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).

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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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