Lecture Notes on Partial
Differential Equations
Dr. E. Natarajan
IIST Lecture Notes Series-2
Government of India
Department of Space
Indian Institute of Space Science and Technology
Valiamala P.O, Thiruvananthapuram-695547
December, 2012
.
Lecture Notes on Partial
Differential Equations
Dr. E. Natarajan
Assistant Professor, Department of Mathematics
Indian Institute of Space Science and Technology,
Valiamala P.O, Thiruvananthapuram, India.
[For internal circulation only.]
Published by :
Government of IndiaDepartment of Space
Indian Institute of Space Science and TechnologyDeemed to be University under Section 3 of the UGC Act, 1956
Valiamala P.O, Thiruvananthapuram-695547
ii
Acknowledgement
This lecture notes is based on my teaching B. Tech students of IIST
for the course MA 221 for several semesters. The main purpose of
this notes is to give students material which is mostly self explanatory
with deep conceptual backend suited for undergraduates.
I thank Dr.K.S.Dasgupta Director, IIST who motivated me to pre-
pare this material. I thank Library and Information services, IIST for
support. I would thank all the members of Department of Mathemat-
ics specially Shri. Abdul Karim.
Hope this will go into the minds of young students and not into
the unread shelf of library.
iii
Contents
1 Modeling partial differential equations 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Vibrating string problem . . . . . . . . . . . . . . . . . 2
1.3 Heat equation . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Well-posed PDE . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Solution of PDE’s by Fourier transform . . . . . . . . . 7
2 Second order linear PDE’s 11
2.1 Classification . . . . . . . . . . . . . . . . . . . . . . . 11
2.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . 11
2.2 Canonical forms . . . . . . . . . . . . . . . . . . . . . . 13
2.2.1 Canonical form of hyperbolic PDE . . . . . . . 14
2.2.2 Canonical form of parabolic PDE . . . . . . . . 16
v
2.2.3 Canonical form of elliptic PDE . . . . . . . . . 17
3 Method of Separation of variables 19
3.1 Heat equation . . . . . . . . . . . . . . . . . . . . . . . 19
3.2 Wave equation . . . . . . . . . . . . . . . . . . . . . . . 22
3.3 Mathematical justification of the method . . . . . . . . 25
4 First order partial differential equations 27
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2 Method of Characteristics . . . . . . . . . . . . . . . . 29
4.2.1 Transversality condition . . . . . . . . . . . . . 31
4.3 Lagranges Method . . . . . . . . . . . . . . . . . . . . 35
5 Tutorial 1 39
6 Tutorial II 43
vi
Chapter 1
Modeling partial differential
equations
1.1 Introduction
Partial differential equations is a relation between an unknown func-
tion and its partial derivatives
F (x1, x2, ..., xn, ux1, ux2, ..., ux11) = 0 (1.1.1)
where x1, x2, ..., xn are independent variables and u(x1, x2, ..xn) depen-
dent variable uxi =∂u
∂xi
• Order of the partial differential equation is the order of the high-
est partial derivative. utt − uxx = f(x, t) is a second order PDE
whereas ut + uxxxx = 0 is a fourth order PDE.
1
ψ(x,t)
x
Figure 1.1: Vertical displacement of string
• An equation is called linear if in Eq(1.1.1) F (.) is a linear func-
tion of the unknown function u and its partial derivatives. For
example x2ux + x y uy + sin(x2 + y2) u = x3 is a linear equation
whereas uxx2 + uyy
2 = 0 is a non-linear equation. An equa-
tion is quasilinear if it is linear in the highest partial derivative
uxx + uyy = |∇u|2 u and semilinear if it is non-linear in its un-
known function uxx + uyy = u3.
1.2 Vibrating string problem
Let ψ(x, t) measures the vertical displacement of the string from equi-
librium. Length of this piece
√
∆x2 +∆ψ2 where ∆ψ = ψ(x+∆x, t)−ψ(x, t) with Density ρ (mass per unit length) is the mass of the string
particle.
ρ
√
∆x2 +∆ψ2 = ρ∆x
√
1 +O(
∆ψ
∆x
)2
(1.2.1)
2
• Displacement of the string is slight, so terms of order 2 for ψ
and∂ψ∂x
are neglected.
Mass ≈ ρ∆x and acceleration in the vertical direction is∂2ψ
∂t2.
Force acting on the string particle is Tension T . This acts parallel to
the string i.e., tangentially. Tension is assumed to be uniform. This
gives
T (x, t) = T.i+ ψ
′
(x) j√
1 + (ψ′(x))2(1.2.2)
Vertical component, ↑ T (x, t) =T ∂ψ∂x
1 + (ψ′(x))2≈ T
∂ψ
∂x
Net vertical tensile force is, ↑ T (x + ∆x, t)− ↑ T (x, t) ≈ T∂2ψ
∂x2+
O(∆x2)
Using Newton’s second law
T∂2ψ
∂x2∆x+O
(
∆x2)
= ρ∆x∂2ψ
∂t2(1.2.3)
Dividing by ρ∆x and letting ∆x → 0 we get∂2ψ
∂t2=
T
ρ
∂2ψ
∂x2= c2
∂2ψ
∂x2
where c =
√
T
ρ(dimensions of velocity).
Let f be twice-differentiable then ψ(x, t) = f(x−ct) satisfies the wave
equation. It is the wave propagating to the left at speed c and f(x+ct)
is the wave propagating to the right at speed c so the solution will be
ψ(x, t) = f(x− ct) + f(x+ ct) where f and g are arbitrary.
In order to determine the motion completely ψ(x, 0) = F (x) and
3
Dirichlet Neumann Robin
Figure 1.2: Dirichlet, Neumann and Robin boundary condition for left
end of the string
∂ψ∂t
(x, 0) = G(x) are given as initial displacement and initial ve-
locity.
ψ(x, t) =F (x− c t) + F (x+ c t)
2+
1
2 c
∫ x+ct
x−ctG(ξ) dξ (1.2.4)
This is known as the D’Alembert’s form of the solution. This says that
to determine ψ(x, t) we need to know F and G within ±c t distance.
