Let D be the open unit disk and let u(x,y) be a real valued function on D

Question

Let D be the open unit disk and let u(x,y) be a real valued function on D.let D be harmonic on D.and $u(x,y)=1-2y^2$on the boundary of D.how can i find the function U(x,y)? i am not able to find any solution of this problem. please help me

Answer 1

There are a couple of things that you can do to find the solution. When I am looking for a harmonic function on the unit ball, I initially think of Poisson's Integral Formula. This is okay, but the integral that you have to solve is pretty difficult (or at least it was for me).

So then, our next option is separation of variables. But since the boundary condition is on a ball, this suggests using the polar form of Laplace's equation $$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta \theta}=0$$ I will give here a "quick and dirty" separation of variables calculation, and you can fill in the details. Assuming that $u(r,\theta)=R(r)\Theta(\theta)$, and following the usual procedure, we get ODEs for $R$ and $\Theta$: $$r^2R''+rR'-\lambda R=0, \quad \Theta''+\lambda\Theta=0$$ For $R$, this is just an Euler equation which means that $R(r) = Ar^{\sqrt{\lambda}} + Br^{-\sqrt{\lambda}}$. For $\Theta$ we get $\Theta(\theta) = C\cos(\sqrt{\lambda}\theta)+D\sin(\sqrt{\lambda}\theta)$. According to the boundary condition, we need $\Theta(\theta) = 1-2\sin^2(\theta) = \cos(2\theta)$. This gives $\sqrt{\lambda}=2, C=1, D=0$. We want the solution to be defined at $0$, so we choose $B=0$, and the solution becomes $$u(r,\theta) = R(r)\Theta(\theta) = Ar^2\cos(2\theta)$$ Again, we must choose $A=1$ to satisfy the boundary condition. Now we just need to convert back into rectangular coordinates. Again using the identity $\cos(2\theta)=1-2\sin^2\theta$ we have $$u=r^2-2r^2\sin^2\theta = x^2+y^2-2y^2 = x^2-y^2$$

Answer 2

$$U(x,y)=x^2-y^2$$ is harmonic and is equal to $1-2y^2$ on the unit circle.