Lecture Notes on Mathematical Physics

Andrea GiustiDIFA - Universita di Bologna & Collegio Superiore

Introduction to Partial Differential Equations

Fall 2014

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Course Description

ObjectivesThe aim of the course consist in providing a general understanding of the methods of solutionsfor the most important PDE that arise in Mathematical Physics.At the end of the course, the students should be able to:

• use the method of characteristics to study first-order equations;

• classify a second order PDE (as elliptic, parabolic or hyperbolic);

• have a basic understanding of the concept of Well-Posed problem for PDE;

• use few standard methods (separation of variables, Green’s functions,...) to solve someelementary exercises.

Outline of Course

1. Introduction & Basic Definitions;

2. First Order PDEs

• Linear & Semi-linear;

• Characteristics.

3. The Diffusion (aka Heat) Equation

• Physical Properties;

• Well-posedness & Boundary conditions (Dirichlet, Neumann and Mixed BC);

• Separation of variables & Basics of Fourier Analysis;

• Uniqueness & the Energy Method;

• The Fundamental solution.

4. The Wave Equation

• Physical Properties;

• Well-posedness & Uniqueness;

• d’Alembert’s formula;

• The General Solution (1 + 1 dimensions);

• Causality;

• Uniqueness & the Energy Method.

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Homework Assignments & Form of AssessmentHomework is perhaps the most important component of this course: it provides you with regularfeedback on whether or not you are keeping up with the material, and it challenges you to creativelyapply what you have already learned. There will be an assignment almost every two weeks.Homework assignments will typically be posted on the course website.For your own benefit, I encourage you to solve all the proposed exercises because they will be themain topic of the exam, together with a subject of your choice within the topics discussed duringthe course.

Lecture Notes & BooksEverything that will be covered in this course is contained in these lecture notes (at least, I hopeso...). I have also included a number of non-examinable appendices.In principle, no reading is required beyond these notes. However, in practice, you might some-times find it helpful to consult additional resources for complementary viewpoints and additionalexplanations. For this purpose, I recommend the following textbooks:

• S. Salsa - Partial Differential Equations in Action: From Modelling to Theory.Springer, 2010;

• A. C. King, J. Billingham and S.R. Otto - Differential Equations: Linear, Nonlinear, Ordi-nary, Partial. Cambridge University Press, 2003;

• P. J. Collins - Differential and Integral Equation - Oxford University Press, 2006.

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Contents

1 Introduction 51.1 What is a PDE? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 The Mathematical Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 PDEs and Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Linear Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 How to solve PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.6 First Order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.6.1 Characteristic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6.2 Quasilinear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.6.3 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.7 Appendix: General Method of Characteristics . . . . . . . . . . . . . . . . . . . . 21

2 Transport Equation 232.1 Initial-value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.2 Non-homogeneous Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Introduction to Second Order PDEs 25

4 Heat (aka Diffusion) Equation 284.1 The Fundamental Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.2 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.3 Energy Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Wave Equation 345.1 Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.1.1 Causality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Energy Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

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Chapter 1

Introduction

NotationsWe will be studying functions u : Rn −→ R, such that u = u(x1, . . . , xn) = u(x), (that can beeasily generalized to the case of vector fields u : Rn −→ Rm) and their partial derivatives. Herex1, . . . , xn are the standard Cartesian coordinates on Rn. We will also use the alternate notationu(x), u(x, y), u(x, y, z), etc. and u(r, θ, φ) for spherical coordinates on R3.Sometimes, we will also consider a time coordinate t, in which case (t, x1, . . . , xn) denotes thestandard Cartesian coordinates on R1+n.We also use lots of different notation for partial derivatives:

∂u

∂xk≡ ∂xku ≡ ∂ku ≡ uxk k ∈ 1, . . . , n

and so on also for derivatives of higher order.

1.1 What is a PDE?

Definition 1 (PDE). A Partial Differential Equation in a single unknown function u : Rn → Ris an equation involving u and its partial derivatives. All such equations can be written as

F (u, ux1 , . . . , uxn , ux1x1 , . . . , uxk1 ···xkN ;x1, . . . , xn) = 0 k1, . . . , kN ∈ 1, . . . , n (1.1)

for some function F .Here N is called the order of the PDE; indeed N is the maximum number of derivatives appearingin the equation.

Remark. The previous definition can also be restated in a more compact formalism, as follows.For a given domain Ω ⊂ Rn, a function u : Ω → R and a real function F ∈ C1, the equation

F (Dαu, . . . , Du, u,x) = 0 (1.2)

with α = (α1, . . . αn) multi-index such that:

Dα :=∂|α|

∂xα11 · · · ∂xαn

n

, |α| = α1 + · · ·+ αn

is called PDE of order α.

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Example 1. Consider the PDE:

−∂2t u+ (1 + cosu) ∂3

xu = 0 u = u(t, x)

this is a clear example of a third-order nonlinear PDE.

Example 2. Consider the PDE:

−∂2t u+ ∂2

xu+mu = 0 u = u(t, x) , m ∈ R+0

this is a clear example of a second-order linear PDE.

The last kind of PDEs are usually called constant coefficient linear PDE because u and itsderivatives appear linearly, i.e. first power only, and are multiplied only by constants.

Remark. Few useful definitions:

1. In application, xk are are often space variables and a solution may be required in some regionΩ. In this case there will be some conditions to be satisfied on the boundary ∂Ω; these arecalled Boundary Conditions (BCs);

2. Also in applications, one of the independent variables can be time, then there will be someInitial Conditions (ICs) to be satisfied;

3. A linear equation is one in which the equation and any boundary or initial conditions donot include any product of the dependent variables or their derivatives; an equation that isnot linear is a nonlinear equation;

4. A nonlinear equation is semilinear if the coefficients of the highest derivative are functionsof the independent variables only.

Example 3. Consider u = u(x, y),

xux + x3y2uy = u5 (Ex.a)

xuxx + x(y + 1)uyy + uux = 0 (Ex.b)

More generally, we can say that a semilinear PDE has the following form:∑|α|=N

aα(x)Dαu+ a0(DN−1u, . . . , Du, u,x) = 0 (1.3)

5. A nonlinear PDE of order N is quasilinear if it is linear in the derivatives of order N withcoefficients depending only on the space (and time) coordinates and derivatives of order< N .

Example 4. Consider u = u(x, y),

[1 + (uy)2]uxx − 2uxuyuxy + [1 + (ux)

2]uyy = 0

More generally, we can say that a quasilinear PDE has the following form:∑|α|=N

aα(DN−1u, . . . , Du, u,x)Dαu+ a0(DN−1u, . . . , Du, u,x) = 0 (1.4)

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1.2 The Mathematical Problem

Suppose that we are interested in some physical system. A very fundamental question is:“Which PDEs are good models for the system?”.A major goal of modeling is to answer this question. Obviously, there is no general recipefor answering this question. In practice, good models are often the end result of confrontationsbetween experimental data and theory.The aim of this course is to discuss some important physical systems and the PDEs that arecommonly used to model them.Now let’s assume that we have a PDE that we believe is a good model for our system of interest.Then, the primary goals of PDE are to answer the following questions:

1. Does the PDE have any solutions? (aka Existence Problem)

2. What kind of data do we need to specify in order to solve the PDE?

3. Are the solutions corresponding to the given data unique? (aka Uniqueness Problem)

4. What are the basic qualitative properties of the solution?

5. Does the solution contain singularities? If so, what is their nature?

6. What happens if we slightly vary the data? Does the solution then also vary only slightly?

7. What kinds of quantitative estimates can be derived for the solutions?

Finally, the last three questions represent the main concern of the so called Well-posedness Prob-lem.

