Author Topic: P4  (Read 1258 times)

Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
P4
« on: February 15, 2018, 07:06:24 PM »
Consider the PDE  with boundary conditions:
\begin{align}
&(u_x -\alpha u_{tt})|_{x=0}=0,\tag{2}\\
&(u_x +\beta u_{tt})|_{x=L}=0\tag{3}
\end{align}
where  $c>0$, $\alpha>0$, $\beta>0$ and $a$ are constant. Prove that the energy $E(t)$ defined as

E(t)= \frac{1}{2}\int_0^L \bigl( u_t^2 + c^2u_{x}^2 +au^2)\,dx +\frac{\alpha c^2}{2}u_t(0,t)^2+
\frac{\beta c^2}{2}u_t(L,t)^2
\tag{4}

does not depend on $t$.

Jilong Bi

• Jr. Member
• Posts: 8
• Karma: 9
Re: P4
« Reply #1 on: February 15, 2018, 09:11:09 PM »
fist take derivative with respect to t,$$\frac{1}{2}\int_0^L(2u_tu_{tt}+c^22u_xu_{xt}+2auu_t)dx+\frac{ac^2}{2}2u_t(0,t)u_{tt}(0,t)+\frac{bc^2}{2}2u_{t}(L,t)u_{tt}(L,t)$$ By the boundary condition given, $$u_{tt} = c^2u_{xx}-au$$ $$\Rightarrow \int_0^L(c^2u_tu_{xx}-auu_t+c^2u_xu_{xt}+auu_t)dx+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ $$\Rightarrow c^2\int_0^L(u_tu_{xx}+u_xu_{xt})dx+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ Since $$u_tu_x$$ take derivative with respect to x is equal to $$u_tu_{xx}+u_xu_{xt}$$ $$\Rightarrow c^2 u_t(L,t)u_x(L,t) - c^2u_t(0,t)u_x(0,t)+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ By another boundary condition $$u_x(0,t) = au_{tt}(0,t)$$ $$\Rightarrow c^2 u_t(L,t)u_x(L,t) - c^2au_t(0,t)u_{tt}(0,t)+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$ And for the last boundary condition $$-u_x(L,t)=bu_{tt}(L,t)$$$$\Rightarrow -c^2b u_t(L,t)u_{tt}(L,t) - c^2au_t(0,t)u_{tt}(0,t)+ac^2u_t(0,t)u_{tt}(0,t)+bc^2u_{t}(L,t)u_{tt}(L,t)$$  last step: $$-c^2b u_t(L,t)u_{tt}(L,t) +cb^2u_{t}(L,t)u_{tt}(L,t)- c^2au_t(0,t)u_{tt}(0,t)+ca^2u_t(0,t)u_{tt}(0,t)$$  All terms cancel ,so the equation equal to 0 and the energy does not depend on t.

Victor Ivrii

This problem describes the string with the masses on its ends (forces are $\pm c^2u_x$ on the left/right ends and acceleration $u_{tt}$)