Show that the initial-boundary value problem

\begin{align} & {{u}_{tt}}={{u}_{xx}}\text{ }(x,t)\in \left( 0,l \right)\times \left( 0,T \right),\text{ }T,l>0 \\ & u\left( x,0 \right)=0,\text{ }x\in \left[ 0,l \right] \\ & {{u}_{x}}\left( 0,t \right)-u\left( 0,t \right)=0,\text{ }{{u}_{x}}\left( l,t \right)+u\left( l,t \right)=0,\text{ }t\in \left[ 0,T \right]\\ \end{align}

has zero solution only.

My attempt :

By separation of variables $\frac{{{X}''}}{X}=\frac{{{T}''}}{T}=\lambda .$

For case $\lambda =0$ or $\lambda <0$ , it is easy to show $X=0$ . Hence $u$ is zero.

For case $\lambda >0$ , let $\lambda ={{\omega }^{2}},\text{ }\omega >0.$

$\Rightarrow X\left( x \right)={{c}_{1}}\cos \omega x+{{c}_{2}}\sin \omega x,\text{ }{X}'\left( x \right)=-\omega {{c}_{1}}\sin \omega x+\omega {{c}_{2}}\cos \omega x$


$\begin{align} & {X}'\left( 0 \right)-X\left( 0 \right)=\omega {{c}_{2}}-{{c}_{1}}=0, \\ & {X}'\left( l \right)+X\left( l \right)=\left( -\omega {{c}_{1}}\sin \omega l+\omega {{c}_{2}}\cos \omega l \right)+\left( {{c}_{1}}\cos \omega l+{{c}_{2}}\sin \omega l \right)=0\text{ ,} \\ \end{align}$

We get $\tan \omega l=\frac{2w}{{{w}^{2}}-1}.$

Let ${{\alpha }_{n}}$ be the nth root of $\tan \omega l=\frac{2w}{{{w}^{2}}-1}.$

$\Rightarrow \lambda =\alpha _{n}^{2},\text{ }{{X}_{n}}\left( x \right)={{\alpha }_{n}}{{C}_{2}}\cos {{\alpha }_{n}}x+{{C}_{2}}\sin {{\alpha }_{n}}x$

$\Rightarrow {{T}_{n}}\left( t \right)={{A}_{n}}\cos \left( {{\alpha }_{n}}t \right)+{{B}_{n}}\sin \left( {{\alpha }_{n}}t \right)$

$\Rightarrow {{u}_{n}}\left( x,t \right)=\left[ {{a}_{n}}\cos \left( {{\alpha }_{n}}t \right)+{{b}_{n}}\sin \left( {{\alpha }_{n}}t \right) \right]\left( {{\alpha }_{n}}{{C}_{2}}\cos {{\alpha }_{n}}x+{{C}_{2}}\sin {{\alpha }_{n}}x \right)$

$\Rightarrow u\left( x,t \right)=\sum\limits_{n=1}^{\infty }{{{u}_{n}}\left( x,t \right)}.$

Since $u\left( x,0 \right)=0=\sum\limits_{n=1}^{\infty }{{{a}_{n}}\left( {{\alpha }_{n}}{{C}_{2}}\cos {{\alpha }_{n}}x+{{C}_{2}}\sin {{\alpha }_{n}}x \right)}$ , we get ${{a}_{n}}=0.$

Thus, $u\left( x,t \right)=\sum\limits_{n=1}^{\infty }{{{b}_{n}}\sin \left( {{\alpha }_{n}}t \right)\left( {{\alpha }_{n}}{{C}_{2}}\cos {{\alpha }_{n}}x+{{C}_{2}}\sin {{\alpha }_{n}}x \right)}$

Now I have problem to make $u\left( x,t \right)$ equal to zero. Help please.



Use energy method by showing that for

$$E(t) := \int_0^{L} (u_t^2 + u_x^2) dx$$

  1. $\forall t \in [0,T], E'(t) = 0$

  2. $E(t)$ is constant.

  3. $E(0) = 0$

  4. $E(t) = 0 \ \forall t \in [0,T]$

  5. $u_t^2 + u_x^2 = 0$

Based on Pinchover and Rubinstein:

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