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I have the heat equation: $$u_t (x,t) -ku_{xx}(x,t)=0 \quad 0<x<L,0<t$$ $$u(x,0)=\phi(x) \quad 0 \leq x \leq L$$

I want to show that if $\phi(x)=0$ then using the maximum/ minimum principle to show that this will be the trivial solution. I know that we are trying to show that $u(x,t)=0$ whenever $0\leq x\leq L$ and $0\leq t$. I am really unsure how to do this.

I know the maximum principle says that the maximum is attained either at t=0 or on one of the sides $x=0$ or $x=L$.

How can I say that on the sides $x=0$ or $x=L$ it is also zero.

wroobell
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user147825
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  • Do you have boundary conditions for $u$? – littleO Mar 18 '16 at 10:19
  • No, I just I have the initial condition nothing else. – user147825 Mar 18 '16 at 10:23
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    Often with the heat equation we solve "initial / boundary value problems" (aka IBVPs). I think that without boundary conditions for $u$, the solution to the PDE is not uniquely determined, and not guaranteed to be $0$. – littleO Mar 19 '16 at 03:01
  • Take a look at this post: http://math.stackexchange.com/questions/1277769/growth-of-tychonovs-counterexample-for-heat-equation-uniqueness . You're not going to get what you want. There the OP cites the famous non-uniqueness result of Tychonov, which is a solution of the heat equation with $0$ initial data on the real line. – Disintegrating By Parts Mar 19 '16 at 20:55

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