Types of boundary condition for the string tied at left boundary:
• Dirichlet boundary condition ψ(0, t) = c
• Neumann boundary condition∂ψ∂x
(0, t) = d
• Robin boundary condition αψ(0, t) + β∂ψ∂x
(0, t) = l
4
Figure 1.3: Heat transfer in a uniform rod
1.3 Heat equation
Consider heat flowing through a uniform thin rod. Experiment shows
that ∆Q = c ρ u(x, t)∆V for a uniform cross section ∆V = A∆x
between small portion x ∈ [a, b] where c is the concentration, ρ is
the density, u(x, t) is the temperature and ∆V is the small volume
element.
Q(x, t) =
∫ b
a
c ρ u(x, t)Adx
dQ
dt=
∫ b
a
c ρ∂u(x, t)
∂tA dx
This gives the rate at which heat accumulates in this segment. Ac-
cording to Newton’s Law of Cooling the rate of change of the temper-
ature of an object is proportional to the temperature gradient. Also
heat energy flows from warmer region to cooler region. This gives
dQ
dt= k [Rate in − Rate out] A (1.3.1)
dQ
dt= k
[
∂u
∂x(b, t)− ∂u
∂x(a, t)
]
A (1.3.2)
In the integral form this givesdQ
dt=
∫ b
a
k A∂2u
∂x2dx. Equating similar
terms we get∫ b
a
(
c ρ∂u(x, t)
∂t− k
∂2u
∂x2
)
Adx = 0 (1.3.3)
5
Under suitable conditions it reduces to∂u
∂t= k
∂2u
∂x2where k =
k
ρ cis
the thermal diffusivity.
1.4 Well-posed PDE
Any physical phenomena can be modeled and it gives rise to partial
differential equation. Suppose motion of a pendulum is modeled as
a differential equation, definitely basic problem has a sure motion
whereas the modeled equation need not have solution. This doesn’t
contradict with the physics of the problem but it says that the modeled
equations are not sufficient enough to talk about the physics of the
problem.
Jacques Hadamard came up with a set of conditions PDE has to
satisfy so that the equation can undoubtedly talk about the physics
of the problem and the PDEs satifying those conditions are known as
Well-posed problems. They are
(i) Existence of a solution
(ii) Uniqueness of the solution (there exists only one solution)
(iii) Solution is stable (continuous dependence of data)
We will illustrate the third point with an example.
ut = k uxx, −∞ < x <∞ (1.4.1)
u(x, 1) = c sin
(
x
c√k
)
(1.4.2)
6
where c is small. This can be solved by taking u(x, t) = c sin
(
x
c√k
)
g(t)
with g(1) = 1. ( Try it!). The solution is u(x, t) = c e(1−t)
c2 sin
(
x
c√k
)
.
Note that when c → 0 when 0 < t < 1 the solution u(x, t) → ∞.
A small change in the initial condition leads to a large change in the
solution thus killing the stability of the problem.
1.5 Solution of PDE’s by Fourier transform
Heat conduction in an infinite rod
ut = α2 uxx, −∞ < x <∞, 0 < t <∞u(x, 0) = f(x),−∞ < x <∞
Note: This problem can be solved by separation of variables if f(x)
is defined in finite interval or even if f is defined in infinite interval
provided if it is periodic. Suppose f is defined in infinite interval
and aperiodic like f(x) = e−|x|. Let us take fourier transform of heat
equation with respect to x.
F{ut} = α2 F{uxx}
∫ ∞
−∞
∂u
∂te−i ω x dx = α2 (i ω)2 u
d
dt
∫ ∞
−∞u(x, t) e−i ω x dx = −α2 ω2 u
Under the assumption (u→ 0, ux → 0, x→ ±∞)
du
dt+ α2 ω2 u = 0
7
The solution is u(x, t) = A(ω)e−α2 ω2 t. We also need to transform the
initial condition. F (u(x, 0)) = F (f(x)) ⇒ u(ω, 0) = f(ω). Using this
in the solution we get A = f(ω). Finally we get
u(ω, t) = f(ω) e−α2 ω2 t
u(x, t) = f(x) ∗ 1
2α√π t
e−x2
4α2 t
u(x, t) =1
2α√π t
∫ ∞
−∞f(ξ) e−
(x−ξ)2
4α2 t dξ
Case 1: If f is constant F then u(x, t) = F which violates the assumption
u→ ∞, x→ ±∞.
Case 2: f(x) =
F, x > 0
0, x < 0= F H(x)
u(x, t) =F
2
[
1 + erf
(
x
2α√t
)]
, where erf(x) =2√π
∫ x
0
e−ξ2
dξ.
In a more formidable way our solution can be written as
u(x, t) =
∫ ∞
−∞f(ξ)K(ξ − x; t) dξ
where
K(ξ − x; t) =e−
(x−ξ)2
4α2 t
2α√πt
is called as Kernel of the integral. If K(ξ − x; t) → δ(ξ − x) as t → 0
then
limt→0
u(x, t) = limt→0
∫ ∞
−∞f(ξ)K(ξ − x; t) dξ = f(x)
Let f(x) = δ(x− x0). Then
u(x, t) =
∫ ∞
−∞δ(ξ − x0)K(ξ − x; t) dξ
= K(x0 − x; t) = K(x− x0; t)
8
If initial temperature distribution is δ(x − x0) then K(x − x0; t) is
the solution to the heat equation. Diffusion process smoothes out the
solution.
Example 1.5.1. ut = uxx, −∞ < x <∞, t > 0, u(x, 0) = F (x), −∞ <
x <∞. By solving this problem using Fourier transform we get
u(x, t) =1√4 π t
∫ ∞
−∞F (y) e−
(x−y)2
4 t dy
where e(x−y)2
4 t > 0 ∀x, y. For t however small u depends on F (x), −∞ <
x < ∞ which means the speed of propagation is infinite. But energy
cannot propagate faster than speed of light. This model of heat equa-
tion has a flaw.
