Let $\Omega :x^2+y^2<4$ in $\Bbb R^2$ with boundary $\partial \Omega$. If $u(x,y)$ is the solution of the Dirichlet problem $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0, \hspace{1cm} (x,y)\in \Omega$$ and $u(x,y)=1+2x^2$ for all $(x,y)\in \partial \Omega$ then find the value of $u(0,1)$.
By solving the Laplace equation I am not getting anything. How to use the condition $u(x,y)=1+2x^2$ in $\partial \Omega$ ?
Any help please..
On
You can use Poisson's representation formula: $$ u(r, \theta)=\frac{1}{2 \pi}\int_0^{2 \pi}\dfrac{4-r^2}{4-4r \cos(\theta-\phi)+r^2} h(\phi) \, d\phi, $$
where $h(\phi)$ is the given boundary data.
For the point $(0,1)$ you use $r=1$ and $\theta=\frac{\pi}{2}$, leading to the integral $$ \frac{1}{2\pi} \int_0^{2\pi}\dfrac{3(5+4 \cos(2\phi))}{5-4 \sin \phi} \, d\phi = 4. $$
Let's write the boundary condition in polar coordinates.
$$ u = 1 + 2(2\cos \phi)^2 = 5 + 4 \cos 2\phi \ \ \ \ \text{when} \ \ r = 2.$$
The following is a solution of Laplace's equation: $$ u = 5 + r^2\cos2\phi$$
This solution satisfies the above boundary condition. By the uniqueness theorem for the Laplace equation, this is in fact the unique solution satisfying our boundary condition.