PDE: Guess non-polynomial solution of $\Delta u=\frac{\delta^2u}{\delta x^2}-\frac{\delta^2u}{\delta y^2}=0$

Question

I need to verify that the function $$u(x,y)=a\cdot ln(x^2 + y^2)$$ is a solution of the two-dimensional Laplace equation $$\Delta u=\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} \qquad (1)$$ and later I have to use the result to guess one non-polynomial (non-trivial) solution of the equation $$\Delta u=\frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=0 \qquad (2)$$ and verify my result. I did first part of the problem where I calulated first and second derivative of u and inserted it into equation. Later I made a guess that the solution for (2) can be: $$u(x,y)=a\cdot ln(x^2 - y^2) \qquad (3)$$ and I calulcated derivaties and so on. My question is, is my guess correct?

Answer 1

Your guess

$$u(x,y)=a\cdot ln(x^2 - y^2) $$

does satisfy the $$\frac {\partial ^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=0 $$

Thus your guess is as good as gold.

Answer 2

The function $$u(x,y)=a\cdot ln(x^2 - y^2)$$ is a solution of the equation $(2)$, as you successfully guessed. $$\Delta u=\frac{\partial^2u}{\partial x^2}-\frac{\partial^2u}{\partial y^2}=0 \qquad (2)$$

Note that a lot of solutions can be found on the general form : $$u(x,y)=f(x+y)+g(x-y)$$ with any differentiable functions $f$ and $g$.

For example, taking $\quad \begin{cases}f(x+y)=a\:\ln(x+y) \\g(x-y)=a\:\ln(x-y) \end{cases}\quad $ $$u(x,y)=a\:\ln(x+y)+a\:\ln(x-y)=a\ln((x+y)(x-y))=a\ln(x^2-y^2)$$ is solution of Eq.$(2)$.

But other functions of $(x^2-y^2)$ are not solution of Eq.$(2)$.

Other example, with $f(x+y)=\sin(x+y)$ and $g(x-y)=\sin(x-y)$ we obtain $u(x,y)=2\cos(x)\sin(y)$ which is solution of Eq.$(2)$.

One can found many other solutions with other functions $f$ and $g$ , even with $f\neq g$, including a lot of polynomial solutions in case of polynomial functions $f$ and $g$.