9
10
Chapter 2
Second order linear PDE’s
2.1 Classification
Let us consider the general form of second order linear PDE’s in two
variables
A(x, y) uxx + 2B(x, y) uxy + C(x, y) uyy = Φ(x, y, ux, uy, u) (2.1.1)
where the left hand side of Eq (2.1.1) denotes the principal part of the
PDE.
2.1.1 Definitions
(a) Equation 2.1.1 is said to be hyperbolic if B2 − AC > 0
(b) Equation 2.1.1 is said to be parabolic if B2 − AC = 0
(c) Equation 2.1.1 is said to be elliptic if B2 − AC < 0
11
What do these classification do? Why is it needed?
These classifications can be related with equation of general conic
section
a x2 + 2 b xy + c y2 + d x+ e y + f = 0 (2.1.2)
(a) Equation 2.1.2 represents eqn of hyperbola if b2 − ac > 0
(b) Equation 2.1.2 represents eqn of parabola if b2 − ac = 0
(c) Equation 2.1.2 represents eqn of ellipse if b2 − ac < 0
The need for this classification is any second order PDE’s in two
variables can be reduced to either hyperbolic, parabolic or elliptic.
Once the behaviour of the solutions of hyperbolic,parabolic and elliptic
PDEs are known then classification to one of the three forms helps in
understanding the problem apriori.
Examples:
utt = c2 uxx where A = 1, B = 0, C = −c2 gives B2 − AC > 0
then the equation is said to be hyperbolic.
ut = k uxx where A = 0, B = 0, C = −k gives B2−AC = 0 then
the equation is said to be parabolic.
uxx + uyy = 0 where A = 1, B = 0, C = 1 gives B2 − AC < 0
then the equation is said to be elliptic.
12
2.2 Canonical forms
Any second order PDE in two variables can be reduced to canonical
form of hyperbolic, parabolic and elliptic PDE’s under suitable trans-
formation. Let us convert the PDE Equation 2.1.1 using the coordi-
nate transformation ξ(x, y), η(x, y). Also take w(ξ, η) = u(x(ξ, η), y(ξ, η))
ux = wξ ξx + wη ηx
uy = wξ ξy + wη ηy
Similarly compute uxx, uyy and uxy in terms of wξξ, wξη and wηη. After
substitution of these terms into the Equation 2.1.1 gives
a(ξ, η)wξξ + 2 b(ξ, η)wξη + c(ξ, η)wηη = Ψ (wξ, wη, w, ξ, η)
where
a(ξ, η) = Aξx2 + 2B ξx ξy + Cξy
2
b(ξ, η) = Aξx ηx +B (ξx ηy + ξy ηx) + C ξy ηy
c(ξ, η) = Aηx2 + 2B ηxηy + C ηy
2
We need a justification that the form of the PDE remains invariant
even after the coordinate transformation i.e., hyperbolic remains as
hyperbolic, parabolic remains parabolic and vice versa. It can be
observed that(
a b
b c
)
=
(
ξx ξy
ηx ηy
) (
A B
B C
) (
ξx ηx
ξy ηy
)
Taking the determinant on both sides gives
b2 − a c = (ξxηy − ξyηx)2 (B2 − AC
)
13
where (ξxηy − ξyηx)2 is nothing but the square of the Jacobian J2.
The canonical form of the PDE after transformation reduces to
(i) wξη = G(wξ, wη, ξ, η) if it is hyperbolic.
(ii) wξξ = G(wξ, wη, ξ, η) or wηη = G(wξ, wη, ξ, η) if it is parabolic.
(iii) wξξ + wηη = G(wξ, wη, ξ, η) if it is elliptic.
2.2.1 Canonical form of hyperbolic PDE
If A = C = 0 then uxy =Φ(ux, uy, u, x, y)
2Bin which case there is no
need of transformation.
Let us assume A 6= 0 and in order to make b2 − a c > 0 we take
a = c = 0. This gives
Aξx2 + 2B ξx ξy + C ξy
2 = 0
Aηx2 + 2B ηx ηy + C ηy
2 = 0
After factorizing
1
A
(
Aξx +(
B −√B2 − AC ξy
))
(
Aξx +(
B −√B2 + AC ξy
))
= 0
14
Aξx +(
B +√B2 − AC
)
ξy = 0
Aξx +(
B −√B2 − AC
)
ξy = 0
Aηx +(
B +√B2 −AC
)
ηy = 0
Aηx +(
B −√B2 −AC
)
ηy = 0
We get totally two equations for ξ and two equations for η but we
need one solution for ξ and one for η. We take
Aξx +(
B +√B2 − AC
)
ξy = 0
Aηx +(
B −√B2 −AC
)
ηy = 0
The above equations are itself PDE in terms of ξ(x, y). On the
characteristic curves ξ(x, y) = constant we get dξ = 0 ⇒ ξx dx +
ξy dy = 0. Similarly dη = 0 ⇒ ηx dx + ηy dy = 0. Equating the like
terms gives
dy
dx= −ξx
ξy=B +
√B2 −AC
A
dy
dx= −ηx
ηy=B −
√B2 −AC
A
Solving the above two ordinary differential equations we get two curves
ξ(x, y) = C1 and η(x, y) = C2. Thus we get the coordinate transfor-
mations by which we can reduce the PDE’S to canonical form.
Example 2.2.1. Determine the canonical form of utt − c2 uxx = 0
B2 − AC = c2 > 0 so the eqn is hyperbolic.
15
We getdx
dt= c and
dx
dt= −c
This gives ξ(x, y) = x− ct and η(x, y) = x+ ct. Using this we get the
canonical form wξ η = 0 ⇒ w(ξ, η) = f(ξ) + g(η).
So u(x, y) = f(x− ct) + g(x+ ct).
2.2.2 Canonical form of parabolic PDE
In order to make b2−a c = 0 we need either b = c = 0 or b = a = 0 Let
us take c = 0 because b = 0 will be automatically forced by parabolic
condition.
