Minimization and Constraints

of

Partial Differential Equations

Author: Cathal Ormond - 07616520Supervisor: Dr. John Stalker

Contents

1 Preliminaries 11.1 Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Euler-Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 Travelling Wave Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.7 The Beam Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Energy and Momentum 92.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Noether’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Euler-Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Minimization of Energy 133.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Beam Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Conclusions 184.1 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

A Proofs 19A.1 Noether’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Bibliography 23

1

Abstract

We seek to examine solutions of certain linear and non-linear partial differential equations whichwill minimize the PDE’s energy when the momentum is constant. In particular, we will look atthe Wave Equation and the Beam Equation, where the non-linearities are functions of the solutionalone. We will begin by an introduction to some of the basics of minimization problems and to thecalculus of variations. To find explicit values of the energy and momentum for our chosen PDEs,we will apply Noether’s Theorem to spatial and time symmetries. The version of the theorem thatI am presenting is a simplification of the one as detailed in [4], which depends on Lie groups. Oncewe have determined the values of our energy and momentum, we will investigate which solutionsminimize the energy whose momenta are constant.

Chapter 1

Preliminaries

Before we begin our extremization, we need to make some basic definitions and introduce somenotation. Many of the concepts used below will be familiar in terms of mechanics (especiallyNewtonian and Lagrangian mechanics), but I will be defining them in more abstract, mathematicalterms.

1.1 Calculus of Variations

The calculus of variations deals mostly with “finding functional forms for which integral formsassume maximum or minimum values”[2]. We will try to determine the value of u = u(x) on [a, b]for which

I(u) =

∫ b

aF (x, u(x), u′(x)) dx

is a minimum, where F is a given functional form. As we do not (yet) know the relationshipbetween x and u, we wish to find the path of integration which minimizes our above integral. Wewill usually assume that F has specific properties, i.e smoothness, or at least existence of partialderivatives up to a certain order.

Definition 1.1.1 (First Variation). Given a functional form I, we define the First Variation ofI as

δI(u) =d

dεI(u+ εv)

∣∣∣∣ε=0

where v is an arbitrary function, usually assumed to be differentiable with compact support.

The theory behind this definition is that if u is the function which minimizes the value of I,then we must have that for any other function v and ε ∈ R

I(u) ≤ I(u+ εv)

when ε is sufficiently small. Considering f(ε) = I(u+ εv) as a real valued function, the derivativeof f must vanish at the point ε = 0. The case for a maximum is analogous. Hence, to find thefunction which make our functional a maximum or minimum, we can consider the first variation ofthe functional, and determine the functions for which the first variation vanishes.

1

1.2. EULER-LAGRANGE EQUATIONS CHAPTER 1. PRELIMINARIES

Lemma 1.1.2 (Fundamental Lemma of The Calculus of Variations). Suppose that f and is a Ck

function on [a, b] and that ∫ b

af(x)g(x) dx = 0

for all Ck functions g on (a, b) which vanish at a and b. Then f must be identically zero on (a, b).

Proof. Let f be a function as above, and let h(x) be some smooth function which is positive on(a, b) and vanishes at a and b, e.g. h(x) = (x−a)(b−x), if a and b are both finite. If the endpointsare infinite, a Gaussian function would also suffice, i.e. h(x) = e−x

2. Then, we let g(x) = h(x)f(x),

which is a Ck function on (a, b) which vanishes at its endpoints. We have that

0 =

∫ b

af(x)g(x) dx =

∫ b

af(x)2h(x) dx

As the integrand is non-negative for all values of x ∈ (a, b), we must have that it is identicallyzero on (a, b). Since h(x) is non-negative, we must have that f(x) is identically zero on (a, b), asrequired.

1.2 Euler-Lagrange Equations

The Euler-Lagrange Equations sometimes provide a simpler method of determining the extremalvalues of a functional form, although in essence they use the same method.

Theorem 1.2.1. Let u : R→ R be an extremum of the integral

I(u) =

∫ b

aL(x, u(x), u′(x)) dt

where L = L(x, u, u′) is some functional with continuous first-order derivatives. Then, u mustsatisfy the Euler-Lagrange Equation for L, namely

∂L

∂u=

d

dx

(∂L

∂u′

)Proof. Suppose that u(x) is an extremal of I, that ε > 0 and that v(x) is a differentiable functionwhich vanishes on its boundary. We define

w(x) = u(x) + εv(x)

We will compute the first variation of I at w. We have

d

dεI(w) =

∫ b

a

dL

dε(x,w,w′) dx

=

∫ b

a

[∂L

∂x

dx

dε+∂L

∂w

∂w

∂ε+∂L

∂w′∂w′

∂ε

]dx

=

∫ b

a

[∂L

∂wv(x) +

∂L

∂w′v′(x)

]dx

2

1.2. EULER-LAGRANGE EQUATIONS CHAPTER 1. PRELIMINARIES

Letting ε→ 0, we have that w → u, so we see that

d

dεI(w)

∣∣∣∣ε=0

= δI(u) =

∫ b

a

[∂L

∂uv(x) +

∂L

∂u′v′(x)

]dx

which we know must vanish, as u was assumed to be an extremal of L. Now, we integrate thesecond term in our integral by parts to see that

0 = δI(u) =

∫ b

av(x)

