# Finding Solutions to a PDE with Polar Change of Variable

• geoffrey159
In summary, the polar change of variable allowed the problem to be brought to a fundamental pde. By isolating the partial derivatives with respect to x and y, it was possible to find the function ##f## which was a solution to the pde. This solution had the form ##f(x,y) = \lambda(\sqrt{x^2+y^2}) e ^{c\arctan(y/x)}##.
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 ?

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

## 1. What is a PDE?

A PDE, or partial differential equation, is a type of mathematical equation that involves multiple variables and their partial derivatives. It is commonly used to describe physical phenomena in fields such as physics, engineering, and finance.

## 2. What is a polar change of variable?

A polar change of variable is a mathematical transformation that converts a PDE from its original Cartesian coordinates to polar coordinates. This change of variable can make it easier to solve certain types of PDEs, particularly those that involve circular or symmetric domains.

## 3. How does a polar change of variable help in solving PDEs?

A polar change of variable can help in solving PDEs by simplifying the equations and making them easier to solve. In polar coordinates, the PDE may have fewer terms and may be separable, meaning it can be split into simpler equations that are easier to solve.

## 4. What are the key steps in using a polar change of variable to solve a PDE?

The key steps in using a polar change of variable to solve a PDE are:
1. Convert the PDE from Cartesian coordinates to polar coordinates using the appropriate transformation formulas.
2. Substitute the new variables into the PDE and simplify the resulting equation.
3. Apply any necessary boundary conditions to the PDE.
4. Solve the simplified PDE using separation of variables or other appropriate methods.

## 5. What types of PDEs can benefit from a polar change of variable?

A polar change of variable can benefit PDEs that involve circular or symmetric domains, as well as those that have Laplace's equation or the Helmholtz equation as their governing equation. These types of PDEs are commonly found in fields such as electromagnetism, fluid dynamics, and heat transfer.

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