As I understand it, your system does have a great deal of symmetry. I'm assuming the problem is cylindrical, i.e. the picture you have drawing is valid at all values of $z$, if we make the $z$ co-ordinate out of the page. So, I'm assuming your problem definition is $\vec{H}(t) = k\,t\, \hat{z};\forall\,t>0;\, r<r_0$ where $r_0$ is the radius of the magnetized region and the initial conditions are $H(r,z,0)=E(r,z,0)=0$.
By symmetry, you argue that the fields depend on $r$ and $t$ only. So it remains to find out the $r,\,t$ dependence and the directions of the fields. Use the integral forms: these are as good to work with as the differential forms for a problem like this. You can work out the governing equations easily from these symmetries, but you're going to need either series of Bessel functions (which you may not yet be familiar with) or numerical analysis to solve them.
Consider the radial components $E_r(r,t)$ and $B_r(r,t)$. Apply Gauss's laws to an arbitrary radius cylinder centered on the circular region. There is no electric nor magnetic charge in side this surface, and the fluxes through it per unit length are $2\,\pi\,r\,E_r(r,t)$ and $2\,\pi\,r\,H_r(r,t)$ so these must both be nought. The radial components of all fields vanish.
Now consider a circular loop concentric with the field system. Work out the flux of $\vec{B}$ through it and the line integral of $\vec{E}$ around it and apply Faraday's law:
$$-2\,\pi\,\mu_0\,\frac{\mathrm{d}}{\mathrm{d}\,t} \int_0^r\,u\,H_z(u,\,t)\,\mathrm{d}\,u=2\,\pi\,r\,E_\phi(r,\,t)\tag{1}$$
where $E_\phi$ is the component of $\vec{E}$ tangential to the loop. You may have heard this called the azimuthal component. Now do the same for the flux of $\vec{E}$ through the loop and the line integral of $\vec{H}$ around it and apply Ampère's law:
$$2\,\pi\,\epsilon_0\,\frac{\mathrm{d}}{\mathrm{d}\,t} \int_0^r\,u\,\,E_z(u,\,t)\,\mathrm{d}\,u=2\,\pi\,r\,H_\phi(r,\,t)\tag{2}$$
Lastly, we must apply Faraday's and Ampere's laws around a square loop lying in a radial plane stretching from the center of the problem $r=0$ out to the radius $r$. If you do this you get:
$$\mu_0\,\frac{\mathrm{d}}{\mathrm{d}\,t} \int_0^r\,H_\phi(u,\,t)\,\mathrm{d}\,u=E_z(r,\,t)-E_z(0,\,t)\tag{3}$$
$$-\epsilon_0\,\frac{\mathrm{d}}{\mathrm{d}\,t} \int_0^r\,E_\phi(u,\,t)\,\mathrm{d}\,u=H_z(r,\,t)-H_z(0,\,t)\tag{4}$$
Now the usual method is to differentiate all these equations with respect to $r$; this will yield exactly the same results that you would have gotten with the differential form Maxwell equations:
$$-\mu_0\,\,r\,\frac{\partial}{\partial\,t}\,H_z(r,\,t)=r\,\frac{\partial}{\partial\,r}\,E_\phi(r,\,t) + E_\phi(r,\,t)\tag{1a}$$
$$\epsilon_0\,\,r\,\frac{\partial}{\partial\,t}\,E_z(r,\,t)=r\,\frac{\partial}{\partial\,r}\,H_\phi(r,\,t) + H_\phi(r,\,t)\tag{2a}$$
$$\mu_0\,\frac{\partial}{\partial\,t} H_\phi(r,\,t)=\frac{\partial}{\partial\,r}\,E_z(r,\,t)\tag{3a}$$
$$-\epsilon_0\,\frac{\partial}{\partial\,t} E_\phi(r,\,t)=\frac{\partial}{\partial\,r}\,H_z(r,\,t)\tag{4a}$$
and you now differentiate (1a) and (2a) with respect to $t$ and thensubstitude (3a) and (4a) into the time-derviatives of (1a) and (2a) to get two, second order differential equations in $E_z$ and $H_z$. These are the cylindrical wave equations for $E_z$ and $H_z$. Notice how $(H_\phi,\,E_z)$ and $(E_\phi,\,H_z)$ are independent pairs - neither of $H_\phi,\,E_z$ has any bearing on $E_\phi,\,H_z$ nor contrariwise. The wave equation for $H_z$ is at last:
$$\mu_0\,\epsilon_0\,r\,\frac{\partial^2}{\partial\,t^2}\,H_z(r,\,t)=r\,\frac{\partial^2}{\partial\,r^2}\,H_z(r,\,t) + \frac{\partial}{\partial\,r}\,H_z(r,\,t)\tag{5}$$
and our task is now to match up solutions of this one with your initial conditions.
This is actually highly nontrivial to solve properly. I'll sketch how it is done. One usually builds solutions up as superpositions of solutions that vary sinusoidally with time, and uses Fourier analysis to do so. Outside the central "excitation" region i.e. where $r>r_0$, when the solution varies sinusoidally with with, i.e. $H_z(r,\,t) = h_z(r)\,e^{i\,\omega\,t}$, the solution to (5) is an outwardly propagating wave:
$$H_z(r,\,t) = H^{(1)}_0\left(\frac{\omega\,r}{c}\right)\,e^{i\,\omega\,t}\tag{6}$$
where $H^{(1)}_0$ is the Hankel function; you probably haven't met this one yet, but for large distances it looks like a wave of the form $e^{i\,k\,r}/\sqrt{r}$, so your system is radiating waves travelling at the speed of light $c$. The disturbance takes time $r/c$ to reach a radius $r$. The general solution is a superposition of this over all frequencies:
$$H_z(r,\,t) =\int_{-\infty}^\infty A(\omega)\, H^{(1)}_0\left(\frac{\omega\,r}{c}\right)\,e^{i\,\omega\,t}\,\mathrm{d}\,\omega\tag{7}$$
and the task reduces to one of finding the superposition function $ A(\omega)$ to match your problem.
