Math 222a (Partial Differential Equations 1) Lecture Notes Izak Oltman last updated: December 11, 2019 These are the lecture notes for a first semester graduate course in partial differential equations taught at UC Berkeley by professor Sung-Jin Oh during the fall of 2019. Contents 1 Introduction to PDE’s 3 1.1 Overview of PDE’s .................................... 3 1.2 Examples ......................................... 3 1.3 Basic Problems ...................................... 5 1.4 First Order (scalar) PDEs ............................... 6 1.5 Method of Characteristics ............................... 10 1.6 Existence of Characteristic Solutions ......................... 14 1.7 noncharactersitic condition for kth order system .................. 20 2 Distributions 25 2.1 Motivation ......................................... 26 2.2 Basic Definitions ..................................... 27 2.3 Operations ......................................... 30 2.4 convergence of distributions .............................. 32 2.4.1 Approximation Method ............................. 35 2.5 Fundamental Solutions ................................. 37 2.5.1 Uniqueness .................................... 37 3 Laplace Equation 40 3.1 Cauchy Riemman Equation ............................... 43 3.2 Boundary Value Problem ................................ 45 3.3 Green’s Function ..................................... 48 4 Wave Equation 55 4.1 1 dimension ........................................ 55 1
Transcript
Math 222a (Partial Differential Equations 1) LectureNotes
Izak Oltman
last updated: December 11, 2019
These are the lecture notes for a first semester graduate course in partial differentialequations taught at UC Berkeley by professor Sung-Jin Oh during the fall of 2019.
1. basic issues (existence, uniqueness, regularity, ... ) will be demonstrated by means ofexamples (Evans chapters 2,3,4)
2. theory of distributions and the Fourier transform (Evans chapter 4 and notes)
3. Sobolev spaces (Evans Chapter 6)
1.1 Overview of PDE’s
At a rudimentary level, a partial differential equation (PDE) is a functional equation, that
involves partial differentiation∂
∂xα ∂α. The unknown depends on several variables. PDEs
are ubiquitous, they arise from many fields in science and mathematics.
1.2 Examples
Here are some fundamental examples of PDEs:
Example 1.2.1 (Laplace Equation). u Rd R or C, with ∆u 0, where ∆ is the
Laplace operator defined as ∆ Pdj1 ∂
2j
Example 1.2.2 (Wave Equation). ju 0 where j ∂20 ∆ (the d’Alembertian) and
u R1d R or C. Here we have define x0 t as the time coordinate.
Both of these are scalar equations. The Laplace equation could model temperature distri-bution at equilibrium or electrostatics which can measure the electric potential where thereis no charge. The wave equation models wave propagation (at finite speed). This could beused to model elastic waves or acoustic waves or electromagnetic waves.
The Laplace equation is the prototypical example of elliptic PDEs while the wave is theprototypical example of hyperbolic PDEs
Example 1.2.3 (Heat Equation). u R1d R,C has the form ∂t ∆u 0
Example 1.2.4 (Schrodinger equation). u R1d C with i∂t ∆u 0
The heat equation is a parabolic PDE while the Schrodinger equation is a dispersivePDE.
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1. Introduction to PDE’s
Example 1.2.5 (Cauchy-Riemann). u, v R2 R, we have:
∂xu ∂yv 0
∂yu ∂xv 0
The Cauchy-Riemman equations give necessary and sufficient conditions for a pair offunctions u, v to give us that uz ivz is holomorphic with z x iy. Also note that uand v also satisfy the Laplace equation.
Example 1.2.6 (Maxwell Equations (vacuum)). Given E R13 R3 and B R13
Note that F is linear for all these examples. So all equations stated so far are linear PDEs.As we can write the operator of the thing equal to zero, we have a linear homogeneousPDE. If instead we had Fu f then we have a linear inhomogeneous. In the case wehave ∆u f we then have what is called the Poisson equation.
There are many nonlinear PDE examples:
Example 1.2.7 (Minimal Surface Equation). u Rd R with:
This is to find a surface with minimal area that satisfies something. This is an exampleof an elliptic PDE.
Example 1.2.8 (Korteng-de Vries (KDV) Equation). u R11 R with:
ptu ∂3xu 6u∂x 0
This is an example of a dispersive nonlinear PDE.
Here are nonlinear systems of PDEs
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1. Introduction to PDE’s
Example 1.2.9 (Ricci Flow). g a Riemannian metric (positive definite symmetric n nmatrix) with:
∂tg 2RicgWhere Ric g1∂2g g1∂gg1∂gExample 1.2.10 (Einstein Equation). g a Lorentzian metric which is a 1 3 1 3symmetric metric with signature ,,, (similar to diag1,1,1,1):
Ricg 0
1.3 Basic Problems
Sometimes there is no solution to a PDE, sometimes there is no unique solution, so we mayhave to impose more information. One way to do this is to impose boundary values, whichis the Boundary Value Problem (BVP)
Example 1.3.1 (Dirichlet Problem). Given U `open,bdd Rd, u U R with
¢¦¤∆u f in U
u g on ∂U
Example 1.3.2 (Neumann Problem). Given U and u as above, but:
¢¦¤∆u f in U
ν Du g on ∂U
where ν is the outer normal to ∂U , and we require RU fdx R∂U gdANote that translated solutions are still solutions, therefore we say that solutions are
equivalent if u v c
Example 1.3.3 (Initial Value Problem (IVP)). For the heat equation:
¢¦¤∂tu ∆u f x > 0,ª Rd
u g x > t 0Rd
For Schrodinger
¢¦¤i∂tu ∆u f x > R Rd
u g x > t 0 Rd
For the wave equation:
¢¦¤ju f x > R Rd
u g x > t 0 Rd
∂tu h x > t 0 Rd
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1. Introduction to PDE’s
Definition 1.3.1. We say a boundary value problem or initial value problem is wellposed(Hadaward) if it satisfies
1. existence
2. uniqueness
3. continuous dependence on dataa
a PDE that fails this is called illposed
The regularity of a solution (if not we have a singularity). We also have asymptotics anddynamics
Let’s introduce notation. First the multi-index notation: given α α1, . . . , αd then wedefine:
∂α Dα ∂α1
1 ∂α22 ∂αdd
We call SαS Pαi the order of α. The order of a PDE is the order of the highest order ∂α
that appears in the equation.
Nonlinear PDEs are classified into three categories:
Here are three examples of first order PDEs (some of which are not in the text)
1. Positive example: ∂tat, x∂xu f where u R11 R, f R11
R, a R11 R
such that supx>R Sat, xS supx S∂xaS BMt @ª, and f, a > Cª
2. Inviscid Burger equation ∂tu u ∂xu 0 with u R11 R.
This is an example of a nonlinear first order scalar PDE. Something goes wrong withthis equation.
athat is the map from the data to the solution is continuous
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1. Introduction to PDE’s
3. Lewy–Nirenberg example: ∂t it∂xu f with u, f R11 C.
It turns out that there exists a function f > Cª such that there is no C1 solution uexists in any neighborhood of 0,0If we identify C by R2, this is also an example of a linear first-order system of PDEs.
Let’s begin with a linear first order scalar PDE.
First suppose a 0 (this is known as a transport equation), then we have ∂tu f , so wecan integrate both sides to get:
ut, x S fdt CxWe could also have initial value problem:
¢¦¤∂tu f
u0, x u0xThen we have ut, x u0x R t0 ft, xdt
Now let’s supose a is a general function which obeys the boundedness condition.
Theorem 1.4.1. Given u0 R R smooth, and f R11 R smooth. Then there exists a
unique solution u R11 R to the IVPa
∂tu a∂xu f
u0, x u0xProof. (method of characteristics)
The idea is to try to make a change of variables to reduce this to the simpler case. Inother words, we want a change of variables t, x( s, y such that:
∂s ∂t a∂x (1)
We have:
∂s ∂t
∂s∂t
∂x
∂s∂x
∂y ∂t
∂y∂t
∂x
∂y∂x
anote that this u is a global solution
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1. Introduction to PDE’s
To get (1), we want:∂t
∂s 1 and
∂x
∂ss, y ats, y, xs, y. The first equality gives us
that s t. So we would like to find:
∂x
∂ss, y as, xs, y
This is a first order ODE. Now since SaS is bounded for each t, this allows us to concludethat xs, y is always global (in s)a. Let’s allow x0, y y.
In our equation we had ∂sus, y fs, y therefore us, y u0y R s0 fs, ydsThe last step is to show that x( y is invertible. We have:
∂s∂yx ∂xat, xs, y∂yx @Mt∂yxby comparison of ODEs, we have:
S∂yxS C e R t0 Mtdt∂yx0, y A 0
Therefore x is strictly increasing or strictly decreasing as we increase y, therefore we havethat x( y is invertible.
Remark 1.4.1. This proof is an instance of the methods of characteristics. Because thexs, y are called the characteristics and the ode involving x is called the characteristic ode.
Now let’s turn the the Inviscid Burger equation:
¢¦¤∂tu u∂xu 0 t, x > R2
u0, x u0x t, x > 0 R
Theorem 1.4.2. For any smooth and compactly supported u0 R R, then there does notexist a smooth global solution u R11
R with this data.
It turns out that there always exists a local solution, that is there exists δ A 0 such thaton 0, δ R, there exists a unique smooth solution u.
Proof. Suppose that u R11 R exists. Now let’s apply the method of characteristics to
the operator ∂t u∂x.So we have ∂sx us, xs, y and us, y u0y which is just constant. Visualy things
are pretty clear.
Or we could use:
0 ∂x∂tu u∂xu ∂t∂xu u∂x∂xu ∂xu2
After a little, we get ∂sxx2 0, which is known as the Ricatti equation, which blows up infinite time. This can be solved via separation of variables.
athis is an ODE result, and requires the additional assumption that M is locally bounded. Then we applyPicard’s theorem and prove by contradiction on a saturated solution
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1. Introduction to PDE’s
Now let’s turn to Lewy-Nirenberg’s example:
∂tu it∂xu f
Theorem 1.4.3. There exists a smooth function f R11 C R2 with the property that
there does not exist any C1 solution u to this PDE on any neighborhood of 0,0Proof. (requires complex analysis). Notation: let Br t, x > R11 t2 r2 @ r2 with ∂Br
the boundary. Take a smooth function f with the properties:
1. ft, x ft, x2. for some sequence rn 0 (ex. 2n) we want f SBrnδnBrnδn 0 with RBrn fdtdx x 0
Assume that such a C1 solution to the PDE, u, exists on some neighborhood of 0,0. Pick
n large enough so that Brn ` U . Then let ut, x 1
2ut, xut, x, then observe that u
is again a C1 solution on Brnδ. But observe u is odd.
From now on u u. On the one hand, we have:
SBrn
∂tu it∂xu SBrn
f x 0
By the divergence theorem, the left side is just:
SBrn
∂tu it∂xu S∂Brn
tº
t2 x2xº
t2 x2
uitu
dσ
Therefore uS∂Brn x 0. On the other hand, it turns out uS∂Brn 0.
By continuity, and the fact that f is zero on a bunch of thin annuli. On each annulus,we have that f 0 ∂tu it∂xu. Let’s call the right half of the annulus U, that is:
U rn δ2
@ t2 x2@ rn δ2, t A 0
On U, change the variables to t, x ( s, y 1
2t2, x so 0
1
t∂tu i∂xu ∂s i∂yu
which is the Cauchy - Riemann operator. So now:
U s, y ` R11
rn δn2@ 2s y2
@ rn δn2We can then apply the Schwartz reflection to get a contradiction.
Lecture 3 (9/5)
Today we will cover section 3.2 from Evan’s book. Some notation things: we will use dinstead of n for the spacial dimension. We will also us xj as the jth coordinate (instead ofxj).
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1. Introduction to PDE’s
1.5 Method of Characteristics
We will be interested in the general, fully nonlinear, first order, scalar, PDE: F x,u,Du 0.So F U RRd
R and u U R. And we will subject this to a boundary condition u gon Γ ` ∂U . And we will assume that F and g are smooth. The coordinates in the domainof F will be written as x, z, p1, . . . , pd x, z,p.
Our first goal is to derive the characteristics of ODEs. To do this, we want to find thesolution ux by finding a suitable curve γ that connects x to x0 > Γ on which the evolutionof u can be computed.
Recall that the simplest example is ∂tu 0 with Γ 0R, then the curves are just thevertical straight lines.
Let’s parameterize our curve γ ` U with xs such that s > I. Then let the function valuealong this curve be zs uxs. We would also like to keep track of the partials, by thesymbols pis ∂iuxs
Assume that a smooth solution u exists. Then we have:
∂spis pis Qj
∂j∂iuxsxjsFurthermore, we have:
0 ∂xiF x,ux,Dux ∂xiF x,ux,Dux ∂zF x,ux,Dux∂iu Q
j
∂pjF x,ux,Dux∂i∂ju (2)
So now we have second order terms in terms of first order terms. Now we will set:
Then the collection of the three ODEs involving z, x, p are called the characteristic ODEs.
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1. Introduction to PDE’s
Theorem 1.5.1. If u is a smooth solution to F x,u,Du 0. Then on a curve x satisfying(3):
zs uxs pis ∂iuxswill obey (4) and (5)
Example 1.5.1. Consider when F is linear. Suppose
0 F x,u,Du Qj
bj∂ju cu
where bj bjx and c cx.In this case F x, z, p Pj b
jpj cz. Then if we plug in our characteristic equations, weget that
xjs bjxszs Q
j
bjpj cxszsRemark 1.5.1.
1. The p equation is not needed
2. there is a hierarchy of the ODEs
Example 1.5.2 (Linear F).
¢¦¤x1ux2 x2ux1 u in U
u g on Γ
With U x1, x2 > R2, x1 A 0, x2 A 0 and Γ x1,0 > R2 ` ∂U .
Therefore F x1p2 x2p1 z, so b x2, x1, and c 1. If we use the characteristicequations, we get that x1 x2 and x2 x1, and z z.
The solution is x1s x0 coss and x2s x0 sins with x0 the initial condition. Wealso have zs esz0 gx0es.
Based on our domain, 0 @ s @ π~2. So our path that we know the solution for is a quartercircle. And we know that solution on this quarter circle is gx0es.
Now given x1, x2 > U , we want to find the quarter circle that hits this point. That is we
want to find x0, s such that x1, x2 x0 cos s, x0 sin s. Then we get x0
»x12 x22,
and s arctanx2
x1. So the result is:
ux1, x2 g »x12 x22 exparctanx2
x1
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1. Introduction to PDE’s
Remark 1.5.2. We recover the method of characteristics for ∂t at, x∂xu 0 by thelinear case.
Example 1.5.3. Consider when F is quasilinear, that is:
F x,u,Du Qj
bjx,u∂ju cx,uIn this case we have:
F Qj
bjx, zpj cx, zPlug in our expressions, to get:
¢¦¤xj bjxs, zsz Pj pj∂pjF cxs, zs
Remark 1.5.3.
1. There is no need to introduce p
2. there is no hierarchy of ODEs.
Example 1.5.4 (Burger’s Equation). Recall this equation was: ∂x1u u∂x2u 0. Thenwe have F p1 zp2, then x1 1 and x2 z and z 0
Example 1.5.5. Consider
¢¦¤∂1u ∂2u u2 in U
u g in Γ
with U x1, x2 > R2, x2 A 0 and Γ x1,0 > R2. So we have b1 1, b2 1 and c z2.Therefore: x1 x2 1 and z z2.
So we have x1, x2 x0 s, s.For the other one, we have:
dz
z2 ds therefore
1
zs 1
z0 s therefore zs z01 sz0 .
Then since z0 gx0, then zs gx01 gx0s .
Finally, given x1, x2 > U , we need to find x0 and s so that x1, x2 x0 s, s. But
then x0 x1 x2, s x2. Therefore ux gx1 x21 gx1 x2x2
Example 1.5.6. Here will have a fully nonlinear F given as:
¢¦¤∂x1u∂x2u u in U
u g on Γ
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1. Introduction to PDE’s
Now let U x1 A 0 and Γ 0, x2 > R2. Now we have:
xj ∂pjF x, z, ppj ∂zF x, z, ppj ∂xjF x, z, pz Q
j
pj∂pjF x, z, pThen in this case we have: F p1p2 z, x1 p2, x2 p1, p1 p1, p2 p2 and z 2p1p2.
Then we have:
¢¦¤p1s p01es
p2s p02es
Therefore: z 2p01p02e2s, so that z z0 p01p02e2s 1. Lastly, x1 p02es 1so that x1 p02es 1 and x2 x0 p01es 1.
And z0 gx0, p02 gx0, p1 gx0gx0 .
In the case where gx0 x20, then we have that z0 x2
0, p01 1
2x0, p02 2x0. Now
given x1, x2 > U we want to find x0, s such that x1, x2 x1x0s, x2
x0s. Plugging inwhat we know about these already, we get:
x0 x2
1
4x1
es 1
xx2
1
4x1 x2
14x
1
x2 14x
1
4x2 x1
4x2x1
Then we get u and we simplify, to get: ux x1 4x22
16
Lecture 4 (9/10)
Today we will continue the material in Evans section 3.2.
Recall what we did last time with precise logical assumptions. We were looking at thescalar first order PDE: F x,u,Dx 0 with F U RRd
R, with the boundary conditionu g on Γ ` ∂U with U a domaina
Then we derived the characteristic equations. For this we supposed that there exist a C2
solution u to our PDE. From this we looked for a trajectory xs > U, zs > R, ps > Rd suchthat uxs zs, pis ∂iuxs and xs is chosen so that we have a closed system ofODEs for x, z, ps.
aopen and connected set in Rd
– 13 –
1. Introduction to PDE’s
The worst term is pis ∂j∂iuxsxjs. To simplify this, we differentiate F in xi toget:
∂xiF ∂xiF x,ux,Dux ∂zF x,ux,Dux∂iu ∂pjF x,ux,Dux∂i∂juxTo simplify everything, we choose xj ∂j∂iuxsxjs. Then we get: pis ∂zF x, z, ppi∂xiF x, z, p. And lastly, we get that z Pi pi∂piF x, z, p.a.
