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)).


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}


(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.


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

By choosing


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).

About baotangquoc

Lecturer School of Applied Mathematics and Informatics Hanoi University of Science and Technology No 1, Dai Co Viet Street, Hanoi
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