Let us assume A 6= 0. This gives
Aηx2 + 2B ηx ηy + C ηy
2 = 0
Substituting C =B2
Aand factorizing we get
Aηx2 + 2B ηx ηy +
B2
Aηy
2 = 0
1
A(Aηx +B ηy)
2 = 0
Aηx +B ηy = 0 ( ∵ A 6= 0)
On the characteristic curve η = const we have dη = 0 ⇒ ηx dx +
ηy dy = 0. This gives the transformation η(x, y) from −ηxηy
=dy
dx=B
A.
Now we have the freedom to choose the transformation ξ(x, y) such
that ξxηy − ξyηx 6= 0. Thus we can reduce the PDE to its canonical
form.
16
Example 2.2.2. Determine the canonical form of x2 uxx− 2 x y uxy+
y2 uyy + xux + y uy = 0.
B2 − AC = 0 so the eqn is parabolic.
We getdy
dx= −y
x⇒ xy = const
This implies η(x, y) = xy. Now let us take ξ(x, y) = x so that ξx ηy −ξy ηx = x 6= 0. You can choose any ξ satisfying above condition.
The canonical form is ξ2wξξ + ξ wξ = 0. Solving this finally we get
w(ξ, η) = ln(ξ) g(η) + f(η) ⇒ u(x, y) = ln(x) g(xy) + f(xy).
2.2.3 Canonical form of elliptic PDE
For the elliptic case we need to make b2 − a c = 0, we can make
a = c and b = 0. This gives
Aξx2 + 2B ξx ξy + C ξy
2 = Aηx2 + 2B ηx ηy + C ηy
2
Aξx ηx +B (ξxηy + ξyηx) + C ξy ηy = 0
This is in itself coupled PDE in ξ(x, y) and η(x, y). Let us try to
convert these two equations in complex form by taking Φ = ξ + i η.
Then we can express the above two equations as
A (ξx2 − ηx
2) + 2B (ξx ξy − ηx ηy) + C (ξy2 − ηy
2) = 0
Aξx iηx +B (ξx iηy + ξyiηx) + C ξy iηy = 0
In terms of Φ,
17
AΦx2 + 2BΦx Φy + C Φy
2 = 0,
Factorizing gives(
AΦx +(
B + i√AC − B2
)
Φy
)
(
AΦx +(
B − i√AC −B2
)
Φy
)
= 0
On Φ = constant we have dΦ = 0 ⇒ Φx dx+ Φy dy = 0.
Equating −ΦxΦy
=dy
dxto the factorization above gives
dy
dx=B + i
√AC − B2
A
dy
dx=B − i
√AC − B2
A
This gives ϕ(x, y) and χ(x, y) both are complex functions. From this
we can extract real valued function ξ(x, y) and η(x, y). We will try to
understand by an example
Example 2.2.3. Determine the canonical form of uxx + x2 uyy = 0
B2 − AC = −x2 < 0 so the eqn is elliptic
We have
dy
dx=i x
2,dy
dx= −i x
2
After solving these two ODES we get
ϕ(x, y) =x2
2+ i y and χ(x, y) =
x2
2− i y
Taking ξ =ϕ+ χ
2and η =
ϕ− χ
2 igives ξ =
x2
2and η = y. Using
this ξ and η the canonical form reduces to wξξ + wηη = −wξ2ξ
.
18
Chapter 3
Method of Separation of
variables
3.1 Heat equation
Let us consider the heat conduction problem
ut − k uxx = 0 0 < x < L, t > 0 (3.1.1)
u(0, t) = u(L, t) = 0, t ≥ 0 (3.1.2)
u(x, 0) = f(x), 0 ≤ x ≤ L (3.1.3)
Eq (3.2) and (3.3) implies f(0) = f(L) = 0. Let us assume the solution
u(x, t) = X(x) T (t). This gives XT′
= k X′′
T .
T′
k T=X
′′
X= −λ gives
d2X
dx2= −λX, 0 < x < L,
dT
dt= −λ k T, t > 0
19
First let us solve the ODE forX(x). We need to find the boundary con-
dition for X(x) at x = 0 and x = L. Using u(0, t) = 0 ⇒ X(0) T (t) =
0 ⇒ X(0) = 0 and u(L, t) = 0 ⇒ X(L) T (t) = 0 ⇒ X(L) = 0.
d2X
x2+ λX = 0, 0 < x < L (3.1.4)
X(0) = X(L) = 0 (3.1.5)
This is an eigenvalue problem with eigenvalue λ and eigenfunction Xλ.
(λ < 0) thenX(x) = a cosh(√−λ x)+b sinh(
√−λ x) withX(0) = X(L) =
0 ⇒ X(x) = 0.
(λ = 0) then X(x) = a+ b x with X(0) = X(L) = 0 ⇒ X(x) = 0.
(λ > 0) then X(x) = a cos(√λ x) + b sin(
√λx) with X(0) = X(L) =
0 ⇒ a = 0 and sin√λL = 0 ⇒ λ =
(nπ
L
)2
, n = 1, 2, 3..
Solving ODE for T (t) gives Tn(t) = e−k (nπL )
2t, n = 1, 2, 3, .... For
each λn the product of Xn(x) ∗Tn(t) is a solution so by the superposi-
tion principle we get u(x, t) =∞∑
n=1
An sinnπx
Le−k (
nπL )
2t. In order to de-
termine the constants An we need to use the condition u(x, 0) = f(x).
This gives∞∑
n=1
An sinnπx
L= f(x), Multiplying both sides by sin
mπx
L
and integrating from 0 toL gives Am =2
L
∫ L
0
sinmπx
Lf(x) dx, m =
1, 2, 3, ...
Let us solve a problem with some simple f(x).