1.6 Existence of Characteristic Solutions
Goal: We would like to develop local existence theory for smooth solutions to our PDE.That is, given data near a point x0 > Γ, then under certain conditions, show that there existsa smooth solution to our PDE in a small neighborhood V ? x0.b
Let’s start by considering a simpler case. Let U Rd xd A 0 and Γ ` xd 0
Let’s say we have the data x, z, p0 at x0.
¢¨¦¨¤
x0 x0 > Γ
z0 gx0pi0 ∂iux0 ∂igx0 for i 1, . . . , d 1
F x0, z0, p0 0
(6)
So we have 2d 1 equations for 2d 1 for x, z, p (2d 1 variables). These are known as thecompatibility conditions.
Assume that we start with x0, z0, p0 that satisfy (6) (such a triple is called admis-sible). Let x0 > W ` Γ. Parametrize the points in W by y y1, y2, . . . , yd1,0. Now thegoal is to find the characteristics:
xy, szy, spy, s
With
¢¨¦¨¤
xy,0 yzy,0 gypiy,0 ∂igy for i 1, . . . , d 1
pdy,0 F xy,0, zy,0, py,0 0
(7)
To do this, we will use the implicit function theorem.
aThis can fail in many ways, one to keep in mind is if different trajectories collidebwe mean local in space and time. And we were from now on refer to this as local existence.
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1. Introduction to PDE’s
Theorem 1.6.1 (Implicit Function Theorem). Given F Rm Rn Rm which is C1 and
suppose that F y0, p0 0 is satisfied for some y0, p0. If ∂yF is invertible (as a matrix) thenthere exist neighborhoods U ? y0 and V ? p0 then there exists a C1 function p U V suchthat py0 p0 and F y, py 0 for all y > U .
Moreover, if F y, p 0 for some y > U and p > V then p pyTherefore (7) is solvable if ∂pdF x0, z0, p0 x 0. In this case, this is known as nonchar-
acteristic.
In summary: we began with x0, z0, p0 with z0 gx0 and p0 ∂1g, . . . , ∂d1gx0, p0dwhich is admissible. And ∂pdF x0, z0, p0 x 0 noncharactersitic. Then we have:
¢¦¤xy,0 yzy,0 gypy,0
With xys, zy, s, py, s are given by the charactersitic equations.
Now, given x > U near x0, we want to find y, s such that x xys.Lemma 1.6.1. Under the noncharactersitic assumption, the map y, s( xy, s is invert-ible near x0. Denote the inverse by x( yx, sxProof. Inverse function theorem. Compute using the characteristic equation that x satisfies:
∂yjxk ∂sxk Uyxns0
Id1
Fp1x0, y0, p0
Fpn1x0, z0, p00 Fpnx0, z0, p0
Compute the determinant which is just xdx0,0 ∂pnF x0, z0, p0 x 0
Theorem 1.6.2. Under the above assumption, ux zyx, sx where y, sx aredefined with x > V . Then this gives us a solution to the PDE
¢¦¤F x,u,Du 0 in V
u g on Γ 9 V `W
Consider the quasilinear 1st order PDE:
Qj
bjx,u∂ju cx,u 0
F Qj
bjx, zpj cx, zThen the noncharacteristic condition requires ∂pdF x0, z0, p0 bdx0, z0 x 0. So in thequasilinear case, the noncharacteristic assumption is equivalent to the existence of a unique
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1. Introduction to PDE’s
choice p0d from everything else.
In the fully nonlinear case, take F ∂xu2 ∂yu2 1. Then F p21 p
12 1. Then given
p01 1
2, then we have two choices of p02
Now let’s prove Theorem 1.6.2.
Proof. step 1 veryify that F xy, s, zy, s, py, s 0. Compute:
∂sF x, z, p Qi
∂xiFxi ∂zF z Q
j
∂jF pj
Then apply the characterisitc equations to get that the whole thing is zero. Then sinceF xy,0, zy,0, py,0 0 by preparation, we are good.
step 2 We know that xyx, sx x and zyx, sx ux. So it remains to showthat ∂iux piyx, sx.
pizyx, sx Qj
∂yjz∂iyj ∂sz∂is
And recall that ∂sz Pi pi∂piF . So we need to compute ∂yjz. From zy, s uxy, s ,then we need to prove:
∂yjz Qi
piy, s∂yjxiDefine:
rjs ∂yjz Qi
piy, s∂yjxiy, srj0 0 by preparation. Then by a long computation, we get that:
∂srj ∂zFr
j
From this we get that some equation is:
Qj
Qk
pk∂yjxk∂xky
jQ
k
pk ∂pkF ´¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¶∂sxk
∂xis Qk
pkQj
∂yjxk∂xiy
j ∂sx
k∂xis
Qk
pkδki pi
Lecture 5 (9/12)
For problem number 4 of the homework (Evans section 3.3.5 problem 6(b)). Takeux, t gx0, x, tJ0, x, t (take this as the definition of u and verify that it is a solution).
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1. Introduction to PDE’s
Review: in the case Γ ` xd 0 U xd A 0. We wanted to solve:
¢¦¤F x,u,Du 0 in U
u g on Γ
Then we used the method of characteristics, picking x0 > Γ
¢¦¤∂sxi ∂piF x, z, p∂sz Pi pi∂piF x, z, p∂spi ∂zF x, z, ppi ∂iF x, z, p
We are interested in solving for this on an open set U . To solve this we needed initial values:
¢¦¤x0 x0
z0 gx0pi0 p0i
For p0 we need to make a choice. For p0i ∂igx0 with i 1, . . . , d1. But for the normaldirection, we need to figure this out such that it solves F x0, z0, p0 0. Such a triple iscalled admissible.
Noncharacteristic x0, z0, p0 means that ∂pdF x0, z0, p0 x 0. This allows us to solvefor pdy,0 for y >W ? x0 and W ` Γ (unique continuous choice with pdx0,0 p0d.
We also could invert y, s( xy, s for s very small and y close to x0.
Now define ux zyx, sx. Then the claim is that this u is a solution to the bound-ary valued problem in V . And this is smooth.
Now let’s extend this to general U and Γ (Ref: Evans Appendix C)
Definition 1.6.1 (Ck Boundary). Given U open bounded in Rd. We say that ∂U is Ck atsome x0 > ∂U if there exists some r A 0 and γ Rd1
R in Ck so that:
U 9 Bx0, r´¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¶ball of radius r centered at x0
xd A γx1, . . . , xd1
And ∂U 9Bx0, r xd γx1, . . . , xd1So now we would like to flatten the boundary. Define the coordinates y1, . . . , yd so that:
¢¦¤yi xi i 1, . . . , d 1
yd xd γx1, . . . , xd1Then U 9Bx0, r ` yd A 0 and ∂U 9Bx0, r ` yd 0.
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1. Introduction to PDE’s
Note this is invertible by:
¢¦¤xi yi i 1, . . . , d 1
xd yd γy1, . . . , yd1If U or ∂U is Ck for all k, we say it is Cª
Suppose we have the general problem:
¢¦¤F x,u,Du 0 in U
u g on Γ
With U a Cª domain. Then there exists x ( yx, smooth, invertible in a neighborhoodV ? x0. So if vy uxy, then v satisfies a PDE of the same form.
F xy, vy,Dvy∂y∂x
xy 0
Then we can write the above operator as Gy, vy,Dvy 0. Then applying our theory ofwhat we have done so far to our equation Gy, v,Dv 0 then we solve the original problem.
In this case, the noncharacteristic condition now becomes:
Qj
νj∂U∂pjF x0, z0, p0 x 0
Where ν∂U is the unit outer normal vector to ∂U
Theorem 1.6.3. Local existence near a noncharacteristic boundary value holds for generalCª domains.
Let’s now talk about uniqueness. Suppose that we already have two solutions u1, u2that are C2 solving:
¢¦¤F x,u,Du 0 in U
u g on Γ
does u1 u2?
The idea is the method of characterics from Γ to x in U . But we should be careful aboutthe normal derivative νj∂U∂ju 0
Example 1.6.1. Consider ∂1u2 1 on RC0 and u 1 on u. We have two solutionsu 1 x and u 1 x. This has to do with the choice of p01 Pj u
j∂Up0j.
So for uniqueness, we require that the normal derivatives also agree: P∂j∂U∂ju1
P∂j∂U∂ju2.
– 18 –
1. Introduction to PDE’s
Then for all x > U such that there exists a characteristic xs, zs, ps with x0 > Γ, z0 ux0, p0 Dux0 such that xs x and xs > U for all s > 0, s. Then we haveu1x u2x by uniqueness of ODEs.
What about continuous dependence? Continuous dependence does hold, but it is prettysubtle.
Example 1.6.2. Consider Burger’s equation
¢¦¤∂tu u∂xu 0 in t A 0u g on t 0
Suppose that g > C2 and YgYC2 B A a then there exists a neighborhood V 0, T R withT only dependent on A (in particular independent of g) such that there exists a unique C2
solution u on V .
Now let’s write g ( ug is continuous in C2, in the sense that if gn g in C2Γ thenugn ug in C2V .
However, there exist counterexamples h1n , h
2n such that:
Yuh1n uh2
n YC2V C nYh1n h
2n YC2Γ
i.e. g ( ug cannot be Lipschitz C2 C2.
Terry Tao wrote a good blog on quasilinear well-posedness.
Here we will be looking at Evans section 4.6. We would like to study the extension ofthe notion of noncharacteristic bounded data to higher order systes of ODEs.
Let’s begin with the noncharaceristic condition for a quasilinear, 1st order, scalar PDE.Recall, we were looking at (with Γ ` xd 0):
F x,u,Du Qj
ajx,u∂ju a0x,u 0
In this case x0, p0, z0 became our characteristic if and only if adx0, z0 x 0.
Note that given this, we can uniquely write:
F x, z, p Qj
ajx, zpj a0x, zx0, z0 is noncharacteristic, therefore this is equivalent to:
If we are interested, we can determine ∂αux0 under he noncharacteristic assumption.Because from the boundary data g we got z0, p01, . . . , p0d1 and from nonchar we gotp0d ∂dux0. Now suppose we want ∂i∂jux0. This is easy if i j 1,2, . . . , d 1. Butalso, for y close to x0, we get ∂i∂dux0 from i 1, . . . , d 1. To compute the second normalderivative, we differentiate the equation in ∂d to get:
0 Qj
ajx,u∂j∂du bx,u, ∂ubut if we evaluate at x0, then we know everything but ajxp, z0∂2
du. In other words wecan write adx0, z0∂2
0x0 things that only depend on x0, ux0, ∂jux0 j 1, . . . , d, otherthings. But the thing on the left is nonzero, so we can solve for the thing on the left.
So the noncharacteristic assumption of x0, z0 is equivalent to determining all derivativesof u at x0.
zLecture 6 (9/17)
1.7 noncharactersitic condition for kth order system
Reference: Evans section 4.6. The things that are important
4.6.1: noncharacteristic cauchy data for quasilinear kth order systems
4.6.3: Cauchy - Kovalewski theorem
Consider the quasilinear K-th order system of PDEs. We have u U®Rd
RN2, with bαAB with A the row index and B the column index.
And C defined similarly. (I don’t really understand the indicies).
The Cauchy data is data for u, ∂νu, . . . , ∂k1νk1
u on Γ ` ∂U .
More simply, we could have Γ ` xd 0, with Cauchy data for u:
u, ∂du, ∂2du, . . . , ∂
k1d u g0, g1, . . . , gk1
on Γ.
– 20 –
1. Introduction to PDE’s
If we assume that we have a smooth solution u, one question we will ask is whether wecan determine all derivatives Dnux0 with x0 > Γ from the Cauchy data and the equation?
The answer is to come up with a noncharacteristic condition for the Cauchy data near x0.
Given the Cauchy data: g0, . . . , gk1 near x0 > Γ. Then we can compute any Dkux0 fork B k except ∂kdux0. We can get this from the PDE. Moving things around we get:
b0,...,0,k∂kdux0 QSαSk
αx0,...,0,dbαDαu cx0
We need that b0,...,0,d is invertible, in which case we have that:
∂kdux0 b0,...,0,d1
Q
SαSkαx0,...,0,d
bαDαu cx0
We want to this again and again. Note that b0,...,0,d is invertible around x0, therefore we cancompute ∂kdux in a neighborhood of x0 (in Γ). Therefore we can compute all of Dk1uxexcept ∂k1
d ux0.Now take ∂d of the equation, keep ∂k1
d u on the left hand side and the rest on the righthand side. Then we get:
b0,...,0,d∂k1d ux0 known
but we can invert and get the something known. The same condition (that b0,...,0,d is in-vertible) allows us determine ∂k1
d ux0 and ∂k1d x in a neighborhood of x0.
Definition 1.7.1 (noncharacterisitic condition). g0, . . . , gk1 is noncharacteristic atx0 if b0,...,0,d evaluated at x0, ux0, . . . ,Dk1ux0 is invertiblea.
Theorem 1.7.1. If g0, . . . , gk1 is noncharacterisitic at x0, then all derivatives of u at x0
can be determined.
We can generalize this to general domains. Given U a smooth domain, which means thatfor all x0 > ∂U , after renaming and reorienting the coordinate axis, there exists r A 0 suchthat:
∂U 9Bx, r xd γx1, . . . , xd1with γ a smooth function. Then we can take yi xi for i 1, . . . , d 1 and yd xd γx1, . . . , xd1.
athe imput to b can be computed from the Cauchy data
– 21 –
1. Introduction to PDE’s
Recall we can also define the unit normal to the boundary ∂U as ν∂U .
One case of this is if aij is a positive definite matrixa. For example if aij is the identitymatrix, then our PDE becomes ∆u 0 and all boundaries are noncharacteristic (this is theproperty of ellipticity).
Another example would be:
aij 1 00 I
Then we get the wave equation ju 0. This is characteristic if and only if ∂0w2
Pdi1∂iw2 0 (these are characteristic hypersurfaces).
Lecture 7 (9/19)
Recall our discussion on the noncharacterisitic condition for
¢¦¤PSαSk bαx,u, . . . ,Dk1uDαu cx,u, . . . ,Dk1u 0 in U
u, ∂νu, . . . , ∂k1νk1
g0, . . . , gk1 on Γ ` ∂U(8)
aPaijξiξj A 0
– 23 –
1. Introduction to PDE’s
We say that the boundary value is noncharacteristic at some point x0 > Γ implies thatwe can compute (or determine) all of the derivatives of u at x0 (Dβux0) if and only if thematrix PSαSk bαx0, ux0, . . . ,Dk1ux0να is invertible.
This motivates the following theorem. Can we construct a solution of u near x0 by theknowledge of all Dαux0? The answer is yes, provided that everything can be written aspower series. Equivalently, if everything is real analytic.
Definition 1.7.2 (real-analytic). We say that f U RN is real analytic if for all x0 > U ,there exists r A 0 and coefficients aα > RN where α are multi-indicies such that:
fx Qα
aαx x0αfor Sx x0S @ rRemark 1.7.1. If f is real analytic, then we have that:
Dαfx0 α!aα α1!α2!αd!
Definition 1.7.3 (real-analytic boundary). We say ∂U (or U) is real analytic if for allx0 > ∂U there exists r A 0 such that ∂U 9 Bx0, r xd γx1, . . . , xd1 (after a suitablerelabeling and reoriented of the coordinates), with γ real analytic
Theorem 1.7.2 (Cauchy–Kovaleskaya Theorem). Consider the boundary valued prob-lem (8) such that bα, c, g0, . . . , gk1, ∂U are real-analytic. And suppose the boundary valuesare noncharacterisitc at x0 > Γ. Then there exists a neighborhood V ? x0 such that thereexists a unique, real-analytic solution u to the boundary valued problem on V 9U .
Moreover, this solution:
ux Q 1
α!Dαux0x x0α
near x0
Proof. Really clever proof, see the text.
Remark 1.7.2 (about Cauchy–Kovaleskaya theorem). At first, you might think that thistheorem looks like a very general existence and uniqueness statement for PDEs. However,real-analytic functions are too restrictive to be useful.
Suppose that f was real-analytic on Rd, and suppose that f 0 on B0,1, then f 0 onRd.
The dependence of u on the boundary data in the Cauchy–Kovaleskaya theorem is veryweak. This would make small disturbances of g’s lead to large disturbances for u.
ux1, x2 sinkx1 sinhkx2is a solution to the boundary value problem. Now everything is analytic, so by the theorem,this is the unique solution.
Now let’s perturb the initial data. Let’s have the initial data now 0, ek1~2k sinkx0, thenwe get:
u ek1~2
sinkx1 sinhkx2Now as we send k ª, then our data goes to 0,0 in any Cn-norm.
However, let’s look at the solution with x2 A 0, then we have:
ek1~2
sinkx1 sinhkx2 kªÐÐ
1
2ekx
2
ek1~2
sinkx1but that thing blows up in the uniform norm.
2 Distributions
We now are turning to the second part of the course. We have two goals
1. We would like to study several fundamental linear 2nd order scalar PDEs (Laplace,Wave, Heat, Schrodinger). (chpt 2.2, 2.4, 2.3)
2. Understand the theory of distributions. This is the “completion of differential calcu-lus”. We will also like to study Fourier analysis (chpt 4ish). Check the bcourse site forlecture notes on these topics.
– 25 –
2. Distributions
2.1 Motivation
Schwartz was the founder of the theory of distributions, but it was independently constructedearlier. Let’s consider a motivating example from electrostatics.
