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Heat Equation - Maximum Principle proof

  1. May 16, 2010 #1
    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!
     
    Last edited: May 16, 2010
  2. jcsd
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