[∂L

∂u− d

dx

∂L

∂u′

]dx

as v vanishes at on its boundary. We apply Lemma 1.1.2 above to see that

∂L

∂u− d

dx

∂L

∂u′= 0

Example 1.2.2. Possibly the most basic example one could consider for this problem is findinga function which gives the shortest distance between two points on the plane, a and b. We mayalready guess that the function which minimizes this distance should be given by a straight line.The arc-length of any function y = y(x) which passes through these two points is a functional formgiven by

l(y) =

∫ b

a

√1 +

(dy

dx

)2

dx

If y minimizes this integral, then by above, it must satisfy the Euler-Lagrange Equation, so if welet L(x, y, y′) =

√1 + (y′)2 , then

∂L

∂y=

d

dx

∂L

∂y′⇔ 0 =

d

dx

(y′√

1 + (y′)2

)

Hence, the rightmost value in brackets is independent of x, so we have

d

dx

(y′(x)√

1 + (y′(x))2

)= 0 ⇔ y′(x)√

1 + (y′(x))2= c

⇔ y′(x) =c√

1− c2= A

⇔ y(x) = Ax+B

Hence, as we might have already guessed, the function which minimizes the distance between pointsin the place is linear. A more interesting yet related example is Brachistochrone problem: to findthe curve of “fastest descent” between two points, i.e. the shortest curve a point traces whenstarting at rest and accelerated by gravity from one point in the plane to another.

3

1.3. LAGRANGIAN CHAPTER 1. PRELIMINARIES

1.3 Lagrangian

When we consider a PDE in the general case, we will assume that any solution is a mappingu : Rn × [0,∞) → R. We will use the notation that u = u(x1, x2, . . . , xn, t), where x1, x2, . . . , xnrepresent our spatial coordinates, and t represents our time coordinate. The Lagrangian of agiven partial differential equation will be a smooth functional of u, its derivatives and its coordi-nates, where any solution of the PDE must satisfy the Euler Lagrange Equations for L. We willalso assume that the Lagrangian is invariant under transformation of coordinates.

The Lagrangian will be the main functional with which we will concern ourselves. It representsthe dynamics of the PDE we will be considering, in particular the energy and momentum. As anexample, we will consider the Lagrangian of the function u : R2 → R given by

L(x, y, u, ux, uy) =1

2u2x +

1

2u2y

It turns our that our PDE depends on spatial coordinates, not time coordinates, so we use y insteadof t. Applying the Euler Lagrange Equations, we have

∂L

∂u=

d

dx

(∂L

∂ux

)+

d

dx

(∂L

∂ux

)⇔ 0 =

d

dxux +

d

dyuy

⇔ 0 = uxx + uyy

which gives us the Laplace Equation. Historically, many partial differential equations were derivedby considering various Lagrangians under the Euler-Lagrange Equations, as was indeed the casefor Laplace’s Equation.

1.4 Lagrange Multipliers

A familiar technique in extremization problems with constraints is the method of Lagrange Multi-pliers. The following explanation is adapted from [5]. We will consider a finite dimensional examplefor convenience, but it turns out that the methods is still true for infinite dimensional spaces, whichis what we will be considering later on.

We wish to find extremal functions for the functional F (u1, u2, . . . , un) subject to the constraintthat f(u1, u2, . . . , un) = 0. We begin by noting that the first variation of f must vanish, so

δf =n∑i=1

∂f

∂uiδui = 0 (1.4.1)

Also, we wish to find functions which minimize F , so its first variation must also vanish:

δF =n∑i=1

∂F

∂uiδui = 0 (1.4.2)

Now, if the δui terms were all linearly independent, then ∂F∂ui

would have to vanish for each i =1, 2, . . . , n. However, this cannot be true because of equation 1.4.1. Hence, we will, without loss

4

1.4. LAGRANGE MULTIPLIERS CHAPTER 1. PRELIMINARIES

of generality, we will eliminate δun in terms of each other term, relabelling the terms if necessary.Now, we can combine equations 1.4.1 and 1.4.2 to note that

δF − λδf =n∑i=1

(∂F

∂ui− λ ∂f

∂ui

)δui = 0 (1.4.3)

where λ is some undetermined factor. As we wished to eliminate δun, we can choose λ as

∂F

∂un− λ ∂f

∂un= 0

Hence, equation 1.4.3 becomes

δF − λδf =

n−1∑i=1

(∂F

∂ui− λ ∂f

∂ui

)δui = 0

and so the coefficient of δui must vanish, so

∂F

∂ui− λ ∂f

∂ui= 0

for all i = 1, 2, . . . , n. Thus, the extrema of F may be determined by the above n equations.