Example 2.1.1. Given E R3 R3 the electric field and ρ R3
A basic problem in electrostatics is to determine E (or ϕ) from ρ. Let’s solve:
∆ϕ ρ
First consider the case when ρ is the sum of point charges at xk with charges qk. By linearity,it is enough to find:
ϕ Q q1ϕ0x xkwhere ϕ0 is the potential of the point charge with charge q 1 at the point 0. Then we cansay that ρ could be written as a continuous sum of point charges:
ρ S ρydyThen we can say that ϕ is the continuous sum of point-charge electric potentials. So:
ϕ S ρyϕ0x ydySo all that remains to find is the electric potential of a point charge, ϕ0. Note by rotationalsymmetry of the problem, ϕ0 must be radial: ϕ0 ϕ0r. By the Gauss equation, we have:
r2. So we know normalize so that ϕ0 0 as r ª and then we see
that:
ϕ0 1
4π
1
r
1
4πSxSAn alternative derivation of this comes from Evans chapter 2.2.
Lecture 8 (9/24)
– 26 –
2. Distributions
2.2 Basic Definitions
I would like to point out that the word distribution to define these objects doesn’t reallymake a lot of sense. I prefer the word ‘generalized functions’, I think that makes more sense.
Today will the real beginning of our discussion on distribution theory. Let’s motivate thedefinition of a distribution.
Let’s try to make sense of the notation of point charges qi at points xi > R3. Let u be thecharge distribution function of these charges. We would like:
1. u 0 outside xi2. RU u Pxi>U qi
However, u cannot be a function that satisfies this. Rather than viewing u as a function onR3, it is more natural to interpret u as a functional on the set of sets. By this we mean:
uU Qxi>U
qi
Given two open disjoint subsets U,V , then uU 8 V uU uV . If we interpret thesehas integrals, we have (not rigoruously):
S uχU χV S uχU S uχV
where χ is the characteristic function. It is therefore natural to interpret u as a linear func-tional on the space of characteristic functions.
So we want to view u as a linear functional on the space of nice functions. The nicest wecan think of is the space of smooth, compactly supported functions: Cª
0 R3.Here is a rough definition of a distribution:
Definition 2.2.1. A distribution u on a set U ` Rd is such that u Cª
0 U R a linearfunctional which is also “continuous”
Notation: Given ϕ > Cª
0 U, we will write:
uϕ `u,ϕewhich mimics the L2 inner product.
We call the space Cª
0 U the space of test functions.
How “flexible” is this space? There exists one nontrivial element:
ψ R R
– 27 –
2. Distributions
with the property ψ > Cª such that suppψ x ψx x 0 ` 0,ª. We can set:
ψ
¢¦¤e1~x x A 0
0 x B 0
Then ψ works. Now consider ϕ Rd R by writing ϕx ψ1 x12 xd2, then
suppϕ ` B0,1.Definition 2.2.2 (Convolution). Given f a locally integrable function, ϕ > Cª
0 Rd, thenwe write:
f ϕx S fyϕx ydyRemark 2.2.1. Note that even if f is nondifferentiable, then f ϕ is Cª, because:
∂xjf ϕ ∂xj S fyϕx ydy S fy∂xjϕx ydyWhere the second equality relies on f being locally integrable and ϕ smooth.
Remark 2.2.2. If f > CRd and ϕ > Cª
0 Rd. Then suppf ϕ ` supp f suppϕ. Wheresupp f suppϕ x y > Rd x > supp f, y > suppϕ
In particular, if f > C0Rd and ϕ > Cª
0 Rd, then f ϕ > Cª
0 RdLemma 2.2.1 (Approximation by mollification). Let f > CkRd and ϕ > Cª
0 Rd with
R ϕ 1. Then given δ A 0, define:
ϕδx 1
δdϕx~δ
Therefore suppϕδ ` B0, δ and R ϕδ 1. Then f ϕδuniformlyÐÐÐÐÐ f as δ 0. Moreover, up
to k derivatives uniformly converge:
supRd
SDαf ϕδ fS 0
for all α, SαS B k.
Proof. When SαS 0 and k 0. Then we have:
f ϕδx fx S fyϕδx ydy fx S fx yϕδydy S fxϕδydy S ϕδyfx y fxdy S fx δz fxϕzdz
Then as δ 0, we use continuity of f to get that this integral goes to zero.
For higher k with SαS A 0, you let the derivatives fall on f , then proceed with the sameproof.
– 28 –
2. Distributions
Now what is the topology that we should take?
Definition 2.2.3 (Convergence of Test Functions). We say that a sequence ϕj > Cª
0 Uconverges to ϕ > Cª
0 U (uj u) if there exists a compact set K ` U such that suppϕj `K(for all but finitely many) and suppϕ `K and supx>K SDαϕjxDαϕxS 0 as j ª forall multi-index α
Remark 2.2.3 (Optional). This isn’t a Frechet space or a Banach space
Remark 2.2.4 (Optional). This is the strongest topology such that every Cª
0 K embedsinto this space for every compact subset K. Where Cª
0 K is the space of functions insuppϕ ` K endowed with ϕj ϕ if and only if supx>K SDαϕjx DαϕxS 0 for all α.This is a complete and metrizable set (Frechet).
Definition 2.2.4 (Distribution). u Cª
0 U U is a distribution if it is linear andcontinuous in the sense that if ϕj is a sequence of test functions which converge to ϕ in theabove sense, then uϕj converges to uϕ.Lemma 2.2.2. Let u Cª
0 U R be a linear functional. Then u is a distribution (i.e.continuous) if and only if (“boundedness”) for all compact K ` U there exists N A 0 andC CK,N A 0 such that for all ϕ > Cª
0 U, suppϕ `K:
S `u,ϕe S B CK,N QSαSBN
supx>K
SDαϕxSProof. If you have boundedness, then you have continuity. We can ask (by linearity) if`u,ϕj ϕe 0. But ϕjϕ 0 in Cª
0 U there exist K compact such that supx>K SDαϕjxϕxS 0. By boundedness, there exists a N and C such that
S `u,ϕj ϕe S B C QSαSBN
supx>K
SDαϕjx ϕxS 0
For the other direction, suppose that u doesn’t have boundedness. Then we get a compactK ` U such that for all N , there exists ϕN with S `u,ϕNe S C N PSαSBN supx SDαϕxS. Thenset ψNx ϕNxN PSαSBN supx SDαϕxS1. Then SDαψN S B 1~N forall SαS B N . ThenψN 0 in Cª
0 U but not because the above.
Lecture 9 (9/26)
Recall that we defined the notation of a distribution.
Definition 2.2.5 (distribution). A distribution u is a linear functional ϕ ( `u,ϕe onCª
0 U that is continuous.
i.e. if ϕj ϕ in the sense that there exists a compact set K ` U such that suppϕj, suppϕ `
K, we have:
supx>K
SDαϕjx DαϕS 0
as j ª for all α. Then `u,ϕje `u,ϕe.– 29 –
2. Distributions
Lemma 2.2.3. A linear functional u Cª
0 U R is a distribution if and only if it is“bounded” in the sense that for all compact subsets K ` U there exists an N and CK,N A 0such that for all suppϕ `K, we have that S `u,ϕe S B CK,N supx>K supSαSBN SDαϕxSWe will call the space of all distributions DUDefinition 2.2.6 (Order of a distribution). We say that a distribution u > DU is saidto be of order B N if there exists N such that for all compact sets K there exists CK,N A 0such that S `u,ϕe S B CK,N supx>K supSαSBN SDαϕxS
The order of a distribution is the minimum among such N .
Definition 2.2.7 (Support of a Distribution). Given an open subset V ` U we say thatu vanishes on V if for all ϕ > Cª
0 U, suppϕ ` V implies that `u,ϕe 0.
Then set Vmax V u vanishes on V . Then set suppu V cmax
Example 2.2.1. Any continuous function u defines a distribution by `u,ϕe R uϕ for allϕ > Cª
0 U.The order of such a distribution is zero and the support definitions agree
Example 2.2.2. Locally integrable functions are distributions. Recall u a function on U islocally integrable if for all compact subsets K ` U we have that RK SuS @ª.
This is a distribution, the order is 0 and suppu suppuduExample 2.2.3. The signed Borel measure on U , BU, defines a distribution (just integrateagainst it). The order is 0.
Example 2.2.4. The simplest example of order N is `u,ϕe Dαϕ
2.3 Operations
Now let’s discuss the basic operations for distributions.
The basic principal (the adjoint method): take an operation P on a function u, we willtry to compute P so that R Puϕ R uP ϕ for all test functions ϕ. Then given a distributionu > DU, we define Pu by `Pu,ϕe `u,P ϕeExample 2.3.1. Consider multiplication of u > DU by f > CªU. When u > Cª
0 US fuϕ S ufϕ
Following this we say that `fu,ϕe `u, fϕe
– 30 –
2. Distributions
Example 2.3.2 (Differentiation). Start with u > Cª
0 U and ϕ > Cª
0 U then
S Dαuϕ 1SαSS uDαϕ
So define: `Dαu,ϕe 1SαS `u,DαϕeNote that all distributions are differentiable. So distribution theory extends differential
calclus. Moreover, all distributions are locally Dαf for f continuous functions.
Example 2.3.3 (convolution). Let’s first consider u > DU and v > Cª
0 U. What is uv?
If v,ϕ > Cª
0 U, then recall:
u vx S uyvx ydySo naturally:
`u v,ϕe S u vxϕx U uyvx ydyϕxdx S uyS vx yϕxdx´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
vϕy
dy
So `u v,ϕe `u, v ϕeBecause v > Cª
0 U, then u v is not just a distribution, it is a smooth function.
Proof. Consider u vx `u, vx e. By continuity of u and the fact that the mapx( vx is continuous in the space of test functions.
So if xn x, then vxn y vx y in Cª
0 U. This means that u vxn u vxie. as a function, u vx is continuous. Note that:
S u vxϕx `u v,ϕeit is easy to show smoothness.
Let’s discuss more examples of distributions.
Example 2.3.4 (Delta Function). Delta distribution δx0 > DU where `δx0 , ϕe ϕx0.
This is equivalently given by an atomic measure of mass 1 supported x0
Example 2.3.5 (Heaviside Function). Heaviside function H R R with:
Hx ¢¦¤1 x A 0
0 x B 0
Then H δ0. Because
`H , ϕe `H,ϕe S ª
0ϕxdx ϕ0
– 31 –
2. Distributions
Remark 2.3.1. We can generalize this computation to give a distribution theoretic proof ofthe divergence theorem.
Remark 2.3.2. Trying to make sense of 1~x as a distribution. This isn’t a L1loc function.
However, does there exists a distribution u such that u 1~x in R 0. Take log SxS anddifferentiate it as a distribution (this is known as taking the principal value of 1~x.
`1~x,ϕe S ª
ª
ϕ
xdx S
ª
ª
log SxSϕxdx S L
0log SxSϕ ϕ0 S 0
Llog SxSϕ ϕ0
SL
L
ϕx ϕ0x
dx Sε
L
ϕx ϕ0x
dx Sε
ε
ϕx ϕ0x
dx SL
ε
ϕx ϕ0x
dx
(Yeah I know, a bunch of those bounds should be negative) So we then send this to infinity.The middle term ends up going to zero so we define:
`PV 1~x, eϕ limε0
S ε
L
ϕx ϕ0x
SL
ε
ϕx ϕ0x
To show the middle term actually goes to zero we note that:
Sε
ε
ϕx ϕ0x
dx Sε
εS
1
0ϕxσdσdx
Then we can bound ϕ and send ε to zero to get that this integral goes to zero.
Lecture 10 (10/1)
Today we will continue on our discussion of distribution theory. Recall that distributions(DU) are defined as linear functionals on Cª0 U that are continuous in the sense of con-vergence on Cª0 U.
We discussed basic operations on distributions: multiplication by a test function, differ-entiation, convolution with a test function.
2.4 convergence of distributions
Let’s now discuss sequences of distributions.
Definition 2.4.1 (Convergence in the sense of distributions). Given a sequence un >DU, we say that un u > DU in the sense of distributions if for all ϕ > Cª0 U,`un, ϕe `u,ϕeRemark 2.4.1. This is the same of weak convergence of linear functionals our space.
Theorem 2.4.1 (Sequential Compactness). Let un > DU and suppose that for allϕ > Cª
0 U, `un, ϕe is convergent. Then the claim is that the linear functional u defined by
`u,ϕe lim `un, ϕe (9)
is a distribution (i.e. it is continuous on Cª0 U).
– 32 –
2. Distributions
Proof. (optional) It’s clear that u as defined above is linear. So it suffices to check continuity.
Let K be any compact subset of U . Then u obeys (9) for ϕ > Cª
0 K. Note that thisspace is nice (for a fixed K). Convergence is just ϕn ϕ when for all N , SαS B N , we havesupx>K SDαun DαuxS 0. So define the seminorm ρNϕ supSαSBN supx>K SDαϕxS.
Then let:
dϕ,ψ QN
2NPNϕ ψ
1 PNϕ ψthen we have convergence of ϕn if and only if dϕn, ϕ 0. So Cª0 K, d defines thesame topology as above and d is invariant in the sense that dϕ v,ψ v (.ϕ,ψ for all
ϕ,ψ, v > Cª0 K, and d is complete.
These must be proven. But once these are shown, then un is the family of linearfunctionals on Cª0 K which is a complete metric space (compatible with the vector spacestructure). Then we can apply the boundedness principal (or the Banach-Steinhaus theo-rem). If `un, ϕe is bounded for each ϕ, then un is equicontinuous.
By equicontinuity, we have that uSCª0 K is continuous and K was arbitrary. Then sinceCª0 U is the strongest topology such that Cª0 K Cª0 U imbedds continuously. It followsthat u is continuous on Cª0 UExample 2.4.1. Consider a sequence of functions un > L1
locU. In this case you can connectthe notion of pointwise convergence and the convergence in the sense of distributions via thedominated convergence theorem.
That is, if un u in the pointwise sense, and there exists some function g > L1loc such
that SunS B g almost everywhere on U , then un u in the sense of distributions
Proof. We just need to show that R unϕ R uϕ for all ϕ > Cª0 U. But this follows imme-diately from the dominated convergence theorem.
Example 2.4.2. Let h > Cª with h 1 on 1,ª and supph ` 0,ª. Then if we takelimδ0 hx~δ this converges to the heaviside function pointwise and in the sense of distri-butions.
Example 2.4.3. Consider the sequence:
unx ¢¦¤einx x A 0
0 x B 0
Then unx does not have a pointwise limit for 0 @ x x 2πZ. But this sequence does have alimit in the sense of distributions, and it is zero.
But we can control (bang) as its absolute value is bounded by1
n Rª
0 S∂xϕxSdx which goes tozero.
Example 2.4.4. Take the same un but normalize it by n. Then the limit of limnª nun 1
iδ0
in the sense of distributions (check).
Example 2.4.5 (approximation of the identity). Let ϕδ be the standard mollifer withsupport contained in a ball of radius δ. Then we saw that ϕδ u u as δ 0.
Now if we take the same ϕδ, then ϕδ δ0 as δ 0 in the sense of distributions.
Proof. Fix ψ > Cª0 Rd, then:
`ϕδ, ψe S δdϕδ1xψxdx S ϕzψδzdzthen apply dominated convergence theorem.
Remark 2.4.2. δ0 u u
Proposition 2.4.1. u > DRd take ϕ > Cª0 Rd with R ϕ 1, ϕδx δdϕx~δ, thenϕδ u u in the sense of distributions.
Proof.
`ϕδ u,ψe `u,ϕδ ψewith:
ϕδ ψ S ϕδx yψxdx S ϕδzϕz ydz δ0
ÐÐ ψyrepeat for Dαϕδ ψ, then ϕδ ψ ψ in Cª0 U. Then by continuity of u, we get`u,ϕ0
ψe `u,ψeRemark 2.4.3. ϕδu is a smooth function. Therefore every distribution can be approximatedby smooth functions.
Take u > DU, then there exist a sequence uj > Cª0 U such that uj u in the sense ofdistributions.
Here is a construction. We would like to construct a sequence Kj of compact subset ofU which is nicely nested such that:
1. Kj `K0j1
2. jKj U
now construct χj > Cª0 U such that χj 1 on Kj but suppχj `Kj1. Then we can computeϕδ χju then set uj be a sequence of these then we are done.
– 34 –
2. Distributions
2.4.1 Approximation Method
An operation P on smooth (smooth and compactly supported) functions is generalized tou > DU by taking a sequence of continuous functions uj > CªU such that uj u in thesense of distributions and define Pu limPuj
Example 2.4.6. Prove that if u > DU such that ∂ju 0 then u is a constant. This istrivial with the approximation method:
Let uδ ϕδu, this converges to u distributionally. But we also know that ∂juδ ϕδ∂iu 0,therefore uδ 0, therefore u must be zero.
Example 2.4.7 (Differentiation of the characteristic function). Recall last time wecomputed that H δ0.
Now let U ` Rd be an open domain. Define 1U χUx as the characteristic function.Then the claim is that if U has a C1 boundary then we have:
∂jχU ν∂UjdSwith dS the Euclidean surface element. So `ϕ, dSe R∂U ϕS∂UdS.
This actually implies the divergence theorem. Let bj > CªRd, then we have:
Proof. Note that ∂j1U is supported in ∂U . It suffices to compute this on balls that cover∂U . Then B 9 ∂U xd γx1, . . . , xd1 and B 9U xd @ γ1, . . . , xd1. Then we have:
1Ux limδ0
hδ1γx1, . . . , xd1 xdset v ∂1γ, . . . , ∂d1γ,1. Then we can differentiate 1U by differentiating all the hδ’s. Bycomputaion we have:
∂jh hδ1γx1, . . . , xd1 xdδ1vj
now for each x1, . . . , xd1 as δ goes to zero, this converges to the distribution δγx1,...,xd1xd,therefore for each ϕ > Cª0 U with suppϕ > Bα we have:
Today will be the last class where we will discuss distribution theory. We will:
introduce the notion of fundamental solutions
discuss operations between distributions
Let’s consider multiplication of distributions. Take u, v > DU. What does uv look like?This doesn’t always work, for example 1~ºx and 1~ºx are locally integrable, therefore theydefine distributions, however, their multiplication gives 1~x which isn’t locally integrable.
