7More partial differential equations

IB Methods

7.1 Well-posedness

Recall that to find a unique solution of Laplace’s equation

∇

2

φ

= 0 on a bounded

domain Ω

⊆ R

n

, we imposed the Dirichlet boundary condition

φ|

∂Ω

=

f

or the

Neumann boundary condition n · ∇φ|

∂Ω

= g.

For the heat equation

∂φ

∂t

=

κ∇

2

φ

on Ω

×

[0

, ∞

), we asked for

∂|

∂Ω×[0,∞)

=

f

and also φ|

Ω×{0}

= g.

For the wave equation

∂

2

φ

∂t

2

=

c

2

∇

2

φ

on Ω

×

[0

, ∞

), we imposed

φ|

∂Ω×[0,∞)

=

f

,

φ|

Ω×{0}

= g and also ∂

t

φ|

Ω×{0}

= h.

All the boundary and initial conditions restrict the value of

φ

on some co-

dimension 1 surface, i.e. a surface whose dimension is one less than the domain.

This is known as Cauchy data for our partial differential equation.

Definition

(Well-posed problem)

.

A partial differential equation problem is

said to be well-posed if its Cauchy data means

(i) A solution exists;

(ii) The solution is unique;

(iii)

A “small change” in the Cauchy data leads to a “small change” in the

solution.

To understand the last condition, we need to make it clear what we mean by

“small change”. To do this properly, we need to impose some topology on our

space of functions, which is some technicalities we will not go into. Instead, we

can look at a simple example.

Suppose we have the heat equation

∂

t

φ

=

κ∇

2

φ

. We know that whatever

starting condition we have, the solution quickly smooths out with time. Any

spikiness of the initial conditions get exponentially suppressed. Hence this is

a well-posed problem — changing the initial condition slightly will result in a

similar solution.

However, if we take the heat equation but run it backwards in time, we get

a non-well-posed problem. If we provide a tiny, small change in the “ending

condition”, as we go back in time, this perturbation grows exponentially, and

the result could vary wildly.

Another example is as follows: consider the Laplace’s equation

∇

2

φ

on the

upper half plane (x, y) ∈ R × R

≥0

subject to the boundary conditions

φ(x, 0) = 0, ∂

y

φ(x, 0) = g(x).

If we take g(x) = 0, then φ(x, y) = 0 is the unique solution, obviously.

However, if we take g(x) =

sin Ax

A

, then we get the unique solution

φ(x, y) =

sin(Ax) sinh(Ay)

A

2

.

So far so good. However, now consider the limit as A → ∞. Then

g(x) =

sin(Ax)

A

→ 0.

for all x ∈ R. However, at the special point φ

π

2A

, y

, we get

φ

π

2A

, y

=

sinh(Ay)

A

2

→ A

−2

e

Ay

,

which is unbounded. So as we take the limit as our boundary conditions

g

(

x

)

→

0,

we get an unbounded solution.

How can we make sure this doesn’t happen? We’re first going to look at first

order equations.