Example 1.4.1. We’ll consider a basic minimization problem. Suppose that we wish to findthe extrema of F (u, v) = u − v subject to the fact that u(x)2 + v(x)2 = 1 for all x ∈ R, whereu, v : R→ R. For convenience, we’ll let f(u, v) = u2+v2−1 = 0. We then have that the derivativesof the function

Λ(u, v, λ) = F (u, v)− λf(u, v) = u− v − λ(u2 + v2 − 1)

must all vanish. This gives us:

∂Λ

∂u= 0 ⇔ 1− 2λu = 0

∂Λ

∂v= 0 ⇔ −1− 2λv = 0

∂Λ

∂λ= 0 ⇔ u2 + v2 − 1 = 0

The final equation is nothing more that the original constraint. If we solve for λ for the first twoequations and equate the results, we have that v(x) = −u(x) for all x ∈ R. Now, we recall ourinitial conditions, and substitute for u to see that

2u(x)2 = 1 ⇔ u(x) = ± 1√2

for all x. Hence, the extrema of F are constant functions given by:

u(x) =1√2

v(x) = − 1√2

or u(x) = − 1√2

v(x) =1√2

5

1.5. TRAVELLING WAVE SOLUTIONS CHAPTER 1. PRELIMINARIES

for all x ∈ R. We can see that when u is positive, F is a maximum with F (u, v) =√

2 , and when vis positive, F is a minimum with F (u, v) = −

√2 . To test this, let u(x) = sinx and v(x) = − cosx.

We have that (sinx)2 + (− cosx)2 = 1, so our initial constraint is satisfied. By trigonometricformulae, we can write F in the form

F (u, v) = sinx+ cosx =√

2 sin(x+

π

4

)Hence, as sinx ∈ [−1, 1] for all x ∈ R, we have that F (u, v) ∈ [−

√2 ,√

2 ], which agrees with whatwe determined above.

1.5 Travelling Wave Solutions

Suppose we have some solution to a partial differential equation u = u(x, t), where we impose somegeneric initial condition:

u(x, 0) = ϕ(x)

Suppose that our solution also solvesut = vux

for some constant v ∈ R We will consider two methods of funding solutions to this initial valueproblem.

1.5.1 Method of Characteristics

Suppose that we can parametrize the coordinates in terms of some variable s. We then solve for

dt

ds= 1

dx

ds= v

du

ds= 0

under initial data. These lines are called characteristics. Solving these ODEs, we see that t = s,x = vs+x0 and u(s) = u(x(s), t(s)) = u0. This gives the relation that x0 = x−vt. As u is constantwith respect to s, we have that

u(x, t) = u(x(0), t(0))= u(x0, 0)= ϕ(x0)= ϕ(x− vt)

as required.

1.5.2 Symmetry Reduction

The characteristics may not always exist, however. Even if they do, they may cross each other andhence solutions like the above may not exist. However, if our solutions are symmetry invariant, wecan still solve the above problem. Suppose that our solutions are translation invariant, i.e.

u(x+ α1s, t+ α2s) = u(x, t)

6

1.6. THE WAVE EQUATION CHAPTER 1. PRELIMINARIES

for α1, α2 ∈ R Taking s = − tα1

, we have u(x, t) = u(x − λt, 0) = ϕ(x − λt), where λ = α2α1

. Now,substituting this back into our original equation, we must have

ut = vux ⇔ −λϕ(x− λt) = vϕ(x− λt)

Assuming that our solutions is not initially the zero solution, we have that v = −λ. In practice,we will not concern ourselves with the change of sign, as our v term is squared in each PDE weconsider, so we can simply replace v by −v.

In either case, whether characteristics exist or not, our solutions are of the form

u(x, t) = ϕ(x− vt)

We call these solutions Travelling Wave Solutions.

1.6 The Wave Equation

The one dimensional wave equation will be our first basic model for our minimization problem. Afunction u = u(x, t) is said to satisfy the (linear) wave equation if

∂2u

∂t2= c2

∂2u

∂x2

for some non-zero constant c ∈ R. We will use the following notation also

u(x, 0) = ϕ(x) ut(x, 0) = ψ(x)

The wave equation has the following Lagrangian

I(u) =

∫ ∞−∞

1

2(ut)

2 − 1

2(cux)2 dx

Later, we will also study a non-linear version of the wave equation, namely

∂2u

∂t2− c2∂

2u

∂x2= f(u)

The Lagrangian of this non-linear equation is given by

I(u) =

∫ ∞−∞

1

2(ut)

2 − 1

2(cux)2 + F (u) dx

where F is some antiderivative of u. We will sometimes assume that c = 1 for convenience.

1.7 The Beam Equation

The Beam Equation forms a more interesting example, being a higher order PDE, although we willconsider simplified version. A function u = u(x, t) is said to satisfy the Beam Equation if

∂2u

∂t2+ a2

∂u4

∂x4= 0

7

1.7. THE BEAM EQUATION CHAPTER 1. PRELIMINARIES

for some non-zero constant a ∈ R. Again, we will adopt the following notation also

u(x, 0) = ϕ(x) ut(x, 0) = ψ(x)

The beam equation has the following Lagrangian

I(u) =

∫ ∞−∞

1

2(ut)

2 − 1

2(auxx)2 dx

Again, we will study a non-linear version of the beam equation, namely

∂2u

∂t2+ a2

∂4u

∂x4= f(u)

where the Lagrangian is given by

I(u) =

∫ ∞−∞

1

2(ut)

2 − 1

2(auxx)2 + F (u) dx

again where F is some antiderivative of u.

8

Chapter 2

Energy and Momentum

2.1 Introduction

In this section, we will develop methods for determining the energy and momentum of a given PDE.The first method we will use is Noether’s Theorem, which states that certain symmetries of PDEsgive conserved quantities. The statement for Noether’s theorem below is a simplified version (aswe only need to consider one kind of symmetry), and assumes that our Lagrangian has no higherorder derivatives than first order derivatives. For the Beam Equation, our Lagrangian has higherorder derivatives, and so this version of Noether’s Theorem does not apply. However, we can usemethods of calculus of variations to determine what our energy and momentum should be.