20
Example 3.1.1. Consider the problem
ut − uxx = 0, 0 < x < π, t > 0
u(0, t) = u(π, t) = 0, t ≥ 0
u(x, 0) =
x, 0 ≤ x ≤ π2
π − x, π2≤ x ≤ π
We get the solution
u(x, t) =
∞∑
n=1
An sin(nx) e−k n2 t
Using the condition of u(x, 0) we can obtain An =4
π n2sin
nπ
2. The
final solution is
u(x, t) =4
π
∞∑
n=1
(−1)n+1
(2n− 1)2sin ((2n− 1) x) e−(2n−1)2 t
Our solution is an infinite series of functions. So in order for the
solution to be well defined we need to discuss the convergence of the
solution. We can use Weierstrass M-test to check the convergence
of series of functions
∞∑
n=1
un(x, t) where we try to bound un(x, t) <
Mn then if the series of real numbers
∞∑
n=1
Mn converges then the real
series
∞∑
n=1
un(x, t) converges uniformly. In general∂
∂x
∞∑
n=1
un(x, t) 6=∞∑
n=1
∂
∂xun(x, t) may not be equal so the convergence of
∞∑
n=1
∂un
∂tand
∞∑
n=1
∂2un
∂x2should also be verified. I will leave it as an exercise.
21
3.2 Wave equation
Let us consider the problem
utt − c2 uxx = 0, 0 < x < L, t > 0
ux(0, t) = ux(L, t) = 0, 0 ≤ x ≤ L
u(x, 0) = f(x), ut(x, 0) = g(x), 0 ≤ x ≤ L
Compatibility conditions
f′
(0) = f′
(L) = g′
(0) = g′
(L) = 0
Let u(x, t) = X(x) T (t) which gives X T′′
= c2X′′
T
T′′
c2 T=X
′′
X= −λ
From this we get two ordinary differential equations
X′′
+ λX = 0, T′′
+ λ c2 T = 0
This is an eigenvalue problem. The boundary condition gives ux(0, t) =
0 ⇒ X′
(0) T (t) = 0 ⇒ X′
(0) = 0. Similarly X′
(L) = 0.
(λ < 0) thenX(x) = a cosh(√−λ x)+b sinh(
√−λ x) withX
′
(0) = X′
(L) =
0 ⇒ b = 0 and λ =(nπ
L
)2
which cannot happen since λ is neg-
ative.
(λ = 0) then X(x) = a + b x with X′
(0) = X′
(L) = 0 ⇒ X0(x) = const.
(λ > 0) then X(x) = a cos(√λ x) + b sin(
√λx) with X
′
(0) = X′
(L) =
0 ⇒ b = 0 and sin√λL = 0 ⇒ λ =
(nπ
L
)2
, n = 1, 2, 3....
Xn(x) = an cos(nπ x
L
)
.
22
Solving the equation T′′
+ λ c2 T = 0
(λ > 0) then Tn(t) = αn cos(√
λn c2 t) + βn sin(√
λn c2 t), n = 1, 2, 3., .
(λ = 0) then T′′
= 0 ⇒ T0(t) = α0 + β0 t.
Finally u(x, t) = X0(x) T0(t) +∞∑
n=1
Xn(x) Tn(t)
⇒ u(x, t) =A0 +B0 t
2+
∞∑
n=1
(
An cos
(
cπnt
L
)
+Bn sin
(
cπnt
L
))
cosnπ x
L, n = 1, 2, 3, ..
Example 3.2.1. Consider the problem
utt − 4 uxx = 0, 0 < x < 1, t > 0
ux(0, t) = ux(1, t) = 0, t ≥ 0
u(x, 0) = cos2(π x), 0 ≤ x ≤ 1
ut(x, 0) = sin2(π x) cos(π x), 0 ≤ x ≤ 1
Using the solution of the previous problem
u(x, t) =A0 +B0 t
2+
∞∑
n=1
(An cos(2nπ t) +Bn sin(2nπ t)) cos(nπ x)
Using the conditions of intial displacement and intial velocity we get,
u(x, 0) =A0
2+
∞∑
n=1
An cos(nπ x) = cos2 π x =1 + cos(2 π x)
2
23
This gives A0 = 1, A2 =1
2, An = 0, ∀n 6= 0, 2
ut(x, 0) =B0
2+
∞∑
n=1
Bn (2nπ)cos(nπ x) =
sin2(π x) cos(π x) =cosπ x
4− cos 3 π x
4
This gives B1 =1
8 π, B3 = − 1
24 π, Bn = 0, ∀n 6= 1, 3
u(x, t) =1
2+
1
8 πsin(2 π t) cos(π x) +
1
2cos(4 π t) cos(2 π x)−1
24 πsin(6 π t) cos(3 π x).
Example 3.2.2.
utt − uxx = cos(2 π x) cos(2 π t), 0 < x < 1, t > 0
ux(0, t) = ux(1, t) = 0, t ≥ 0
u(x, 0) = cos2(π x), 0 ≤ x ≤ 1
ut(x, 0) = 2 cos(π x), 0 ≤ x ≤ 1
We have Xn(x) = cos(nπ x), λn = (nπ)2, n = 0, 1, 2, .... Let us
assume the solution as
u(x, t) =T0(t)
2+
∞∑
n=1
Tn(t) cos(nπ x)
Substituting this expression in the PDE and using the initial con-
ditions finally we get the solution
u(x, t) =1
2+
(
cos(2 π t)
2+t+ 4
4 πsin(2 π t)
)
cos(2 π x).
24
3.3 Mathematical justification of the method
The method of separation of variables has limitations and the points
to be noted before applying the method are as follows.
(a) In the partial differential equation Lu = f(x, y)(
Ex : Forwave equationL =∂2
∂t2+
∂2
∂x2
)
L must be separable.
∄φ(x, y) such thatL(X(x) Y (y))
φ(x, y)X(x) Y (y)= F (x)G(y) for some F
and G functions of x and y.
(Ex: L(u) =∂2u
∂x2+∂2u
∂x∂y+∂2u
∂y2, substituting u(x, y) = X(x) Y (y)
L(X(x) Y (y)) = X′′
Y + X′
Y′
+ X Y′′
cannot be written as
F (x) +G(y). So the PDE is not separable.
(b) All initial and boundary conditions must be on lines x = const
and y = const. (Ex: Suppose a domain not rectangular with
sides parallel to x and y axes.)
