diffusion equation with time and space derivatives in boundary conditions

Question

I think the system I'm working with can be modeled with the following initial boundary value problem based on the 1D diffusion equation.

partial differential equation

$u_t - Du_{xx} = 0$

boundary conditions

$u_t(0,t) = \beta D u_x(0,t)$

$u(L,t) = 0$

initial condition

$u(x,0) = \phi(x)$

Where $u$ is the dependent variable, $0<x<L$ is location, $t$ is time, $D$ is the diffusion coefficient, $\beta$ is a constant, and $\phi$ is some arbitrary function of $x$ specifying the initial value of $u$ over the domain.

What solution methods exist for this sort of IBVP? Would the BC at $x=0$ be considered a Neumann boundary condition? The methods I've read about so far don't deal with any scenarios where both time and space derivatives are present in the BC.

Answer 1

Starting from @Leon Avery's answer , you have to solve $$\cos (\omega )-\frac{\omega }{\beta }\, \sin (\omega )=0$$ Admitting that you are concerned by the first solution in $\omega$, rewrite the equation as $$\omega \tan(\omega)=\beta$$ and have a look at this question of mine where is proposed, as a quite good approximation

$$\frac{\pi^2 \omega^2\left(\left(\frac{\pi ^2}{3}-\frac{17}{4}\right) \omega^2 +\pi^2\right) }{\omega^4-\frac{17 \pi ^2 }{4}\omega^2+\pi ^4 }=\beta$$ which reduces the problem to a quadratic equation in $\omega^2$.

Answer 2

This is tractable by separation of variables. Real solutions of the diffusion equation are of the form

$$ u(x,t) = \sum e^{-D\omega_i^2 t}(a_i cos(\omega_i x) + b_i sin(\omega_i x)) $$

The boundary conditions can be satisfied nontrivially only with certain values of $\omega_i$. Specifically, letting $L = 1$, the first BC gives

$$ -a\omega = b\beta \\ b = -\frac{a\omega}{\beta} $$

By the way, I don't think this BC is any of the usual types (Dirichlet, Neumann, Robin), because it involves the time derivative, which none of these do. The second BC then becomes

$$ cos(\omega) - \frac{\omega}{\beta}sin(\omega) = 0 $$

In general there are a (countably) infinite number of values of $\omega$ that satisfy that condition. You can satisfy yourself of that by plotting $cos(\omega)-\frac{\omega}{\beta}sin(\omega)$ vs $\omega$. These are the values $\omega_i$ can take on in the general solution. I suspect that for general $\beta$ you will need to find them numerically. Then, of course, you need to find the values of $a_i$ such that the initial condition is satisfied:

$$ \phi(x) = \sum a_i\left(cos(\omega_i x) - \frac{\omega_i}{\beta}sin(\omega_i x)\right) $$

Unless you can find some nice orthogonality condition that these eigenfunctions satisfy, you may be stuck with solving a system of linear equations to approximate the $a_i$.

Addendum:

Following @Claude_Leibovic's observation that the condition on $\omega$ reduces to

$$ \omega\tan(\omega) = \beta \\ \tan(\omega) = \frac{\beta}{\omega}, $$

I note that there is an easily visualized graphic solution to this constraint. Plot $\tan(\omega)$ and $\beta/\omega$ against $\omega$ -- the points where they cross are the desired solutions.

Now, note that $\tan(\omega)$ and $\beta/\omega$ are odd functions of $\omega$. If $\omega$ is a solution, $-\omega$ is one, too. Thus, it suffices to find the positive solutions. Picturing $\tan(\omega)$ and $\beta/\omega$, it is evident that, for positive $\beta$, there is one solution in $(0,\pi)$ and one solution in $(2\pi,3\pi), (4\pi,5\pi),...$, i.e. $(2n\pi,(2n+1)\pi)$ for every non-negative integer $n$. For $\beta < 0$, there is no solution in $(0,\pi)$, and there is one in each of $(\pi,2\pi), (3\pi,4\pi), ...$.

Often one is not really interested in a full solution, but only in knowing how rapidly the initial condition decays to nothing. The long-term behavior of the solution is dominated by the eigenfunction corresponding to the smallest eigenvalue $\omega_{min} := \min(\left|\omega\right|)$. $\omega_{min}$ can be approximated as described in @Claude_Leibovic's answer. This eigenfunction decays with time constant $\tau = \frac{1}{D\omega_{min}^2}$.