2.2 Noether’s Theorem

For this version of Noether’s Theorem, some background knowledge of Lie groups and symmetrygroups is required. The particular symmetries we will be considering are translation of coordinates,so we need some generator for the group of translations. Since much of this background material issomewhat beyond the scope of this Research Assignment, we will state, possible quite rashly, thatthe group of the translation (x, t) 7→ (x, t) + (α1, α2) is generated by the vector field

v = α1∂

∂t+ α2

∂

∂x

A fuller justification of this can be found in [4].

Theorem 2.2.1 (Noether’s Theorem (Simplified)). Suppose that the Lagrangian L depends onlyon first order derivatives, and that the vector field that generates the group of symmetries is of theform

v =n∑i=1

ζi(x, u)∂

∂xi

Then,

Pi = ζiL− ∂L

∂ui

n∑j=1

ζj∂u

∂xj

is constant with respect to xi for i = 1, 2, . . . , n.

9

2.2. NOETHER’S THEOREM CHAPTER 2. ENERGY AND MOMENTUM

We will not provide a proof for this, as it quite complicated. A particular version is proved inthe appendix, and a proof for the general case may be found in [4].

2.2.1 The Wave Equation

We will consider what implications Noether’s theorem has for the wave equation. Specifically,we will determine what the energy and momentum are for both the linear and non-linear waveequation.

Example 2.2.2 (Linear Wave Equation). The Lagrangian density for the wave equation

L(x, t, u, ux, ut) =1

2u2t −

1

2(cux)2

depends only on first order derivatives of solutions to the wave equation, so we apply the previoustheorem to see that two quantities are conserved:

P1 = α1L−∂L

∂ut

(α1∂u

∂t+ α2

∂u

∂x

)= −1

2α1u2t − 1

2α1c2u2x − α2uxut

P2 = α2L−∂L

∂ux

(α1∂u

∂t+ α2

∂u

∂x

)= 1

2α2u2t + 1

2α2c2u2x + α1c

2uxut

Note that the above are the densities for the conserved quantities. Letting α2 = 0, we have timetranslation invariance and letting α1 = 0, we have spatial translation invariance. In either case, wehave that

E(t) =1

2

∫ ∞−∞

u2t + c2u2x dx

P (t) =

∫ ∞−∞

uxut dx

(2.2.1)

are conserved. This is the energy and momentum of the wave equation. If we let t = 0, our initialenergy and momentum are given by

E =1

2

∫ ∞−∞

ψ(x)2 + c2(ϕ′(x))2 dx

P =

∫ ∞−∞

ψ(x)ϕ′(x) dx

(2.2.2)

Example 2.2.3 (Non-Linear Wave Equation). The Lagrangian density for the non-linear waveequation is given by

L(x, t, u, ux, ut) =1

2u2t −

1

2(cux)2 + F (u)

It also depends only on first order derivatives of solutions to the wave equation, so:

P1 = α1L−∂L

∂ut

(α1∂u

∂t+ α2

∂u

∂x

)= −1

2α1u2t − 1

2α1c2u2x − α2uxut + α1F (u)

P2 = α2L−∂L

∂ux

(α1∂u

∂t+ α2

∂u

∂x

)= 1

2α2u2t + 1

2α2c2u2x + α1c

2uxut + α2F (u)

10

2.3. EULER-LAGRANGE EQUATIONS CHAPTER 2. ENERGY AND MOMENTUM

This gives us that our energy E and momentum P are given by

E(t) =

∫ ∞−∞

1

2u2t +

1

2c2u2x + F (u) dx

P (t) =

∫utux dx

(2.2.3)

Again, under initial data, this becomes

E =

∫ ∞−∞

1

2ϕ(x)2 +

1

2c2(ϕ′(x))2 + F (ϕ) dx

P =

∫ ∞−∞

ψ(x)ϕ′(x) dx

(2.2.4)

2.3 Euler-Lagrange Equations

As mentioned above, this method fails for the beam equation. However, we can apply a strongerversion of Noether’s theorem (again in [4]) and use the Euler-Lagrange equations for the BeamEquation to determine the energy and momentum. The idea is that if we multiply the Euler-Lagrange equations by ut, and express this as the divergence of some scalar field, then the firstcoordinate of this field (i.e. the coordinate which is differentiated with respect to t in the divergence)will be our conserved quantity for the time translation symmetry. For the momentum, we multiplythe Euler-Lagrange equations by ux, and again express this as the divergence of some scalar field.The first coordinate of this field will be our conserved quantity for the spatial translation symmetry.

As an example of this, we return briefly to the wave equation. The Lagrangian density is asabove, so we have the Euler-Lagrange equations for the as:

utt − uxx = 0

So, for the energy, we wish to express(utt − uxx)ut

as the divergence of some scalar field. One can see by direct computation that

∇ ·(

1

2u2t +

1

2ux,−utux

)= (utt − uxx)ut

Hence, the first coordinate is our conserved density, so the energy is given by

E(t) =1

2

∫ ∞−∞

u2t + u2x dx

which agrees with what we worked out above.

