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I'm trying to solve a set of exercises in order to prepare myself for a test. This question verses about energy method on partial differential equations and I would like to ask for help on that, and, if possible, a refference on energy methods and maximum principles. I'm a begginer on partial differential equations.

Show that there is at most one solution to the problem $$\begin{cases}u_t=\alpha^2u_{xx}+g,\textrm{ in }(0,L)\times(0,\infty)\\ u(0,t)=u(L,t)=0,t\geqslant 0\\ u(x,0)=u_0(x),\textrm{ in }[0,L]\\ u\in C^2([0,L]\times(0,\infty))\cap C([0,L]\times[0,\infty)) \end{cases}$$ if $u$ is a continuous, differentiable by parts, $u_0(0)=u_0(L)=0$, $g\in C((0,L)\times(0,\infty))$, using: (a) the maximum principle; (b) the energy method. Obtain the candidate for a solution, of the form $u(x,t)=\sum_{n=1}^{\infty}c_n(t)\sin\left(\frac{n\pi x}{L}\right)$.

Thanks in advance! P.S.: Oscar Niemeyer lives forever!

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    Livessssss Foooreeeeeeveeerrrrrrrrrrr!!!!!!!!2012-12-07
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    Foreverrrr!!!! :)2012-12-07
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    FALSE! Niemeyer is DEAD now.2012-12-08
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    It was just a tribute from an admirer, only. But the question of PDE? Thank you2012-12-08

1 Answers 1

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The existence is guaranteed by Theorem (Theorem 2, page 50, Evans, L.C. Partial Differential Equations). Let $ T>0$ arbitrary and suppose $w=u_{1}-u_{2}$. Then $$\begin{cases}w_{t}=\alpha^2w_{xx}, \textrm{ in }(0,L)\times(0,T]\\ w(x,0)=0, x\in(0,L)\\ w(x,t)=0, \{0,L\}\times[0,T] \end{cases}$$ By the principle of maximum

max $w$ in$[0,L]\times[0,T]=$max $w$ in $\{0,L\}\cup\{0,T\}=0$. Then, $w=0$ in $[0,T], \forall T>0\Longrightarrow u_{1}=u_{2}$.

Item b), I do not know :(

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    For b), note that $$\frac{d}{dt} \int_0^L w^2 = \int 2ww_t = \int_0^L 2\alpha^2 ww_{xx} = -2\alpha^2 \int_0^L (w_x)^2 \le 0.$$2012-12-11
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    Good morning Sam. First thanks for listening. I understand, this is going to help me in other things. But you used Green's identity, right? This is energy method? thank you2012-12-12
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    I don't know if I would call it "Green's identity", the third equality is by integration by parts, using the fact that the boundary terms vanish. I believe this is the energy method, yes. If we define the energy functional by $E[w] = \int_0^L w^2$, then this inequality says that the energy can never increase for solutions of your PDE. So the point is that you want to find some functional $E$, which satisfies $E[w]=0$ iff $w=0$ and such that $E[w]$ is non-increasing for all solutions of the problem at hand. This immediately gives uniqueness.2012-12-12
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    So the problem of using the energy method to prove uniqueness basically consists in finding the right energy functional for your problem. At least this is how I interpret it!2012-12-12
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    @Sam Why can switch derivative and integral in first equality?2015-11-17