How to solve $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y} = \gamma f$

Question

Let $f:\mathbb{R}^2\to \mathbb{R}$ be a $\mathcal{C}^1$ function. I would like to solve the partial differential equation: $$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y} = \gamma f,$$ where $\gamma$ is a fixed real number.

I tried doing it with a polar change of variables: let $g(r,\theta) = f(r\cos(\theta),r\sin(\theta))$. Then I find $$r\frac{\partial g}{\partial r} = \gamma g.$$

First, is it correct? And then how to go from there on?

Answer 1

First of all $$r\frac{\partial g}{\partial r} = \gamma g\implies \dfrac{\dfrac{\partial g}{\partial r}}{g}=\dfrac{\gamma}{r}.$$ Integrating with respect to $r$ we have

$$\ln \dfrac{g(r,\theta_0)}{g(r_0,\theta_0)}=\gamma\ln \dfrac{r}{r_0}.$$

Thus it is

$$g(r,\theta_0)=g(r_0,\theta_0)\left(\frac{r}{r_0}\right)^{\gamma}.$$

Finally, since $\theta_0$ is arbitrary we have

$$g(r,\theta)=g(r_0,\theta)\left(\frac{r}{r_0}\right)^{\gamma}.$$

Answer 2

This looks to me like an ideal candidate for "separation of variables". Look, at first, for a solution of the form f(x,y)= X(x)Y(y). Then the equation becomes $xY\frac{dX}{dx}+ yX\frac{dY}{dy}= \gamma XY$. Dividing both sides by XY, $\frac{x}{X}\frac{dX}{dx}+ \frac{y}{Y}\frac{dY}{dy}= \gamma$. Since one term is a function of x only while the other is a function of y only, in order to sum to a constant each must be a constant. That is, we must have $\frac{x}{X}\frac{dX}{dx}= \alpha$ and $\frac{y}{Y}\frac{dY}{dy}= \beta$ where $\alpha+ \beta= \gamma$.

To solve $\frac{x}{X}\frac{dX}{dx}= \alpha$ write it as $\frac{X}{dX}= \alpha\frac{dx}{x}$ and integrate: $ln(X)= \alpha ln(x)+ C= ln(x^{\alpha})+ C$. Then $X= C'x^{\alpha}$ where $C'= e^C$.

Doing the same thing with Y gives $Y= D'y^{\beta}$ and the product XY is $Ex^{\alpha}y^{\beta}$ where again $\alpha+ \beta= \gamma$.

So how do we determine $\alpha$, $\beta$, and, for that matter $\gamma$ itself? Well, that depends upon what boundary conditions are given- and you didn't give any!

Answer 3

$$r\frac{\partial g}{\partial r} = \gamma g.\quad\text{is correct.}$$ In this equation no $\frac{\partial}{\partial\theta}$ appears. So, $\theta$ can be considered as a parameter and the equation becomes an ODE : $$r\frac{dg}{dr} = \gamma g$$ $$\frac{dg}{g} = \gamma \frac{dr}{r}$$ $$g=c\:r^{\gamma}$$ where $c$ is a constant with respect to $r$, but can be function of all the parameters involved in the ODE. Thus : $$g(r,\theta)=c(\theta)\:r^{\gamma}$$ where $c(\theta)$ is an arbitrary function of $\theta$. $$f(x,y)=h\left(\frac{y}{x}\right) (x^2+y^2)^{\gamma/2}$$ where $h\left(\frac{y}{x}\right)=c\left(\tan^{-1}(\frac{y}{x})\right)$ is an arbitrary function because $c(\theta)$ is an arbitrary function.

The function $\:h\:$ has to be determined according to some boundary conditions.