So in general, this is impossible, but if for all x > U , at least one of u or v is smooth, thenu v is well defined.
Definition 2.4.2 (Singular Support). Let u > DU. We say that u is smooth on anopen subset V ` U if `u,ϕe R uϕ for some u > Cª and for all ϕ > Cª0 U with suppϕ ` V .We define the singular support as:
singsuppu U V ` U an open subset in U on which u is smoothIntuitively, this is the collection of points where the distribution fails to be a smooth
function
Proposition 2.4.2. If the singular supports of u and v are disjoint, then their product iswell-defined.
Proof. Construct a smooth function χ such that χ 1 on singsuppU with suppχ ` U
singsuppu. Then for all ϕ > Cª0 U we have ϕ χϕ 1 χϕ. So suppχϕ ` U singsuppuand supp1 χϕ ` U singsupp v.
Then `uv,ϕe `uv,χϕe `uv, 1 χϕe `v, uχϕe `u, v1 χϕeRemark 2.4.4. The approximation method is more flexible than the adjoint method.
The converse of Proposition 2.4.2 is false as the following example shows:
Example 2.4.8. In R2, let u dSx0 and v dSy0. So now suppu singsuppu x 0and supp v singsupp v y 0. Now singsuppu 9 singsupp v 0 but uv is well-defined.
We have u limδ0 δ1ψδ1x, ψ > Cª
0 R with R ψ 1. and v defined similarly. Thenwe have uv“” limδ δ2ψx~δψy~δ δ0
Now let’s discuss the convolution of two distributions. In general you cannot define this.
Proposition 2.4.3. If u and/or v (distributions) have compact support, then u v is welldefined and moreover u v v u and suppu v ` suppu supp v
Proof. welldefined: recall when v is a function `u v,ϕe `u, v ϕe, where:
v ϕy S vx yϕxdx `v,ϕx ewithout loss of generality, suppose that supp v is compact. Then since ϕ > Cª0 Rd thenv ϕ > CªRd even when v > DU and suppv ϕ ` supp v suppϕ. Thereforev ϕ > Cª0 Rd therefore `u v,ϕe `u, v ϕe is well-defined.
– 36 –
2. Distributions
2.5 Fundamental Solutions
By the previous proposition, for any u > DRd we can always make sense of u δ0 δ0 uwhich is by definition just u. If u > Cª0 Rd, then ux `δx, ue `δ0x , ue δ0 ux.We can write:
`δ0x , ue S δ0x yuydy“” uxExample 2.5.1. If V is a finite dimensional vector space, then we can write v Pi viei ifei span V .
Now suppose that Pu v where P is a linear operator from U to V , then if we areinterested in the existence of solutions to this equation follows if we know Pui ei. Then ifv > V with v P viei then u P viui solves Pu v
So the above is like saying δ0x span the whole space Cª0 RdLet’s consider P as a linear scalar partial differential operator of the form:
Pu QSαSBk
aαxDαu
And aα is a real-valued smooth function. Then Pu makes sense for u > DRdDefinition 2.5.1 (Fundamental Solution). Given y > Rd, we say Ey is a fundamentalsolution for P if it solves the equation PEy δy δ y
The application is that if we would like to solve Pu v we expect that u“” R vyEyxto be a solution (in analogy to the finite dimensional vector space example)
vx S vyδ0x ydySo we have (hopefully)
P S vyEyxdy S vyδ0x ydy vxbut these equalities need to be checked in a case-by-case basis.
2.5.1 Uniqueness
Example 2.5.2. Let U and V be finite dimensional vector spaces and P be a liner operator.Let Pu v. Let ϕ > V then:
`u,P ϕe `Pu,ϕe `v,ϕe (10)
With P V U the adjoint. Suppose there exist a set of vectors θi that span U. If for
all θi we can find Ei > V such that P Ei θi, then by (10) (replacing ϕ with Ei)
au, θif av,Eiffor all i. If our dual space could seperate points, then this would give uniqueness. Because ifu1, u2 both solve Pu v, then we must have that `u1, θie `u2, θie for all θi > U.
– 37 –
2. Distributions
Now given P define the formal adjoint as:
Pv Q
SαSBK1SαSDαaαu
note that if one of u or v is smooth an compactly supported, then `Pu, ve `u,P ve (thisfollows by integrating by parts).
Consider the fundamental solutions E
yx to P E
y δ y, then you expect a charac-terization of u in Pu v:
S vxE
yxdx S PuxE
yxdx ? S uxP E
yxdx S uxδ0x ydx uyExample 2.5.3. Let P ∂x on R. We know that
d
dxH δ0 so
d
dxHx y δ0x y δy
therefore H is a fundamental solution for ∂x.
So let’s suppose we would like to solve ∂xu u. Then by our formula we have:
ux S vyHx ydy S x
ª
vydyThe adjoint operator is just P ∂x therefore the solutions we are looking for with
P E
yx δ0x y. Just let E
yx Hy x. Then compute:
uy ? S vxHy xdx S y
ª
vxdxwhen v > Cª0 R, we have:
S ∂xuHy xdx S ux∂xHy xdx uySuppose we are interested in solving ∂xu v on a, b.
S 1a,bx∂xuxHy xdx S ux∂x1a,bxHy xdxThen since ∂x1a,b δa δb we get that the above becomes:
uy uaHy a ubHy bbecause y @ b. And so we get that uy uaR ya vdx and we have the fundamental theoremof calculus.Lecture 12 (10/8)
Today we will cover the Laplace equation (Evans 2.2).
Recall we were discussing the fundamental solution. If P is a linear operator of the form:
P QSαSBk
aαDα
– 38 –
2. Distributions
we say that Ey is a fundamental solution for P at y > Rd if PEy δy δ0x y.Once we have this, we can prove existence. Given a nice function f , then if we formally
write:
uf S fyEyxdyThen we expect that Puf f .
We also get uniqueness, we look at the adjoint of P which is characterized by:
`Pu, ve `u,P veafter integrating by parts, we get that:
Pv Q
SαSBk1αDαaαv
Then the fundamental solution can be defined as P Ex δx. So now given a nice u, wehave:
ux `u, δxe `u,P Exe `Pu, ExeThen we can recover Pu (which gives us uniqueness). This formula above we will call therepresentation formula for u.
We also have uniqueness for a boundary valued problem. This is the same idea where wewrite for x > U :
But these extra terms give us the contributions of the boundary values of u.
Remark 2.5.1. If you consider constant coefficient P. This is the case where aα are con-stants. There are two simplifications that occur.
1. if PE0 δ0 then Eyx E0x y is a fundamental solution for P at PE0 y δ0 y δy. So we get translational invariance.
2. P v PSαSBk1SαSaαDαv Pvx. Therefore if we set Exy E0x y thenP
yEx PE0x y δ0x y δxSo we will now only consider the constant coefficient P . Then Eyx E0x y andExy E0x y. Then we have that:
uf S fyE0x ydy f E0xSo for uniqueness, Exy E0x y:
`Pu, Exe S PuyE0x ydy Pu E0x– 39 –
3. Laplace Equation
3 Laplace Equation
Now we will consider P ∆. Let’s begin by deriving fundamental solutions to the Laplacian.That is solve ∆E0 δ0 in Rd for all d C 2. We already did this. We will use rotationalinvariance of ∆. That is given a rotation matrix R on Rd, then we have:
∆URx ∆uRxThen we can look for E0 E0r. So we would like a∆E0,1B0,rf aδ0,1B0,rf 1.Integrating by parts the thing on the left, we get:
And the first term can be shown to be zero by the divergence theorem(possibly?).
Lecture 13 (10/15)
Day of the midterm Lecture 14 (10/17)
Today we will spend 20-30 minutes discussing the midterm and review what we have doneso far in the course.
General Theory
1. For first-order scalar PDE’s we studied the method of characteristics. The idea is thatwe can always reduce PDE’s to first order ODE’s which then we can use ODE resultsto get existence and uniqueness. This required noncharacteristic boundary data. Inthe quasilinear case, the noncharacteristic boundary data is equivalent to the conditionthat we are able to compute all derivatives of the solution at the boundary point. Weextended this to general quasilinear k-th order systems (see Holmgien’s theorem aboutan application). We then discussed Cauchy Koweleskaya, but it isn’t the end due toHadamard’s example. He emphasized well-posedness (which asks not only existenceand uniqueness, but also continuous dependence).
2. Then we turned to constant coefficient 2nd order scalar PDE’s. We are studying thefour equations:
¢¨¦¨¤
∆u 0
ju 0
∂t ∆u 0
i∂t ∆u 0
this can be studied on a case by case basis (like in Evans), but we can study them atthe same time using distribution theory. This allows us to make use of the fundamentalsolution: Pu δy and P u δx. We gen existence for Pu f . And we get uniqueness:if u is nice (eg compact support) then ux E0 Pu, u is only defined on Ω soux E0 Pu R∂Ω
What’s to come: for the next 2-3 weeks we will focus on these equations, then we willdiscuss Sobolev spaces. We wil be studying A-priori estimates. We will be studying:
¢¦¤Pu f in Ω
u g on Γ
– 42 –
3. Laplace Equation
we first assume a solution exists, then we prove estimates on them. For existence we wouldlike `u,P ϕe `f,ϕe. This is something we can define if P is injective. Then the linearfunctional `u,P ϕe is well-defined in spanP ϕ ϕ > Cª
0 U. This strategy favors L2 typespaces, in which we can find solutions u > L2, and all derivatives will stay in L2. A Sobolevspcae is Hk u Dαu > L2, SαS B k.
3.1 Cauchy Riemman Equation
Let’s apply what we have learned to the Cauchy Riemman equation for f u iv:
∂xu ∂yv 0
∂yu ∂xv 0
This is equivalent to ∂x i∂yf 0. Note that if this is true, then each component of f isharmonic as ∂x i∂y∂x i∂y ∂2
x ∂2y ∆. Now in 2 d the fundamental solution is:
∆ 1
2πlog r δ0
Decomposing ∆, we get the fundamental solution ∂x i∂y 1
2πlog r
1
2πx~r2 iy~r2
1
2πz
Theorem 3.1.1. E0 1
2π
1
zis a fundamental solution for ∂x i∂y
Theorem 3.1.2. If ∂x i∂yf 0 for f > DU then f > CªUProof. Let z0 > U , let χ be 1 near z0 and zero outside a small ball centered at z0. Then:
∂x i∂yχf f∂x i∂yχThen since χf is compactly supported, we can use the representation formula to get that:
But this last term vanishes near z0, if we let that term be F , then we have (if F is a function):
S E0z wF wdwxdwywhich is smooth. If F isn’t smooth, then we use approximation
Corollary 3.1.1 (Morera’s Theorem). If f is C1 on U and RΓ fzdz for all closed curvesΓ, then f is holomorphic (i.e. f is a Cª solution to CR equation)
– 43 –
3. Laplace Equation
Proof. We say that f is a distribution solution if and only if `f, ∂x i∂yϕe 0 for all ϕ >
Cª0 U complex valued. Where the inner product is now defined as `u, ve R uzvzdxdy(which isn’t real valued anymore). We have that if:
0 S fz∂x i∂yϕzdxdyfor all ϕ then f is smooth.
Lemma 3.1.1.
S hz∂x i∂yχΩdxdy 1
i S∂Ωhzdx idy
By hypothesis, we have that `f, ∂x i∂yχΩe 0. Let ϕ χΩ be an approximation,then we can approximate ϕ be step functions in L1. Then `∂x i∂yf,χΩe 0 implies that`∂x i∂yf,ϕe 0 for all ϕ > Cª0 UTheorem 3.1.3 (Cauchy Integral formula).
fz0 1
2π S∂Ω
fzz z0
dz
Proof. (uniqueness in the case of u on Ω). We have:
fz0 fz0χΩz0 S δz z0χΩz0fzdz S ∂x i∂yE0z0 zχzfzdzthen we integrate by parts, when the differential operator falls on f we get zero, when itfalls on χ we use the lemma to get:
1
2π S1
z0 z∂X i∂yχΩzfzdxdy 1
2πi S∂Ω
dz
z z0
fzwhere there was a sign error.
Lecture 15 (10/22)
Today we will continue the discussion of the Laplace equation. We were looking at:
∆u f
what we did last time was we computed the (radial) fundamental solutions for ∆. That is∆E0 δ0 which had the form:
E0x ¢¦¤
1
2πlog SxS d 2
1
dαdd 2 SxSd2 d C 3
with αd the volume of the d-dimensional unit ball.
Recall that we proved:
– 44 –
3. Laplace Equation
existence of a solution uf to ∆u f for f a compactly supported continuous functionon Rd where:
ufx E0 fx uniqueness due to the representation formula. When u is compactly supported:
E0 ∆u u combined with the regularity (smooth, analytic) of E0 outside x 0 leads to smooth-
ness (and analycity) of u near x > U , ∆u 0 in U
derivative estimate: given ∆u 0 in U and Bx, r ` U then:
SDαuxS B Cαr1SαS SBx,r
SuSdy Liouvile theorem: if u is a harmonic bounded function on Rd, then u is constant.
in d C 3, any bounded solution to ∆u f in Rd is of the form u uf c where c issome constant.
see the notes for d 2
We said that u has to be a compactly supported distribution for: u E0∆u to hold.If u is a distribution solution to ∆u 0 in U . Then we truncate and approximate u bydistributions with compact support.
3.2 Boundary Value Problem
Now will try to apply the representation formula for a boundary valued problem. Supposewe are solving for u in U . So we have for x > U :
Remark 3.2.1. The same proof applies to ∆u x 0. This leaves to a nice proof of Jensen’sformula in complex analysis.
Theorem 3.2.2 (Maximal Principal). Let U be a bounded C1-domain (open and con-nected). Let u > C2U 9C0U with ∆u 0 in U then:
1. (weak maximal principal)
maxU
u max∂U
u
2. (strong maximal principal): if x0 > U and ux0 maxU u, then u is constant.
Proof. It suffices to prove the strong maximal principal. Let M maxU u. Let Bx, r ` U .Then:
M ux0 1S∂Bx, rS S∂Bx,rudSy BM
this means that u M on ∂Bx, r (because we have R∂BM u´¹¹¹¹¸¹¹¹¹¹¶C0
dSy). We can do this
for all balls in U , but this implies that u M in Bx, r. Then by connectedness, we aredonea
alet Z x > U ux M. This is open by the aboe computation and closed as it is the preimage of aclosed set, therefore it is U
– 46 –
3. Laplace Equation
Corollary 3.2.2 (Uniqueness of Dirichlet Problem). Given U a bounded C1 domain.The Dirichlet problem is:
¢¦¤∆u f in U
y g on ∂U
note that ν Du is not prescribed. If f and g are continuous and u > C2U 9CU then uis unique.
Proof. If u and u solve this, then u u is harmonic with boundary value 0. Then by themaximal principal we get that u u 0 (show B and C).
Remark 3.2.2. This uniqueness theorem is a global phenomenon!
Remark 3.2.3. The uniqueness that comes from the maximal principal implies the existenceof a representation formula.
Proof. (sketch) Given x > U with ∆u 0. If we take uS∂U > C∂U and map it to ux > R.Then this map is bounded by the maximal principal. Then we look at the linear subspaceV uS∂U of ∆u 0 in U,u > C2U 9CU. Then we use Hahn-Banach.
Note that this is almost useless, as we do not get what the formula is, just that it exists.
Theorem 3.2.3 (Harnack’s inequality). Given u > C2U 9CU, ∆u 0, u C 0. Thenon any bounded domain V ` V ` U :
supV
u B C infVu
Where C depends only on U and V
Proof. We will apply the mena value property for balls:
ux 1SBx, rS SBx,ruydy
this follows by averaging1S∂Bx, rS R∂Bx,r udS for 0 @ r @ r.
Now take r P distV , ∂U. Then take x, y > V such that Sx yS @ r. Then we can write:
ux 1SBx, rS SBx,rudz B
1SBx, rS SBy,2rudy
2dSBy,2rS SBy,2rudy B 2duy
using that Bx, r ` By,2r. This tells us:
2duy B ux B 2duythen we cover U with balls of radius r. So if we jump N balls, we get:
2dNuy B ux B 2dNuysince V is compact, we know that N is bounded and we are done.
– 47 –
3. Laplace Equation
Lecture 16 (10/24)
Recall our discussion of the Laplace equation. We got uniqueness of a solution to the Dirchletproblem for a bounded domain U
¢¦¤∆u f in U
u g on ∂U
If f, g are continuous and u > C2U 9CU, then we have uniqueness.
3.3 Green’s Function
Definition 3.3.1 (Green’s function). Given a domain U and some y > U , we say thatG, y > DU is a Green’s function for U at y if
¢¦¤∆xG, y δ0x y in U
Gx, y 0 x > ∂U
Remark 3.3.1. This definition is opposite to Evan’s definition.
Note that Gx, y E0x y is a harmonic function. So one way to find Gx, y is to takeGx, y E0x y hyx such that h solves:
¢¦¤∆hyx 0 in U
hyx E0x y for x > ∂U
So, existence theory for the Dirichlet problem implies the existence of a Green’s function.
Now, if you have a Green’s function, you have existence of a solution uf to:
¢¦¤∆uf f in U
uf 0 on ∂U
where we set:
ufx S fyGx, ydyThis has to be justified, but morally speaking, one can solve:
¢¦¤∆u 0 in U
u g on ∂U
by working with v u g. We find a Cª extension g > CªU such that gS∂U g. Then:
¢¦¤∆v ∆u ∆g ∆g in U
v 0 on ∂U
– 48 –
3. Laplace Equation
So if we solve v and put u v g, then u solves the original problem.