(c) Linear operator boundary conditions x = const involve no par-
tial derivative of u w.r.t x. Similarly y = const involve no partial
derivative of u w.r.t y. (Ex: Suppose at x = 0 the condition is
given as∂u
∂x+∂u
∂y= 0 ).
25
26
Chapter 4
First order partial differential
equations
4.1 Introduction
In this chapter we will see the methods to solve first order PDES which
are Quasilinear.
a(x, y, u) ux + b(x, y, u) uy = c(x, y, u) (4.1.1)
Let us start solving this PDE
Example 4.1.1.
ux = c0 u+ c1(x, y), u(0, y) = y (4.1.2)
Eventhough this a PDE since there is no term with uy this can be
looked as a first-order ODE and then solved by method of integrating
27
X
Y
Characteristic curves
Initial Curvey = const
y=const
factors.dy
dx+ P (x) y = Q(x)
u(x, y) e−c0 x =
∫ x
0
c1(ξ, y) e−c0 ξ dξ + T (y). By using the condition
u(0, y) = y. We find that T (y) = y. So this problem has a unique
solution.
In solving the above problem by fixing y = const we see that the
PDE gets converted to set of ODE’s on each y = const line. So we
have solved the set of ODES’s. These are solutions of ODE’s on the
line y = const. We then call the line as characteristic curve and the
solution as characteristic solution.
Now it is not always true that a first-order PDE with given condi-
tion will have a solution which is unique. Let us see few examples on
28
this issue.
Example 4.1.2. Consider the PDE ux = c0 u with u(x, 0) = 2 x.
The solution is u(x, y) = ec0 x T (y) by using the condition we see
that T (0) = 2 x e−c0 x which is impossible. So the PDE has no solution.
Example 4.1.3. Consider the same PDE ux = c0 u with u(x, 0) =
2 ec0 x. we see that T (0) = 2 which means there are infinitely many
functions T (y) satisfying the above condition. So the PDE has in-
finitely many solutions.
4.2 Method of Characteristics
Consider the quasi-linear first-order partial differential equation
a(x, y, u) ux + b(x, y, u) uy = c(x, y, u)
The initial condition is given on some curve Γ(s) : (x0(s), y0(s), u0(s))
in the parametric form where s ∈ (α, β). This can be written as
(a, b, c) . (ux, uy,−1) = 0 where (ux, uy,−1) is normal to the surface
u(x, y) − u = 0 on the (x, y, u) plane. Therefore (a, b, c) lies in the
tangent plane. So we get the equations
dx
dt= a(x(t), y(t), u(t))
dy
dt= b(x(t), y(t), u(t))
du
dt= c(x(t), y(t), u(t))
We call this set of ODE’s as characteristic equation. Solving this
set of ODE’s gives characteristic curves (x(t, s), y(t, s), u(t, s)) where
29
Initial curve
Characteristic curve
x
y
u
each curve emanates from different points on Γ(s) for the given initial
condition x(0, s) = x0(s), y(0, s) = y0(s), u(0, s) = u0(s).
Example 4.2.1. Solve the PDE ux + uy = 2, u(x, 0) = x2
The set of characteristic equations are
dx
dt= 1,
dy
dt= 1,
du
dt= 2
where x(0, s) = s, y(0, s) = 0, u(0, s) = s2. Solving this set we get
x(t, s) = t + s, y(t, s) = t, u(t, s) = 2 t + s2. This gives u(x, y) =
2 y+(x− y)2. To see the characteristics in the x− y plane we can use
the ODEdy
dx=b(x, y)
a(x, y)which gives y = x+ c for the current problem.
30
4.2.1 Transversality condition
In order for the PDE to have a unique solution near the initial curve
Γ(s) the transformation x(t, s), y(t, s) has to be invertible. That is the
jacobian should be non-zero on the points of initial curve Γ.
∂(x, y)
∂(t, s)=
∣
∣
∣
∣
∣
xt yt
xs ys
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
a b
(x0)s (y0)s
∣
∣
∣
∣
∣
6= 0 Geometrically this means
projection of Γ on the x − y plane is tangent at this point to the
projection of the characteristic.
Example 4.2.2. Solve the equation ux = 1 subject to u(0, y) = g(y).
dx
dt= 1,
dy
dt= 0,
du
dt= 1
where x(0, s) = 0, y(0, s) = s, u(0, s) = g(s). Solving this set we
get x(t, s) = t, y(t, s) = s, u(t, s) = t + g(s). This gives u(x, y) =
x + g(y). If the initial condition is changed to u(x, 0) = h(x). we get
where x(0, s) = s, y(0, s) = 0, u(0, s) = h(s). Solving for this condi-
tion we get x(t, s) = t+s, y(t, s) = 0, u(t, s) = t+h(s). The transver-
sality condition
∣
∣
∣
∣
∣
1 0
1 0
∣
∣
∣
∣
∣
= 0. The solution has a problem with unique-
ness. The characterisitc curves for this equation isdy
dx= 0 ⇒ y = c.
The initial curve Γ(s) is the x axis which is also the characteristic
curve on y = 0. So the initial curve Γ and the characteristic curve
coincides. So we loose the uniqueness near the initial curve.
Example 4.2.3. Solve the equation ux + uy + u = 1 subject to u =
sin x, on y = x+ x2, x > 0.
dx
dt= 1,
dy
dt= 1,
du
dt= 1− u
31
−5 −4 −3 −2 −1 0 1 2 3 4 5−10
−5
0
5
10
15
20
25
30
Initial curve
characteristic curves
where x(0, s) = s, y(0, s) = s + s2, u(0, s) = sin s. Solving this set
we get x(t, s) = t+ s, y(t, s) = t+ s+ s2, u(t, s) = 1+ (sin s− 1) e−t.
