Laplace equation on unit disc

Question

Let $u(x,y)$ be the solution of$$ u_{xx}+u_{yy}=64$$ in unit disc $\{(x,y): x^2+y^2<1\}$ and such that $u$ vanishes on the boundary of disc then find $u(\frac{1}{4},\frac{1}{√2})$ What I tried i know how to solve Laplace's equation so i make a transformation $ v = u -32x^2$

With this my problem converted into $v_{xx}+v_{yy}=0, v= -32x^2$ on boundary of disc Then i shifted to polar coordinates because of unit disc also i know the solution of Laplace equation in polar coordinates is $$ v(r,\theta)= a_0 + \sum_{n=1}^{\infty} a_n r^n cos(n\theta)+\sum_{n=1}^{\infty} b_n r^n sin(n\theta)$$

Boundary condition becomes $ v (1,\theta)= - 32 cos^2\theta= -16( 1+cos2{\theta})$ I tried to match this conditions with solution i got $$a_0= -16 = a_2$$ other coefficients becomes zero .so solution becomes $$ v(r,\theta)= -16-16 r^2 cos2(\theta)$$ how to calculate $u(\frac{1}{4},\frac{1}{√2})$ from there Please help .

Answer 1

One of the common tricks in solving Laplace's equation is to exploit the symmetry, as the shape of the domain is usually what determines the solution. In this case, your domain is the unit disc, which is radially symmetrical. As such, it would make more sense to change to polar coordinates instead.

Letting

$$x = r\cos \theta \qquad \qquad y = r\sin \theta$$

the problem becomes

$$\frac 1r \frac{\partial}{\partial r}\bigg(r\frac{\partial u}{\partial r}\bigg) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 64 \qquad \qquad u(1,\theta) = 0$$

where now $u = u(r,\theta)$. Since the problem is radially symmetrical, we expect that $u$ only depends on $r$, so that the problem simplifies to

$$\frac 1r \frac{\partial}{\partial r}\bigg(r\frac{\partial u}{\partial r}\bigg) = 64 \qquad \qquad u(1) = 0$$

where now $u = u(r)$. The general solution is

$$u(r) = 16r^2 + A\ln (r) + B$$

We require the solution to be at least continuous in the domain, so we must have $A=0$ in order to avoid the blowup at $r=0$. We then have

$$u(1) = 0 \implies B = -16$$

It follows that

$$u(r) = 16(r^2-1)$$

Switching back to Cartesian coordinates, we find that

$$u(x,y) = 16(x^2+y^2-1) \qquad u \bigg(\frac 14, \frac{1}{\sqrt 2}\bigg) = -7$$

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EDIT:

Just realised that you actually had the correct answer. You just have to undo all of the transformations.

In Cartesian coordinates,

$$v(x,y) = -16-16r^2\cos(2\theta) = -16-16r^2 (\cos^2 (\theta) - \sin^2 (\theta)) = -16-16x^2+16y^2$$

and $u$ is calculated via

$$u(x,y) = v(x,y) + 32x^2 = -16+16x^2+16y^2 = 16(x^2+y^2-1)$$