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Let $a \in R$ and consider a sufficiently regular solution of:

$u_{tt}+au_t-u_{xx}=0$, $t>0, x \in ]0,1[$

$u(0,t)=u(1,t)=0$, $t>0$

$u(x,0)=\phi(x), u_t(x,0)=\psi(x)$, $x \in [0,1]$

Define $E(t)=\int_0^1 u_t^2+u_x^2 dx$.

Show that $E(t)=E(0)-a\int_0^t \int_0^1 u_t^2 (x,s) dxds$.

My idea:

We have: $E'(t)=\int_0^1 u_tu_{tt}+u_xu_{xx} dx=\int_0^1 -au_t^2 + u_{xx}u_t+u_xu_{xt} dx=\int_0^1 -au_t^2+ (u_xu_t)_x dx $

Integrating:

$E(t)-E(0)=\int_0^t E'(t) =\int_0^t \int_0^1 (-au_t^2+ (u_xu_t)_x dx) ds$

$=-a\int_0^t \int_0^1 u_t^2 (x,s) dxds + \int_0^t \int_0^1 (u_xu_t)_x dxds$

Then, I tried to show that $\int_0^t \int_0^1 (u_xu_t)_x dxds=0$ and obtain the result, but I don't get any result. Is this the right way?

Renato Moreira
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1 Answers1

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Differentiating $$E(t)=\int_0^1 (u_t^2+u_x^2)\, dx$$ with respect to $t$ yields $$E'(t)=2\int_0^1 ( u_t u_{tt}+ u_x u_{xt})\, dx$$ Your formula is missing $2$ and has $u_{xx}$ there. After substituting $u_{tt}=-au_t + u_{xx}$ we get $$E'(t)=-2a\int_0^1 u_t^2 \,dx +2\int_0^1 (u_t u_{xx} + u_x u_{xt}) \,dx$$ The second integral is zero, because the integrand is the $x$-derivative of $u_t u_x$, which vanishes at both endpoints. Thus, $$E'(t)=-2a\int_0^1 u_t^2 \,dx$$ which integrates to $$E (t)=E(0)-2a\int_0^t\int_0^1 u_t^2 \,dx\,ds$$

  1. Does the extra factor of 2 resolve or modify the question raised in the comments?

  2. Comments by Euler....IS_ALIVE are acknowledged.

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