1.3 PDEs and Physics

The first revolution of Ordinary Differential Equations with Leibniz and Newton, with thedifferential calculus but most importantly the first differential equation describing a physical law:the fundamental principle of dynamics of Newton. Combined with the universal law of gravita-tion, it lead to the development of mathematical celestial mechanics, and later to ballistic and soon... But first of all this was an immense conceptual revolution: express in terms of differentialequations physical laws in order to predict the evolution of a system.This novel idea then spread everywhere in science, and enter a second higher stage with Euler,Fourier and the birth of partial differential equations for modelling mathematically the evolutionof continuum systems. Then PDEs accompanied each new field which emerged in modern sciencesince then.Here you can find a few examples:

The incompressible Navier-Stokes equations of Fluid Mechanics: ∂tu + (u · ∇)u +1

ρ∇p = ν∆u

∇ · u = 0

(1.5)

with u = u(t,x) is the velocity field of the fluid, ρ is the density of the fluid, p is the pressure, ν

is the viscosity and ∆def= ∇ · ∇ is the Laplacian differential operator.

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The Maxwell Equations of Electromagnetism (in vacuum):

∇ · E = 0 ∇ ·B = 0

∇× E + ∂tB = 0 ∇×B− 1

c2∂tE = 0

with E = E(t,x) ∈ R3 and B = B(t,x) ∈ R3.

The Boltzmann Equation of kinetic theory:

∂tf + v · ∇f =

∫R3

∫S2

f(v′)f(v′∗)− f(v)f(v∗) B(v− v∗; σ) dσ d3v∗

where f = f(t,x,v) ≥ 0 is integrable with total unit mass, B is the collision kernel and

v′ =v + v∗

2+ σ|v− v∗|

2, v′∗ =

v + v∗2− σ |v− v∗|

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The Schrodinger Equation of Quantum Mechanics:

i~ ∂tψ = − ~2

2m∆ψ + V ψ

where ψ = ψ(t,x) ∈ C and V = V (t,x) is a potential.

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1.4 Linear Partial Differential Equations

Before we dive into a specific model, let’s discuss a distinguished class of PDEs that arerelatively easy to study. The PDEs of interest are called linear PDEs. Most of this course willconcern linear PDEs.

Definition 2. A linear differential operator L is a differential operator such that:

L(au+ bv) = aLu+ bLv , ∀a, b ∈ R , ∀u, v functions (1.6)

Remark. The notation was introduced out of convenience and laziness. The definition is closelyconnected to the superposition principle.

Definition 3 (Linear PDEs). A Partial Differential Equation is linear if it can be written as:

Lu = f(x) (1.7)

for some linear operator L and some function f : Ω ⊂ Rn → R of the coordinates.If f = 0 then we say that the PDE is homogeneous, otherwise we say that it is inhomogeneous.

Here is an incredibly useful property of linear PDEs.

Theorem 1. (Superposition Principle).

u1, . . . , uM such that

Lui = 0 , ∀i = 1, . . . ,M

(Solutions of the homogeneous equation)

=⇒

M∑i=1

ci ui(x) is also a solutions

of the homogeneous equation

with c1, . . . , cM ∈ R

(1.8)

Proof. Left for exercise (Exercise 1).

This shows that the set of all solutions to Lu = 0 is a vector space when L is linear. Moreover,we can immediately deduce that the solutions of inhomogeneous and homogeneous linear PDEsare closely related. Indeed,

Proposition 2. Let Sh be the set of all solutions to the homogeneous linear PDE, i.e.

Shdef= u : Ω ⊂ Rn → R | Lu = 0 (1.9)

and let u∗ be a “fixed” solution to the inhomogeneous linear PDE:

Lu = f(x) (1.10)

Then the set S of all solutions to the inhomogeneous is the translation of Sh by u∗, i.e.:

S def= uh + u∗ | uh ∈ Sh (1.11)

Proof. Assume that Lu∗ = f and let w 6= u∗ be such that Lw = f , then L(u∗ − w) = f − f = 0so that u∗ − w ∈ Sh. Thus, w = u∗ + (w − u∗) and so w ∈ S by definition.On the other hand, if w ∈ S, then w = u∗ + uh for some uh ∈ Sh. Therefore, Lw = L(u∗ + uh) =f + 0 = f . Thus, w is a solution to the inhomogeneous equation (1.10).

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1.5 How to solve PDEs

Before we start with studying the most important PDEs, I would like to express the following(fundamental) remarks concerning solution of these kind of problems. Firstly, there is no generalrecipe that works for all PDEs! We will develop some tools that will enable us to analysesome important classes of PDEs.In the second instance, usually, we do not have explicit formulas for the solutions to thePDEs we are interested in! Instead, we are forced to understand and estimate the solutionswithout having explicit formulas.The two things that you typically need to study a PDE are: the PDE (obviously) and some“data”.

1.6 First Order PDEs

Method of Characteristics for Linear and Semilinear Equations

We consider linear first order PDE in two independent variables:

a(x, y)ux + b(x, y)uy + c(x, y)u = f(x, y) (1.12)

where a, b, c, f ∈ C(Ω) in some region Ω ⊂ R2 and we assume that a, b 6= 0 for the same (x, y).

Remark. Indeed, we could consider semilinear first order equation such as

a(x, y)ux + b(x, y)uy = κ(x, y, u)

instead of a linear equation as the theory of the former does not require any special treatment ascompared to that of the latter.

Now, the key to the solution of the equation (1.12) is to find a change of variables:

ξ ≡ ξ(x, y) , η ≡ η(x, y) (1.13)

which transforms (1.12) into the simpler equation

wξ + h(ξ, η)w = F (ξ, η) (1.14)

where w = w(ξ, η)def= u(x(ξ, η), y(ξ, η)).

We shall define this transformation so that it is one-to-one, at least ∀(x, y) ∈ D ⊂ Ω Then, onD we can, in theory, solve for x and y as functions of ξ and η.To ensure that we can do this, we require that the Jacobian of the transformation does not vanishin D:

J = det

∣∣∣∣ξx ξyηx ηy

∣∣∣∣ = ξxηy − ξyηx 6= 0, ∞ ∀(x, y) ∈ D

We begin looking for a suitable transformation by computing derivatives via the chain rule:

ux =∂w

∂ξ

∂ξ

∂x+∂w

∂η

∂η

∂x= wξξx + wηηx

uy =∂w

∂ξ

∂ξ

∂y+∂w

∂η

∂η

∂y= wξξy + wηηy

(1.15)

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If we now substitute these into equation (1.12) we get

a (wξ ξx + wη ηx) + b (wξ ξy + wη ηy) + cw = f (1.16)

We can rearrange this as

(a ξx + b ξy)wξ + (a ηx + b ηy)wη + cw = f (1.17)

This is close to the form of equation (1.12); if we can choose η = η(x, y) such that

a ηx + b ηy = 0 (1.18)

Provided that ηy 6= 0 we can express this required property of η as

ηxηy

= − ba

(1.19)

Supposing that we can define a new variable η which fulfils this constraint, what is the equationdescribing the curves of constant η?Setting η = const. = k, then

dη = ηx dx+ ηy dy = 0 =⇒ dy

dx= −ηx

ηy=b

a(1.20)

So, the equation η(x, y) = k defines the solutions of the following Ordinary Differential Equation(ODE):

dy

dx=b(x, y)

a(x, y)(1.21)

This equation is called the Characteristic Equation of the linear equation (1.12). Its solutioncan be written in the form g(x, y, η) = 0 (where η is the constant of integration) and defines afamily of curves in the plane called characteristics or characteristic curves of (1.12).Characteristics represent curves along which the independent variable η of the new coordinatesystem (ξ, η) is constant.

