Partial differential equation, known solution

Question

Knowing that any solution of $y \frac{\partial ^2 u}{\partial y^2} + 2 \frac{\partial u}{\partial y} = \frac{2}{x}$ is of the form $u(x,y) = \frac{\varphi(x) + y^2}{xy} + \psi(x)$,

where $\varphi, \ \psi $ are $C^2$,

find the solution of

$(1) \ \ y \frac{\partial ^2 u}{\partial y^2} + 2 \frac{\partial u}{\partial y} = \frac{2}{x}, \ u(x,x)=g(x)$

and of

$(2) \ \ y \frac{\partial ^2 u}{\partial y^2} + 2 \frac{\partial u}{\partial y} = \frac{2}{x}, \ u(y^3, y)=g(y)$.

Here $g: \mathbb{R} \rightarrow \mathbb{R}$ is a given function.

I'm having problems solving this.

I've solved a similar problem $\frac{\partial u}{\partial x} = 2x \frac{\partial u}{\partial y}, \ u(4,y)=g(y)$, ($g$ is given) where the solution was $u(x,y)=\varphi(y+x^2)$ and I just plugged $\varphi$ to the equation: $\varphi(y+16)=u(4,y)=g(y), \ \ \varphi(y)=g(y-16), \ \ u(x,y)=g(x^2+y-16)$.

But here it doesn't seem to work. Could you explain to me how to solve it?

Thank you!

is this two different problems ((1) and (2)) or must both conditions be satisfied simultaneously? — Julián Aguirre, Oct 07 '14 at 10:07
These are two different problems. I'll edit the question to make it less confusing. — Bilbo, Oct 07 '14 at 10:10

score 1 · Accepted Answer · answered Oct 07 '14 at 10:24

This is a second order equation. The general solution depends on two arbitrary functions ($\phi$ and $\psi$.) If you impose only one condition, you will get infinite answers depending on one arbitrary function. Consider (1). You want $$ u(x,x)=\frac{\phi(x)+x^2}{x^2}+\psi(x)=g(x). $$ You can choose $\psi$ arbitrarily and let $$ \phi(x)=x^2(g(x)-\psi(x)-1). $$

On the other hand, if you impose both conditions (1) and (2), you will get a system of two equations with the two unknowns $\phi$ and $\psi$, and you will find (in most cases) a unique solution $u$.

Partial differential equation, known solution

1 Answers1