11

2.3. EULER-LAGRANGE EQUATIONS CHAPTER 2. ENERGY AND MOMENTUM

2.3.1 Beam Equation

We will perform the same calculations for the linear and non-linear beam equation. The Lagrangiandensity for the non-linear beam equation is given by

1

2u2t −

1

2(auxx)2 + F (u)

so the Euler-Lagrange equations are

utt + a2uxxxx + f(u) = 0

We wish to find some scalar field A such that

∇ ·A = (utt + a2uxxxx + f(u))ut

The scalar field is give by A = (12u2t + 1

2a2u2xx + F (u), a2uxxxut − uxxutx) as

∇ ·A =∂

∂t

(12u

2t + 1

2a2u2xx + F (u)

)+

∂

∂x

(a2uxxxut − uxxutx

)= (uttut + a2uxxuxxt + f(u)ut) + (a2uxxxxut + a2uxxxutx − uxxxutx − uxxutxx)

= (utt + a2uxxxx + f(u))ut

Thus, our energy is given by

E(t) =

∫ ∞−∞

1

2u2t +

1

2a2u2xx + F (u) dx

We now wish to find some scalar field B such that

∇ ·B = (utt + uxxxx + f(u))ux

The scalar field is give by B = (utux, uxuxxx − 12u

2t + 1

2a2u2xx + F (u)) as

∇ ·B =∂

∂t(utux) +

∂

∂x(a2uxuxxx − 1

2u2t − 1

2a2u2xx + F (u))

= (uttux + utuxt) + (a2uxxuxxx + a2uxuxxxx − ututx − a2uxxuxxx + f(u)ux)

= (utt + a2uxxxx + f(u))ux

Hence, the momentum is given by

P (t) =

∫ ∞−∞

utux dx

Letting t = 0, our energy and momentum under initial data are given by

E =

∫ ∞−∞

1

2ψ(x)2 +

1

2a2ϕ′′(x)2 + F (ϕ(x)) dx

P =

∫ ∞−∞

ψ(x)ϕ′(x) dx

(2.3.1)

12

Chapter 3

Minimization of Energy

3.1 Introduction

Theorem 3.1.1. The integral

I(x) =

∫ ∞−∞

F (t, xi, x(1)i , . . . , x

(k)i ) (3.1.1)

where i = 1, 2, . . . , n and x(j)i =

djxidtj

, is stationary for weak variations when the following n

equations are satisfied

∂F

∂xi=

k∑j=1

(−1)j+1 dj

dtj

(∂F

∂x(j)i

)for i = 1, 2, . . . , n

Proof. We will transform each function xi to yi = xi + εiqi, where qi = qi(t) has compact support,and each εi is linearly independent and independent of t for all i = 1, 2, . . . , n. Firstly note that

∂y(l)j

∂εi= δji q

(l)j

by direct computation. We will compute the first variation of I(x) for each εi:

δI(x)i =d

dεiI(x+ εq)

∣∣∣∣εi=0

=

∫ ∞−∞

∂F∂t

dt

dεi+

k∑j=0

(n∑l=1

∂F

∂y(j)l

dy(j)l

dεi

) dt

∣∣∣∣∣∣ε=0

=

∫ ∞−∞

k∑j=0

∂F

∂y(j)i

q(j)i

dt

∣∣∣∣∣∣ε=0

=

∫ ∞−∞

k∑j=0

∂F

∂x(j)i

q(j)i

dt

13

3.2. THE WAVE EQUATION CHAPTER 3. MINIMIZATION OF ENERGY

Now, we integrate each term by parts a sufficient number of times so that our integral is in termsof qi only, and not its derivatives. As qi has compact support for all i, each of the boundary termswill vanish except for a change in sign, so we have:

δI(x)i =

∫ ∞−∞

∂F∂xi−

k∑j=1

(−1)j+1 dj

dtj

(∂F

∂x(j)i

) qi dtBy 1.1.2, and given that each variation must vanish if x is an extremal, we must have that

∂F

∂xi−

k∑j=1

(−1)j+1 dj

dtj

(∂F

∂x(j)i

)= 0

for all i = 1, 2, . . . , n, as required.

3.2 The Wave Equation

3.2.1 Non-Linear Equation

We wish to examine solutions to the non-linear wave equation which minimize the energy and haveconstant momenta. The energy and momentum formulae can be are given above in equation 2.2.4.We will use the method of Lagrange multipliers, and consider the function

Λ(x, ϕ, ψ, ϕ′, ψ′) =

∫ ∞−∞

1

2ψ(x)2 +

1

2c2ϕ′(x)2 + F (ϕ(x))− λψ(x)ϕ′(x) dx+ λk (3.2.1)

Applying the theorem 3.1.1, we must have that:

∂F

∂ϕ− d

dt

∂F

∂ϕ′= 0 ⇒ f(ϕ)− c2ϕ′′(x) + λψ′(x) = 0 (3.2.2)

∂F

∂ψ− d

dt

∂F

∂ψ′= 0 ⇒ ψ(x)− λϕ′(x) = 0 (3.2.3)

Taking the derivative of the second equation, we have that

ψ′(x) = λϕ′′(x) (3.2.4)

Substituting this into the first equation, we get

f(ϕ) = (c2 − λ2)ϕ′′(x) (3.2.5)

For the non-linear equation, we can assume that f 6= 0, so the second derivative of ϕ doesn’t vanish,and λ 6= ±c. Returning to what equation 3.2.3 actually means in terms of u(x, t), we have

ut(x, 0) = λux(x, 0)

We can solve this by considering section 1.5.1, and so the functions which minimize the energymust be travelling wave solutions, i.e. u(x, t) = ϕ(x − λt). They must also, however, satisfy theabove ODE initially.