So we have that “homogeneous and nontrivial boundary value” can be solved by solving“inhomogeneous and trivial boundary value”
It turns out that existence to Dirichlet problem is equivalent to the existence of a Green’sfunction.
Remark 3.3.2. It turns out that G is a Cª function in U U y xTheorem 3.3.1 (uniqueness and symmetry of Green’s functions). Let U be a boundedC1 domain and suppose there exists a Green’s function Gx, y. Then (for x x y, x, y > U:
1. If Gx, y is also a Green’s function, then Gx, y Gx, y2. Gx, y Gy, x
Proof.
Gx, y S δ0x zGz, ydz S 1Uzδ0x zGz, ydz S ∆zGz, xGz, y1Uzdz
d
Qj1S ∂jGz, x∂jGz, y1Uzdz d
Qj1S ∂jGz, xGz, y∂y1Uzdz
d
Qj1S Gz, x∂2
jGz, y1Uzdz d
Qj1S Gz, x∂jGz, y∂j1Uzdz
d
Qj1S ∂jGz, xGz, y∂j1Uzdz
S Gz, xδ0z, ydz´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Gy,x
0 0
Remark 3.3.3. We required that ∂G to be continuous up to ∂U
So now if we take G G, then we see the symmetry property for any Greens’ function.Then we use this to see that Gx, y Gy, xRemark 3.3.4. Gx, is a solution to the adjoint problem.
¢¦¤∆Gx, δxGx, 0
Revised definition of Greens’ function:
– 49 –
3. Laplace Equation
Definition 3.3.2. A Greens function is a solution to:
¢¦¤∆G, y δy in U
G, y 0 on ∂U
and G, y > C1U yNow let’s turn to construction of Green’s functions on some domains. Here we will use atechnique called the method of image charges.
The idea is that the fundamental solution E0x, y models the electric potential of a unitpoint charge which is situated at y. To ensure that on the boundary that the potential iszero, we simply pretend there is an opposite charge on the other side of the boundary. Sowe would get Gx, y E0x y E0x y, where y is y reflected across the boundary
Example 3.3.1. Let U B0,1.If we look for Sx yS Sx yS then we get hyperplanes. Recall that for d C 3 we have:
E0X 1
dαdd 2 1SxSd2
so we get that
1
λd2E0x E0λx
for λ A 0
So now we can consider the hypersurface for R A 1 P x > Rd Sx yS RSx yS.
On the plane, this is the circle.
In this case for y > U :
Gx, y E0x y E0SySx ywhere y
ySyS2Proof. (that this is the Green’s function) Let y > B0,1. It suffices to check that Sx yS SySSx yS for x > ∂B0,1 if and only if SxS2 1.
But this is just:
SxS2 2x y SyS2 SyS2SxS2 2x y SyS2but the right side is:
SyS2SxS2 2x ySyS2 Sy2SSy4S SyS2SxS2 2x y 1
– 50 –
3. Laplace Equation
In Summary we have the following Greens functions:
1. For Rd x > Rd xd A 0, then if y > Rd
let y y1, . . . , yd1,yd. Then we set:
Gx, y E0x y E0x y2. For B0,1 we have y > B0,1 y
ySyS2 then:
Gx, y E0x y E0SySx yTheorem 3.3.2 (Representation formula for the Dirichlet problem). Let u > C2U9CU, then we have that:
First assume that u > CªU. Then the first and third term is zero by Gx, y 0 for y > ∂U .For general u, you can approximate u in C2U 9CU.Theorem 3.3.3 (Poisson integral formula). Given g > C∂U (for today U is the halfplane or the unit ball). If we define:
ux S νy DyGx, ygydythen u > C2U 9CU and u g on ∂U and ∆u 0
Lecture 17 (10/29)
Recall that we were discussing the Green’s function for the Laplace equation. It solved:
¢¦¤∆G, y δ y in U
G, y 0 on ∂U
For a domain U ` Rd with y > U . We found that G is unique and that Gx, y Gy, x. Sowe also have: ¢¦¤
∆Gx, δx in U
Gx, 0 in ∂U
which is the definition in Evan’s.
– 51 –
3. Laplace Equation
Theorem 3.3.4 (Poisson Integral Formula). Given U a C1-domain and u > CªU,then:
ux SUGx, y∆uydy S
∂Uνy DyGx, y´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Poisson integral kernel
uydSy
Remark 3.3.5. Note that this representation formula does not have a term with the normalof u.
Remark 3.3.6. ∆u 0 is what is usually called Poisson integral formula.
Remark 3.3.7. If we want to solve:
¢¦¤∆u g in U
u h in ∂U(11)
If a solution exists, then:
ux SUGx, y∆uydy S
∂Uν DyGx, yudS
In fact, we can define:
ux SUGx, ygydy S
∂UνgDyGx, yhydSy
it turns out that ux is indeed the solution to the Dirichlet problem (11), but we will notjustify this here.
It isn’t hard to show that ∆u g, it is difficult to show that ux xx0>∂UÐÐÐÐÐ hx0
We had examples of Green’s functions using the method of image charges.
Example 3.3.2. When U Rd xd A 0. In this case, we had:
Gx, y E0x y E0x ywith y y1, . . . , yd1,ydExample 3.3.3. When U B0,1, we used the fact that if y > B0,1 and let y SyS2y,then we have that:
∂B0,1 x > Rd Sx yS SySSx yS
for any y > B0,1. This suggested that we take:
Gx, y E0x, y E0SySx yif y 0, then we set Gx, y E0x, y 0 E0x, y
– 52 –
3. Laplace Equation
Given an integral operator:
Tfx S Kx, yfydythen K is the kernel of the integral operator. By the Schwartz kernel theorem, any linearT Cª
0 X DY is of the form Tf `Kx, , feNow we would like to compute the Poisson kernel: νyDyGx, y for the two above
examples.
Example 3.3.4. Swap x and y, set:
Gx, y E0y x E0y xνy Dy on ∂U xd 0 is just ∂yd, so we have:
∂ydE0y x ∂yd Sy xSE
0Sx yS 1
dαd yd xdSy xS 1Sy xSd1
where we used that:
E
0x 1
dαd 1
rd1
If we restrict to y y,0, then we get:
1
αdd xdSy xSdThen we compute:
∂ydE0y x 1
dαd xdSy xSdthen restrict to y y1,0, to get
1
dαd xdSy xSd . So our Poisson kernel becomes:
νyDyGx, y 2xd
dαd 1Sy xSdthere was a sign error, the final answer is right. So the Poisson integral formula for harmonicu in Rd
is:
ux S 2xd
dαd 1Sy xSdhydyRemark 3.3.8. ux hx0 as x x0 if h > Cª∂U
– 53 –
3. Laplace Equation
Example 3.3.5. For the ball, we have:
Gx, y E0y x E0SxSSy xSand again we will use that:
E
0r 1
dαd 1
rd1
then we have that on ∂B0,1:νy Dy Q
α
yα∂α
Then we have:
Qα
yα∂yαE0y x Qα
yα∂yα Sy xS 1
dαd 1Sy xSd1
SyS2 y xdαd 1Sy xSd
And we have:
Qα
yα∂yαE0SxSSy xS Qα
yα∂yαSxSSy xS 1
dαd 1SxSd1Sy xSd1
1
dαd 1SxSd2
Pα yαyα xαSy xS 1Sy xSd1
1
dαd 1SxSd2
SyS2 yxSy xSdthen if we restrict both to ∂B0,1, then we get:
1 y x
dαd 1Sy xSdfor the first term, and the second term becomes something else, and after we add them up,we get:
SxS2 1
dαd 1Sy xSdNote that if u is harmonic in B0,1, we get:
ux S∂B0,1
νy DyGx, yhydSy S∂B0,1
1 SxS2dαd hySy xSddSy
– 54 –
4. Wave Equation
4 Wave Equation
The wave equation is an equation on ϕ R1d R where the coordinates of the domain arex0, x1, . . . , xd with x0 t and our operator is:
jϕ 1
c2∂2t ∆ϕ 0
this is called the d’Alembertian operator. This is the prototypical hyperbolic equation.We always take c 1 in math courses. We have three goals:
1. find an explicit fundamental solution for j. We will figure out that there are a lot ofsymmetries of j.
2. derive an (explicit) representation formula for:
¢¦¤jϕ f, in R1d
t A 0
ϕ g, on ∂R1d
∂tϕ h on ∂R1d
(12)
3. Prove existence of a solution to (12)
4.1 1 dimension
Let’s consider the domain R11. Then we have:
j ∂2t ∂
2x ∂t ∂x∂t ∂x
So let’s take advantage of this factoring by finding the fundamental solution. Let’s make achange of coordinates u t x, v t x. Then we get that du dt dx, dv dt dx threfore
dt 1
2du dv and dx
1
2du dv this means that ∂u
∂t
∂u∂t
∂x
∂u∂x
1
2∂t
1
2∂x and
∂v 1
2∂t
1
2∂x.
Then if E is our fundamental solution, we have:
jE0 4∂u∂vE0u, v 2δ0u, vwhere we had to use the approximation method to compute δ with a change of variables. Sowe just need to solve:
∂u∂vE0 1
2δ0uδ0v
So that:
E0 1
2Hu c1Hv v2
– 55 –
4. Wave Equation
The forward fundamental solution E is defined so that E is a fundamental solution andsuppE ` t C 0. This forces c1 c2 0. Therefore the fowards fundamental solution hasthe form:
E 1
2HuHv 1
2Ht xHt x
Lecture 18 (10/31)
Let’s rectify what was messed up last time.
Theorem 4.1.1 (Change of Variables for Distributions). Suppose that Φ is a diffeo-morphism (smooth change of coordinates) from X1 to X2 (open subsets in Rd. And supposethat u > DX2, then we can define u XΦ, by for all ϕ > Cª0 X1:
`u XΦ, ϕe du, 1SdetDΦSϕ XΦ1iRemark 4.1.1. u( u XΦ is continuous and linear in u.
Proof. First assume that u > Cª0 X2,then let y1, . . . , yd Φx1, . . . , xd, then we have:
S uΦxϕxdx S uyϕΦ1y W∂x1, . . . , xd∂y1, . . . , yd Wdy
du,ϕΦ1y 1SdetDΦySiNow since Φ is a diffeomorphism, we have that we still get a test function for the input ofu
Corollary 4.1.1. δ0 XΦ 1SdetDΦSδΦ10
Let’s now return to the wave equation:
jϕ ∂2t ∆ϕ 0
this is the sign choice that people usually use in general relativity. For R11, we got:
E0 1
2Ht x c1Ht x c2
Then we introduced something called the forward fundamental solution, which requiredthat suppE ` t C 0. This is the relevant choice for solving the wave equation because wewould like to know what happens in the future. This requires jE δ0 and E 0 beforeseeing δ0. Then we get c1 c2 0, and get the unique forward fundamental solution:
E 1
2Ht xHt x– 56 –
4. Wave Equation
Our goal today is to derive a representation formula using E. In 1 1 dimension, we getthat for all ϕ > CªRd
, for all t, x > Rd
then, we get d’Alembert’s formula
ϕt, x 1
2 St
0S
xtsxts
jϕs, ydyds 1
2ϕ0, x t ϕ0, x t 1
2 Sx6
xt∂tϕ0, ydy
Note that there is a sign change in Evans due to the choice of sign of j.
Let’s turn to the derivation of the reresentation formula for any dimension (given thatwe know the forward fundamental solution). Then we will go towards a way of derivingfundamental solutions for all dimensions.
Let E be a forward fundamental solution to j in R1d. In other words we get that:
¢¦¤jE δ0
suppE ` t C 0Assume in addition that for all bounded interval
suppE 9 t > I (13)
is compact. It turns out (13) is required for the following
Lemma 4.1.1. Let u > DR1d, and suppu ` t > L,ª, then u E is well-defined.
Proof. (sketch) Suppose that u is a Cª function and E is a function. Let’s make sense ofu Et, x, this is just:
u Et, x U us, yEt s, x y´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶compactly supported
dsdy
Proof. (real) Given an interval I a, b with a, b > R 8 ª, let χI be a smooth functionthat is 1 on I and 0 in s B a1 and s C b1. Then we will consider χIuE for all bounded I.
Now for ϕ > Cª0 Rd, we have:
`χIu E, ϕe ? `u E, χIϕe du, E χIϕ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
need to show is a test function
iActually we will prove that χL,ªEχIϕ. For functions, we had:
f gy S fx ygxdxthen we had suppf g ` supp f supp g.
– 57 –
4. Wave Equation
Now suppϕχI ` t > I, then we would like to show that supp f g ` L,ª. NowE χM,ªE 1 χM,ªE. Now:
suppχM,ªE χIϕ ` t CM t > I
Yeah...I got a little lost, this is one of those things that is easier to just work out by yourself.
Theorem 4.1.2 (Uniqueness of forward fundamental solutions). Let E be a forwardfundamental solution with the property (13) and let E
be any forward fundamental solution
to:
¢¦¤jE
δ0
suppE
` t C 0
Then E
E
Proof. We have:
E δ0 E jE
E0 jE
E E
jE E
δ0 E
Theorem 4.1.3. Let ϕ > CªRd, t, x > Rd
. Then:
ϕt, x E δtϕδt0 ∂tEt ϕ∂t0 Et jϕ1tC0
Proof. We have:
ϕ1tC0 δ0 ϕ1tC0 jE ϕ1tC0 ∂2
tE ϕ1tC0 ∆E ϕ1tC0 ∂tEt ∂tϕ1tC0 ∂tE ϕδt0 E ∆ϕ1tC0 E ∂2
t ϕ1tC0 E ∂tϕδt0 E ∆ϕ1tC0 ∂tE ϕδt0
Looking at the representation formula and pretend everything is a function, we have:
E jϕ1tC0t, x S ª
0S jϕs, yEt s, x ydyds
Then if you believe that suppE ` SxS B t, then you are integrating over the triangle belowthe thing.
Let’s specialize to d 1, then E 1
2HtxHtx, then let’s look at the representation
formula. Then we have:
E ψt, x 1
2 S S ψs, yHt s x yHt s x ydsdy 1
2 U ψs, ydsdy
– 58 –
4. Wave Equation
where we are integrating over s, y t s C x y, t s C x y, s C 0. This is telling usthat x t s C y C x t s. So we get:
1
2 St
0S
xtsxts
ψs, ydydsThen if we apply this expression to the last term we get:
E jϕ 1
2 St
0S
xtsxts
jϕdyds
and we get that:
E ∂tϕδt0 1
2 Sxt
xt∂tϕ0, ydy
and:
∂tE ϕδt0 ∂t1
2 Sxt
xtϕ0, ydy 1
2ϕ0, x t ϕ0, x t
Lecture 19 (11/5)
We were discussion the wave equation:
jϕ 0
In R11 we were able to compute the fundamental solution Et, x 1
2HtxHtx. In
R1d, given a forward fundamental solution E with the condition that for all bounded inter-vals suppE 9 t, x t > I is compact. Then we were able to construct the representationformula for the Cuachy problem:
¢¦¤jϕ f in R1d
ϕ,∂tϕ g, h in 0 Rd(14)
then:
ϕt, x E ∂tϕδt0 ∂tE ϕδt0 E jϕ1R1d
If everything is a function, then we have:
E f S Et s, x yfs, ydsdythen we are just integrating over the inverted cone in R1d. This is known as “finite speedof propagation”.
– 59 –
4. Wave Equation
Remark 4.1.2. In the case of wave equations, we can take the representation formula towrite down the solution to the initial value problem. For t, x > R1d
, define
ϕt, x E hδt0 ∂tE gδt0 E f1R1d
Then the claim is that for nice enough f, g, h (for instance all smooth) then ϕt, x does solvethe initial value problem (14).
The goal today is to find the foward fundamental solution in R1d for d C 1. To do this,we will exploit the symmetries of j to narrow down the possible candidates for E.
Symmetries of j, we would like to find L R1d R1d such that:
jϕ XL jϕ XLOne example is would be translations. If x0 > R1d, then we have:
jϕt, x x0 jϕt, x x0This is not helpful for E because jE δ is not invariant under translations.
We are instead interested in symmetries that fix 0,0. These are called the Lorentztransformations. These are:
1. (rotations), R > SOda. Then jϕt,Rx jϕt,Rx.2. (reflections) L t, x1, . . . , xd t, x1, . . . , xd or we would reflect a spacial coordinate.
Then since all derivatives are second order, then this is invariant.
3. (Lorentz boosts) Lorentz boost with speed v and direction x1 is the formula:
t, x1( t vx1º1 v2
,x1 vtº
1 v2
Any other Lorentz transform is given by a composition of the above. Another way tocharacterize these is L is a Lorentz transformation if and only if it preserves:
t2 x12 xd2
Another property of the Lorentz transformation is the behavior of j under scaling. For allλ A 0, given the scaling transformation t, x( t~λ,x~λ, then we have:
jϕt~λ,x~λ 1
λ2jϕt~λ,x~λ
Definition 4.1.1 (Homogeneity). We say a function ϕ Rn 0 R is homogeneous of
degree a if ϕx~λ 1
λaϕx
a RRt RtR I and detR 1
– 60 –
4. Wave Equation
Example 4.1.1. If n 1 and ϕ xa for x A 0. Then this is homogeneous of degree a.
Remark 4.1.3. If ϕ is homogeneous of degree a, then jϕ is homogeneous of degree a 2
For distributions consider:
`ϕ~λ, ψe S ϕx~λψxdx λdS ϕyψλydy aϕ,λdψλfDefinition 4.1.2. For ϕ > DRn or DRn 0 is homogeneous of degree a if for allψ > Cª
0 Rn or Cª
0 Rn 0 we have:
`ϕ,λnψλe λa `ϕ,ψeExample 4.1.2. δ0 on Rn, then the degree is n.
So if we are looking for jE δ0 and j lowers the degree by 2 and δ0 has degree 1 d,it is reasonable to look for a homogeneous E of degree d 1
Example 4.1.3. Consider ϕx xa for x A 0 and 0 for x @ 0. Then ϕ is a homogeneousfunction of degree a on R 0. When a A 1, then ϕ can be extended as a homogeneousdistribution on the whole R.
