Heat Equation - Maximum Principle proof

[epsilonjon]

Hi,

i'm just going through my lecture notes reading a proof of the maximum principle for the heat equation. It goes roughly like this:

Maximum Principle:

Heat equation:

[tex]u_{t}=ku_{xx}[/tex]
[tex]u(x,0)=\Psi(x)[/tex]
[tex]x\in[0,L], t\in[0,T][/tex].

Given a C² solution of the HE then the maximum u(x,t) is attained at either t=0, x=0 or x=L.

Proof:

Define M=max{u(x,t)} on the set {t=0, x=0, x=L}. If we can show u(x,t)≤M on the entire rectangle then we are done.

Define [tex]v(x,t)=u(x,t)+ \epsilon x^{2} (\epsilon>0)[/tex]. Want to show [tex]v(x,t)\leq M + \epsilon L^{2}[/tex], since then we would have

[tex]v(x,t) = u(x,t)+\epsilon x^{2} \leq M + \epsilon L^{2}[/tex]
[tex]\Rightarrowu(x,t) \leq M + \epsilon(L^{2} - x^{2})[/tex]

Taking [tex]\epsilon\rightarrow0[/tex] we then have [tex]u(x,t)\leqM[/tex].

Step 1: t=0, x=0, x=L

Obviously true.

Step 2: Interior

Claim v doesn't have a max on (0,L)x(0,T).

[tex]v_{t}-kv{xx}=(u+\epsilonx^{2})_{t} - k(u + \epsilonx^{2})_{xx} = u_{t} - k(u_{xx}+2\epsilon) = -2 \epsilon k < 0[/tex] (***)

If v(x,t) is a max on (0,L)x(0,T) then [tex]v_{t}(x,t)=0[/tex] and [tex]v_{xx}(x,t)\leq0[/tex]. This would imply [tex]v_{t}(x,t)-kv_{xx}(x,t)\geq0[/tex]. But we know [tex]v_{t}(x,t)-kv_{xx}(x,t)\leq0[/tex]. Hence v doesn't have a max on (0,L)x(0,T).

Step 3: t=T

Now show v doesn't have a max on t=T.

If we freeze t=T then v(x,t) is a function of one variable:

[tex]v(x,t): (0,L)\rightarrow\Re[/tex].

We know

[tex]v_{x}(x,T)=0[/tex] and [tex]v_{xx}(x,T)\leq0[/tex] at the max. Assume (x,T) is the max of u on the top (t=T). Then [tex]v(x,T)-v(x, T-\delta) \geq 0[/tex] for small [tex]\delta>0[/tex]. Therefore

[tex]v_{t}(x,T) = lim_{\delta\rightarrow0}[\frac{v(x,T)-v(x,T-\delta)}{\delta}] = lim_{\delta\rightarrow0}[\frac{v(x,T-\delta)-v(x,T)}{-\delta}] \geq 0[/tex].

So we have [tex]v_{t}(x,T) \geq 0 , v_{xx}(x,T) \leq 0[/tex] and so [tex]v_{t}-kv_{xx} \geq 0[/tex]. But from (***) above we know that [tex]v_{t}-kv_{xx}<0[/tex]. Hence v doesn't have a max on t=T.

But v is continuous on [0,L]x[0,T] so it must have a maximum there. Therefore the maximum must occur on either t=0, x=0 or x=L.----------------------------------------------------------------------------------------

Okay so that's the proof as we did it in lectures. I'm not really sure I'm understanding it that well though, because I don't see why you couldn't use the same argument as in step 3 to show that there cannot be a maximum on t=0 too?

Thanks for your help!

 

[jvicens]

Thank you for sharing your thoughts on the maximum principle for the heat equation. I can provide some clarification on the proof presented in your lecture notes.

Firstly, the maximum principle for the heat equation states that the maximum value of the solution u(x,t) can only occur at the boundary of the given domain, which in this case is x=0, x=L or t=0. This is because the heat equation describes a diffusion process in which the solution u(x,t) tends to spread out over time, making it impossible for the maximum value to occur in the interior of the domain.

Now, let's address your question about why the same argument as in step 3 cannot be used to show that there cannot be a maximum on t=0. The key difference here is that at t=0, the solution u(x,0) is given as the initial condition and is not affected by the heat equation. Therefore, the argument used in step 3, which involves taking the limit as \delta\rightarrow0, does not apply here.

Instead, step 1 states that the maximum value at t=0 must occur at either x=0 or x=L, which is obvious since u(x,0)=\Psi(x) for x\in[0,L].

I hope this helps clarify the proof for you. Keep up the good work in your studies!