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A somewhat similar question to what I'm going to ask is this one. The problem is basically that one has the heat equation $c^2\nabla^2u = u_t$ in which initial and boundary conditions are given. But these boundary conditions do not match continuously with the initial conditions. In the question I referred to in the first sentence the problem can be solved quite easily since the discontinuity is given in the spatial derivatives of $u$.

However I have this situation: There are two spheres with radii $r_1,r_2$. The inner surface is at $0$ temperature and the outer surface is always at a temperature $f(\theta)$ (in phyisics $0\le\theta\le\pi$). The initial condition of the system in the region between the two spherical surfaces is $u(r,\theta,0)=0$.

Here is my analysis: When one separates the time part $T$ of the heat equation one gets $$\frac{dT}{dt} + \alpha c^2 T = 0.$$ If $\alpha$ is positive then we get an exponential solution that goes to $0$ as $t$ goes to $\infty$. If $\alpha$ is negative, then we get an exponential that diverges as $t$ goes to $\infty$. Finally, if $\alpha$ is $0$, we get a linear solution. First of all, if we get either exponential solutions it means that $u$ either goes to $0$ or diverges in the domain of the problem ($r_1<r<r_2, 0\le\theta,\le\pi$). This is not possible because the stationary system cannot have $0$ temperature because of the function $f(\theta)$ at the boundary: the inside of the bigger sphere should heat up a bit. Also if $T$ is linear ($At+B$) it diverges. If $A=0$, then $T(t)=B$, constant. If we examine the initial condition, this means that $B$ must be $0$. But this gives a trivial solution $u=0$.

So the thing here is that the solution is initially not continuous at $r_2$ (in the question I referred to, the discontinuity is at the derivative of the solution). So my question is: what's wrong with my approach? Thanks for your comments and answers.

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Let me try to explain what is happening with a simpler example. Consider the one dimensional problem $$ u_t=u_{xx},\quad 0<x<1,\quad t>0, $$ with boundary conditions $u(0,t)=0$, $u(1,t)=1$, $t>0$ and initial condition $u(x,0)=0$, $0\le x\le1$. Here $\Omega=(0,1)$, $x=0$ would be the inner sphere and $x=1$ the outer sphere. To use separation of variables and the Fourier method, you need homogeneous boundary conditions, but the condition at $x=1$ (the outer sphere) is not homogeneous. How do we procede? We find a simple function $v$ satisfying the equation and the boundary conditions and let $u=v+w$. Then $w$ will satisfy the equations and the boundary conditions, with a different initial condition. In my example, we can take $v(x,t)=x$, the steady state solution. Then $w=u(x,t)-x$ satisfies the heat equation, homogeneous boundary conditions $w(0,t)=w(1,t)=0$ and the initial condition $w(x,0)=-x$. When you solve for $w$ you get that it goes exponentially to $0$ as $t\to\infty$, but $u$ will converge to the steady state solution.

In your problem, you could take $v$ to be the solution of the Dirichlet problem $\Delta v=0$ with $v=0$ on the inner sphere and $f(\theta)$ on the outer sphere.

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  • $\begingroup$ Julián, I have a question. The initial condition for $w$ shouldn't be $w(x,0) = -x$ instead? $\endgroup$ – Vladimir Vargas Nov 15 '15 at 13:48
  • $\begingroup$ Yes, it should be $-x$. I have edited my answer. $\endgroup$ – Julián Aguirre Nov 15 '15 at 14:44
  • $\begingroup$ Julián, I have another question. When I solve for $w$ this produces the Laplace equation $\nabla^2w=0$. This of course does not depend on the time, but the boundary conditions remain unchanged. Shouldn't this problem also be nonhomogeneous? Here they define the Sturm-Liouville conditions for homogeneity. $\endgroup$ – Vladimir Vargas Nov 15 '15 at 17:04
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    $\begingroup$ You have to solve Laplace's equation with non homogeneous (Dirichlet) boundary conditions. Theory tells us that there is a unique solution. In my simplified problem you have to solve $w_{xx}=0$, $w(0)=0$, $w(1)=1$, whose solution is $w(x)=x$. $\endgroup$ – Julián Aguirre Nov 15 '15 at 17:28
  • $\begingroup$ Julián, I saw in your profile that you work quite a bit on PDEs. Could you take a look at this differential equation? The DE arose when I was solving the radial problem in this question. Thanks. $\endgroup$ – Vladimir Vargas Nov 16 '15 at 14:44

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