Example 5. Consider the linear first order equation:

x2 ux + y uy + xy u = 1 (1.22)

In this case we have:

a(x, y) = x2 , b(x, y) = y , c(x, y) = xy , f(x, y) = 1

then the characteristic equation is given by

dy

dx=b(x, y)

a(x, y)=

y

x2(1.23)

by separation of variables we get:

ln y +1

x= k , ∀y > 0 , ∀x 6= 0 k ∈ R (1.24)

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This is an integral of the characteristic equation describing curves of constant η and so we choose

η(x, y) = ln y +1

x(1.25)

Now, if we choose, for example, ξ(x, y) = x we have the Jacobian:

J = ηy =1

y6= 0

as required.Since ξ = x,

η(x, y) = ln y +1

x= ln y +

1

ξ=⇒ y = exp

(η − 1

ξ

)(1.26)

Now we apply the transformation

ξ = x , η = ln y +1

x

with w(ξ, η) = u(x, y) and we have

ux = wξξx + wηηx = wξ − x−2wη = wξ −wηξ2

uy = wξξy + wηηy = 0 + wη1

y= exp

(1

ξ− η)wη

(1.27)

Then the PDE becomes (prove it!)

wξ +exp (η − 1/ξ)

ξw =

1

ξ2(1.28)

Result that concludes our example.

Equivalent set of ODEs

The point of this transformation is that we can solve equation (1.12).Considering

wξ + h(ξ, η)w = F (ξ, η) (1.29)

as a Linear First Order ODE in ξ, with η carried along as a parameter.Thus we use an integrating factor method, i.e. defining

H ≡∫ ξ

h(ξ′, η) dξ′

we haveeH wξ + h(ξ, η) eH w = F (ξ, η) eH (1.30)

thus∂

∂ξ

(eH w

)= F (ξ, η) eH (1.31)

Now we integrate with respect to ξ, since η is being carried as a parameter, the constant ofintegration may depend on η:

eH w = g(η) +

∫F (ξ, η) eH dξ (1.32)

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in which g is an arbitrary differentiable function of one variable.Now, the general solution of the transformed equation is

w(ξ, η) = e−H(ξ,η) g(η) + e−H(ξ,η)

∫ ξ

F (ξ′, η) eH(ξ′,η) dξ′ (1.33)

We then obtain the general solution of the original equation by substituting back ξ(x, y), η(x, y):

u(x, y) = eα(x,y) [β(x, y) + g(η(x, y))] (1.34)

A certain class of first order PDEs (linear and semilinear PDEs) can then be reduced to a set ofODEs. This makes use of the general philosophy that ODEs are easier to solve than PDEs.

1.6.1 Characteristic Curves

We investigate the significance of characteristics which, defined by the ODE

dy

dx=b(x, y)

a(x, y)(1.35)

represent a one parameter family of curves whose tangent at each point is in the direction of thevector n = (a, b). Note that the left-hand side of equation

a(x, y)ux + b(x, y)uy = κ(x, y, u) (1.36)

is the derivation of u in the direction of the vector n, i.e. ∂u∂n

= n · ∇u = aux + buy = κ(x, y, u).Their parametric representation can be expressed as

γ : R −→ R2 ; γ(s) = (x(s), y(s)) (1.37)

where x(s) and y(s) satisfy the pair of ODEs

dx

ds= a(x, y) ;

dy

ds= b(x, y) (1.38)

Then we havedu

ds= ux

dx

ds+ uy

dy

ds= a(x, y)ux + b(x, y)uy = κ(x, y, u) (1.39)

which shows the variation of u along the curves. The one parameter family of characteristic curvesis parametrised by η, i.e. each value of η represents one unique characteristic.The solution of equation (1.36) then reduces to the solution of the family of ODEs

du

ds= κ(x, y, u) (1.40)

along each characteristics.The parametric characteristic equations (1.38) have to be solved together with equation (1.40),called the compatibility equation, to find a solution to semilinear equation (1.36).

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The Cauchy Problem

Consider a curve Γ in R2 such that

Γ(σ) = (x0(σ), y0(σ)) (1.41)

The Cauchy problem consist in finding a solution of the equation

F (ux, uy, u, x, y) = 0 (1.42)

in a neighbourhood of Γ∗ (graph of Γ) such that

u = u0(σ) on Γ∗ (1.43)

called Cauchy Data on Γ.

Remark. .

1. u can only be found in the region between the characteristics drawn through the endpointof Γ;

2. Characteristics are curves on which the values of u combined with the equation are notsufficient to determine the normal derivative of u;

3. A discontinuity in the initial data propagates onto the solution along the characteristics.These are curves across which the derivatives of u can jump while u itself remains continuous.

Existence & Uniqueness

Why do some choices of Γ in R2 give a solution and other give no solution or an infinite numberof solutions?It is due to the fact that the Cauchy data (initial conditions) may be prescribed on a curve Γwhich is a characteristic of the PDE. To understand the definition of characteristics in the contextof existence and uniqueness of solution, return to the general solution:

u(x, y) = eα(x,y) [β(x, y) + g(η(x, y))] (1.44)

Consider the Cauchy data, u0, prescribed along the curve Γ whose parametric form is given by

Γ(σ) = (x0(σ), y0(σ))

and suppose thatu0(x0(σ), y0(σ)) = q(σ)

Then, If Γ is not a characteristic, the problem is well-posed and there is a unique function gwhich satisfies the condition

q(σ) = exp α(x0(σ), y0(σ)) [β(x0(σ), y0(σ)) + g(η(x0(σ), y0(σ)))] (1.45)

If, on the other hand, Γ(σ) = (x0(σ), y0(σ)) is the parametrisation of a characteristic (i.e. η = k),the relation between the initial conditions q and g becomes

q(σ) = exp α(x0(σ), y0(σ)) [β(x0(σ), y0(σ)) +G] (1.46)

where G = g(k) = cost.; the problem is ill-posed.The functions α(x, y) and β(x, y) are determined by the PDE, so the latter equation places aconstraint on the given data function q(x). Ifq(σ) is not of this form for any constant G, thenthere is no solution taking on these prescribed values on Γ. On the other hand, if q(σ) is of thisform for some G, then there are infinitely many such solutions, because we can choose for g anydifferentiable function so that g(k) = G.

