Trinh @ Bath

Lecture 33: Uniqueness of solutions I

In this lecture, we'll prove uniqueness for two variants of the heat equation. The different variants you should consider as as follows (all posed on x[0,L]x[0,L]): ut=κuxx,u(x,0)=0, with either Dirichlet conditions [D] u=T0(x=0),u(L,t)=T1(x=L), or Neumann conditions [N] ux=G0(x=0),ux=G1(x=L), or mixed conditions [M] A0u+B0ux=H0(x=0),A1u+B1ux=H1(x=L).

Section 19.1: Uniqueness for zero Dirichlet heat equation

Let's show uniqueness of solutions to the heat equation with Dirichlet boundary conditions.

Let u, v be distinct solutions and set w=uv. Then w satisfies wt=κwxx,w(0,t)=T0T0=0,w(L,t)=0,w(x,0)=0,

Essentially this is a heat problem where the system begins at zero heat, has BCs with zero heat everywhere; it is sensible that the only solution is trivial. To prove it, we can manipulate the equation in the following way.

wt=κwxxwwt=κwwxxt12w2=κwwxxtL012w2dx=κL0wwxxdx

Let's call the quantity E(t)=L012w2dx, the energy.

Now use the fact that x(wwx)=w2x+wwxx to re-write the RHS. We then get after integrating once, E(t)=κwwx|L0κL0w2xdx=κL0w2xdx where the second equality occurs because w(0,t)=0=w(L,t) [*]. So in the end, we have the fact that E(t)=κL0w2xdx0, so the energy is always decreasing. But note that E(0)=0 since w(x,0)=0. Finally, note that E(t) is always 0 by its form (the integral of a squared quantity). So the energy is always decreasing, begins from zero, and can never be negative. We have the three statements:

  1. E(t)0 for all time,
  2. E(0)=0,
  3. E(t)0 for all time,

and you would conclude that it has to remain at its initial value, and therefore E(t)0 for all time. Looking at the form of the integrand, you would conclude that the only way this occurs is if the integrand is itself zero, or w2(x,t)=0 for all x[0,L] and for all t0. So w=uv0 and thus u(x, t) \equiv v(x, t)$ and the solutions must be the same.

Section 19.2: Uniqueness for other BCs of the heat equation

The proof of uniqueness for the Neumann condition [N] is done identically to the above, except in the step marked with [*] we use the fact that wx(0,t)=0=wx(L,t). But you would arrive at the same conclusions as above.