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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?

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    $\begingroup$ 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? $\endgroup$ Commented Jan 27, 2023 at 3:23
  • $\begingroup$ @Aruralreader Nope, no other assumptions given in the question unfortunately $\endgroup$ Commented Jan 27, 2023 at 12:05
  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Commented Jan 29, 2023 at 10:25
  • $\begingroup$ @KurtG. Yep, it does satisfy $B_t - \alpha B_{yy} = 0$ and the condition at $y=0$. $\endgroup$ Commented Jan 29, 2023 at 16:10
  • $\begingroup$ @Aruralreader, I checked that too. Then what is OP's question about? $\endgroup$
    – Kurt G.
    Commented Jan 29, 2023 at 18:34

1 Answer 1

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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{equation} \begin{aligned} -\omega c(\omega, y) &= \alpha s_{yy}(\omega, y), \\ \omega s(\omega, y) &= \alpha c_{yy}(\omega, y). \end{aligned} \tag{2} \end{equation} 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*}

Addendum in hindsight

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:

$$B(t, y) = u(y)\cos\omega_0 t + v(y)\sin\omega_0 t.$$

Working with this form to begin with gets you to the same result with a lot less pain, and maybe a bit of physical-reasoning satisfaction.

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  • $\begingroup$ 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)? $\endgroup$ Commented Jan 31, 2023 at 0:18
  • $\begingroup$ @VJ123. It's simply the real form of the (inverse) Fourier transform, so you're starting with a rather broad basis. $\endgroup$ Commented Jan 31, 2023 at 0:25
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    $\begingroup$ 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? $\endgroup$ Commented Jan 31, 2023 at 1:07
  • $\begingroup$ @VJ123. Yes, that’s the case. $\endgroup$ Commented Jan 31, 2023 at 1:11
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    $\begingroup$ @VJ123: Too long? Not a single moment working through a problem in mathematics is wasted. Be patient and persistent. $\endgroup$ Commented Jan 31, 2023 at 3:40

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