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1.6.2 Quasilinear Equations

We consider first a special class of nonlinear equations where the nonlinearity is confined tothe unknown function u. The derivatives of u appear in the equation linearly. Such equations arecalled quasilinear. More generally, we can say that a Quasilinear PDE has the following form:∑

|α|=N

aα(DN−1u, . . . , Du, u,x)Dαu+ a0(DN−1u, . . . , Du, u,x) = 0 (1.47)

Trying to be coherent with the previous sections, we are going to discuss in detail only the case ofquasilinear PDEs in two dimensions. Indeed, let us consider the most general quasilinear equationof the first order:

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u) (1.48)

where a, b, c ∈ C1(D), D ⊂ R3, and a2 + b2 6= 0.One can readily verify that the method of characteristics developed in the previous sections alsoapplies to the quasilinear case (1.48) as well.Namely, each point on the initial curve Γ, defined as

Γ : ]α, β [−→ R3 , Γ(σ) = (x0(σ), y0(σ), u0(σ)) (1.49)

is a starting point for a characteristic curve.The characteristic equations are now

dx

ds= a(x, y, u) ,

dy

ds= b(x, y, u) ,

du

ds= c(x, y, u) (1.50)

supplemented by the initial condition

x(s = 0, σ) = x0(σ) , y(0, σ) = y0(σ) , u(0, σ) = u0(σ) (1.51)

The problem consisting of (1.48) and the latter initial conditions is called the Cauchy Problem forquasilinear equations.The main difference between the characteristic equations for the linear/semilinear case, i.e.

dx

ds= a(x, y) ,

dy

ds= b(x, y) ,

du

ds= κ(x, y, u)

and the set (1.50) is that in the former case the first two equations are independent from the thirdequation and of the initial conditions. In the quasilinear case, this uncoupling of the characteristicequations is no longer possible, since the coefficients a and b depend upon u. We can also pointout that in the semilinear case, the equation for u is always linear, and thus it is guaranteed tohave a global solution (provided that the solutions x(t) and y(t) exist globally).

Finally, let us show the method for a few simple cases.

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Example 6. Solve the Cauchy Problem ux + uy = 2

u(x, 0) = x2 (1.52)

The characteristic equations and the parametric initial conditions are then given by

x = 1 , y = 1 , u = 2

x(0, σ) = σ , y(0, σ) = 0 , u(0, σ) = σ2 (1.53)

where we have made use of the convention g ≡ dg/ds.Now, it is simple to integrate the previous ODEs:

x(s, σ) = s+ f1(σ) , y(s, σ) = s+ f2(σ) , u(s, σ) = 2s+ f3(σ) (1.54)

Upon substituting into the initial conditions, we find

x = s+ σ , y = s , u = 2s+ σ2 (1.55)

We have thus obtained a parametric representation of the so called integral surface.To find an explicit representation of the surface u as a function of x and y we need to invert therelations x = x(s, σ) and y = y(s, σ) simultaneously (i.e. the Jacobian J must be non-vanishingin the region in which we want to extend our solution).In the current example the inversion is straightforward:

s = y , σ = x− y (1.56)

Thus the explicit representation of the integral surface (the solution) is given by

u(x, y) = 2y + (x− y)2 (1.57)

result which concludes our discussion.

Example 7. Solve the Cauchy Problemux + uy + u = 1

u|Ω = sinx , Ω = (x, y) ∈ R2 | x > 0 , y = x+ x2(1.58)

The characteristic equations and the associated initial conditions are given by

x = 1 , y = 1 , u = 1− ux(0, σ) = σ , y(0, σ) = σ + σ2 , u(0, σ) = sinσ

(1.59)

respectively.Now, if we compute the Jacobian

J = det

∣∣∣∣xs xσys yσ

∣∣∣∣ = 2σ 6= 0 (1.60)

Thus we anticipate a unique solution at each point where σ 6= 0. Since we are limited to theregime x > 0, i.e. σ > 0, we indeed expect a unique solution.The parametric integral surface is then given by

x(s, σ) = σ + s

y(s, σ) = σ + σ2 + s

u(s, σ) = 1− (1− sinσ)e−s(1.61)

16

In order to invert the mapping, we substitute the equation for x into the equation for y to obtain

σ =√y − x

in particular, the sign of the square root was selected according to the condition x > 0.Now it is easy to find

s = x−√y − x

whence the explicit representation of the solution is given by

u(x, y) = 1− (1− sin√y − x) exp

(√y − x− x

)(1.62)

Notice that the solution exists only in the domain

D = Ω ∩ y − x ≥ 0

and, moreover, it is not differentiable at the origin of R2.

1.6.3 Conservation Laws

Let u = u(x, t) a function representing a certain physical quantity for which we can define theflow ΦS through the surface S ⊂ D, with D a suitable spatial domain.Considering the one-dimensional case, we can choose D = [x1, x2] ⊂ R and the surface is given bythe boundary of the interval. A Conservation Law for a function of such kind can be definedin the following integral form

d

dt

∫ x2

x1

u(ξ, t) dξ = f(u(x1, t), x1, t)− f(u(x2, t), x2, t) ≡ −∆xf(u, x, t) (1.63)

where the RHS (right-hand side) represents the flux of the quantity u through the boundary ofthe domain.If we consider the classical example of the Traffic Flow, given a street starting at point x1 andending at point x2, the LHS (left-hand side) would represent the time variation of the the totalnumber of cars between points x1 and x2.Assuming u ∈ C1(D; t ≥ 0) we see that∫ x2

x1

ut(ξ, t) dξ = −∆xf(u, x, t) (1.64)

and, therefore,

1

x2 − x1

∫ x2

x1

ut(ξ, t) dξ = −f(u(x2, t), x2, t)− f(u(x1, t), x1, t)

x2 − x1

(1.65)

then, the limit as x1 → x2 and assuming f ∈ C1, we get

ut(x, t) + [f(u, x, t)]x = 0 (1.66)

which represents the differential form for a Conservation Law.Now, if we express explicitly the spatial differentiation (chain rule) we can easily get

ut + fu ux + fx = 0 (1.67)

17

If we then assume that f(u, x, t) = f(u), thus we reach the well known one-dimensional non-linear wave equation

ut + c(u)ux = 0 (1.68)

with c(u) ≡ f ′(u).

Consider then the Cauchy Problemut + c(u)ux = 0 Ω = (x, t) | x ∈ R , t ≥ 0u(x, 0) = φ(x) φ ∈ C1(R)

(1.69)

The characteristic equations and the associated initial conditions are given by

t = 1 , x = c(u) , u = 0

x(0, σ) = σ , t(0, σ) = 0 , u(0, σ) = φ(σ)(1.70)

respectively.Then, integrating the former ODEs, taking into account the initial conditions, one can find

t(σ, s) = s

x(σ, s) = s c(u) + σ

u(σ, s) = φ(σ)

(1.71)

Now, defining F (σ) ≡ c(u(σ, s)) = c(φ(σ)) we get

t(σ, s) = s

x(σ, s) = s F (σ) + σ(1.72)

Now, if we compute the Jacobian

J = det

∣∣∣∣xs xσts tσ

∣∣∣∣ = −[sdF (σ)

dσ+ 1

]≡ −[tF ′(σ) + 1] (1.73)

Thus J 6= 0 in a neighbourhood of s = 0 so we can actually find a unique regular solution for theCauchy Problem in a neighbourhood of the initial curve (Γ∗ = (x, t) : x ∈ R , t = 0).On the other hand, there may exists some values for s (and thus also for t) such that J = 0, whichmeans that the solution becomes singular. These values, if they exist, are the root of the equation

J = −[tF ′(σ) + 1] = 0 (1.74)

so they satisfy the following relation

t = − 1

F ′(σ), σ ∈ R (1.75)

Thus, we can state that, in general, the regular solution for the Cauchy Problem (1.69) isdefined for 0 ≤ t < tc <∞, where tc, if it exists, is such that:

tc :=1

max |F ′(σ)|(F ′(σ) < 0) (1.76)

The latter is then called Critical Time.

