# Diffusion equation with time-dependent boundary condition

The bottom surface (located at $$y=0$$) of a semi-infinite, electrically conducting slab is subject to the slowly oscillating magnetic field $$B = B_0 \cos(\omega t)$$.

I am trying to find an expression for the magnetic field within the conductor as a function of $$y$$ and $$t$$ using the 1D diffusion equation: $$\frac{\partial B}{\partial t} = \alpha \frac{\partial^2 B}{\partial y^2}$$

I know that the solution is given by: $$B = B_0 \exp(-\frac{y}{\delta})\cos(\omega t -\frac{y}{\delta})$$ where $$\delta = \sqrt{2 \alpha/\omega}$$.

However, I can't quite figure out how to get there due to the time-dependent boundary condition.

I have tried separation of variables but with not much luck, since the solution I am looking for can't be expressed as a product $$Y(y)T(t)$$.

Could someone please tell me how to get to the solution?

• The solution you're looking at points to the initial state of the medium being $B(0, y) = B_0e^{-y/\delta}\cos(y/\delta)$. This seems pretty specific. Are there other assumptions about the physics? Commented Jan 27, 2023 at 3:23
• @Aruralreader Nope, no other assumptions given in the question unfortunately Commented Jan 27, 2023 at 12:05
• From a physical point of view I find it odd that a magnetic field obeys a heat equation. Anyway. Are you sure the solution $B=B_0\,e^{-\frac{y}{\delta}}\cos(\omega t-\frac{y}{\delta})$ solves the equation? This answer might be very useful. Commented Jan 29, 2023 at 10:25
• @KurtG. Yep, it does satisfy $B_t - \alpha B_{yy} = 0$ and the condition at $y=0$. Commented Jan 29, 2023 at 16:10
• @Aruralreader, I checked that too. Then what is OP's question about? Commented Jan 29, 2023 at 18:34

With slight changes in notation, we look for bounded solutions of $$B_t = \alpha B_{yy}$$ in the half-plane $$y>0$$ subject to $$B(t, y=0) = B_0\cos\omega_0 t$$.

Suppose $$B$$ is a solution. Proceeding formally, assume $$\begin{equation*} B(t, y) = \frac{1}{\pi}\int_0^\infty c(\omega, y)\cos\omega t\, d\omega + \frac{1}{\pi}\int_0^\infty s(\omega, y)\sin\omega t\, d\omega \tag{1} \end{equation*}$$ where \begin{align*} c(\omega, y) &= \int_{-\infty}^\infty\, B(t, y)\cos\omega t\, dt, \\ s(\omega, y) &= \int_{-\infty}^\infty\, B(t, y)\sin\omega t\, dt \end{align*} are the Fourier cosine and sine transforms of $$B(t, y)$$ with respect to $$t$$. If $$B$$ satisfies $$B_t = \alpha B_{yy}$$ then substitution into (1) and rearrangement shows these transforms must be related as \begin{aligned} -\omega c(\omega, y) &= \alpha s_{yy}(\omega, y), \\ \omega s(\omega, y) &= \alpha c_{yy}(\omega, y). \end{aligned} \tag{2} Computing $$c_{yyyy}$$ and $$s_{yyyy}$$ we see that $$c$$ and $$s$$ satisfying this system also satisfy $$c_{yyyy} + (\omega/\alpha)^2 c = 0$$ and $$s_{yyyy} + (\omega/\alpha)^2 s = 0$$. Each has four linearly independent solutions \begin{align*} & \exp(-\sqrt{\omega/2\alpha}y)\cos(\sqrt{\omega/2\alpha}y), \\ & \exp(-\sqrt{\omega/2\alpha}y)\sin(\sqrt{\omega/2\alpha}y), \\ & \exp(\sqrt{\omega/2\alpha}y)\cos(\sqrt{\omega/2\alpha}y), \\ & \exp(\sqrt{\omega/2\alpha}y)\sin(\sqrt{\omega/2\alpha}y). \end{align*} The third and fourth forms are unbounded as $$y\rightarrow\infty$$ so we ignore them and make use of the first and second only: $$\begin{equation*} c(\omega, y) = \exp(-\sqrt{\omega/2\alpha}y)\biggl(c_1(\omega)\cos(\sqrt{\omega/2\alpha}y) + c_2(\omega)\sin(\sqrt{\omega/2\alpha}y)\biggr), \end{equation*}$$ and similarly $$\begin{equation*} s(\omega, y) = \exp(-\sqrt{\omega/2\alpha}y)\biggl(s_1(\omega)\cos(\sqrt{\omega/2\alpha}y) + s_2(\omega)\sin(\sqrt{\omega/2\alpha}y)\biggr). \end{equation*}$$ Note that (2) implies a relation between $$c_1, c_2$$ and $$s_1, s_2$$, specifically $$s_1(\omega) = - c_2(\omega)$$, $$s_2(\omega) = c_1(\omega)$$.

With these results, at $$y = 0$$ we have \begin{align*} B(t, 0) &= \frac{1}{\pi}\int_0^\infty\, c(\omega, 0)\,\cos\omega t\, d\omega + \frac{1}{\pi}\int_0^\infty\, s(\omega, 0)\,\sin\omega t\, d\omega \\ &= \frac{1}{\pi}\int_0^\infty\,c_1(\omega)\cos\omega t \, d\omega + \frac{1}{\pi}\int_0^\infty\,s_1(\omega)\sin\omega t \, d\omega. \end{align*} The condition $$B(t, 0) = B_0\cos\omega_0 t$$ suggests $$c_1(\omega) = \pi B_0\delta(\omega - \omega_0)$$ and $$s_1(\omega) = 0$$. Thus $$c_2(\omega) = 0$$, $$s_2(\omega) = \pi B_0\delta(\omega - \omega_0)$$. So \begin{align*} c(\omega, y) &= \pi B_0\exp(-\sqrt{\omega/2\alpha}y)\cos(\sqrt{\omega/2\alpha}y)\delta(\omega - \omega_0) \\ s(\omega, y) &= \pi B_0\exp(-\sqrt{\omega/2\alpha}y)\sin(\sqrt{\omega/2\alpha}y)\delta(\omega - \omega_0). \end{align*} Substitution into (1) and a bit of algebra gives $$\begin{equation*} B(t, y) = B_0\exp(-\sqrt{\omega_0/2\alpha}y) \cos(\sqrt{\omega_0/2\alpha}y - \omega_0 t). \end{equation*}$$

The above approach works for a large class of driving functions at $$y = 0$$, however, with $$B_0\cos\omega_0 t$$, things are far simpler. From a physics perspective you might argue the driver is $$\omega_0$$-periodic, so look for $$\omega_0$$-periodic solutions. That is:
$$B(t, y) = u(y)\cos\omega_0 t + v(y)\sin\omega_0 t.$$
• Thank you very much for your response. This looks really promising. Might I ask why we assume that $B(y,t)$ takes the form given in (1)? Commented Jan 31, 2023 at 0:18
• Ah I see, so we're essentially saying that $B(y,t)$ is equal to the inverse Fourier transform of the Fourier transform of $B(y,t)$ if I understand correctly? Commented Jan 31, 2023 at 1:07