Eliminating 3rd term in $\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2} + 3u = 0$ through change of variables?

Question

I want to eliminate the third term in this pde

$$\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2} + 3u = 0$$

Here's what I thought of, is this ok?

Let $a = \lim \limits_{k \to \infty} kx$

Let $b = \lim \limits_{k \to \infty} ky$

Then

$\frac{d}{dx} = \frac{d}{da}\frac{da}{dx} = \lim \limits_{k \to \infty} k\frac{d}{da}$

$\frac{d^2}{dx^2} = \frac{d}{da}\frac{da}{dx}\frac{d}{da}\frac{da}{dx} = \lim \limits_{k \to \infty} kx\frac{d}{da}\lim \limits_{k \to \infty} k\frac{d}{da} = \lim \limits_{k \to \infty} k^2\frac{d^2}{da^2}$

Similarly,

$\frac{d^2}{dy^2} = \lim \limits_{k \to \infty} k^2\frac{d^2}{db^2}$

Substituting,

$\lim \limits_{k \to \infty} k^2\frac{d^2u}{da^2} + k^2\frac{d^2u}{db^2} + 3u = 0$

$\lim \limits_{k \to \infty} \frac{d^2u}{da^2} + \frac{d^2u}{db^2} + \frac{3u}{k^2} = 0$

$\frac{d^2u}{da^2} + \frac{d^2u}{db^2} = 0$

So is that all ok?

Answer 1

$$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} + 3u = 0$$ Let $u=v+u_p$ where $u_p$ is a particular solution, so that $\frac{\partial^2v}{\partial x^2} + \frac{\partial^2v}{\partial y^2} = 0$ , which will be the expected equation.

$$\frac{\partial^2u}{\partial x^2} + \frac{d^2u}{\partial y^2} + 3u = \frac{\partial^2v}{\partial x^2} + \frac{\partial^2v}{\partial y^2} + \frac{\partial^2 u_p}{dx^2} + \frac{\partial^2u_p}{\partial y^2} + 3u_p = 0$$ So, it is necessary that: $$\frac{\partial^2u_p}{\partial x^2} + \frac{\partial^2u_p}{\partial y^2} + 3u_p = 0$$ It seems that there is no pregress compared to the first equation. Do not think so ! Because in the first equation we are locking for ALL solutions, while now we are loocking for only ONE solution, doesn't matter which one.

For example, we can decide that $u_p$ be function of $x$ only. Then $$\frac{d^2u_p}{dx^2} + 3u_p = 0$$ This is a simple ODE easy to solve. I let you continue to find $u_p$.

Alternatively, we could dicide that $u_p$ be function of $y$ only. This would also solve the problem, apparently leading to a different general solution $u=v+u_p$. But, in fact, both are the same presented on different forms.