18

From the characteristic equations we can also deduce that

t(σ, s) = s

x(σ, s) = t c(u) + σ

u(σ, s) = φ(σ)

(1.77)

We can also recover an implicit expression for the integral surface, indeed

u(x, t) = φ[x− c(u) t] , x ∈ R , t ∈ [0, tc[ (1.78)

Now, if we calculate the derivatives with respect to x and t we get the following expressions (Proveit!):

ut = φ′(σ)σt = − F (σ)

1 + tF ′(σ)φ′(σ)

ux = φ′(σ)σx =φ′(σ)

1 + tF ′(σ)

(1.79)

Remark. It is worthy to notice that if we multiply both sides of the former Cauchy Problem bythe term c′(u) (supposing c′(u) 6= 0) we get a new problem, namely

vt + v vx = 0 Ω = (x, t) | x ∈ R , t ≥ 0v(x, 0) = g(x)

(1.80)

with v(x, t) = c(u(x, t)) and g(x) = c(f(x)).The letter differential equation is known as the inviscid Burgers’ Equation.

Example 8. Find the regular solution for the Cauchy Problemut + uux = 0

u(x, 0) = A sinx , x ∈ [0, 2π] , t > 0 , A > 0(1.81)

The initial curve is then given by

Γ(σ) : x0(σ) = σ , t0(σ) = 0 , u0(σ) = A sinσ (1.82)

The characteristic equations and the parametric initial conditions are then given by

t = 1 , x = u , u = 0

x(0, σ) = σ , t(0, σ) = 0 , u(0, σ) = A sinσ(1.83)

from which we gett(σ, s) = s

x(σ, s) = σ + As sinσ

u(σ, s) = A sinσ

(1.84)

Thus,x = σ + (A sinσ)t (1.85)

and thenu(x, t) = A sin(x− ut) (1.86)

19

In order to calculate the critical time we should notice that

ux =A cosσ

1 + tA cosσ, ut = −A

2 sinσ cosσ

1 + tA cosσ(1.87)

and then

tc =1

max |A cosσ|(cosσ < 0) =⇒ tc =

1

A(1.88)

indeed the cosine gets its maximum negative value, in x ∈ [0, 2π], in x = π.

20

1.7 Appendix: General Method of Characteristics

We finish by describing the general method for solving a first-order equation in n variables.Consider the first-order, nonlinear equation,

F (x, u,∇u) = 0 , x ∈ Rn , ∇ =n∑i=1

ei ∂xi (1.89)

In the case of two spatial variables, we prescribed initial data on a curve Γ in R2. Now wemust prescribe data on an (n− 1)-dimensional manifold Γ in Rn.

Remark. An m-dimensional manifold is a surface which can be represented locally as the graphof a function.

Our Cauchy problem is F (x, u,∇u) = 0 x ∈ Rn

u|Γ = φ(1.90)

First, we parametrize Γ by the vector σ = (σ1, . . . , σn−1) ∈ U ⊂ Rn−1 so that Γ(σ) = (x 10 (σ), . . . , x n

0 (σ)).By letting z(s) = u(x(s)) and pi(s) = uxi(x(s)), we rewrite our equation as

F (x, z,p) = 0 (1.91)

We then define the 2n+ 1 characteristic equations by

dxids

= Fpi ,dz

ds=

n∑i=1

pi Fpi ,dpids

= −Fxi − pi Fz (1.92)

for i = 1, . . . , n.Our initial conditions are given by

xi(σ, 0) = x i0 (σ) , z(σ, 0) = φ(σ) , pi(σ, 0) = ψi(σ) (1.93)

where the functions ψi(σ), i = 1, . . . , n are determined by solving the following equations.First, we need,

F [x 10 (σ), . . . , x n

0 (σ), φ(σ), ψ1(σ), . . . , ψn(σ)] = 0

Second, we need∂u

∂σi(σ, 0) =

n∑k=1

∂u

∂xk

∂xk∂σi

for i = 1, . . . , n− 1.But, u(σ, 0) = φ(σ), xi(σ, 0) = x i

0 (σ) and uxi = pi. Therefore, this equation becomes

∂φ

∂σi=

n∑k=1

ψk∂x k

0

∂σi

Therefore, our system of n equations for the n unknown functions ψi(σ) are given by

φσi =n∑k=1

ψk∂x k

0

∂σi, i = 1, . . . , n− 1

F [x 10 (σ), . . . , x n

0 (σ), φ(σ), ψ1(σ), . . . , ψn(σ)] = 0

(1.94)

21

Again, functions ψi may not exist or may not be unique, but if they do exist, we can find a uniquesolution of (1.92) satisfying the initial conditions (1.93) for that choice of ψi.In order to guarantee that we can invert the function x = x(σ, s) near the manifold Γ we willassume our initial data is noncharacteristic. That is, defining Ψ(σ) = (ψ1(σ), . . . , ψn(σ)), we saythat Γ(σ), φ(σ),Ψ(σ) is noncharacteristic if

N · ∇pF = 0 (1.95)

where N = N [Γ(σ)] is the normal vector to the (n− 1)-dimensional manifold Γ.

In summary, for noncharacteristic boundary data (Γ, φ,Ψ), we can find a local solution of (1.90) bysolving the characteristic equations (1.92) with initial conditions (1.93) and letting u(x) = z(σ, s).More precisely, ∀σ ∈ U ⊂ Rn−1, let (x(σ, s), z(σ, s), p(σ, s)) be the unique solution of (1.92), (1.93).By the noncharacteristic assumption on the initial data, we can invert the function x = x(σ, s)near s = 0. That is, we can find functions f, g such that σ = f(x) and s = g(x). Now, let

u(x) ≡ z(σ, s) = z(f(x), g(x))

for x near Γ∗.

22

Chapter 2

Transport Equation

One of the simplest PDEs is the transport equation with constant coefficients, i.e.

ut + b · ∇u = 0 , in Ω = ]0,+∞[ × Rn (2.1)

where b ∈ Rn and u : Ω −→ R, u = u(t,x).

Now, which function u solve (2.1)? To answer this question let us suppose for the momentthat we are given some smooth solution u and try to compute it. To do so, we first must recognizethat the PDE (2.1) states that a particular directional derivative of u vanishes. We exploit thisinsight by fixing any point (t,x) ∈ Ω and defining

z(s) := u(t+ s,x + sb) s ∈ R (2.2)

We can then calculate

dz

ds= u(t+ s,x + sb) + b · ∇u(t+ s,x + sb) = 0 (2.3)

where the second equality is due to (2.1). Thus, z is a constant function of s, and consequently∀(t,x) ∈ Ω, u is constant on the line through (t,x) with the direction (1,b) ∈ R1+n. Hence, if weknow the value of u at any point on each such line, we know its value everywhere in Ω.

