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Solving a pde

  1. Nov 27, 2015 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

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

    3. 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 ?
     
  2. jcsd
  3. Nov 27, 2015 #2

    Samy_A

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    Science Advisor
    Homework Helper

    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.
     
  4. Nov 27, 2015 #3
    Thank you
     
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