Finding Solutions to a PDE with Polar Change of Variable

[geoffrey159]

Homework Statement

Find all ${\cal C}^1(\mathbb{R}_+^\star \times \mathbb{R},\mathbb{R})$ solutions to the pde $x\frac{\partial f}{\partial y} - y \frac{\partial f}{\partial x} = cf$, where $c$ is a constant. Use a polar change of variable.

Homework Equations

Trying to bring the problem to a fundamental pde by a change of variable.

The Attempt at a Solution

We define the polar change of variable $(x,y) = \phi(r,\theta) = (r \cos \theta, r\sin \theta)$, which is known to be a ${\cal C}^1$-diffeomorphism from $\mathbb{R}_+^\star \times ]-\frac{\pi}{2},\frac{\pi}{2}[ \to \mathbb{R}_+^\star \times \mathbb{R}$, and we define the function $F = f\circ\phi$ which is of class ${\cal C}^1(\mathbb{R}_+^\star \times ]-\frac{\pi}{2},\frac{\pi}{2}[,\mathbb{R})$ if and only if $f$ is of class ${\cal C}^1(\mathbb{R}_+^\star \times \mathbb{R},\mathbb{R})$.

Assume that $f$ is a solution, then $F$ has all its first partial derivatives defined and by the chain rule: $F_r = \cos\theta (f_x\circ \phi )+ \sin\theta( f_y\circ \phi)$ and $F_\theta = -r\sin \theta (f_x\circ \phi )+ r\cos\theta (f_y\circ \phi)$.

Isolating the partial derivatives with respect to $x$ and $y$, we have :$f_x\circ \phi = \cos\theta F_r -\frac{\sin\theta}{r} F_\theta$ and $f_y\circ \phi = \sin\theta F_r + \frac{\cos \theta}{r} F_\theta$.

Finally, $f$ is a solution if and only if $F$ is a solution to $F_\theta - c F = 0$ which has solution $F(r,\theta) = \lambda(r) e^{c\theta}$, where $\lambda$ is a function of class ${\cal C}^1(\mathbb{R}_+^\star,\mathbb{R})$. Returning to $f$, the solutions have the form $f(x,y) = \lambda(\sqrt{x^2+y^2}) e ^{c\arctan(y/x)}$

Is that correct ?

 

[Samy_A]

Redid all the steps and can't find an error. Plugging in your result in the PDE shows the function satisfies the PDE. So yes, looks correct.

 

[geoffrey159]

Thank you