14

3.3. BEAM EQUATION CHAPTER 3. MINIMIZATION OF ENERGY

3.2.2 Linear Equation

If we set f(u) = 0, we have the linear wave equation. This changes only equation 3.2.2 above,giving

0 = c2ϕ′′(x)− λψ′(x)

which is nothing more that equation 3.2.3 with derivatives taken on both sides. If we substituteour travelling wave solution into the wave equation, we have that

0 = utt − c2uxx = λ2ϕ′′(x− λt)− c2ϕ′′(x− λt) = (λ2 − c2)ϕ′′(x− λt)

Hence, λ ought to be the velocity c for the travelling wave up to a sign.

3.3 Beam Equation

The Beam equation provides us with a more interesting results, as its Lagrangian depends on higherorder derivatives. As well as that, as long as solutions satisfy our constraint on momentum, it isalso possible to compute explicit solutions to the linear Beam Equation.

3.3.1 Non-Linear Equation

As with the wave equation, we consider the same problem:

Λ(x, ϕ, ψ, ϕ′, ψ′) =

∫ ∞−∞

1

2ψ(x)2 +

1

2a2ϕ′′(x)2 + F (ϕ(x))− λψ(x)ϕ′(x) dx+ λk (3.3.1)

where the energy and momentum are derived in the previous chapter. Applying theorem 3.1.1, wemust have that:

∂F

∂ϕ− d

dt

∂F

∂ϕ′+d2

dt2∂F

∂ϕ′′= 0 ⇒ f(ϕ) + λψ′(x) + a2ϕ(4)(x) = 0 (3.3.2)

∂F

∂ψ− d

dt

∂F

∂ψ′= 0 ⇒ ψ(x)− λϕ′(x) = 0 (3.3.3)

Again, as with the wave equation, we can differentiate the latter to see that ψ′(x) = λϕ′′(x), whichwe substitute into the former to see that

f(ϕ(x)) = −(a2ϕ(4)(x) + λ2ϕ′′(x))

We cannot use the methods of characteristics to solve the equation, as characteristics do not exist.Instead, we recall that solutions are translation invariant, so we have by section 1.5.2 above thatu(x, t) = ϕ(x− vt). However, unlike the wave equation, there is not much more we can say aboutthe non-linear beam equation. We cannot infer any relationships between λ, v and a.

15

3.3. BEAM EQUATION CHAPTER 3. MINIMIZATION OF ENERGY

3.3.2 Linear Equation

Letting f(u) = 0, we have the linear beam equation. From above, solutions for the linear beamequation must initially solve the above ODE:

a2ϕ(4)(x) + λ2ϕ′′(x) = 0

This is a linear, homogeneous ODE, and has as an auxiliary equation az4 + λz2 = 0, which hasroots 0, 0± ηi where η = λ

a . Hence, the above ODE has following solution

ϕ(x) = c1eηix + c2e

−ηix + c3x+ c4

= A1 cos(ηx) +A2 sin(ηx) +A3x+A4

where A1 = c1 + c2, A2 = c1 − c2, A3 = c3 and A4 = c4. Now we recall that our solution must alsobe a travelling wave solution, so u(x, t) = ϕ(x− vt). We can determine v by noting that

0 = utt + a2uxxxx

= v2ϕ′′(x− vt) + ϕ(4)(x− vt)

= (v2 − λ2)ϕ′′(x− vt)

As ϕ′′(x) 6= 0, we must have that λ = ±v. We’ll ignore the sign for the moment. This gives

u(x, t) = ϕ(x− λt)

= A1 cos(ηx− ηλt) +A2 sin(ηx− ηλt) +A3(x− λt) +A4

We will now try and determine the value of λ, and hence η, in terms of a. Considering u(x, t) =ϕ(x− λt), we substitute this into our original equation. Hence:

0 = utt + a2uxxxx

= (a2η4 − η2λ2)(A1 cos(ηx− ηλt) +A2 sin(ηx− ηλt))

Hence,0 = a2λ4 − λ6 = λ4(a2 − λ2)

We will assume that λ 6= 0, so we have that the Lagrangian multiplier again is the velocity, up toa sign. Hence η = ±1, and so

u(x, t) = A1 cos(x− λt) +A2 sin(x− λt) +A3(x− λt) +A4

by modifying the constants if necessary.

Now, this particular solution has terms which oscillate infinitely as x and t vary over their do-main. Worse still, the term A3(x − λt) blows up also as x, t become infinite. However, solutionswhich remain finite do indeed exist. For example, for the non-linear equation

utt + (1.69)uxxxx = eu − 1

16

3.3. BEAM EQUATION CHAPTER 3. MINIMIZATION OF ENERGY

Table 3.1: A numerical solutions to the non-linear beam equation for c = 1.3

can be solved numerically. A graph of this solution can be seen below. We will not concern ourselveswith providing these solutions. We can see that this solution does indeed behave like a travelling,but is bounded so must remain finite over its domain. The graph is symmetric around the origin,so the waves travelling in either direction are the same, and hence have the same velocity.