Remark. As well known from Calculus, we have that a directional derivative of a functionu : D ⊂ Rm → R along a certain vector n ∈ Rm is given by:

∂u

∂n= n · ∇u (2.4)

Then we can easily write:

∂u

∂n= n · (∂t, ∇)u = ut + b · ∇u = 0 , n = (1,b) ∈ ]0,+∞[ × Rn (2.5)

as previously stated. /

23

2.1 Initial-value Problem

Let us consider the initial-value problem (IC)ut + b · ∇u = 0 , in Ω = ]0,+∞[ × Rn

u(0,x) = g(x)(2.6)

where b ∈ Rn and g : Rn → R are known.Given (t,x), as above, the line through this point with direction (1, b) is represented parametri-cally by

(t(s),x(s)) = (t+ s,x + sb)

with s ∈ R.This line ”hits” the hyperplane Γ = t = 0 × Rn when s = −t, at the point (0,x − bt). Now,since u is constant along the line (t(s),x(s)) and given that

u(0,x) = g(x) ⇒ u(0,x− bt) = g(x− bt)

we can deduce that:u(t,x) = g(x− bt) t ≥ 0 , x ∈ Rn (2.7)

Remark. It is easy to directly check that if g ∈ C1 then u defined by (2.7) is indeed a solutionof (2.8). On the other hand, if g /∈ C1, then there is obviously no C1 solution of (2.8); but,even in this case, the expression (2.7) certainly provides a strong, and in fact the only reasonable,candidate for a solution.We may thus informally declare u(t,x) = g(x− bt) to be a weak solution of (2.8). /

2.2 Non-homogeneous Problem

Next, let us consider the associated Non-homogeneous problem:ut + b · ∇u = f , in Ω = ]0,+∞[ × Rn

u(0,x) = g(x)(2.8)

As before, given (t,x) ∈ R1+n and defining z(s) := u(t+ s,x + sb) with s ∈ R, then

dz

ds= u(t+ s,x + sb) + b · ∇u(t+ s,x + sb) = f(t+ s,x + sb) (2.9)

Consequently,

z(0)− z(−t) = u(t,x)− g(x− bt) =

∫ 0

−t

dz

dsds =

∫ 0

−tf(t+ s,x + sb) ds =

s → s′ = s+ t ⇒ =

∫ t

0

f(s′,x + (s′ − t)b) ds′(2.10)

and so

u(t,x) = g(x− bt) +

∫ t

0

f(s,x + (s− t)b) ds t ≥ 0 , x ∈ Rn (2.11)

24

Chapter 3

Introduction to Second Order PDEs

Classification & Canonical Form

Consider a general Second Order Linear Equation in two independent variables:

a(x, y)∂2u

∂x2+ b(x, y)

∂2u

∂x∂y+ c(x, y)

∂2u

∂y2+ d(x, y)

∂u

∂x+ e(x, y)

∂u

∂y+ f(x, y)u = g(x, y) (3.1)

which, in the case of a semilinear equation, the coefficients d, e, f, g could be functions of ux, uyand u as well.Now, the question that could arise quite naturally is: can we also make a transformation

(x, y)T−→ (ξ, η) to put the equation into a simpler form?

So, let’s consider the previous coordinate transformation with the prescription:

J =∂(ξ, η)

∂(x, y)6= 0∞

then, by the inverse function theorem, there exist an open neighbourhood D of (x, y) and anotherneighbourhood of U of (ξ, η) such that the transformation:

T |D,U : D −→ U (x, y) 7−→ (ξ, η)

is invertible.As before, we compute chain rule derivations

∂u

∂x=∂u

∂ξ

∂ξ

∂x+∂u

∂η

∂η

∂x∂u

∂y=∂u

∂ξ

∂ξ

∂y+∂u

∂η

∂η

∂y

∂2u

∂x2=

∂

∂x

(∂u

∂x

)=

(∂ξ

∂x

∂

∂ξ+∂η

∂x

∂

∂η

)(∂u

∂ξ

∂ξ

∂x+∂u

∂η

∂η

∂x

)and analogously for uyy , uxy

(3.2)

Then, the equation becomes (Exercise 2, prove the following statements)

Auξξ +B uξη + C uηη + F (uξ, uη, u, ξ, η) = 0 (3.3)

where we have written explicitly only the Principal Part of the PDE, precisely the one involvingthe highest-order derivatives of u.

25

This expression may seems as difficult as the first one; on the other hand, we have the remarkablefreedom of setting A, B or C (depending on the kind of the PDE) to zero by means of a wisechoice of ξ and η.Moreover, It is easy to verify that (Exercise 2)

B2 − 4AC = (b2 − 4ac) J2 (3.4)

So, provided J 6= 0 we see that the sign of the discriminant b2− 4ac is invariant under coordinatetransformations. We can use this invariance properties to classify the equation.

Studying the structure of the Characteristic Equations (which is not of our concern) we haveto distinguish three different cases:

1. If b2 − 4ac > 0 we can find a change of variable (x, y) → (ξ, η) such that it transform theoriginal PDE into

uξη + (lower order terms) = 0 (3.5)

In this case the equation is said to be Hyperbolic and has two families of characteristics.

2. If b2 − 4ac = 0, a suitable choice for ξ still simplifies the PDE, but now we can choose ηarbitrarily and the equation reduces to the form

uηη + (lower order terms) = 0 (3.6)

The equation is then said to be Parabolic and has only one family of characteristics.

3. If b2 − 4ac < 0 we can again apply the change of variables (x, y) → (ξ, η) to simplify theequation, but now this functions will be complex conjugate. To keep the transformationreal, we apply a further change of variables (ξ, η) → (α, β) via

α = ξ + η

β = i (ξ − η)=⇒ uξη = uαα + uββ (3.7)

so, the equation can be reduced to

uαα + uββ + (lower order terms) = 0 (3.8)

In this case the equation is said to be Elliptic and has no real characteristics.

The above forms are called the Canonical Forms of the Second Order Linear/Semilinear equa-tions in two variables.

Remark. • The Characteristic Equation is

dy

dx=b±√b2 − 4ac

2a

• These definitions are all taken at a point x0 ∈ R2; unless a, b and c are all constant, thetype may change with the point x0.

26

Example 9. Here are just a few examples concerning the previous statements.

• The Wave Equation in 1+1 dimension (i.e. (t, x)),

utt − c2 uxx = 0

is clearly hyperbolic, indeed

a = 1 , b = 0 , c = −c2 =⇒ b2 − 4ac = 4c2 > 0

• The Diffusion/Heat equation,ut − κuxx = 0

is clearly parabolic, indeed

a = 0 , b = 0 , c = −κ =⇒ b2 − 4ac = 0

• The Laplace’s equation in 2 dimensions,

uxx + uyy = 0

is clearly elliptic, indeed

a = 1 , b = 0 , c = 1 =⇒ b2 − 4ac = −4 < 0

• The Tricomi’s equation,y uxx + uyy = 0

is respectively hyperbolic, parabolic and elliptic, depending on the sign of y, indeed

a = y , b = 0 , c = 1 =⇒ b2 − 4ac = −4y =

> 0 , if y < 0

0 , if y = 0

< 0 , if y > 0

27

Chapter 4

Heat (aka Diffusion) Equation

The heat equation for a function u(t,x), with x ∈ Rn and t > 0, is

ut −D∆u = 0 , ∆ ≡ div∇ =n∑i=1

∂2xi

(4.1)

Here, the constant D > 0 is the diffusion coefficient and ∆ is the Laplacian operator.In this Chapter we will briefly discuss the main result concerning this important equation, suchas: the fundamental solution, the technique of separation of variables and the energy method.

4.1 The Fundamental Solution

Consider the n-dimensional Heat Equation:

ut −D∆u = 0 , x ∈ Rn , t > 0 (4.2)

with the initial condition (IC)u(0,x) = φ(x) (4.3)

If we recall definition of the Fourier transform for a function f : Rn −→ R, i.e.

f(k) :=

∫Rn

f(x) exp(−ik · x) dnx (4.4)

we can easily prove the following properties:

gt(t,x) = ∂t g(t,k) and ∆g(t,x) = (ik)2 g(t,k) ≡ −k2 g(t,k) (4.5)

where k2 := |k|.Moreover,

(f ∗ g)(x) = f(k) g(k) (4.6)

where the ∗ represents the Fourier convolution integral, i.e.

