## Well posedness of a coupled PDE-ODE system

In this post, I simply give a proof of the well posedness of a “simple” coupled PDE-ODE system (which we mentioned in this post).

The idea is based on “localized method”.

Let $\Omega\subset \mathbb R^n$ is a bounded set with smooth boundary $\partial\Omega$. We consider the following system

$(1) \qquad \begin{cases} L_t - d_L\Delta L = 0, &x\in\Omega,\\ d\frac{\partial L}{\partial\nu} + \beta L = \alpha L, &x\in\partial\Omega, \\ l_t + \alpha l = \beta L, &x\in\partial\Omega,\\ L(0,x) = L_0(x),&x\in\Omega,\\ l(0,x) = l_0(x),&x\in\partial\Omega.\end{cases}$

Theorem: Suppose that $L_0\in L^2(\Omega)$ and $l_0\in L^2(\partial\Omega)$. Then there exists $T>0$ such that system (1) has a unique weak solution $(L^*,l^*)\in C([0,T];L^2(\Omega))\times C([0,T];L^2(\partial\Omega))$.

Proof:

Denote by $X = C([0,T]; L^2(\partial\Omega))$. We construct a mapping $A: X \rightarrow X$ as follows. Take any $l_1 \in X$. We denote by $L_1$ is a unique weak solution of the following reaction diffusion equation with inhomogeneous Robin boundary condition

$(2)\qquad \begin{cases} L_t - d_L\Delta L = 0, &x\in\Omega, \\ d_L\frac{\partial L}{\partial \nu} + \beta L = \alpha l_1, &x\in\partial\Omega,\\ L(0,x) = L_0(x),&x\in\Omega\end{cases}$.

We know that $L_1\in L^2(0,T;H^1(\Omega))$, thus, $L_1|_{\partial\Omega} \in L^2(0,T;L^2(\partial\Omega))$. Hence, solve the ODE

$(3) \qquad \begin{cases} l_t + \alpha l = \beta L_1,&x\in\partial\Omega,\\ l(0,x) = l_0(x), &x\in\partial\Omega\end{cases}$

we have get a unique weak solution $l_2\in X$. Now we define $A: X \rightarrow X$ as

$Al_1 = l_2$

We will prove that $A$ is a contraction mapping. Denote by $l_2 = Al_1, p_2 = Ap_1$ where $l_1, p_1 \in X$. We also denote by $L_1, P_1$ solutions of (2) w.r.t $l_1, p_1$ respectively. For the differences, we write $U = l_2 - p_2, V = L_ 1 - P_1$. Then from (2) and (3), we have

$(4)\qquad \begin{cases} U_t + \alpha U = \beta V, &x\in \partial\Omega,\\ U(0,x) = 0\end{cases}$

and

$(5)\qquad \begin{cases}V_t - d_L\Delta V = 0, &x\in\Omega,\\ d_L\frac{\partial V}{\partial\nu} + \beta V = \alpha(l_1 - p_1), &x\in\partial\Omega,\\ V(0,x) = 0, &x\in\Omega.\end{cases}$

Multiply (4) by $U$ then integrate on $\partial\Omega$, we get

$\frac{d}{dt}\int_{\partial\Omega}U^2dS+\alpha\int_{\partial\Omega}U^2dS =\beta \int_{\partial\Omega}UVdS$

$\leq \alpha\int_{\partial\Omega}U^2dS + \frac{\beta^2}{4\alpha}\int_{\partial\Omega}V^2dS.$

Integrating on $(0,t)$, we get

$(6)\qquad \frac{1}{2}\int_{\partial\Omega}U(t)^2dS \leq \frac{\beta^2}{2\alpha}\int_0^t\int_{\partial\Omega}V(s)^2dSds$

Now, multiplying (5) by $V$ then integrating over $\Omega$, doing some integrations by parts, we find that

$\frac{d}{dt}\int_{\Omega}V^2dx + d_L\int_{\Omega}|\nabla V|^2dx = \int_{\partial\Omega}(l_1-p_1-\beta V)VdS$

$\leq -\frac{\beta}{2}\int_{\partial\Omega}V^2dS + \frac{\alpha^2}{2\beta}\int_{\partial\Omega}|l_1-p_1|^2dS$

Integrate on $(0,t)$, we get, inparticular

$\beta\int_0^t\int_{\partial\Omega}V(s)^2dSds \leq \frac{\alpha^2}{\beta}\int_0^t\int_{\partial\Omega}|l_1(s)-p_1(s)|^2dSds\leq \frac{\alpha^2}{\beta}T\|l_1-p_1\|_X^2$.

Apply this to (6) we have, for all $t\in [0,T]$

$\int_{\partial\Omega}U(t)^2dS \leq \frac{\alpha}{2}T\|l_1 - p_1\|_X^2$.

Thus,

$\|l_2 - p_2\|_X \leq \sqrt{\frac{\alpha}{2}T}\|l_1-p_1\|_X$.

By choosing

$T<\frac{2}{\alpha}$

we see that $A$ is a contraction mapping. Then $A$ has a unique fixed point $l^*$. Insert $l^*$ to (2) to get a unique $L^*$. It is easy to verify that $(L^*, l^*)$ is the solution of the system (1).

The uniqueness of solutions is standard since the system is linear, so we omit it here (or leave it to the readers).