This gives
u(x, y) = 1+(
sin√y − x− 1
)
e−(x−√y−x). This solution is valid in
the domain D = {(x, y)|0 < x < y} ∪
{(x, y)|x ≤ 0 and x+ x2 < y} The transversality condition
∣
∣
∣
∣
∣
1 1
1 1 + 2s
∣
∣
∣
∣
∣
=
2 s 6= 0 when s 6= 0. If you observe clearly the solution u is not differ-
entiable at the origin. The characteristics aredy
dx= 1 ⇒ y = x + c.
We can observe that the Slope of the initial curve at the origin =
Slope of the characteristic at the origin. When choosing s we have
omitted s = −√y − x. So the solution near the curve in the region
{(x, y)|x < 0 and y = x+ x2} gives non-uniqueness of the solution.
Example 4.2.4. Solve the equation −y ux + xuy = u subject to
32
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Initial curve
characteristic curve
u(x, 0) = ψ(x).dx
dt= −y, dy
dt= x,
du
dt= u
where x(0, s) = s, y(0, s) = 0, u(0, s) = ψ(s). Solving this set
we get x(t, s) = s cos t, y(t, s) = s sin t, u(t, s) = et ψ(s). This gives
u(x, y) = ψ(
√
x2 + y2)
etan−1 ( y
x).
The transversality condition
∣
∣
∣
∣
∣
1 s
1 0
∣
∣
∣
∣
∣
= −s 6= 0 when s 6= 0. In
choosing ψ(
√
x2 + y2)
we have assumed x > 0. Each characteristic
intersects the initial curve Γ twice.
Example 4.2.5. Solve the equation ux + 3 y23 uy = 2 subject to
33
u(x, 1) = 1 + x.dx
dt= 1,
dy
dt= 3 y
23 ,
du
dt= 2
where x(0, s) = s, y(0, s) = 1, u(0, s) = 1 + s. Solving this set we
get x(t, s) = t + s, y(t, s) = (t + 1)3 u(t, s) = 2 t + s + 1. This
gives u(x, y) = x + y13 . The transversality condition
∣
∣
∣
∣
∣
1 3
1 0
∣
∣
∣
∣
∣
= −3 6=
y=1
Initial curve
Characteristic curve
0. The characteristic equationdy
dt= 3 y
23 with y(0, s) = 1 is not
Lipchitz continuous at the origin. This gives y = t3 and y = 0 as
solutions. So this does not have unique solution. Hence y = 0 is also
a characterisitic. We also get y = (x − s + 1)3. For each point s
on y = 1 we get different characteristic curve which intersects with
34
another characteristic y = 0. Thus u may not have unique solution
there. Hence the solution u(x, y) = x+ y13 is singular on the x− axis.
4.3 Lagranges Method
Consider the quasilinear PDE a(x, y, u) ux+b(x, y, u) uy = c(x, y, u).
The characteristic equations are
dx
dt= a(x, y, u),
dy
dt= b(x, y, u),
du
dt= c(x, y, u)
From this we can extract two set of equations
dy
dx=b(x, y, u)
a(x, y, u),du
dx=c(x, y, u)
a(x, y, u)
We get two family of curves φ(x, y, u) = α, ψ(x, y, u) = β. So the
general solution to the quasilinear PDE is
F (φ(x, y, u), ψ(x, y, u)) = 0.
Lemma 4.3.1. Let φ = φ(x, y, u), ψ = ψ(x, y, u).
If f(φ, ψ) = 0 then u satisfies
∂u
∂x
∂(φ, ψ)
∂(y, u)+∂u
∂y
∂(φ, ψ)
∂(u, x)=∂(φ, ψ)
∂(x, y)
Proof. We know that f(φ, ψ) = 0. Differentiating with respect to x
and y we get
∂f
∂φ
(
∂φ
∂x+∂φ
∂u
∂u
∂x
)
+∂f
∂ψ
(
∂ψ
∂x+∂ψ
∂u
∂u
∂x
)
= 0
∂f
∂φ
(
∂φ
∂y+∂φ
∂u
∂u
∂y
)
+∂f
∂ψ
(
∂ψ
∂y+∂ψ
∂u
∂u
∂y
)
= 0
35
∣
∣
∣
∣
∣
φx + ux φu ψx + ux ψu
φy + uy φu ψy + uy ψu
∣
∣
∣
∣
∣
= 0
which gives∂u
∂x
∂(φ, ψ)
∂(y, u)+∂u
∂y
∂(φ, ψ)
∂(u, x)=∂(φ, ψ)
∂(x, y)
Theorem 4.3.2. The general solution of PDE a(x, y, u) ux+b(x, y, u) uy =
c(x, y, u) is f(φ, ψ) = 0, where φ(x, y, u) = c1, ψ(x, y, u) = c2 are the
solution curves of
dx
a(x, y, u)=
dy
b(x, y, u)=
du
c(x, y, u)
Proof. Since φ(x, y, u) = c1 and ψ(x, y, u) = c2
dφ = φx dx+ φy dy + φu du = 0
dψ = ψx dx+ ψy dy + ψu du = 0
Usingdx
a(x, y, u)=
dy
b(x, y, u)=
du
c(x, y, u)
a φx + b φy + c φu = 0
aψx + b ψy + c ψu = 0
Solving for a, b, c we get
a∂(φ,ψ)∂(y,u)
=b
∂(φ,ψ)∂(u,x)
=c
∂(φ,ψ)∂(x,y)
The proof can be deduced by using the above lemma.
36
Example 4.3.1. Find the general solution of xux + y uy = u. The
set of equations aredx
x=dy
y=du
u
We get φ(x, y, u) =y
x= c1, ψ(x, y, u) =
u
x= c2. Thus the general
solution is f(y
x,u
x
)
= 0. This can be written as u(x, y) = x g(y
x
)
.
Example 4.3.2. Find the general solution of x2 ux+y2 uy = (x+y) u.
The set of equations are
dx
x2=dy
y2=
du
(x+ y) u
dx
x2=dy
y2gives
y − x
xy= c1 and
dx− dy
x2 − y2=
du
(x+ y) u⇒ x− y
u= c2.
The general solution of the above PDE is f
(
y − x
xy,x− y
u
)
= 0.