(f ∗ g)(x) :=

∫Rn

f(x− y) g(y) dny = (g ∗ f)(x)

Now, taking the Fourier transform of both the equation and the IC we getut + k2D u = 0 , u = u(t,k)

u(0,k) = φ(k)(4.7)

28

Multiplying both sides of the first equation by the integrating factor exp(Dk2t) the equationbecomes

∂t

[eDk

2t u(t,k)]

= 0 (4.8)

Thus,u(t,k) = f(k) exp(−Dk2t) , f(k) arbitrary function (4.9)

Using the initial condition

u(0,k) = φ(k) =⇒ f(k) = φ(k) (4.10)

Now we know thatu(t,k) = φ(k) exp(−Dk2t) (4.11)

Finally, reversing the Fourier transform

u(t,x) =1

(2π)n

∫Rn

dnk u(t,k) exp(ik · x) =

=1

(2π)n

∫Rn

φ(k) exp(−Dk2t+ ik · x) dnk =

= (Γ(t) ∗ φ)(x)

(4.12)

where Γ(t,x) is the so called Fundamental Solution for our problem.

In order to find this function we just need to reverse the Fourier transform of the function Γ(t,k) =exp(−Dk2t) indeed,

Γ(t,x) =1

(2π)n

∫Rn

dnk exp(−Dk2t+ ik · x) =

=n∏a=1

∫ +∞

−∞

dka2π

exp(−Dk2at+ ikaxa) =

=1√

4πDtexp

(− x2

4Dt

) (4.13)

Thus, the Fundamental Solution for the Heat Equation is given by:

Γ(t,x) =1√

4πDtexp

(− x2

4Dt

)(4.14)

end the general solution for the Cauchy problem is then given by:

u(t,x) = (Γ(t) ∗ φ)(x) =

∫Rn

Γ(t,x− y)φ(y) dny (4.15)

29

4.2 Separation of Variables

In this section we introduce a very useful technique, called the method of separations ofvariables, for solving initial boundary value-problems.We consider the heat equation in (1 + 1)D satisfying the initial conditions

ut −Duxx = 0 , x ∈ [0, L] , t > 0

u(0, x) = φ(x) u = u(t, x)(4.16)

We seek a solution u satisfying certain boundary conditions (BC); in the present case we considerthe so called Dirichlet BC, i.e.

u(t, 0) = u(t, L) = 0 , ∀t > 0 (4.17)

In this method, we look for solutions of the form

u(t, x) = X(x)T (t) (4.18)

where X and T are function which have to be determined.Substituting this ansatz into the equation, we get

X(x)T (t)−DT (t)X ′′(x) = 0 (4.19)

where X ′ = Xx and T = Tt, from which we get

T (t)

DT (t)=X ′′(x)

X(x)(4.20)

The left side depends only on t whereas the right hand side depends only on x. Since they areequal, they must be equal to some constant −λ, with λ ∈ R. Thus

T (t) + λDT (t) = 0 , X ′′(x) + λX(x) = 0 (4.21)

which are two ODEs that we can easily solve.The general solution of the first equation is given by

T (t) = A exp(−λDt) , A ∈ R (integration constant) (4.22)

The general solutions of the second equation are as follows:

X(x) =

for λ < 0 , α cosh(

√−λx) + β sinh(

√−λx)

for λ = 0 , αx+ β

for λ > 0 , α cos(√λx) + β sin(

√λx)

(4.23)

In addition, the function X which solves the second equation will satisfy boundary conditionsdepending on the boundary condition imposed on u.The problem

X ′′(x) + λX(x) = 0

X satisfies boundary conditions(4.24)

30

is called the eigenvalue problem, a non-trivial solution is called an eigenfunction associatedwith the eigenvalue λ.So, if we consider the Dirichlet condition:

u(t, 0) = u(t, L) = 0 , ∀t ≥ 0 (4.25)

In this case the eigenvalue problem becomesX ′′(x) + λX(x) = 0

X(0) = X(L) = 0(4.26)

Now, we have to find non-trivial solutions X for this eigenvalue problem.It is straightforward to find that ∀λ ≤ 0 we have that X(0) = X(L) = 0 ⇒ α = β = 0, so λ ≤ 0is not an eigenvalue of the problem.On the other hand, if we consider λ > 0,

X(0) = α = 0

X(L) = β sin(√λL) = 0

(4.27)

Since X must be a non-trivial solution, β 6= 0 and hence sin(√λL) = 0. Consequently, in order

to get a non-trivial solution we have

λn =(nπL

)2

, n ≥ 1 (4.28)

and the corresponding eigenfunctions are given by

Xn(x) = βn sin(nπLx), and Tn(t) = An exp

(−n

2π2

L2Dt

)(4.29)

Thus we have obtained the following sequence of solutions

un(t, x) = Xn(x)Tn(t) = Bn exp

(−n

2π2

L2Dt

)sin(nπLx)

(4.30)

with Bn ≡ Anβn.Finally, thank to the linearity of the Heat Equation, we can obtain the general solution by meansof the superposition principle:

u(t, x) =∞∑n=1

un(t, x) =∞∑n=1

Bn exp(−n2π2Dt/L2

)sin (nπx/L) (4.31)

Then, if we consider the IC, at time t = 0 we have:

u(t, x) =∞∑n=1

Bn sin (nπx/L) = φ(x) (4.32)

31

In order to compute the coefficients Bn, one should note the following results (Prove it!):

2

L

∫ L

0

sin (nπx/L) sin (mπx/L) dx = δnm

2

L

∫ L

0

cos (nπx/L) cos (mπx/L) dx = δnm∫ L

0

sin (nπx/L) cos (mπx/L) dx = 0

(4.33)

Multiplying both sides of Eq. (4.32) by sin (mπx/L), with fixed m, and then integrating over[0, L], we get:

Bm =2

L

∫ L

0

φ(x) sin (mπx/L) dx (4.34)

Remark. Where we have made use of the well known Kroneker delta which is defined as

δnm =

1 , if n = m

0 , if n 6= m

This last remark actually concludes our discussion. /

32

4.3 Energy Method

We next consider the inhomogeneous heat equation with some auxiliary conditions, and usethe energy method to show that the solution satisfying those conditions must be unique. Considerthe following mixed initial-boundary value problem, which is called the Dirichlet problem for theheat equation

ut −Duxx = f , x ∈ [0, L] , t > 0

u(0, x) = φ(x) u = u(t, x) , f = f(t, x)

u(t, 0) = g(x) , u(t, L) = h(x) ∀t > 0

(4.35)

for given functions f, g, h, φ.Arguing from the inverse, let as assume that there are two functions u and v that both solve theinhomogeneous heat equation and satisfy the initial and Dirichlet boundary conditions,with u 6= v.Then their difference, w = u − v, satisfies the homogeneous heat equation with zero initial-boundary conditions, i.e.

wt −Dwxx = 0 , x ∈ [0, L] , t > 0

w(0, x) = 0 w = w(t, x)

w(t, 0) = w(t, L) = 0 ∀t > 0

(4.36)

