I'm looking at a very simple 2 component linear reaction-diffusion system. The equations are:

$$\partial_t u = a \, \partial_{xx} u + bw - cu \\ \partial_t w = d \, \partial_{xx} w - bw + cu$$

with $a,b,c,d>0$.

There is an internal boundary condition for $w$:

$$\partial_x w(t,0^-)=m\left[w\left(t,0^{-}\right)-1\right]\,,\quad m > 0\,;\qquad\qquad \partial_x w\left(t,0^{+}\right)=0.$$

Notably this means that $w$ is generally not continuous at $x=0$.

Past this point the boundary conditions are a bit flexible for mathematical convenience, but the most natural problem I can think of is to have the spatial domain be all of $\mathbb{R}$ and require $u,w \to 0$ as $|x| \to \infty$.

Otherwise it is a standard Cauchy problem: $u(0,x)$ and $w(0,x)$ are given.

Are there closed form solutions to this problem as a function of the initial data? I tried finding stationary solutions but with this decay boundary condition it turns out that either $u$ or $w$ is forced to be negative in the stationary problem, which does not really make physical sense. I haven't seen examples of separation of variables for coupled systems of linear PDE, so I'm not sure how easy that is to carry out.

It is not difficult to see that the Fourier problem can be solved, as it is just

$$\partial_t \hat{u} = (-a |\xi|^2 - c) \hat{u} + b \hat{w} \\ \partial_t \hat{w} = c \hat{u} + (-d |\xi|^2 - b) \hat{w}$$

so the solution is given through the exponential of the matrix $\begin{bmatrix} -a |\xi|^2 - c & b \\ c & -d |\xi|^2 - b \end{bmatrix}$. I'm not sure this is feasible to Fourier-invert, though, nor am I sure how to insert the internal boundary condition into the Fourier problem.


1 Answer 1


Some ideas you can think about:

  1. Separation of variables. Let $u(t,x)=e^{-kt}U(x)$ and $w(t,x)=e^{-kt}W(x)$. You get two ODEs for $U(x)$ and $W(x)$, which seem straightforward to solve and apply BCs to. To get a more general solution, apply the superposition principle.

  2. Reformulate your problem on $\mathbb R$ with jumps at $x=0$ as two coupled one-sided Cauchy problems with no jumps. Let $w_-(t,x)=w(t,-x)$ and $w_+(t,x)=w(t,x)$ for $x>0$, and similarly for $u_\pm(t,x)$. You would require that $(u_+,w_+)$ and $(u_-,w_-)$ each solve your differential equations for $x>0$. For boundary conditions, you would first impose: \begin{align} \partial_x w_-(t,0)&=-m(w_-(t,0)-1),\\ \partial_x w_+(t,0)&=0. \end{align} Second, I would impose continuity of $w,u,$ and $\partial_x u$ at $x=0$: \begin{align} w_-(t,0)&=w_+(t,0),\\ u_-(t,0)&=u_+(t,0),\\ -\partial_x u_-(t,0)&=\partial_x u_+(t,0). \end{align} The reason is because otherwise, $w_+$ and $w_-$ are completely unrelated.

To solve each one-sided system, apply the Laplace transform in $x$. You will get a coupled system of $t$-ODEs which can probably be solved. The inverse transform probably won't be explicitly computable, but numerical evaluations and/or asymptotics would be possible.

  • $\begingroup$ Interesting ideas, I will think about them. Also, $w_+$ and $w_-$ are coupled through their relationship with $u$, which has no mandatory boundary condition (though it should be continuous at $x=0$ if everything makes physical sense). In fact it is precisely this coupling that was the point of making up this system. $\endgroup$
    – Ian
    Jun 30, 2017 at 23:13
  • $\begingroup$ In idea #1, I guess the first $k$ is $c$ and the second one is $b$? Or are they both $b+c$ (the negative of the nonzero eigenvalue of $\begin{bmatrix} -c & b \\ c & -b \end{bmatrix}$)? $\endgroup$
    – Ian
    Jul 1, 2017 at 0:57
  • $\begingroup$ @Ian $k$ is actually an arbitrary constant. Compare this situation with the heat equation: $u_t=a u_{xx}$. If you seek a solution of the form $u(t,x)=e^{-kt}U(x)$, then there are many such solutions. Note however that the range of $k$ will be restricted by the boundary condition $|u(t,x)|\le C$ as $|x|\to\infty$ (i.e. you don't want $U(x)$ to grow exponentially in $x$). Of course, the situation with $W$ is a little different than with $U$, since other BCs are involved. $\endgroup$
    – user254433
    Jul 1, 2017 at 1:24
  • $\begingroup$ Oh, I see, it's a separation constant, in this case continuous-valued because the domain is infinite. Nice. $\endgroup$
    – Ian
    Jul 1, 2017 at 1:51

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