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I have problems checking that the energy $E(t)=\frac{1}{2}\int_I(u_t^2+c^2u_x^2)dx$ on an open interval $I\subset \mathbb R$, such that $u(0,x)=0$ and $u_t(0,x)=0$ for $x\in\mathbb R\setminus I$ is monotonically decreasing.

I think the best way is to show $E'\le0$

Therefore $E'(t)=\int_I(u_tu_{tt}+c^2u_xu_{xt})dx=\int_I(u_tu_{xx}c^2+c^2u_xu_{xt})dx=c^2\int_I(u_tu_{xx}+u_xu_{xt})dx=c^2\int_I(\frac{\partial(u_tu_x)}{\partial x})dx$

Could you explain why this should be decreasing ?

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I changed $u(0,x=0)$ to $u(0,x)=0$; please confirm if this is what you meant. Is the PDE considered on $I$ or on $\mathbb R$? If the former, do you have boundary conditions as well? If the latter, are the initial conditions for $u,u_t$ imposed only on $I$ or on all of $\mathbb R$? –  user53153 Jan 13 '13 at 15:56
    
Yes I meant that, it is considered on I. the initial conditions are imposed for all $x\in \mathbb R $ without $I$ –  Voyage Jan 13 '13 at 16:09
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But with those initial conditions the solution is $u\equiv 0$. Which has $E(t)\equiv 0$. Problem solved? // Maybe you should write down the wave equation you are using; I've been assuming $u_{tt}=c^2 u_{xx}$. –  user53153 Jan 13 '13 at 16:15
    
Hmm, how can the initial conditions be changed such that $E(t)\le0$ ? This is the second part of an exercise, I alreadey showed in the first part that from -\infty to \infty the energy is constant. –  Voyage Jan 13 '13 at 16:17
    
My guess is that the initial conditions are imposed on $\mathbb R\setminus I$. –  user53153 Jan 13 '13 at 16:20

1 Answer 1

up vote 2 down vote accepted

You calculated $E'(t)=u_t(b) u_x(b)-u_t(a)u_x(a)$ where $[a,b]=I$. Using d'Alembert's formula you can find $$ u_tu_x=\frac{1}{4c}\left\{(cg'(x+ct)+h(x+ct))^2 - (cg'(x-ct)+h(x-ct))^2 \right\} $$ When $x=b$, the first square vanishes and we get $u_tu_x\le 0$. The opposite happens when $x=a$.

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Could you explain me, why the first square vanishes if x=b?? –  Voyage Jan 13 '13 at 21:30
    
@Voyage If $x=b$, then $x+ct>b$. Hence, $x+ct$ falls outside of the interval $I$, where the initial data $g,h$ are identically zero. –  user53153 Jan 13 '13 at 21:33
    
Thanks, but where do you set x=b? –  Voyage Jan 13 '13 at 21:34
    
@Voyage Your calculation ended with $\displaystyle \int_I \frac{\partial (u_tu_x)}{\partial x}\,dx$. The integral of derivative evaluates to $u_tu_x\Big|_{x=a}^{x=b}$, which is what I wrote in my answer. –  user53153 Jan 13 '13 at 21:37
    
So it is decreasing and increasing at the same time, which is just valid if it is constant, which is a stronger claim then just decreasing... or have I overlooked something that permits it to be strictly decreasing? –  Stefan Jun 20 at 0:33

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