Now define the following energy

E(t) = E[w(t)] :=1

2

∫ L

0

w2(t, x) dx (4.37)

which is always positive, and decreasing, if w solves the heat equation. Indeed, differentiatingthe energy with respect to time, and using the heat equation we get

dE

dt=

∫ L

0

wwt dx = D

∫ L

0

wwxx dx (4.38)

Integrating by parts in the last integral gives

dE

dt= Dwwx|L0 −D

∫ L

0

(wx)2 dx = −D

∫ L

0

(wx)2 dx ≤ 0 (4.39)

since the boundary terms vanish due to the boundary conditions.Now, due to the initial condition,

E(t = 0) = 0

But then using the fact that the energy is a non-negative decreasing quantity, we get

0 ≤ E(t) ≤ E(t = 0) = 0

Hence,

E(t) =1

2

∫ L

0

w2(t, x) dx = 0 , ∀t ≥ 0 (4.40)

which implies that the non-negative continuous integrand must be identically zero over the inte-gration interval, i.e

w ≡ 0 , ∀x ∈ [0, L] , ∀t ≥ 0 =⇒ u = v , ∀x ∈ [0, L] , ∀t ≥ 0 (4.41)

which finishes the proof of uniqueness.

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Chapter 5

Wave Equation

In this first part we will solve the wave equation on the entire real line x ∈ R. This correspondsto a string of infinite length.The wave equation, which describes the dynamics of the amplitude u(t, x) of the point at positionx on the string at time t, has the following form

utt − c2uxx = 0 (5.1)

As we saw, the wave equation has the second canonical form for hyperbolic equations.By means of the characteristic equations, one can then rewrite this equation in the first canonicalform, which is

uξη = 0 (5.2)

This is achieved by passing to the characteristic variables

ξ = x+ ct , η = x− ct (5.3)

In particular, we can show the equivalence between (5.1) and (5.2) , let us compute the partialderivatives of u with respect to x and t in the new variables using the chain rule.

ux = uξ + uη

ut = cuξ − cuη(5.4)

Now, if we realize that, for u ∈ C2, the second order linear operator of the wave equation (theD’Alembert operator) factors into two first order operators

2 := ∂2t − c2∂2

x = (∂t − c∂x)(∂t + c∂x) (5.5)

which, under the coordinate transformation (x, t)→ (ξ, η) becomes

2 = −4c2 ∂ξ∂η (5.6)

as we expected.Now, Eq. (5.2) can be treated as a pair two successive ODEs. Integrating first with respect tothe variable η, and then with respect to ξ, we arrive at the solution

u(ξ, η) = f(ξ) + g(η) (5.7)

Thus, the general solution is given by

u(t, x) = f(x+ ct) + g(x− ct) (5.8)

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5.1 Initial Value Problem

If we consider the following Cauchy Problem for the wave equation in (1 + 1)Dutt − c2uxx = 0 , x ∈ R , t > 0

u(0, x) = φ(x) u = u(t, x)

ut(0, x) = ψ(x)

(5.9)

where φ and ψ are arbitrary functions of single variable.The solution to this problem is easily found from the general solution (5.8); indeed, all we needto do is find f and g from the initial conditions.From the first condition:

u(0, x) = φ(x) =⇒ f(x) + g(x) = φ(x) (5.10)

and from the second one:

ut(0, x) = ψ(x) =⇒ cf ′(x)− cg′(x) = ψ(x) (5.11)

where f ′ = fx.To solve this system, we first integrate both sides of the latter equation, from 0 to x to get rid ofthe derivatives on f and g, and rewrite the equations as

f(x) + g(x) = φ(x)

f(x)− g(x) =1

c

∫ x

0

ψ(s) ds+ f(0)− g(0)(5.12)

We can solve this system by adding the equations to eliminate g, and subtracting them to eliminatef . This leads to the solution

f(x) =φ(x)

2+

1

2c

∫ x

0

ψ(s) ds+f(0)− g(0)

2

g(x) =φ(x)

2− 1

2c

∫ x

0

ψ(s) ds− f(0)− g(0)

2

(5.13)

Finally, substituting this expressions for f and g back into the general solution (5.8) we get

u(t, x) =φ(x+ ct) + φ(x− ct)

2+

1

2c

∫ x+ct

x−ctψ(s) ds (5.14)

This is the well known d’Alembert’s solution for the Cauchy Problem of our concern.

35

5.1.1 Causality

The value of the solution to the former Cauchy problem at a certain point (t0, x0) can bededuced from d’Alembert’s formula.

u(t0, x0) =φ(x0 + ct0) + φ(x0 − ct0)

2+

1

2c

∫ x0+ct0

x0−ct0ψ(s) ds (5.15)

We can see that this value depends on the values of φ at only two points on the x axis, x0 + ct0and x0 − ct0, and the values of ψ only on the interval [x0 − ct0, x0 + ct0].For this reason, the interval [x0 − ct0, x0 + ct0] is called interval of dependence for the point(t0, x0). Sometimes the entire triangular region with vertices at (0, x0 − ct0), (0, x0 + ct0) and(t0, x0) is called the domain of dependence, or the past history of the point (t0, x0). Thesides of this triangle are segments of characteristic lines passing through the point (t0, x0). Thus,we see that the initial data travels along the characteristics to give the values at later times.An inverse notion to the domain of dependence is the notion of domain of influence of thepoint (0, x0). This is the region in the x − t plane consisting of all the points, whose domain ofdependence contains the point x0. The region has an upside-down triangular shape, with the sidesbeing the characteristic lines emanating from the point (0, x0).This also means that the value of the initial data at the point x0 affects the values of the solutionu at all the points in the domain of influence. Notice that at a fixed time t0, only the pointssatisfying x0 − ct0 ≤ x ≤ x0 + ct0 are influenced by the point (0, x0).

5.2 Energy Method

The conservation of energy provides a straightforward way of showing that the solution to theCauchy Problem

utt − c2uxx = 0 , x ∈ R , t > 0

u(0, x) = φ(x) u = u(t, x)

ut(0, x) = ψ(x)

(5.16)

is unique.Arguing from the inverse, let as assume that this problem has two distinct solutions, u and v.Then their difference, w = u− v, satisfies the following Cauchy Problem

wtt − c2wxx = 0 , x ∈ R , t > 0

w(0, x) = 0 w = w(t, x)

wt(0, x) = 0

(5.17)

Now, if we define the following energy

E(t) = E[w(t)] :=1

2

∫ ∞−∞

[(wt)

2 + c2(wx)2]dx (5.18)

Hence the energy associated with the solution w at time t = 0 is

E(t = 0) = 0 (5.19)

due to the initial conditions.

36

At the same time

dE

dt=

∫ ∞−∞

[wtwtt + c2wxwtx

]dx

.=

∫ ∞−∞

[wtt − c2wxx

]wt dx = 0 ∀t ≥ 0 (5.20)

where.= stands for ”up to boundary terms”. So, we have been able to show that E(t) is conserved.

But since the integrand in the expression of the energy is non-negative, the only way the integralcan be zero, is if the integrand is uniformly zero. That is,

∇w = (wt, wx) = 0

This implies that w = const. , ∀(t, x) ∈ R+ × R, but since w(0, x) = 0 ⇒ const. = 0. Thus,

w(t, x) = u(t, x)− v(t, x) = 0

which is in contradiction with our initial assumption of distinctness of u and v. This implies thatthe solution to the former Cauchy problem is unique.

37