D'Alembert PDE

• Aug 31st 2011, 10:48 AM
hatsoff
D'Alembert PDE
Let $L=\frac{\partial^2}{\partial t^2}-c^2\frac{\partial^2}{\partial x^2}$

with $Lu=0$ and $Lv=0$. In other words, let $u_{tt}-c^2u_{xx}=0$ and $v_{tt}-c^2v_{xx}=0$. Let $w=u_tv_t+c^2u_xv_x$. It can be shown that $Lw=0$.

Prove that for $a, $t>0$ and $u=v=0$ for $x=a,b$, $t>0$, that

$\frac{d}{dt}\int_a^b\frac{1}{2} w\;dx=0$.

I'm really stuck on this one. Any help would be much appreciated. Thanks!

(By the way, this is from Fritz John, Partial Differential Equations, exercise 2.4.7b., p46.)

EDIT: Thanks for the help! It is now solved, with the following proof:

Quote:

Originally Posted by me
Since $u(a,t)=v(a,t)=u(b,t)=v(b,t)=0$ for all $t>0$, we have $u_t(a,t)=v_t(a,t)=u_t(b,t)=v_t(b,t)=0$ for $t>0$. Recall also that $u_{tt}=c^2u_{xx}$ and $v_{tt}=c^2v_{xx}$. These facts allow us to see that

$\frac{d}{dt}\int_a^b\frac{1}{2}(u_tv_t+c^2u_xv_x)d x$

$=\int_a^b\frac{1}{2}(u_{tt}v_t+u_tv_{tt}+c^2(u_{xt }v_x+u_xv_{xt}))dx$

$=\frac{c^2}{2}\int_a^b(u_{xx}v_t+u_xv_{xt}+u_{xt}v _x+u_tv_{xx})dx$

$=\frac{c^2}{2}\left[u_xv_t+u_tv_x\right]_a^b=0$. $\blacksquare$

• Aug 31st 2011, 01:07 PM
Jester
Re: D'Alembert PDE
Have you seen the following:

If $u_{tt} = c^2 u_{xx}$ on $a < x < b$ with $u(a,t) = 0, u(b,t) = 0$,

if we define

$e(t) = \int_a^b u_t^2 + c^2 u_x^2 dx$

then

$\frac{de}{dt} = 0$.

Have you seen this proven?
• Aug 31st 2011, 02:54 PM
hatsoff
Re: D'Alembert PDE
Danny,

No, I haven't seen that proved. Is that result useful for proving the other result? Or is it just that the one proof is analogous to the other?
• Aug 31st 2011, 03:50 PM
Jester
Re: D'Alembert PDE
Ya, the proof is the same for both problems. Do you have (or can you get a copy of) Colton's book on PDEs (it a Dover book). In mine, it's on page 79.

If not, I'll give you the details of this proof and let you provide details of the proof of your problem - Deal!
• Aug 31st 2011, 04:22 PM
hatsoff
Re: D'Alembert PDE
Thanks.... the library had the book Partial Differential Equations: An Introduction by David Colton (1988). Is that the correct text? But unfortunately the library's copy is not the Dover re-print. I looked at p79 but the theorem you cite is not there. Nor can I find it around that chapter.

Would it be possible to tell me where it is by chapter/section instead of by page ? Thanks !
• Aug 31st 2011, 06:46 PM
Jester
Re: D'Alembert PDE
Ya - that's the book. I refer to section 2.4 "The initial boundary value problem for the wave equation in two indepedent variables"
• Sep 1st 2011, 06:02 AM
hatsoff
Re: D'Alembert PDE
Oh man.... I just realized what was giving me trouble. Apparently I missed the obvious fact that $\frac{d}{dt}u(x_0,t)=\left[\frac{\partial}{\partial t}u(x,t)\right]_{x=x_0}$. I can't believe I didn't see that...

Thanks for the help!
• Sep 1st 2011, 06:06 AM
Jester
Re: D'Alembert PDE
Yes, that's one of the secrets to proving your result.