17

Chapter 4

Conclusions

We conclude the following about solutions of the above PDEs:

Linear Non-Linear

Wave EquationTravelling Wave Travelling Waveλ = ±c u(x, t) = f(ϕ) = (c2 − λ2)ϕ′′(x)

Beam EquationTravelling Wave Travelling Wave

λ = ±a f(ϕ(x)) = −(a2ϕ(4)(x) + λ2ϕ′′(x))

We can see that every solutions which minimizes the energy for constant momentum must be atravelling wave solution. As well as that, when our PDE is linear, the Lagrange multiplier is nothingmore than the wave velocity. We could conjecture from these results that that this is true for PDEsin general, but this is quite beyond the scope of this project.

4.1 Acknowledgements

I would like to sincerely thank my supervisor Dr. John Stalker for his help and guidance duringthis research assignment, without whom this project would have long ago become quite the comedyof mathematical errors. Also, I would like to thank Dr. Paschalis Karageorgis for providing thegraph for a numerical solution to the above non-linear version of the beam equation.

18

Appendix A

Proofs

A.1 Noether’s Theorem

To give at least some justification for using it, we will provide a proof for Noether’s Theorem. Thisparticular version of Noether’s Theorem is adapted from [1].

Theorem A.1.1 (Noether’s Theorem). Assume that a Lagrangian L = L(t, xi, xi) for i = 1, 2, . . . , nis invariant (up to an exact differential) under a transformation of the form

t 7→ t = t(t, xi, αs) xj 7→ xj = xj(t, xi, αs) (A.1.1)

for s = 1, 2, . . . , r, where each αs is an independent parameter of the transformation. What wemean by “up to an exact differential” is that

L(t, xi, xj)dt = L(t, xj , xj)dt+ dϕs(t, x

i)αs (A.1.2)

Then, the following expression is conserved(n∑i=0

∂L

∂xixi − L

)ξs −

(n∑i=0

∂L

∂xiζis

)+ ϕs (A.1.3)

for each s = 1, 2, . . . , r, where we adopt the notation x0 = t and, for convenience,

ξs =∂t

∂αs

∣∣∣∣αs=0

ζis =∂xi∂αs

∣∣∣∣αs=0

(A.1.4)

Proof.

Stage 1: Preliminaries

Consider a 1-parameter family of curves parametrised by u, say

xi = Xi(t, u) (A.1.5)

and let Γ be a given curve corresponding to the initial parameter value u = u0. we can assumethat these curves lie in some 2-dimensional subspace (denoted V ) of Rn+1 defined by

xi = ϕi(u, v) t = ψ(u, v) (A.1.6)

19

A.1. NOETHER’S THEOREM APPENDIX A. PROOFS

We will assume that Xi, ϕi and ψi are C2 functions chosen such that

t = ϕ(u0, t) (A.1.7)

Let the points p1 = (t1, xi1) and p2 = (t1, . . . , x

i2) be fixed on Γ and correspond to the parameters

(u0, v1) and (u0, v2) on V respectively. Let C(u) denote a curve for which u is constant. Let p′1 andp′2 be the points on the curve C(u) corresponding respectively to (u, v1) and (u, v2). For arbitrarydisplacements δt and δxi, we have

δt =∂ψ

∂uδu+

∂ψ

∂vδv δxi =

∂ϕi

∂uδu+

∂ϕi

∂vδv (A.1.8)

For the displacement of the point (δt, δxi), we have

δu = (u− u0) δv = 0 (A.1.9)

which gives us (ignoring the higher powers of δu and δv)

δt = (u− u0)∂ϕ

∂u

∣∣∣∣u=u0

δxi = (u− u0)∂ψi

∂u

∣∣∣∣u=u0

(A.1.10)

By above, this means that the displacement is also given by

δxi =∂Xi

∂t

∣∣∣∣u=u0

δt+ (u− u0)∂Xi

∂u

∣∣∣∣u=u0

= (Xi)u=u0δt+ δ∗xi (A.1.11)

where, for convenience, we define

δ∗xi = (u− u0)∂xi

∂u

∣∣∣∣u=u0

(A.1.12)

Similarly, it may be shown thatδ∗xi = (Xi)u=u0δt+ δ∗xi (A.1.13)

Stage 2: The Lagrangian

We are now ready to begin taking our Lagrangian into account. We define the following integral:

I(u) =

∫ p′2

p′1

L(t, xi, xi) dt (A.1.14)

where the integral is integrated along the curve C(u). We will compare it to the integral takingover Γ between p1 and p2. We will ultimately be studying the variation of I, defined to be

δI = (u− u0)∂I

∂u

∣∣∣∣u=u0

(A.1.15)

We cannot differentiate I directly, as the limits of I also depend on u. Therefore, we changecoordinates from t to v. This means that we can write I(u) in the form

I(u) =

∫ v2

v1

[L(ψ(u, v), Xi(ϕ(u, v), u), Xi(ϕ(u, v), u

)] ∂ϕ∂v

dv (A.1.16)

