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).
Let's show uniqueness of solutions to the heat equation with Dirichlet boundary conditions.
Let u, v be distinct solutions and set w=u−v. Then w satisfies wt=κwxx,w(0,t)=T0−T0=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=κwxx⟹wwt=κwwxx⟹∂∂t12w2=κwwxx⟹∂∂t∫L012w2dx=κ∫L0wwxxdx
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|L0−κ∫L0w2xdx=−κ∫L0w2xdx where the second equality occurs because w(0,t)=0=w(L,t) [*]. So in the end, we have the fact that E′(t)=−κ∫L0w2xdx≤0, 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:
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 t≥0. So w=u−v≡0 and thus u(x, t) \equiv v(x, t)$ and the solutions must be the same.
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.