Let $u(x,y) $ be the solution of $\displaystyle \frac{\partial^{2}u}{\partial x^2}+\frac{\partial^{2}u}{\partial y^2}=64$ in the unit disk $\left \{(x,y)|x^2+y^2<1 \right \}$ and such that $u$ vanishes on the boundary of the disk,Then $\displaystyle u\left ( \frac{1}{4},\frac{1}{\sqrt 2} \right )$
Solution I tried-I tried the method $$m^2+1=0$$ $$m^2=-1$$ $$m=\pm i$$ so the C.S is $$\phi_1(y+ix)+\phi_2(y-ix)$$
$$\text{and P.I i solved is }32x^2$$
so complete soltuion is $$\phi_1(y+ix)+\phi_2(y-ix)+32x^2$$
but t seems that this solution will not help me because there is imaginary part in solution,
I also tried solution by taking $u=Ae^{hx+ky}$ in end i get$$h^2=-k^2$$ complete solution is
$$\sum Ae^{hx \pm ihy}+32x^2$$ but i have no idea how can these solution can help me to get desired result
Please Help
Thank you
Cartesian coordinates $(x,y)\quad :\quad \displaystyle \frac{\partial^{2}u}{\partial x^2}+\frac{\partial^{2}u}{\partial y^2}=64$
Polar coordinates : $(\rho,\theta)\quad : \quad \displaystyle\frac{\partial^{2}u}{\partial \rho^2} +\frac{1}{\rho}\frac{\partial u}{\partial \rho}+\frac{1}{\rho^2}\frac{\partial^2 u}{\partial \theta^2}=64$
The given boundary condition is not sufficient to consider the solutions which might be functions of $\theta$. So we assume that $\frac{\partial u}{\partial \theta}=0$ everywhere.
$$\displaystyle\frac{d^{2}u}{d \rho^2} +\frac{1}{\rho}\frac{d u}{d \rho}=64$$ $$\frac{d u}{d \rho}=c_1\frac{1}{\rho}+32\rho$$
$$u=c_1\ln(\rho)+c_2+16\rho^2$$ The condition $u=0$ at $\rho=1$ implies $c_2=-16$. $$u=c_1\ln(\rho)+16(\rho^2-1)$$ Again the wording of the problem seems ambiguous about the boundary conditions. The most likely it is supposed that $u$ is continuous and finite at $\rho=0$ which implies $c_1=0$. $$u=16(\rho^2-1)$$ $$u(x,y)=16(x^2+y^2-1)$$ This satisfies the PDE and the condition $u=0$ on the unit circle.
$u(1/4,\: 1/\sqrt{2})=16(\frac{1}{16}+\frac12-1)=-7$