1

y (a) = ’ (p1 y (a) + p2 y(a) ’ f (a)), (6.9)

p0

and use it to obtain y (a + ) = y (a) + y (a). We now have initial data,

y(a + ), y (a + ), at the point a + , and can play the same game to proceed

to a + 2 , and onwards.

Suppose now that we have the analogous situation of a second order

partial di¬erential equation

‚2•

aµν (xi ) µ ν + (lower orders) = 0. (6.10)

‚x ‚x

in Rn . We are also given initial data on a surface, “, of co-dimension one in

Rn .

n

t2

p

t1

The surface “ on which we are given Cauchy Data.

148 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

At each point p on “ we erect a basis n, t1 , t2 , . . . of normal and tangents,

and the information we have been given consists of the value of • at every

point p together with

‚• def µ ‚•

=n , (6.11)

‚xµ

‚n

the normal derivative of • at p. We want to know if this Cauchy data

is su¬cient to ¬nd the second derivative in the normal direction, and so

construct similar Cauchy data on the adjacent surface “ + n. If so, we can

repeat the process and systematically propagate the solution forward through

Rn .

From the given data, we can construct

‚2• ‚2•

def

nµ tν

= ,

i

‚xµ ‚xν

‚n‚ti

‚2• 2

νν ‚•

def

= ti tj µ ν , (6.12)

‚ti ‚tj ‚x ‚x

but we do not yet have enough information to determine

‚ 2 • def µ ν ‚ 2 •

=nn . (6.13)

‚xµ ‚xν

‚n‚n

Can we ¬ll the data gap by using the di¬erential equation (6.10)? Suppose

that

‚2•

= φµν + nµ nν ¦ (6.14)

0

µ ‚xν

‚x

where φµν is a guess that is consistent with (6.12), and ¦ is as yet unknown,

0

and, because of the factor of nµ nν , does not a¬ect the derivatives (6.12). We

plug into

‚2•

aµν (xi ) µ ν + (known lower orders) = 0. (6.15)

‚x ‚x

and get

aµν nµ nν ¦ + (known) = 0. (6.16)

We can therefore ¬nd ¦ provided that

aµν nµ nν = 0. (6.17)

If this expression is zero, we are stuck. It is like having p0 (x) = 0 in an

ordinary di¬erential equation. On the other hand, knowing ¦ tells us the

6.1. CLASSIFICATION OF PDE™S 149

second normal derivative, and we can proceed to the adjacent surface where

we play the same game once more.

De¬nition: A characteristic surface is a surface Σ such that aµν nµ nν = 0

at all points on Σ. We can therefore propagate our data forward, provided

that the initial-data surface “ is nowhere tangent to a characteristic surface.

In two dimensions the characteristic surfaces become one-dimensional curves.

An equation in two dimensions is hyperbolic, parabolic, or elliptic at at a

point (x, y) if it has two, one or zero characteristic curves through that point,

respectively.

Characteristics are both a curse and blessing. They are a barrier to

Cauchy data, but are also the curves along which information is transmitted.

6.1.2 Characteristics and ¬rst-order equations

Suppose we have a linear ¬rst-order partial di¬erential equation

‚u ‚u

a(x, y) + b(x, y) + c(x, y)u = f (x, y). (6.18)

‚x ‚y

We can write this in vector notation as (v · )u + cu = F , where v is the

vector ¬eld v = (a, b). If we de¬ne the ¬‚ow of the vector ¬eld to be the

family of parametrized curves x(t), y(t) satisfying

dx dy

= a(x, y), = b(x, y), (6.19)

dt dt

then (6.18) reduces to an ordinary di¬erential equation

du

+ c(t)u(t) = f (t) (6.20)

dt

along each ¬‚ow line. Here,

u(t) ≡ u(x(t), y(t)),

c(t) ≡ c(x(t), y(t)),

f (t) ≡ f (x(t), y(t)). (6.21)

If we have been given the initial value of u on a curve “ that is nowhere

tangent to any of the ¬‚ow lines, we can propagate this data forward along

the ¬‚ow by solving (6.20). If the curve “ did become tangent to one of the

150 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

¬‚ow lines at some point, the data will generally be inconsistent with (6.18)

at that point, and no solution can exist. The ¬‚ow lines are therefore play

a role analagous to the characteristics of a second-order partial di¬erential

equation, and are therefore also called characteristics.

6.2 Wave Equation

6.2.1 d™Alembert™s Solution

Let •(x, t) obey the wave equation

‚2• 1 ‚2•

’ 2 2 = 0, ’∞ < x < ∞. (6.22)

‚x2 c ‚t

We begin with a change of variables. Let

ξ = x + ct,

· = x ’ ct. (6.23)

be light-cone co-ordinates. In terms of them, we have

1

x= (ξ + ·),

2

1

(ξ ’ ·).