Find all real solutions to the two dimensional Laplace equation $U_{xx} + U_{yy} =0$ of the form $u=\log p(x,y)$, where $p$ is a quadratic polynomial.
Solution:
Let $p(x,y) = Ax^2 + By^2 +Cxy + D$ be a quadratic polynomial such that $A, B \not= 0$. Then
$$U_{x} = \frac{2Ax + Cy}{\ln(10)(Ax^2 + By^2 +Cxy + D},$$
$$U_{xx} = \frac{2A ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2},$$
$$U_{y} = \frac{2By + Cx}{\ln(10)(Ax^2 + By^2 +Cxy + D)},$$
$$U_{yy} = \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2}.$$
This implies
$$U_{xx} + U_{yy} = \frac{2A \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} + \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} = 0. $$
I feel like I am not doing this right. Is there a simpler way? Thanks. And also how do I find such solutions.
This is an heavy method, but you can continue. $$U_{xx} + U_{yy} = \frac{2A ln(10)(Ax^2 + By^2 +Cxy + D) - ln(10)(2Ax + Cy)^2}{ln(10) (Ax^2 + By^2 +Cxy + D)^2} + \frac{2B ln(10)(Ax^2 + By^2 +Cxy + D) - ln(10)(2By + Cx)^2}{ln(10) (Ax^2 + By^2 +Cxy + D)^2} = 0$$ After simplification : $$2A(Ax^2 + By^2 +Cxy + D)-(2Ax + Cy)^2 + 2B(Ax^2 + By^2 +Cxy + D)-(2By + Cx)^2=0$$ It is easy to see that $A=B=1\:;\:C=D=0$ is solution. Hense : $p(x,y)=x^2+y^2$ $$U(x,y)=log(x^2+y^2)$$
A simpler method consists in solving first the PDE : $$U(x,y)=f(x+iy)+g(x-iy)$$ any functions $f$ and $g$.
In case of $f=g=log$ then $U=log(x+iy)+log(x-iy)$ $$U(x,y)=log(x^2+y^2)$$