I have the heat equation with no external heat term in polar coordinates which I have to solve on the unit disk.
$$ \dfrac{\partial u}{\partial t} = \dfrac{1}{r}\dfrac{\partial}{\partial r}\left(r \dfrac{\partial u}{\partial r}\right) + \dfrac{1}{r^2} \dfrac{\partial^2 u}{\partial \theta^2} \equiv\Delta u$$
The given initial and boundary conditions are the following:
$$u(1,\theta,t)=\sin 3\theta$$
$$u(r,\theta,0)= 0$$
I read about the approach of introducing a new variable $v(r,\theta,t)$ such that:
$$v(r,\theta,t)=u(r,\theta,t)-u_E(r,\theta)$$
where $u_E(r,\theta)$ is the equilibrium temperature distribution and $v(r,\theta,t)$ is supposed to have homogeneous boundary conditions.
I understand that if the variables were in cartesian coordinates, owing to the fact that the equilibrium temperature distrubution turns out to be a linear distrubution when there is no external heat source, we can claim that:
$$v_{xx}=u_{xx}, v_{yy}=u_{yy}, v_{zz}=u_{zz}, v_{t} = u_{t}$$
But I am confused about what are their equivalents in polar coordinates? Although I think in polar coordinates $v_{rr} = u_{rr}$ and $v_{t} = u_{t}$ could be valid assumptions but I don't think $v_{\theta \theta} = u_{\theta \theta}$ is a valid assumption because the $\theta$ coordinate in general is not linear but a sinusoidal function in the solution expression. I don't have good reasons to set $(u_E)_{\theta \theta}=0$ in:
$$v_{\theta \theta} = u_{\theta \theta} – (u_E)_{\theta \theta}$$
So, how do I define the new variables to make the boundary value homogeneous for this nonhomogeneous boundary value problem?
Best Answer
So, generally speaking for this kind of problems seperation of variables works very well.
So in order to end up with the equilibrium solution start with the defining equations: $$ \Delta u_E(r,\theta)=0\\ u_E(1,\theta)=\sin(3\theta) $$ Making the Ansatz $u_E(r,\theta)=R(r)\Theta(\theta)$ we can immediately deduct $\Theta(\theta)=\sin(3\theta)$ and $R(1)=1$ from the boundary condition. Since we would like our equilibrium solution to be smooth at the origin we additionally require $R(0)=0$. (Otherwise the function would jump when going over the origin in a straight line)
Inserting the Ansatz into the differential equation yields $$ \dfrac{1}{r}\left(r R'(r)\sin(3\theta)\right)' - \dfrac{1}{r^2} R(r)9\sin(3\theta) = 0\\ \Leftrightarrow r(rR'(r))'= 9R(r) $$ This linear ODE can be solved by the standard methods available and we end up with $R(r)=r^3$. Thus, the equilibrium solution reads $u_E(r,\theta)=r^3\sin(3\theta)$.