Application of semigroup method

Question

We consider the following Cauchy problem with certain conditions on the boundary: $$\eqalign{ & {u_{tt}}(t,x) + {u_{tt}}(t,0) - {u_{xx}}(t,x) + u(t,L) = 0{\rm{ in ]0}}{\rm{,T[}} \times {\rm{]0}}{\rm{,L[}} \cr & {\rm{\{ u(0}}{\rm{,x)}}{\rm{,}}{{\rm{u}}_t}(0,x)\} = \{ f(x),g(x)\} \cr} $$ My problem is how to transfer the above equation to the Cauchy problem form and how to deal with the term $u(t,L)$ and ${u_{tt}}(t,L)$ : $$\eqalign{ & Y' = AY \cr & Y(0) = {Y_0} \cr} $$ How can I writing this equation under the Cauchy form problem? Thanks.

Answer 1

I'd say that the boundary conditions are relevant for the matter.

For example, let us assume that the boundary conditions are $$u(t,0)=0,\qquad u_{tt}(t,L)+u_x(t,L)=-u_t(t,L).$$

If we write $v(t)=u(t,L)$, then the equation can be rewritten as $$u_{tt}(t,x)=u_{xx}(t,x)-v(t)-v_{tt}(t).$$

But the boundary conditions imply $$v_{tt}(t)=u_{tt}(t,L)=-u_t(t,L)-u_x(t,L)=-v_t(t)-u_x(t,L)$$ and thus the equation becomes $$u_{tt}(t,x)=u_{xx}(t,x)-v(t)+v_t(t)+u_x(t,L).$$

Now, if we write

$$Y=\begin{pmatrix} u\\ u_t\\ v\\ v_t \end{pmatrix}$$ then $$Y'=\begin{pmatrix} u_t\\ u_{tt}\\ v_t\\ v_{tt} \end{pmatrix}= \begin{pmatrix} u_t\\ u_{xx}-v+v_t+u_x(L)\\ v_t\\ -v_t-u_x(L) \end{pmatrix}. $$

So, the problem can be written in the desired form by taking

$$Y=\begin{pmatrix} u\\ \tilde{u}\\ v\\ \tilde{v} \end{pmatrix},\qquad A\begin{pmatrix} u\\ \tilde{u}\\ v\\ \tilde{v} \end{pmatrix}= \begin{pmatrix} \tilde{u}\\ u_{xx}-v+\tilde{v}+u_x(L)\\ \tilde{v}\\ -\tilde{v}-u_x(L) \end{pmatrix},\qquad Y_0=\begin{pmatrix} f\\ g\\ f(L)\\ g(L) \end{pmatrix}. $$

Note that, in this setting, the phase space should have the form $H_1\times H_2\times \mathbb C\times\mathbb C$, where $H_1$ and $H_2$ are functional spaces (probably some Sobolev space), and the domain of the operator should include the condition $v=u(L)$.

These considerations are based in a similar problem that I studied and maybe it gives you some insights on how to deal with the boundary terms (but in my case there were no boundary terms in the equation).

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Edit (in response to the comments).

Let me consider $$\eqalign{(1)\qquad & {u_{tt}} = {u_{txx}} + {u_{xx}}\cr (2)\qquad& {u_{tt}}(L) = - {u_{tx}}(L) - {u_x}(L) \cr (3)\qquad& u(0,t) = {u_t}(0,t) = 0 \cr (4)\qquad & u(L,t) = a,{u_t}(L,t) = b \cr (5)\qquad & u(x,0) = v(x),{u_t}(x,0) = w(x) \cr}$$ Multiplying equation $(1)$ by $\overline{u}_t$, integrating over $[0,L]$, performing integration by parts and using the boundary conditions $(2)$-$(4)$ we get $$\frac{1}{2}\frac{d}{dt}\left(\int_0^L\Big(|u_t(x,t)|^2+|u_x(x,t)|^2\Big)dx+|b(t)|^2\right)=-\int_0^L|u_{tx}(x,t)|^2\;dx.$$ So, the energy is given by \begin{equation} E(t)=\frac{1}{2}\int_0^L\Big(|u_t(x,t)|^2+|u_x(x,t)|^2\Big)dx+\frac{1}{2}|b(t)|^2. \end{equation} We consider the phase space $$H=H_\star^1\times L^2\times\mathbb{C}\quad(\text{where } H_\star^1=\{f\in H^1\mid f(0)=0\})$$ equipped with the norm associated with the energy: $$\|(u,U,b)\|_H^2=\int_0^\ell\Big(|U|^2+|u_x|^2\Big)dx+|b|^2.$$ Then, the problem can be written as \begin{equation} \left\{\begin{aligned} &Y'=AY\\ &Y(0)=Y_0 \end{aligned}\right. \end{equation} where $Y=(u,U,b)$, $Y_0=(v,w,w(L))$ and $A:D(A)\subset H \to H$ is given by $$AY=\begin{pmatrix} U\\ U_{xx}+u_{xx}\\ U(L) \end{pmatrix}$$ with domain $$D(A)=\{Y\in H\mid u\in H^2,\;U\in H_\star^1\cap H^2,\; b=U(L) \}.$$

We have $$\text{Re }\langle AY,Y\rangle_{H}=-\int_0^L|U|^2\;dx\leq 0$$ and thus $A$ is dissipative.

Here, of course, $\langle\cdot,\cdot\rangle_H$ is the inner product that induces the energy norm $\|\cdot\|_H$:

$$\langle(u^*,U^*,b^*),(u,U,b)\rangle_H=\int_0^L \Big(U^*\overline{U}+u^*_x\overline{u}_x\Big)dx+b^*\overline{b}.$$