Example 4.3.3. Find the general solution of PDE u (x+y)+u (x−y) uy = x2 + y2, u = 0on y = 2x. The set of equations are
dx
u (x+ y)=
dy
u (x− y)=
du
x2 + y2
y dx+ x dy − u du
0=x dx− y dy − u du
0
d
[(
xy − u2
2
)]
= 0, d
[
x2 − y2 − u2
2
]
= 0 which gives u2−x2+ y2 =
c1 and 2 xy−u2 = c2. The general solution is f(
x2 + y2 − u2, 2 xy − u2)
=
0. Using the initial condition u = 0 on y = 2x. We get 4 c1 = 3 c2 ⇒7 u2 = 6 xy + 4 (x2 − y2).
37
38
Chapter 5
Tutorial 1
Exercise 5.0.1.
∂2u
∂t2− 9
∂2u
∂x2= 0, −∞ < x <∞, t > 0
u(x, 0) =
1, if |x| ≤ 2
0, if |x| > 2
∂u
∂t(x, 0) =
1, if |x| ≤ 2
0, if |x| > 2(5.0.1)
(i) Find u(0,1
6)?
(ii) Find large time behaviour of the solution limt→∞
u(ξ, t)?
for fixed ξ ∈ R.
Exercise 5.0.2. Using D’Alemberts form of solution for wave equa-
39
tion find the solution of
∂2u
∂t2− c2
∂2u
∂x2= 0, −∞ < x <∞, t > 0
u(x, 0) = e−x2
, −∞ < x <∞∂u
∂t(x, 0) = 0, −∞ < x <∞ (5.0.2)
Exercise 5.0.3. Thermal conductivity of an aluminium alloy is 1.5
W/cm K. Calculate the steady state temperature of an aluminium bar
of length 1m (insulated along the sides) with its left end fixed at 20◦C
and right end fixed at 30◦C.
Exercise 5.0.4. (a) Show that the function
u(x, t) = e− kθ2t
ρc sin(θx) is a solution to the homogeneous heat
equation ρc ∂u∂t
− k ∂2u∂x2
= 0, 0 < x < l, ∀t.
(b) What values of θ will cause u to also satisfy homogeneous Dirich-
let conditions at x = 0 and x = l?
Exercise 5.0.5. Classify the PDE and reduce to its canonical form
(a)∂2u
∂x2− 2 sin x
∂2u
∂x∂y− cos2 x
∂2u
∂y2− cosx
∂u
∂y= 0
(b)
y5∂2u
∂x2− y
∂2u
∂y2+ 2
∂u
∂y= 0, y > 0
(c)∂2u
∂x2+ (1 + y2)2
∂2u
∂y2− 2y (1 + y2)
∂u
∂y= 0
40
Exercise 5.0.6. Consider the equation x∂2u
∂x2−y ∂
2u
∂y2+1
2
(
∂u
∂x− ∂u
∂y
)
=
0
(a) Find the domain where the equation is elliptic and the domain
where it is hyperbolic.
(b) For each of the above two domains, find the corresponding canon-
ical transformation.
Exercise 5.0.7. Using Separation of variables find the solution to the
following PDE’s
(a)
∂2u
∂t2=∂2u
∂x2, 0 < x < π, t > 0
u(0, t) = u(π, t) = 0, t ≥ 0
u(x, 0) = sin3x, 0 ≤ x ≤ π
∂u
∂t(x, 0) = sin 2x, 0 ≤ x ≤ π (5.0.3)
(b) Solve the heat equation∂u
∂t= 12
∂2u
∂x2, 0 < x < π, t > 0 subject
to the following boundary and initial condition
∂u
∂x(0, t) =
∂u
∂x(π, t) = 0, t ≥ 0
u(x, 0) = 1 + sin3 x, 0 ≤ x ≤ π (5.0.4)
Find limt→∞
u(x, t) for all 0 < x < π and explain physical interpre-
tation of your result.
41
(c) Consider the heat equation∂u
∂t− ∂2u
∂x2= 0, x ∈ R, t ≥ 0.
Find the transformation λ(x, t) by solving an ODE of the form
φ(λ).
42
Chapter 6
Tutorial II
Exercise 6.0.8. Solve the PDE subject to the given Cauchy condition
on the Cauchy data curve
(1) 2 ux − 5 uy = 4, subject to the Cauchy condition u(x, 0) = x.
(2) ux + 3 uy = u+ 2, subject to the Cauchy condition u(0, y) = y.
(3) xux + 2y uy = 3 u, subject to the Cauchy condition u(1, y) =
cos y for y > 1 .
(4) ux−2 uy = u−1, subject to the Cauchy condition u = 2y on x =
ky, giving reasons for any restriction that must be placed on k.
(5) Solve by the method of characteristic ux + 2 xuy = 2 xu, given
that u = x2 on the initial curve Γ with the equation y =x2
2.
(6) ut+2 u ux = 0, subject to the Cauchy condition u(x, 0) = tanh x.
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(7) ut − u ux = et, subject to the Cauchy condition u(x, 0) = −x.
(8) u ut+ux = 0, subject to the Cauchy condition u(x, 1) =1
xfor x ≥
1.
(9) ut + u ux = t, subject to the Cauchy condition u(x, 0) = −2x.
Find the domain in the upper half of (x, t) plane where the
solution is valid.
(10) Solve the PDE ut+3 u3 ux = 0 subject to the Cauchy condition
u(x, 0) =
−1, −∞ < x < −1
x, −1 ≤ x ≤ 4
4, 4 < x <∞
Exercise 6.0.9. Solve the following PDE by Lagrange’s method
(1) 3 ux + 2 uy = 0 with u(x, 0) = sin(x)
(2) y ux + xuy = 0 with u(0, y) = e−y2
(3) y ux + xuy = x y, x ≥ 0, y ≥ 0 with u(0, y) = e−y2
for y > 0, u(x, 0) = e−x2
, for x > 0
(4) 2 xy ux + (x2 + y2) uy = 0, u = ex
x−y on x+ y = 1.
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