Radially symmetric solutions of the Dirichlet problem

Question

How do I find the radially symmetric solutions of this Dirichlet problem?

$$\left\{ \begin{array}{l l} u_{tt} + u_{yy} = (x^2+y^2)^2 & \quad \mbox{ on $1< x^2 + y^2< 4\ $,} \\ \quad u(x,y) = 3 & \quad \mbox{ on $ x^2 + y^2 = 4\ $}, \\ \quad u(x,y) = 2 & \quad \mbox{ on $ x^2 + y^2 = 1\ $} \end{array} \right. $$

Answer 1

Solve $$\left\{ \begin{array}{l l} u_{xx} + u_{yy} = (x^2+y^2)^2 & \quad \mbox{ on $1< x^2 + y^2< 4\ $,} \\ \quad u(x,y) = 3 & \quad \mbox{ on $ x^2 + y^2 = 4\ $}, \\ \quad u(x,y) = 2 & \quad \mbox{ on $ x^2 + y^2 = 1\ $} \end{array} \right.$$

• Let $u=v(x^2+y^2)$. Then $$u_x=v'(x^2+y^2)\,2\,x$$ $$u_{xx}=v''(x^2+y^2)\,4\,x^2+2v'(x^2+y^2)$$ $$u_y=v'(x^2+y^2)\,2\,y$$ $$u_{yy}=v''(x^2+y^2)\,4\,y^2+2v'(x^2+y^2)$$ $$u_{xx} + u_{yy}=4v''(x^2+y^2)(x^2+y^2)+4v'(x^2+y^2)$$
• Let $t=x^2+y^2$. We have ODE problem $$4v''(t)t+4v'(t)=t^2,\;v(4)=3,\; v(1)=2. $$
• Solution is $$v=-\frac{3\log(t)}{4\log(4)}+\frac{t^3}{36}+\frac{71}{36}.$$
• Answer: $$u=-\frac{3\log(x^2+y^2)}{4\log(4)}+\frac{(x^2+y^2)^3}{36}+\frac{71}{36}.$$