20

A.1. NOETHER’S THEOREM APPENDIX A. PROOFS

Now, our limits are constant. Differentiation gives us

∂I

∂u=

∫ v2

v1

{[∂L

∂t+

n∑i=1

(∂L

∂xi∂Xi

∂t+∂L

∂xi∂Xi

∂t

)]∂ψ

∂u

∂ψ

∂u+

∂2L

∂u∂v

}dv

+

∫ v2

v1

[n∑i=1

(∂L

∂xi∂Xi

∂u+∂L

∂xi∂Xi

∂u

)]∂ϕ

∂vdv

=

[L∂ψ

∂u

]v2v1

+

∫ v2

v1

[n∑i=1

(∂L

∂xi∂Xi

∂u+∂L

∂xi∂Xi

∂u

)]∂ϕ

∂vdv

(A.1.17)

We note that the first integral reduces to the final form because

∂L

∂v=∂L

∂t

∂ϕ

∂v+

n∑i=1

(∂L

∂xi∂Xi

∂t

∂ψ

∂v+∂L

∂xi∂Xi

∂t

∂ϕ

∂v

)(A.1.18)

and so the integral becomes∫ v2

v1

{∂L

∂v

∂ψ

∂u+ L

∂2ϕ

∂u∂v

}dv =

∫ v2

v1

∂

∂v

[L∂ψ

∂u

]dv (A.1.19)

Stage 3: u = u0

If we let u = u0, this amounts to letting the points p1 and p2 go to p′1 and p′2 respectively. Whenthis is true, we have that t = v and by our construction

∂ψ

∂v

∣∣∣∣u=u0

= 1 (A.1.20)

Substituting this into our above formula for the derivative of I(u) after setting u = u0, we have

∂I

∂u=

[L∂ψ

∂u

∣∣∣∣u=u0

]t2t1

+

∫ t2

t1

n∑i=1

(∂L

∂xi∂Xi

∂u

∣∣∣∣u=u0

+∂L

∂xi∂Xi

∂u

∣∣∣∣∣u=u0

)dt (A.1.21)

Returning to our the variation for I, we see that

δI =

[(u− u0) L

∂ψ

∂u

∣∣∣∣u=u0

]t2t1

+

∫ t2

t1

n∑i=1

(∂L

∂xiδ∗xi +

∂L

∂xiδ∗xi

)dt (A.1.22)

The first term may be simplified by considering δt and δxi given in equation A.1.10. The integralmay be also simplified by integrating the second term in the integral by parts (shifting the derivativefrom the δ∗ term to the other term). Combining these two, we get

δI =

[Lδt+

n∑i=1

∂L

∂xiδ∗xi

]t2t1

+

∫ t2

t1

n∑i=1

[∂L

∂xi− d

dt

(∂L

∂xi

)]δ∗xi dt (A.1.23)

We note, however, that in this expression, δ∗xi assumes that t is constant. Hence, we will replaceeach of these terms as per equation A.1. We can conclude that

δI =

[n∑i=1

∂L

∂xiδxi −

n∑i=1

(∂L

∂xixi)δt− Lδt

]t2t1

+

∫ t2

t1

n∑i=1

[∂L

∂xi− d

dt

(∂L

∂xi

)]δ∗xi dt (A.1.24)

21

A.1. NOETHER’S THEOREM APPENDIX A. PROOFS

Stage 4: The Transformation

We may finally introduce our transformation. By equations A.1.1 and A.1.4, we can see that

δ∗xi = (ζis − xiξs)αs (A.1.25)

We note that we may also write δI as follows

δI =

∫ t2

t1

L(t, xi, xj)dt−

∫ t2

t1

L(t, xj , xj)dt =

∫ t2

t1

dϕs(t, xi)αs = [ϕ(t, xi)]t1t2α

s (A.1.26)

We may now conclude, by combining equations A.1.24, A.1.25 and A.1.26 that∫ t2

t1

n∑i=1

[∂L

∂xi− d

dt

(∂L

∂xi

)](ζis − xiξs)αs dt =[

ϕ(t, xi)−

(n∑i=0

∂L

∂xixi − L

)ξs +

n∑i=0

∂L

∂xiζis

]t1t2

αs(A.1.27)

We finally note that the as the Euler-Lagrange equations must be satisfied, the left hand side of theabove equation must vanish. As each of the αs terms are linearly independent, we may concludethat each of the terms

ϕ(t, xi)−

(n∑i=0

∂L

∂uiui − L

)ξs +

n∑i=0

∂L

∂uiζis (A.1.28)

must be conserved for s = 1, 2, . . . , r, which completes the proof.

22

Bibliography

[1] Hanno Rund, The Hamilton-Jacobi Theory in The Calculus of Variations, D. Van NostrandCompany Ltd., 1966.

[2] Charles Fox, An Introduction to The Calculus of Variations, Dover Publications Inc, 1950.

[3] Lawrence C. Evans, Partial Differential Equations, Graduate Studies in Mathematics, Volume19, American Mathematical Society, 1998.

[4] Peter J. Olver, Applications of Lie Groups to Differential Equations, Graduate Texts in Math-ematics, Volume 107, Springer, 1993.

[5] Cornelius Lanczos, The Variational Principles of Mechanics, Dover Publications, 1986.

23