`ϕ,ψe S ª
0xaψxdx
when a 1, no homogeneous distribution agrees on R.
Lemma 4.1.2. If ϕ is a homogeneous distribution of degree a A .n in Rn 0, then thereexists a unique distribution ϕ > DRn that is homogeneous of degree a and agrees with ϕ inRn 0.
Now we are looking for E.
Motivated by the homogeity of j and δ0 we may assume that E is a homogeneousdistribution on R1d of degree d1. By the previous lemma, it is enough to constructE in R1d 0,0
s2t, x t2 SxS2 is invariant under all Lorentz transformations, we will assume that
E
kinda χ
d12 t2 SxS2
We wanted the forward fundamental solution. So we rather define E as:
E 1t>0,ªχd1
2 t2 SxS2to make sure that this is well-defined, it is good to require that χ
d12 vanishes for s @ 0.
Claim 4.1.1.
jE 0 in R1d 0,0
– 61 –
4. Wave Equation
Theorem 4.1.4. If ϕ > DRn and suppϕ ` 0, then ϕ Pα cαDαδ0 for at most finitely
many cα x 0
Using this theorem, we have that jE Pα cαDαδ0 and jE is homogeneous of degree d1.
This implies that jE c0δ0. So the last thing you have to check is that c0 x 0 One way todo this is to compute, which is possible, but difficult). But this fact is not difficult, you justneed the uniqueness of jϕ 0 in R1d and suppϕ ` R1d
then ϕ 0 (this can be proved by
Fourier transform or integration by parts).
Remark 4.1.4. χd1
2 is a distribution so the composition of it with t2 SxS2 should beunderstood by approximations.
Let’s now prove the claim.
Proof. letting χ χd1
2
j1tC0χd1
2 t2 SxS2 1tC0 j χd12 t2 SxS2 1tA0∂t2tx1t2 SxS2 Q
j
∂j2xjx1t2 SxS2
lots of computation. We have to use Euler’s identity:
sχs d 1
2χs
this is because χ is a homogeneosus distribution of degreed 1
2 1
d 1
2
Lecture 20 (11/7)
Recall that we were aiming to derive the fundamental solution for j in R1d. what wewere doing was looking at symmetries of j, that is finding L R1d
R1d linear such thats2t, x t2 SxS2 s2Lt,Lx. And we were looking at homogenity of j and δ0. Andlooking at the forward condition suppE ` R1d
. Then our guess was the E is of the form:
E 1R1d
χd1
2 t2 SxS2with χ
d12
homogeneous of degreed 1
2in DR with suppχ
d12
` 0,ª.We computed that jE 0 outside the origin, i.e. jE ` 0,0 and is homogeneous of
degree d 1 which tells us that jE cδ0. Then it is not too difficult to show that c x 0,but refer to the class notes to compute it.
How does χd1
2 look like? Let’s construct a family of distributions, χa
for a > R such
that it is a distribution on R, it is homogeneous of degree a and suppχa` 0,ª.
– 62 –
4. Wave Equation
For a A 1, then there is a obvious candidate:
χaHxxa
To extend this to a B 1, we will differentiate our distribution. Observe that by computation(for a A 1) we have (you have to check this):
d
dxχa aHxxa1
aχa1
Then let’s just define: χa by using:
χa
1
a 1
d
dxχa1
this is good, and will give nice distributions χa
which will happen if a 1,2,3, . . . . That’swhy we need to be a little smarter, let:
χa caHxxa
with ca canceling the possibly zero. If we set:
χa
1
Γa 1Hxxafor a A 1 where:
Γa S ª
0xaex
dx
x
then we wouldn’t have this problem. To see this:
d
dxχa
1
Γa 1aHxxa1
ΓaaΓa 1χa1
Then since Γa 1 aΓa, so the above is just χa1
. So for a B 1, we define:
χad
dxχa1
dk
dxkχak
for k > N large enough so that a k A 1.
Example 4.1.4. χ0Hx. And:
χk
δk10 x
Also we have:
χ1~2
1ºπHxx1~2
this uses the computation that Γ1~2 ºπ. Therefore we have χ1~2k
1ºπHxx1~2k
– 63 –
4. Wave Equation
Remark 4.1.5. Up to a constant, this is the unique distribution with the three desiredproperties.
Remark 4.1.6. jE 2πd12δ0. This means that:
E 1
2πd12
1R1d
χd1
2 t2 SxS2
Proposition 4.1.1 (Properties of E).
1. Finite speed of propagation (weak Huyghen’s principal) This follows from thefact that:
suppE ` t, x > R 0 B SxS B t
With the representation formula that we just derived:
ϕt, x E ∂tϕδt0 ∂tE ϕδt0 E jϕ
with ϕ > Cª0 R1d
and all these convolutions involve integration on inverted coneswhich have finite area at time equal to zero.
2. Strong Huyghens principle when d is odd and C 3. Note that:
E c1R1d
χd1
2 t2 SxS2which is only supported it t, x t SxS which is the boundary of the light cone.
Remark 4.1.7. When d 1, we get E 1
21R11
Ht2 SxS2 1
2Ht xHt x
Remark 4.1.8. IF d 2 then we have:
E 1
2π1R12
1
t2 SxS21~2
If you plug this into the representation formula, you will get Poisson formula for the 2dimensional wave equation.
Remark 4.1.9. For d 3, then we get:
E 1
2π1R13
δ0t2 SxS2the easiest way to compute this is by the approximation method, with
1
λϕs~λ δ0 as λ 0,
so we get:
1
4º
2πtdSC
0
with C
0 t, x t SxS. Plugging this into the representation formula gives you the Kir-choff formula in R13.
Remark 4.1.10. In Evan’s book, for chapter 2.4, there is an alternative derivation of therepresentation formula for j using a very different method (the method of spherical means).
– 64 –
5. Fourier Transform
5 Fourier Transform
Let’s motivate why the Fourier transform takes the form that it does. This will be a brieftwo lecture crash course on the Fourier transform.
The Fourier transform can be thought of as a change of basis in the space of functions.One way to understand fx is a a description of f in the basis that consists of δ0x y forall y > Rd. We can do this because we can write:
fx S fyδ0x ydyThis is not good for understanding ∂x or any Dα. It is reasonable to ask for a basis that di-agonalizes out basis with respect to our linear transformation of differentiation. The Fouriertransform selects a basis eiξxξ>Rd .
An important property of eiξx is that:
∂jeiξx
iξjeiξx
The Fourier transform, Ffξ is the change of basis formula such that:
f S aξeiξxdξwith:
Ffξ aξWe need to compute the transitional matrix m so that:
δ0x y S my, ξeixξdξNote first that:
S my, ξeixξdξ δ0x y S m0, ξeiξxydξ S m0, ξeiξyeiξxtherefore my, ξ m0, ξeiξy.
Note that:
0 xjδ0x S xjm0, ξeixξdξ iS m0, ξ∂ξjeixξdξ iS ∂ξm0, ξeixξdξWe then conclude that m0, ξ c. And we have already dedcued δ0x y R ceiξyeiξxdξ.Now we have:
fx S fyδ0x ydy cS S fyeiξyydyeiξxdξ
– 65 –
5. Fourier Transform
Therefore we have:
Ffξ dS fyeiξydyso all we need to do is to determine the constant c.
We want to determine c, where c m0, ξ c. In other words we want to determine:
δ0x cS eixξdξ
Note that
δ0x δ0x1δ0x2δ0xdAnd note that the above integral can be written as:
cS ex1ξ1dξ1S ex
dξddξd
So if we just compute c1 for
δ0x c1S eixξdξ
in R, then cd cd1
Claim 5.0.1. c1 2π1
Proof.Method 1:
limε0S eεSξSeixξdξ lim
ε0S
ª
0eξεixdξ S
0
ª
eξεixdξ
limε0
1
ε ix
1
ε ix
1
ilimε0
1
x iε
1
x iε
2 limε0
ε
x2 ε2 2πδ0
Lecture 21 (11/12)
Recall our discussion on the Fourier transform. We motivated the Fourier transform as achange of basis from δ0x yy>Rd to a new basis eiξxξ>Rd . And we have:
S fyδ0x ydy fx S aξeiξxdξ
– 66 –
5. Fourier Transform
wo we would like to relate fy with aξ. The heart of the matter is to write:
δ0x cS eixξdξ
we had one way to compute c 2πd. Let’s compute this a different way, using an algebraicmethod.
Let’s start with the knowledge:
δ0x cS eiξxdξ
for d 1. Given any function f , we have:
f0 S fyδ0ydy cU fyeiξydydξLet Âfξ R eiξyfydy. So we have
f0 cS ÂfξdξSome properties:
1. xeiξx i∂ξeixξ
2. ∂xeiξx iξeixξ
So if x ∂xf 0, then we get that i∂ξ ξ Âf 0. This is easy to solve, and we get:
fx f0e 12x2
Let’s normalize f0 1 to get fx e 12x2 , this gives us:
Âfξ Âf0e 12ξ2
but we have:
Âf0 S fydy S e12SyS2dy
º2π
Therefore Âfξ º2πe12SξS2 . Therefore:
1 f0cS Âfξdξ º2πcS e12SξS2dξ 2πc
therefore c c 2π1
Let’s move to a rigorous derivation. Let’s begin with some comments about complexvalued distributions. Let Cª
0 U C be the set of smooth, compactly supported, complexvalued functions in U . We say uj u if and only if Ruj u and Iuj u.
Now given u, v > Cª0 U C, then we let `u, ve R uvdx be the Hermitian product.
– 67 –
5. Fourier Transform
Definition 5.0.1 (Conjugate Linear Complex Distribution). We say a complex valueddistribution, u, is a continuous conjugate linear functional on Cª0 U C almost everywhere,i.e.
uϕ ψ uϕ uψ, ucϕ cuϕThis generalizes uϕ `u,ϕe when u > Cª0 U CRemark 5.0.1. Equivalently, u > DU C if and only if u v iw with v,w distributions.
Definition 5.0.2. Given f > Cª0 Rd C, let:
Ffξ Âfξ S fyeiξydyLet us introduce F in the following way
`a, be2πddξ S abdξ2πd
then define:
`Ff, ae2πddξ `f,FaeWe can compute this by:
`Ff, ae2πddξ U fyeiξyaξ dξ2πd S fyS aξeiξydξ2πddy
Then we will define:
Faf ax S aξeiξx dξ2πd
we will soon prove that F F1
Lemma 5.0.1.
1. if f > L1, then Ff is well-defined and YFfYLª B YfYL1
2. for all x, y > Rd,
Ffxξ eixξFfξFfξ η Feiηfξ
3. if f, ∂jf > L1, then F∂jf iξjFfξ4. if f and xjf > L1, then Fxjf i∂xiFfξ
– 68 –
5. Fourier Transform
Remark 5.0.2. By the Fourier transform, regularity of f is equivalent to decay of FfRemark 5.0.3. Fax 2πdFax
We would now like to build a space such that it is closed under the Fourier transform(note that test functions and L1 functions both won’t work).
Definition 5.0.3. Schwartz class on Rd
SRd C ϕ > CªRd
C ¦α,β > Nn, Yxβ∂αϕYLª @ªwe say that ϕj ϕ in S if and only if:
supx>Rd
SxβDαϕj ϕS 0
for all α,β
By our lemma, we have that S is closed under the Fourier transform, and the adjoint ofthe Fourier transform. Therefore F S
S can act on the dual space of the Schwartz class.
Definition 5.0.4 (Tempered distributions).
SRd
C u continuous, conjugate-linear functions on SRd,C. Where we say u is continuous ϕj ϕ in S if uϕj ϕϕ
So we get F S S by `Fu,ϕe2πddξ `u,Fϕe and F
S S (Defined similiarly,
no time).
Remark 5.0.4. All test funcitons are Schwartz functions but the Gaussian is a Schwartzfunction, but not a distrubution.
Also S ` DRd C, but this is a strict subset.
Remark 5.0.5. Usually, the way to compute Fu when u > S , one approximates u by uj > L1
(in the topology of S ). And
Fuj S ujyeiξydyTheorem 5.0.1 (Fourier Inversion, Plancherel).
1. (Fourier Inversion on S) For f > SRd,C, FFf FFf f2. (Plancherel’s theorem) for f, g > SRd C, we have:
S fgdx S FfFg dξ2πdIn particular:
YfY2L2 YFfY2
L2 dξ
2πd
and F extends uniquely to F L2 L2 by density.
– 69 –
5. Fourier Transform
3. (FI in S ) For u > S Rd C,FFf FFu uProof. It suffices to show that FF id in S or S . Because the other FF id follows fromF
2πdF,
(1) in S, f > S, Ff > F . So we have:
FFf S Ffξeiξx dξ2πd
limε0
S FfξeεSξ1SSξdSeiξx dξ2πd limε0
U fyeiξyeεSξ1SSξdSeiξx dξ2πddy limε0
S d
Mj1S eεSξj Siξjxy2πddξ fydy
S δ0x1 y1δ0xd ydfydy fx
Where we used the following fact that we proved last time
limε0S eεSξSeiξxdξ δ0x2π
(2) (sketch) FF id on S
1, This key point is to justify that:
δ0 F1
There are two steps to show this. First ∂ξj1 0 implies that xjF1 0 for all j, With alittle more argument, this reduces to computing the constant cδ0 F
1. This follows bycomputing:
1 be12 SxS2 , δ0g 1
cbe12 SxS2 ,Fg 1~c
Remark 5.0.6. The Fourier inversion formula and the Plancheral theorem is all aboutfinding a way to justify:
δ0 S eiξxdξ
2π
Remark 5.0.7. The Fourier transform maps convolutions to products (and vice-versa)
Ff g FfFgAnd
a 2πddξ bη S aξbη ξ dξ2πdThen:
F1a 2πddξ bη F1aF1b
– 70 –
5. Fourier Transform
Remark 5.0.8. Let L Rd Rd be an invertible linear map. Then:
Ff XLξ detL1FfL1tξ
Lemma 5.0.2. You can compute the Fourier transform of any Gaussian. If A is a symmetricpositive definite matrix, then:
Fe12 xtAx 2πd~2SdetAS1~2e12 ξtA1ξ
This is the end of our discussion of the Fourier transform.Lecture 22 (11/14)
5.1 Applications
The goal today is to give some applications of the Fourier transform to the study of evolu-tionary PDE’s. The main ideas to get across are:
1. using the Fourier transform to solve homogeneous evolutionary PDE’s. Like PDEs ofthe type ∂t fu 0
2. Duhamel’s principle: using the homogeneous solution to solve the inhomogeneous pde.
3. discuss the Fourier transform of the foward fundamental solution. The main exampleto study is the heat equation
¢¦¤∂t ∆u f in R1d
t A 0
u g in t 0and the Schrodinger equation
¢¦¤i∂t ∆u f in R1d
t A 0
u g in t 0these and the wave equation are worked out in the notes.
Let’s begin by studying the homogeneous equation:
¢¦¤∂t ∆u 0 in R1d
u g on t 0From now on, the Fourier transform only relies of the space fourier transform, and F t willmean the time-spacial Fourier transform.
Letting Âut, ξ Fut, ξ. So we get the new problem
¢¦¤∂t SξS2Âut, ξ 0
Âu0, ξ Âgξ (15)
so solving this first order ODE, we get:
Âut, ξ ÂgξetSξS2– 71 –
5. Fourier Transform
Proposition 5.1.1 (Solvability for the heat equation).
1. (existence) for g > L2, then there exists a solution u > Ct0,ª L2 to (15) such thatYut, YL2 B YgYL2
2. If u,u are solutions to (15), u,u > Ct0,ª, L2, then u ua
5.1.1 Duhamel’s Principle
∂t Atu fwhere At PN
j0 aαt, xDα with Dα not containing ∂t. This is a first-order evolutionaryequation.Suppose that we know how to solve the homogeneous problem. In other words,given a space X of distributions on Rd and there exists the solution operator St, sgwhere g >X and t C s is such that:
∂t AtSt, sg 0
Ss, sg gFor fixed s, St, sg > Cts,ª,X.
Then we can compute:
∂t At1s,ªtSt, sgWe know that: At1s,ªtu 1s,ªtAtu and we have:
∂t1s,ªtu δ0t su 1s,ªt∂tuSo we get:
∂t At1s,ªtSt, sg δ0t sSt, sg δ0t sgSo the affect of initial data is an instantaneous forcing term.
Now given any F t, x R F s, xδ0t sds, this suggests that we want to define
uF t, x S St, sF s, x1s,ªtds S t
ª
St, sF s, xdsThis is called Duhamel’s Formula This computation tells us that:
∂t AtuF S ∂t At1s,ªtSt, sF9s, ds S F s, xδ0t sds F t, xause the formula and Plancherel, then apply ODE uniqueness
– 72 –
5. Fourier Transform
Usually, X will have a norm defined on it, and we will have YSt, sgYX B CYgYX . In thiscase if F > L1
t 0, T ,X (that is YF t, YX is absolutely integrable) then uF > Ct0, t,X.And moreover, if suppF ` 0 B t B T, then
uF St
ª
St, sF sds S t
0St, sF sds
and UF 0, x 0.
Let’s now apply Duhamell’s principle to:
¢¦¤∂t SξS2Âut, ξ 10,ªt Âft, ξÂu0, ξ Âgξ
this tells you that:
ÂuF t, x S t
ª
St, s10,ª Âfsds S t
0etsSξS2 Âfsds
this is supposed to give us the solution to the inhomogeneous equation
¢¦¤∂t SξS2ÂuF 10,ªt ÂfÂuF 0, ξ 0
Then we get that:
Âut, ξ etSξS2Âgξ S t
0etsSξS2 Âfs, ξds
Alternatively we could have done the following: we could compute:
F1etSξS2Âg F1etSξS2 g 1º
4πtd S e
SxyS2
4t gythen we can apply Duhamnels principal directly to ∂t ∆ to get:
ut, x δt,0g S t
0St, sfsds
which solves the inhomogeneous problem
¢¦¤∂t ∆u fu g
Theorem 5.1.1.
