- #1

- 215

- 11

*Solve ##au_{x} + bu_{y} = f(x,y)##, where ##f(x,y)## is a given function. If ##a \neq 0##, write the solution in the form*

$$u(x,y) = (a^{2} + b^{2})^{\frac{-1}{2}} \int_{L} f ds + g(bx - ay)$$

$$u(x,y) = (a^{2} + b^{2})^{\frac{-1}{2}} \int_{L} f ds + g(bx - ay)$$

(from

*Partial Differential Equations An Introduction, 2nd edition*by Walter A. Strauss; pg. 10)

I began by using the change of variables ##x' = ax + by## and ##y' = bx - ay## to reduce the PDE to the following ODE:

$$(a^{2} + b^{2}) u_{x'} = f( \frac{ax' + by'}{a^{2} + b^{2}}, \frac{bx' - ay'}{a^{2} + b^{2}})$$

$$u_{x'} = \frac{f( \frac{ax' + by'}{a^{2} + b^{2}}, \frac{bx' - ay'}{a^{2} + b^{2}})}{a^{2} + b^{2}}$$

This is easily solvable:

$$u(x', y') = \int \frac{f( \frac{ax' + by'}{a^{2} + b^{2}}, \frac{bx' - ay'}{a^{2} + b^{2}})}{a^{2} + b^{2}} dx' + g(y')$$

My difficulty lies in the next step. In using the change of coordinates to return to the unprimed coordinates I'm not precisely sure how ##dx'## is affected. Because ##x'## depends on a, b, x and y I expect it to decompose into some combination of dx and dy which can then be written as a line integral but I'm still not entirely sure how the a and b are encapsulated...

I'm stuck with something like this because I don't understand how to get the a and b terms to come out to give me ##(a^{2} + b^{2})^{\frac{1}{2}}## in the numerator.

Any guidance or explanation would be greatly appreciated. Thanks in advance!

$$u(x,y) = (a^{2} + b^{2}) \int_{L} f ds + g(bx - ay)$$