- #1

geoffrey159

- 535

- 72

## 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 ?