1. (existence) for g > L2 and f > L10, T L2, there exists a solution to the inhomoge-neous equation such that Yut, YL2 B YgYL2 R t0 Yfs, YL2
a
anotation: u > Lpt I;X iff t( Yut, YX > LPt I
– 73 –
5. Fourier Transform
2. (uniqueness) If u and u are in Ct0, T ;L2 and the inhomogeneous equation thenu u
The crux of Duhamels’ formula is:
∂t At1s,ªtSt, sg δ0t sgWe can get the forward fundamental solution via the Fourier transform. We have:
∂t ∆E δ0tδ0xsuppE ` t C 0
This formula tells us that we must have that:
E 10,ªtSt,0δ0xRecall that Fδ0 1. This means that:
Âu ÆSt,0δ0should solve
¢¦¤∂t SξS2Âut, ξ 0
Âu0, ξ 1
but we know that etSξS2 solves this, therefore we get that:
E 10,ªtF1etSξS2 10,ªt 1ºrπt
de
SxS2
4t
Remark 5.1.1. E is smooth outside of the space time origin
Remark 5.1.2. E is analytic outside t 0Remark 5.1.3. For any u > S1R1d with suppu ` t C L , then u E is well-defined.
Now let’s discuss the space time Fourier transform of the forward fundamental solution.Let’s start with:
∂t ∆E δ0
let’s take the Fourier transform to get:
iτSξS2Ft,xE 1
provided that iτ SξS2 x 0, we have
F t,xE 1
iτ SξS2– 74 –
6. Energy Method and Sobolev Spaces
This isn’t quite enough, we also need to use the support property suppE ` t C 0.
We can approximate E by eεtE as ε 0. Now we can try to compute:
∂t ∆eεtE eεt∂t ∆E εeεtE δ0t, x εeεtE
putting these together, we get:
∂t ∆ εeεtE δ0t, xthen take the Fourier transform of this to get:
iτ SξS2 εFt,xeεtEτ 1
for all ε A 0, but note that:
F t,xeεtE 1
iτ YxiS2 εso now:
Ft,xE limε0
Ft,xeεtE limε0
1
iτ SξS2 εRemark 5.1.4. Analyticity in the lower (or upper) half-plane of the Fourier transformcooresponds to the original distribution in t C 0 or t B 0Remark 5.1.5.
Lemma 5.1.1. Let a > H a > C Ia C 0, then:
limε0
F1t 1
τ iε a i10,ªteiat
Assuming this lemma, we see that
E F1t,xFt,xE lim
ε0F
1t,x 1
it iε SξS2 F
1 10,ªteSξS2tLecture 23 (11/19)
6 Energy Method and Sobolev Spaces
Now we will move on to discussion on the energy method and Sobolev spaces. This willfollow Evans chapter 5.
– 75 –
6. Energy Method and Sobolev Spaces
6.1 Energy Method
This is the third general way to trying to understand 2nd order PDEs. This is glorifiedintegration by parts.
Example 6.1.1 (Laplace equation). We will focus on proving a-priori estimates. Bywhich we will assume there is a regular solution to the PDE, then we will prove inequalitiesabout norms of the solution with respect to the data. Let’s look at the Dirichlet problem
¢¦¤∆u f in U
u g on ∂Ω(16)
Proposition 6.1.1. Suppose that U is a C1 bounded domain and u > C2U, f > C0U,g > C0∂U. Then the solution, u, to (16) is unique.
Proof. We have already seen a proof, but here we present a different proof.
Let u1, u2 be two solutions to (16). Then if we let v u2 u1 > C2U, then this solvesthe homogeneous equation:
¢¦¤∆v 0 in U
v 0 on ∂U
The key idea is: multiply the equation by v, integrate over U , then integrate by parts.
Then we get the inequality via rearrangement, Cauchy Schwartz, and Minkowski’s inequality.
Another main thing we do in the energy method is: to apply operators Y to u, then look atP Y u , then apply the process as above.
Notice that for any ∂j, if ∂t∆u f , then ∂t∆∂ju ∂jf , this is because ∂j commuteswith the heat operator. So if we apply the same method to this resulting equation, then wecan get estimates:
Remark 6.1.1. For the wave equation, ju f , a “good” multiplier is ∂tu. This gives usthe physical conservation of energy.a
6.2 Sobolev Spaces
From the energy method, you expect to be able to control something like:
YDαuYL2
One natural question is: can we say other things about the solution from the control of thesekinds of norms?
We know that if f > C1c R, then by the fundamental theorem of calculus
fx S x
ª
f xdx B Yf YL1
Now if x > I (a compact interval) and supp f ` I, then we have:
supISf S B Yf YL1I B Yf YL2ISI S1~2
by Holder’s inequality. Now it turns out we can generalize this to multiple dimensions, whichare known as Sobolev inequalities.
Definition 6.2.1 (Sobolev Space). Given U a domain in Rd, k > ZC0, 1 B p Bª, then welet:
W k,pU u > LpU Dαu > LpU ¦ SαS B kathis multiplier makes sense due to Lagrangian mechanics
– 77 –
6. Energy Method and Sobolev Spaces
where Dα are taken in the sense of distributions.
We associate a norm:
YuYWk,pU QSαSBk
YDαuYLpU
So Sobolev inequalities will look like
YDαuYLq B YuYWk,p
Remark 6.2.1. We want a “function space” framework for converting a-prioi estimates toexistence results.
In linear algebra, given a n m matrix L and we are interested in solving Lu f , thenthe existence of u such that Lu f is equivalent to `f,ϕe 0 for all ϕ such that Lϕ 0(which is equivalent to ImL kerLÙ). It turns out the Sobolev spaces are the idealspace to do the same thing.
Let’s introduce a little more notation. If ϕ > Cª0 U then ϕ > W k,pU for all k, p. Inother words Cª0 U ` W k,pU, and we can consider the completion of Cª0 U in W k,pUwith respect to Y YWk,pU. The resulting space will be denoted W k,p
0 U (the completionof test function in the Sobolev space.
In general, this is a strictly smaller subset, and you should think of these function aselements in the Sobolev space whose boundary value on ∂U is zero.
The case p 2 is of importance (these are the bounds we get from the energy method).We let HkU W k,2U. And we often write Hk
0 U W k,20 U.
Proposition 6.2.1 (Completeness of Sobolev Space, Norm Equivalence To FourierTransform).
1. For all k, p, W k,pU is complete under YYWk,pU. In other words W k,pU is a Banachspace.
2. There is a characterization of HkRd in terms of the Fourier transform, that is:
YuYHkRd Y1 SξSkÂuξYL2ξ
where A B if and only there exists C A 0 such that A B CB and B B CA
Now let’s discuss some basic properties of W k,pU.1. approximation theorem
2. extension theorems
3. trace theorem
– 78 –
6. Energy Method and Sobolev Spaces
Lemma 6.2.1. Given y > Rd, let τy ux ( ux y viewed as a linear operator. Then τyis continuous on Lp as a function of y (implicitly, this means u > Lp). In other words:
limy0
Yτyu uYLp 0
Proof. See Evans appendix.
Corollary 6.2.1. For all u > L,, given a mollifer ϕ, then:
limε0
Yϕε u uYLp 0
Proof. We have:
\S 1
εdϕy~εux ydy ux\
Lp YS ϕzux εzdz S ϕzuxdzYLp YS ϕzux εz uxdzYLpB S ϕzYτεzu uYLpdz
take ε 0 and apply the dominated convergence theorem.
Theorem 6.2.1 (Approximation). By the same token if u >W k,pRd, then ϕε u u inW k,pRd. And we know that ϕε u > CªRdProposition 6.2.2. If u > W k,pU, then there exists uε > CªU such that uε u inW k,pUProof. Consider a sequence of open sets Vn such that Vn is bounded, Vn is compact, Vn ` Vn1
and ª
n1 Vn U . Every open covering has a smooth partition of unity χn subordinate to Vn.
By this we mean that χn > CªRd such that suppχn ` Vn and Pª
n1 χnx 1 and forall x > U , only finitely many terms in the sum are nonzero. And 0 B χn B 1
Then we can split u Pª
n1 χnu which can be defined on Rd. Then the idea is to approxi-mate each χnu >W k,pRd by extending it by 0 outside Vn. Then you take an approximation:
vn ϕεn χnu
such that Yvn χnuY B δnε. Then we let vε Pn vn where δn is chosen so that P δn 1, then
Yvn uYWk,p
ª
Qn1
Yvn χnuYWk,pRd Bª
Qn1
δnε
, which is bounded by ε.
Proposition 6.2.3 (Approximation of elements in CªU). If U is a C1-domain andu >W k,pU, then there exists a sequence uε > CªU such that uε u in W k,pU.Proof. (sketch) It was a picture.
– 79 –
6. Energy Method and Sobolev Spaces
Lecture 24 (11/21)
We will continue the discussion on Sobolev spaces today. Recall we say that u >W k,pUif:
YuYWk,p QSαSBk
YDαuYLp @ªWe then approximated elements of the Sobolev space that were better behaved. If u >
W k,pRd, with ϕ > Cª0 U, R ϕ 1, ϕε εdϕ~ε, then:
ϕε u u
in W k,pRd. Also if u >W k,pU then there exists a sequence uj > CªU such that uj uin W k,pU.
We also have that if u >W k,pU with U a C1 domain, then there is a sequence uj > CªUsuch that uj u in W k,pU.Proposition 6.2.4. If u >W k,pRd, k is a non-negative integer, 1 B p @ª, then χx~Ruu in W k,pRd as R ª (for χ a smooth function that is 1 on B0,1 and 0 outside B0,2)Proof. For k 1 (the others are similar). We need to show that:
YDχ~Ru uYLp Yχ~Ru uYLp 0
as R ª. To show this we right the first term as:
χ~RDu Du Dχ~Ruwhich can both be shown to go to zero by simple estimates.
Corollary 6.2.2. Cª0 Rd is a dense subset of W k,pRd(This can’t be right...we defined W k,p
0 to be the completion of test functions, but by thiscorollary, this means that W k,p
0 W k,p)
Theorem 6.2.2 (Extension). Suppose U is a bounded domain with ∂U a Ck boundary.Let V be an open set such that V a U . Then there exists a linear operator E W k,pU W k,pRd such that:
1. EuUU u
2. suppEu ` V for all u >W k,pU3. YEuYWk,pRd B CYuYWk,pU where C depends on k, p, V,U .
– 80 –
6. Energy Method and Sobolev Spaces
Proof. (sketch)Step 0: it suffices to consider u > CªU (by approximation methods).
step 1: Reduce the result to the following: there exists an extension operator E onU B0,19Rd
and V B0,2, defined for u > CªU and suppu ` B0,19 xd C 0 andYE uYWk,pRd B CYuYWk,pU.
Here we just argue by picture, I did not draw the picture...
Step 2: proof of the claim in part 1. The key is to use reflection. Let x x1, . . . , xd1then define:
ux, xd ¢¦¤ux, xd C 0
ux,xd, xd @ 0
This is a continuous extension. However, can we extend so that Dux,0 Dux,0?Can we do this for all derivatives? The key part is to show that the normal derivatives agree:
∂lxdux,0 ∂lxdux,0For l 0, . . . , k. So let’s introduce α0, . . . , αk, β0, . . . , βk. Then define:
ux, xd ¢¦¤ux, xd, xd C 0
Pkl0αlux,βlxd, xd @ 0
with αi > R and βi A 0. So now if we compute, for xd @ 0:
∂lxdux, xd ∂lxd k
Qm0
αmux,βmxd k
Qm0
αmβml∂lxdux,βmxdSo now if we want the two limits as xd 0 (from both sides), we get the following linearsystem of equations for β and α:
1 α0 αk
1 β0α0 β0αk
1 β0kα0 βkkαkWhich is just the matrix equation:
1
1
β00 βk0
β0k βkkα0
αk
which is just the Vandermond matrix, which has determinant:
M0Bi@jBk
βi βj
– 81 –
6. Energy Method and Sobolev Spaces
so if we choose βi’s to be pairwise distinct, then there exists α0, . . . , αk so that this matrixholds.
So now define:
Eu χu
with χ equal 1 on the support of U and 0 outside a B0,1 9 xd C 1~2. Then you checkthe estimate.
Remark 6.2.2 (Stein). There exists a universal way to extend Sobolev spaces Euniv for allW k,p (k C 0,1 B p @ª which only requires U to be C1.
6.3 Traces
Here we will restrict u to ∂U
Theorem 6.3.1. Let k C 1 and 1 B p @ª, U a C1 domain
1. u > CªU and define:
tru uS∂UThen this extends uniquely as a bounded map from W k,pU to Lp∂U such that:
YtruYLp∂U B CYuYW 1,pU
2. If u >W k,pU and tru 0, then u >W k,p0 U and vice-versa.
Proof. See Evans section 5.5
Remark 6.3.1. In fact, the image of tr leads to the notion of fractional regularity Besovspaces.
It turns out trHkU Hk1~2∂U, with U is Ck (this can be defined with the Fouriertransform)
YuYHsRd Y1 SξSsÂuξYL2ξ
6.4 Sobolev Inequalities
This is all expressions of the form:
YuYW p,q B CYuYWk,pU
or the term on the left could be one of many types of norms.
A key element in this theory is the following inequality
– 82 –
6. Energy Method and Sobolev Spaces
Theorem 6.4.1 (Gagliardo – Nirenberk - Sobolev Inequality). Let u > Cª0 Rd, then:
YuYLd~d1Rd B cYDuYL1RdWe can remember d~d1 through dimensional analysis (or scaling analysis). If we have
ux, define uλx uxλ1. Then we have:
YDuλYL1 S SDuλSdx λd1S SDuySdyWe also have that YuλYLp λd~pYuYLp In order for the inequality to hold for all u and all λ,
we would like p d
d 1.
Proof. Given u > Cª0 Rd, we know that we can write:
ux S xi
ª
∂iux1, . . . , yi®ith component
, . . . , xddyi
This tells us that:
supxi>R
SuxS B S ª
ª
S∂iux1, . . . , xi, . . . , xdSdyiLemma 6.4.1 (Looms-Whitney-Inequality). Let fi be functions on Rd such that:
fix1, . . . ,Âxi, . . . , xd C 0
where Âxi indicates that we are fixing that coordinate (so the function doesn’t depend on thatcoordinate) then:
S f1fddx Bd
Mi1
YfiYLd1Rd1Proof.
S f1fddx1 f1S f2fddx
1B f1Yf2YLd1
x1YfdYLd1
x1
then we integrate this thing and apply Holders inequality again and again and everythingwill end up working out.
computation follows, I couldn’t really read the board, I’ll try to fill this in later.
Now let’s prove the main thing. First for all x > Rd, we have:
SuxS Sux aBx,1
uydyS aB0,1
SuSdy B aB,x1
Sux uyS a SuySdyboth of these can be trivially bounded (see notes). The hard part is bounded the semi-norm.Note that we have:
Sux uyS B Sux aBx,r9By,r
uzdzS SaBx,r9By,r
uzdz uySB a
Bx,r9By,rSux uzSdz a
Bx,r9By,rSuz uySdz
This can be controlled by our previous results, scaling estimates on regions, and Holder’sinequality.
Definition 6.4.2 (Version). We say u is a version of u if u u almost everywhere.
Theorem 6.4.5 (Sobolev Inequality for u >W 1,pU, p A d). given p A d and α 1 d~p1. If u >W 1,p
0 U, then there exists a version u > C0,αU and YuYC0,αU B CYuYW 1,pU
2. Assume also that U is a bounded C1 domain. For any u > W 1,pU there exists aversion u > C0,ªU and the above inequality holds.
Theorem 6.4.6 (General Sobolev inequality for W k,pU). Let k be any non-negativeinteger and 1 B p B ª and assue either U is a domain and u > W k,p
0 U or U is a boundedCk domain and u >W k,pU.
1. let l be a nonnegative integer, 0 B l B k, 1 B q @ª, d~q l C d~p k, then u >W l,qUand YuYW l,qU B CYuYWk,pU
2. let l be a nonnegative integer 0 B l B k, 0 @ α @ 1 if l α C d~p k, then there exists aversion u > C l,αU and YuYCl,αU B CYuYWk,pU
Lecture 26 (12/5)
Today we will wrap up our discussion of Sobolev spaces. We will study compactnessproperties with the main theorem the Relich - Konarchov theorem. We will also look atPoincare’s inequality. And lastly we will look at the dual space of Sobolev spaces, which willnaturally lead to the notation of negative regularity Sobolev spaces.
– 86 –
6. Energy Method and Sobolev Spaces
6.5 Compactness
We want to look at ui `W 1,pU given YuiYW 1,pU B C, then this turns out to be compactin an appropriate larger space.
Definition 6.5.1 (Compact Embedding of Banach Spaces). Let’s say we have X, Y YX and Y, Y YY two normed spaces with Y `X. We say that Y, Y YY embeddes compactlyinto X, Y YX (and say Y bb X if for all bounded sequences in Y has a convergent sub-sequence in X.
Theorem 6.5.1 (Relich–Konoluchov). Let U be a C1 bounded domain, 1 B p @ ª and1 B q @ p, then W 1,pU bb LqU. Recall we have:
YuYLpU B CYuYW 1,pU
and W 1,pU b LpUProof. The key ingredients to this are:
Theorem 6.5.2 (Arezela-Ascoli). Let K be a compact metric space with un ` CK,then given
1. (pointwise boundedness) for all x >K, we have unx is bounded
2. (equicontinuity) for all ε A 0 there exists δ0 such that Sunx unyS @ ε for all n andSx yS @ δ.then there exists a subsequence uni that is convergent in the uniform topology.
