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Studying the maxwell equations I came across the following system of coupled pdes

$$u_t=\frac{v_x-2(1+t)u}{(1+t)^2}\\ v_t=u_x$$

with initial conditions $$u(x,0)=\sin(x)\\ v(x,0)=-\cos(x)/2$$ where $u=u(x,t)$ and $v=v(x,t)$.

I don't expect you to tell me the solution right away..In fact I know the solution. I'm more interested in how to approach such a problem in principle. Which method could work here? I guess it may involve some fourier transformation to get rid of the spatial derivatives followed by a back transformation... BTW:Mathematica and Maple can't do the job.

EDIT:the exact solution is

$$u(t,x)=\frac{1}{\sqrt{(1+t)^3}}\cos(\frac{\sqrt{3}\ln(1+t)}{2})\sin(x)$$ $$v(t,x)=\frac{1}{2\sqrt{1+t}}\left(-\cos(\frac{\sqrt{3}\ln(1+t)}{2})+\sqrt{3}\sin(\frac{\sqrt{3}\ln(1+t)}{2})\right)\cos(x)$$

OD IUM
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1 Answers1

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The first equation can be written as

$$ \big[(1+t)^2 u\big]_t = v_x $$

Rewriting the system with $p = (1+t)^2 u$ gives

\begin{align} p_t &= v_x \\ p_x &= (1+t)^2v_t \end{align}

Equivalently

\begin{align} p_{xx} &= (1+t)^2p_{tt} \\ v_{xx} &= \big[(1+t)^2 v_t\big]_t \end{align}

Conveniently, $\sin x$ and $\cos x$ are eigenfunctions in the $x$ direction, so we may guess a solution of the form

\begin{align} p(x,t) = \sin(x) f(t) \\ v(x,y) = \cos (x) g(t) \end{align}

where $f(0) = 1$, $g(0) = -\dfrac12$. Then we have the second-order ODEs

\begin{align} (1+t)^2 f''(t) + f(t) &= 0 \\ \big[(1+t)^2 g'(t)\big]' + g(t) &= 0 \end{align}

Clearly, $g(t) = f'(t)$, so it's sufficient to solve the first equation as the IVP

$$ (1+t)^2 f''(t) + f(t) = 0, \ f(0) = 1, f'(0) = -\dfrac12 $$

This is a Cauchy-Euler equation, so we know the solution has the form $f(t) = (1+t)^r$. This gives the characteristic equation

$$ r(r-1) + 1 = 0 $$ $$ \implies r = \frac{1 \pm i\sqrt{3}}{2} $$

Thus

$$ f(t) = \sqrt{1+t}\left[ c_1 \cos\left(\frac{\sqrt{3}}{2}\ln(1+t)\right) + c_2 \sin\left(\frac{\sqrt{3}}{2}\ln(1+t)\right) \right] $$

You can do the rest.

Dylan
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