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Partial Differential Equations

  • #1
215
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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

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

(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!

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

Answers and Replies

  • #3
215
11
I haven't tried this particular problem using the method of characteristics. However, I have tried solving several "simpler" (ie. less general) problems using the method of characteristics but personally find the method of characteristics difficult to apply to an inhomogeneous PDE so please bear with me if what follows is utter nonsense.


Beginning with the PDE we wish to solve: ##au_{x} + bu_{y} = f(x, y)##

The method of characteristics tells us that the directional derivative of ##u## in the direction of ##(a,b)## is given by ##f(x,y)##. I'm really not sure where to go from this...everything I can think of saying follows from the single example my text provides in the case of f(x,y) = 0; that is, I would look at lines perpendicular to ##(a,b)## and argue that ##bx - ay = constant## - but I'm unsure whether or not that should be true in this case (and it seems to me like it shouldn't be true).
 
  • #4
20,148
4,218
In applying the method of characteristics, I usually think of a and b as the components of velocity in the x and y directions:
[tex]\frac{dx}{dt}=a[/tex]
[tex]\frac{dy}{dt}=b[/tex]
[tex]ds=\sqrt{(dx)^2+(dy)^2}=\sqrt{(a)^2+(b)^2}dt[/tex]
So,
[tex]\frac{dx}{ds}=\frac{a}{\sqrt{(a)^2+(b)^2}}[/tex]
[tex]\frac{dy}{ds}=\frac{b}{\sqrt{(a)^2+(b)^2}}[/tex]

[tex]{\sqrt{(a)^2+(b)^2}}\frac{du}{ds}=f\left(x_0+\frac{as}{\sqrt{(a)^2+(b)^2}},y_0+
\frac{bs}{\sqrt{(a)^2+(b)^2}}\right)[/tex]

The g function is related to the initial condition, which lies on a line perpendicular to the streamlines.

Chet
 
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  • #5
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Ahh, that method seems a lot cleaner and nicer than what I was going for...

I'm sure it's possible for me to get the same result by using the method I began to employ in my initial post if I were able to express ##dx'## in terms or some combination of ##dx## and ##dy##. I doubt it's as simple as just ##dx' = a dx + b dy## (because I still wouldn't be able to work my way from there to the final result).

I'm just curious to see how I could plow forward with the coordinate method because I expect its possible as the two methods should be equivalent.
 

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