Another key ingredient is a property of mollification called accelerated convergence.
Proposition 6.5.1. Let ϕ > Cª0 Rd with R ϕ 1 and define ϕεx εdϕxε1, thenϕe u u as ε 0 in Lp. It turns out that if you assume that u >W 1,pU, then the rate ofconvergence in ϕε u u can be quantified.
Here is a more general proposition
Proposition 6.5.2. Let k be a positive integer, 1 B p @ª, let ϕ > Cª0 Rd such that R ϕ 1and R xαϕdx 0 for all 1 B SαS B k 1. Then if u >W k,pRd then:
Yϕε u uYLp B Cεk QSαSk
YDαuYLpProof. When k 1:
ux ϕε ux S ϕzux ux εzdz S ϕzS ε
0∂zux tzdz
S ϕzS ε
0z Dux tzdz
– 87 –
6. Energy Method and Sobolev Spaces
then by the Minkowski inequality, we have:
Yux ϕε uxYLp B S Sε
0Yϕzz Dux tzYLpxdtdz
B S Sε
0SϕzzSYDux tzYLpdtdz
B CεYDuYLp
Now let’s prove the RK theorem.
Step 1: It suffices to prove the case when q p. We want to show that if YunYW 1,pU B Cthen there exists a subsequence uni so that Yuni unjYLp 0 as i, j ª .
Indeed, if p @ q, then we have that all the other norms are bounded. If p @ q @ p thenuse Holder’s inequality which states that:
YvYLq B YvYθLpYvY1θLr
with q1 p1θ r11 θStep 2: we would like to prove that there exists a subsequnce uni such that for all ε A 0there exists I with Yuni uniYLpU @ ε if i, i C I.
Fix a bounded subset of Rd (call it V ) and let U ` V . Let’s extend un to elements inW 1,pRd with suppun ` V and
YEunYW 1,pRd B CYunYW 1,pU B C
Now let un Eun. Choose δ and let vn un ϕδ and supp vn ` V another bounded set withV ` V , let δbe such that:
Yun vnYLp B ε~3(where we use the proposition). Then note that vn is smooth, then we have:
vnx S ϕδx yvnydythen by Holder’s inequality, we have that:
YvnYLª B Cδ
and also
YDvnYLª B Cδ
– 88 –
6. Energy Method and Sobolev Spaces
this means that vn has a convergent subsequence by Arzela-Ascoli. So there exists vni sothat Yvni vniYLp @ ε~3, for i, i C I.
So now we have:
Yuni uniYLp B Yuni vniYLp Yvni vni SLp Yuni vniYLp B εnow we must preform a diagonalization argument.
Let’s now consider one nice application of this theorem. This is a key way to estimate uby Du.
Proposition 6.5.3 (Poincare’s Inequality). Let U be a C1 bounded domain, 1 B p @ ª,then:
\u aUu\
LP UB CYDuYLpU
with C only depending on U and p.
Proof. Assume that this fails. Then for all n C 1 there exists un ~ 0 so that:
Yun aUunYLP U C nYDunYLP U
Now let:
vn un `U unYun `U unYLp
Therefore: YvnYLp 1 and ` vn 0, and1
nYvnYLP U C YDvnYLp so that YDvnYLpU 0 as
N ª.
We know that vn is bounded in W 1,p. THerefore there exists vni with vni v in LpU.In particular:
YvYLp limiª
YvniYLp 1
on the other hand:
aUv lim
iªaUvni 0
and finally, YDvnYLp 0 this implies that Dv 0 in the sense of distributions, therefore v isa constant and we get a contradiction.
– 89 –
6. Energy Method and Sobolev Spaces
6.6 Duality
Definition 6.6.1 (Negative Regularity Sobolev Spaces). Let k be a non-negative inte-ger and 1 B p @ª, let:
W k,pU ¢¦¤u > DU ¦ SαS B k,§ gα > LpU, u Q
SαSBkDαgα
£§¥with norm:
YuYWk,pU infgα
uPSαSBkDαgα
QSαSBk
YgαYpLp1~p
Theorem 6.6.1. For all domains U , k a positive integer, 1 @ p @ª:
W k,p0 U W k,pU
– 90 –
Appendices
A Midterm Review
A.1 First order scalar characteristics
We are trying to solve the first order scalar PDE:
¢¦¤F x,u,Du 0 U R Rd
u g Γ ∂U(18)
With x > U ` Rd, u Rd R. Then if we assume there exists a solution, then along any path
x I Rd, we have zs uxs with:
¢¦¤xj ∂pjF
pi ∂zFpi ∂xiF
z Pi pi∂piF
Where in F we replaced u with z and ∂i with pi.
Corollary A.1.1. If F P bj∂ju c is linear or quasilinear, then:
¢¦¤xj bj
z c
A.1.1 Existence
Given our PDE (18), we would like to know when this method works, whether a solutionexists. If our boundary is flat (Γ ` xd 0) then for any characteristic path, we know itmust satisfy compatability conditions:
¢¨¦¨¤
x0 x0 > Γ
z0 gx0pi0 ∂igx0 i 1, . . . , d 1
F x0, z0, p0To apply our method of characteristics, we require the information of pd0 which we canget if ∂pdF x0, z0, p0 x 0 (this is known as the noncharacteristic condition).
Now we would like to show that u has a solution near x0. To do this let’s find a wholebunch of characteristics where the first input indicates where the path starts, and the secondindicates the time on the path:
xy, szy, spy, s
– 91 –
A. Midterm Review
with y y1, y2, . . . , yd1,0. Then in order to find the solution ux we must figure out thecharacteristic that hits x. That is we must determine y and s such that xy, s xa. Itturns out that under the noncharacterisitic assumption, this map is locally invertible, andwe denote the inverse: x( yx, sx, which gives rise to the main theorem:
Theorem A.1.1. Given our PDE 18 and a point x0 > Γ such that ∂pdF x 0 at x0, then thereexists a neighborhood of x0 such that ux zyx, sx solves the PDE for all x in thisneighborhood.
Remark A.1.1. The nonchraracteristic assumption of x0, z0 is equivalent to determiningall derivatives of u (in every direction) at x0.
A.1.2 General Boundary
For a smooth boundary Γ we can do the same thing, but the noncharacterisitc conditionbecomes:
Qj
νj∂U∂pjF x0, z0, p0 x 0
A.1.3 Uniqueness
Given two solutions to such a PDE whose sum of their normal derivatives agree, then theyare the same.
A.2 Noncharacteristic Condition for Kth order systems
Our new quasilinear K-th order PDE is of the form:
. So bα evaluated at each point is aN N matrix. The A,Bth component of this matrix is bαAB. Therefore a very compact(but uninformative) way of writing F would be:
F QSαSk
bαDαu c
But just note that F is a set of N equations. Now the Cauchy data (initial conditions) forthis PDE would be to prescribe normal derivatives. If our boundary is flat, Γ ` xd 0,then our Cauchy data is:
u, ∂du, . . . , ∂k1d u g0, g1, . . . , gk1
Then define the noncharacteristic condition for the Cauchy data for a point x0 > Γ as whenthe matrix:
b0,0,...,0,kx0, ux0, . . . ,Dk1ux0is invertible. If this is the case then we can determine all derivatives of u at x0.
alol, this notation could use work, I literally just figured this out while writing it
– 92 –
A. Midterm Review
A.2.1 General Domains
For a general domain, the noncharacteristic condition requires that:
QSαSk
bανα
is invertible, were ν is the outer unit normal to the boundary.
A.2.2 Power Series
Theorem A.2.1. Given the PDE (19) such that all functions and boundary are real analyticand x > Γ is noncharacteristic, then there exists a unique, local, real-analytic solution u with:
ux Q 1
α!Dαux0x x0α
near x0
A.3 Distributions
Definition A.3.1 (Test Function). A test function on a set U ` Rd is a smooth compactlysupported function on U . The space of test functions on U is denoted Cª
0 RDefinition A.3.2 (Convergence of Test Functions). We say fn f converges in thespace of test functions (with fn, f > Cª
0 U) if there exists a compact set containing thesupport of alla the functions on which all derivatives uniformly converge.
Definition A.3.3 (Distribution). A distribution on a set U ` Rd is a linear functionalb
from Cª
0 U to R which is continuous with respect to convergence of test functions.
Theorem A.3.1 (Boundedness). A linear functional u Cª0 U R is a distribution ifand only if for all compact sets K ` U , there exists a N and C CK,N such that for allϕ > Cª0 U with suppϕ `K:
S `u,ϕe S B C QSαSBN
supx>K
SDαϕxSDefinition A.3.4 (Order of a Distribution). The largest such N in the above theoremis called the order of a distribution.
Theorem A.3.2. If u is a distribution that agrees with 1~x on 0,ª then the order of u isat least 1.
Definition A.3.5 (Convolution). If f > L1loc and ϕ > Cª
0 Rd then:
f ϕx SRdfyϕx ydy
we can also define the starred convolution as:
f ϕx SRdϕyfy xdy
abut finitely many of courseblinear map from a vector space to its base field
– 93 –
A. Midterm Review
Theorem A.3.3. If f > L1loc and ϕ > Cª
0 Rd then f ϕ > Cª
Theorem A.3.4. If f > CRd and ϕ > Cª
0 Rd then suppf ϕ ` supp f suppϕ
Definition A.3.6 (Mollifer). A mollifier is a function ϕ > Cª
0 Rd with R ϕ 1
Theorem A.3.5 (Approximation). If ϕ is a mollifier, let ϕδ δdϕxδ1. Then iff > CkRd then f ϕδ converges uniformly to f (an up to and including k derivativesconverge uniformly).
Theorem A.3.6. If f > L1loc then f defines a distribution.
A.3.1 Operations
We may do the following operations with u a distribution, f > CªU, ϕ, v > Cª0 Rd1. `fu,ϕe `u, fϕe2. `Dαu,ϕe 1SαS `u,Dαϕe3. `u v,ϕe `u, v ϕe
Theorem A.3.7. u v > Cª
Theorem A.3.8. δ0 u u
Theorem A.3.9. If U is an open, connected, bounded subset of Rd with a C1 boundary and1U is the characteristic function (a distribution) then:
∂j1U νjdS∂U
with ν the unit outer normal of ∂U and dS∂U the surface measure.
A.3.2 Convergence of Distributions
Definition A.3.7. un (distributions) converge in distribution to a distribution u if for allϕ > Cª0 Rd then `un, ϕe `u,ϕeTheorem A.3.10. If un u as distributions, then Dαun u as distributions for all α
Theorem A.3.11 (Sequential Compactness). If un are distributions and for all ϕ >
Cª0 U, `un, ϕe converges, then u defined as `u,ϕe limn `un, ϕe is a distribution.
Theorem A.3.12. If ϕ is a moliffier, then ϕδ converges in distribution to δ0
Theorem A.3.13. If ϕ is a mollifier and u a distribution then ϕδ u are smooth functionsthat converge in distribution to u.
– 94 –
A. Midterm Review
A.3.3 Operations between distributions
Definition A.3.8 (Singular Support). The set of points on which a distribution fails toa be a smooth function.
Theorem A.3.14. If two distributions have disjoint singular support, then their product isdefined.
Theorem A.3.15. If u or v have compact support, then u v v u is well-defined(`u v,ϕe `u, v ϕe) and suppu v ` suppu supp v.
A.4 Fundamental Solutions
For this section consider the linear scalar partial differential operator:
Pu QSαSBk
aαxDαu
with aα real valued smooth.
Definition A.4.1 (Fundamental Solution). Given y > Rd, the fundamental solution Eyis such that PEyx δyx δ0x y
Then formally, a solution to this PDE is ux R vyEyxdy.
Definition A.4.2. The formal adjoint to the above linear operator is:
Pv Q
SαSBkaαxDαaαu
which we expect to satisfy `Pu, ve `u,P veRemark A.4.1. Finding fundamental solutions E
yx to the adjoint of P gives us unique-ness, this is because if u solves Pu v, then:
ux `u, δxe `u,P Exe `Pu, Exe `v, ExeWhich expresses u in terms of v.
Theorem A.4.1. If P is constant coefficient (ie all aα are constant) and PE0 δ0, then
1. Eyx E0x y2. P vx Pvx3. Exy E0x y
– 95 –
B. Final Review
A.5 Laplace equation
The fundamental solution to the Laplace equation in Rd is:
E0r ¢¦¤
log r
2πd 2
1
dd 2αdrd2d C 3
Theorem A.5.1. If u > DRd and ∆u 0 then u is smooth and harmonic.
Theorem A.5.2. If u is harmonic, then there exists a constant C CSαS sch that:
SDαuxS B C 1
rdSαS SBx,rSuSdy
Theorem A.5.3. Bounded harmonic functions are constant.
Theorem A.5.4. For f > DRd with compact support with d C 3, then solutions to ∆u fare unique upto addition by a constant.
Theorem A.5.5. If u is harmonic, then:
ux 1S∂Bx, r S∂Bx,rudSy 1SBx, rS SBx,r
uy
B Final Review
B.1 Green’s Functions
Given E0 as the fundamental solution to the Laplace equation, U an open set and u a smoothfunction on U , then we have:
ux SUE0x y∆uydy S
∂UE0x yν Duydy S
∂Uν DyE0x yuydy(20)
Now if u is harmonic and we choose E0 so that it vanishes on the boundary (a Greenfunction) then we can get results like:
Theorem B.1.1 (Mean Value Theorem). if u is harmonic then:
ux 1
dαdrd1 SBx,ruydSy
Concretly, we have that:
Definition B.1.1 (Green’s Function). Given an open set U with y > U , then G, y >C1U y is the Green’s function at y if:
¢¦¤∆xGx, y δ0x y, x > U
Gx, y 0, x > ∂U
– 96 –
B. Final Review
Theorem B.1.2. On an a bounded open set with C1 boundary, Green’s functions are uniqueand symmetric (Gx, y Gy, x).
We can construct Green’s functions for certain domains by the method of image charges.This takes advantage of the symmetries of our domain, which can only be done in specificcircumstances.
By looking at (20) we have the following representation formula for u > C2U 9CU:ux S
UGx, y∆uydy S
∂Uνy DyGx, yuydSy
This leads to
Theorem B.1.3 (Poisson Integral Formula).
ux S∂Uνy DyGx, ygydSy
solves the ODE:
¢¦¤∆u 0 x > U
u g x > ∂U
We even have the stronger formula (with certain conditions):
ux SUGx, ygydy S
∂Uνy DyGx, yhydSy
solves:
¢¦¤∆u g x > U
u f x > ∂U
B.2 Wave Equation
Here we define the d’Alembertian operator j ∂2t ∆ and ask to solve:
¢¦¤jϕ 0 t, x > R1d
ϕ g t, x > ∂R1d
∂tϕ h t, x > ∂R1d
In one dimension, we use change of variables to determine that the foward fundamentalsolution (fundamental solution supported in nonnegative time) is:
Et, x 1
2Ht xHt x
with H the Heaviside function.
– 97 –
B. Final Review
Definition B.2.1 (Forward Fundamental Solution). To solve the wave equation we lookfor forward fundamental solutions E that satisfy
1. jE δ0
2. suppE ` t C 03. suppE 9 t > I is compact for all intervals I.
Theorem B.2.1. Forward fundamental solutions are unique
Theorem B.2.2. For ϕ > CªRd and t, x > Rd
, then:
ϕt, x E ∂tϕδt0 ∂tE ϕδt0 E jϕ1tC0
For the one dimensional case, this leads to
Theorem B.2.3 (d’Alembert’s Formula).
ϕt, x 1
2 St
0S
xtsxts
jϕs, ydyds 1
2ϕ0, x t ϕ0, x t 1
2 Sxt
xt∂tϕ0, ydy
To find the forward fundamental solution for arbitrary dimensions, we take advantage ofthe fact that reflections, rotations, and Lorentz boots are conserved under j. Also bylooking at scaling properties, we know that E should to be homogeneous of degree d 1,ie:
aE, λd1ϕλf λd1 `E, ϕe
B.3 Fourier Transform
The motivation is to preform a change of coordinates in the space of functions to somethingthat behaves well with differentiation. We simply define:
Definition B.3.1 (Fourier Transfom).
Ffxξ 12πd SRdfxeiξxdx Âfξ
And we have the inverse:
Definition B.3.2 (Inverse Fourier Transform).
F1 Âfξx 12πd S Âfξeiξxdξ
Proposition B.3.1. If f > L1, we have Âf is well defined and Y ÂfYLª B YfYL1
Proposition B.3.2. For f > L1:
Ffxξ eixξFfxξFfxξ η Feiηxfxξ
– 98 –
B. Final Review
Proposition B.3.3. Ä∂jf iξj ÂfProposition B.3.4. Äxjf i∂ξ ÂfProposition B.3.5. Åf g ÂfÂgProposition B.3.6. F1f 2πddξ g F1fF1g
The best space to take the Fourier transform is the space of Scwartz functions, whichare like test functions, but are allowed to have non-compact support as long as the decayquickly
Definition B.3.3 (Schwartz Class on Rd).
SRd C ϕ > CªRd
C YxβDαϕYLª @ª ¦ α,β > NnThen F and F1 map S to S. The space of conjugate linear functionals on S are called
tempered distributions and we are allowed to take Fourier transforms (and inverse Fouriertransforms) of these elements (which will be tempered distributions).
Theorem B.3.1 (Fourier Inversion on S). for f > S, we have that F1Ff FF1f
f . The same is true for tempered distributions.
Theorem B.3.2 (Plancherel’s Theorem). For f, g > S, we have the conservation of thetwo inner products:
S fÂgdx S ÂfÂg dξ2πdThere are important integrals to keep in mind to help compute Fourier transforms
1. R eiξxdξ limε0 R eεSξSeiξxdξ δ0x2π2. R e1~2SξS2dξ 2πd~23. Fe1~2x2 º2πe1~2ξ2
