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Simple PDE questions

  1. Sep 21, 2013 #1
    So up to this point we have only learned 2 forms of PDE's to solve:

    Constant Coefficient Equations and Variable Coefficient Equations.

    Questions: Solve:

    1) aUx+bUy + cU = 0

    2) Ux+ UY = 1

    where U = U(x,y)

    Attempt:

    Well for 2) I'm thinking that it doesn't necessarily matter that the constant is 1 even though in the text it is boxed that the equation has to be of the form Ux+ yUy = 0
    but 1 is just another constant. As for 1) I am a little miffed.
     
  2. jcsd
  3. Sep 21, 2013 #2
    In general PDEs are much more difficult to solve than ODEs so in both of these problems I think it prudent to try to reduce them into ODEs using a few "tricks".

    I agree that your first question looks more daunting initially but we'll get to that as soon as we work through the second question (because it's easier and I think it's more enlightening). Before we do that though, we can try to dumb things down and solve some PDEs that are much simpler and work our way into solving these slightly harder ones.

    1) First, what is the general solution to the PDE ##u_{x} = 0## where u is a function of x and y?
    2) Second, what is the general solution to the PDE ##au_{x} + bu_{y} = 0## where u is a function of x and y and a and b are constants not both zero? This is a simple extension of my first question but it is illuminating. Hint: Think about another way you could represent ##au_{x} + bu_{y}## using methods from multivariable calculus (Hint: Think about the gradient.).

    Now that you (hopefully) have the general solution of ##au_{x} + bu_{y} = 0## think about your second question: ##au_{x} + bu_{y} = 1##. What does this mean geometrically?


    If you're interested in a procedure of attack for solving these problems there are two methods that I know about. One is called the geometric method (this is the method I am trying to lead you to by asking my above questions). The second is the coordinate method which I personally find works very nicely (especially for problem of only 2 variables). You might be able to find these in your textbook.

    In fact, you can solve both problems using this method but understanding the geometric method will really help you understand these types of PDEs better. Hopefully this helps!
     
  4. Sep 21, 2013 #3

    Thanks for the suggestion on the plan of attack. W.R.T. this second equation I am working through it via the coordinate method, but that was because I couldn't see where to proceed with the geometric method beyond a certain step. I'll discuss what I have gotten for both ways.

    Geo Method: Looking at it in terms of the geometric method, I could rewrite ##au_{x} + bu_{y} = 1## as (1,1) ° ∇U(x,y) = 1, then doing this I could set it up as a separable one variable equation, but my concern with this is that I tried that but since my original equation did not equal 0 it did not work.

    Coordinate method: I beleive I've gotten closer. I used substitution with x' = ax + by, and
    y' = bx-ay, after performing the necessary simplifiction, I've arrived at:

    (a+b)Ux + (b-a)Uy = 1

    Now I am here trying to figure out a way to eliminate the 1 and turn it to a zero and then hopefully put it in the form of a constant coefficient equation...
     
  5. Sep 21, 2013 #4

    I agree that it is difficult to solve this problem using the geometric method - the important thing is that you understand what's happening in the geometric method. You've rewritten it in the way I hoped you would so let think about that a little bit. What exactly does that expression mean?


    You're getting there with the co-ordinate method. However, Using the substitutions ##x' = ax + by## and ##y' = bx - ay## as you have we can go even farther. Let's find expression for ##u_{x}## and ##u_{y}## in a nifty way.

    ##\frac{∂u}{∂x} = \frac{∂u}{∂x'} \frac{∂x'}{∂x} + \frac{∂u}{∂y'} \frac{∂y'}{∂x}##

    (All we've done is used the chain rule.)

    Simplify that and do the same thing for ##u_{y}## and plug these new expressions in - you should find you get a much simpler result (it should be on ODE).

    You can apply the same procedure for the harder PDE you have.
     
  6. Sep 21, 2013 #5
    Ok, for me, and not to take away the nice job Tsunoyukami is doing, but in my humble opinion, there is only one way to permanently solve your problems with first-order PDEs: You have got to go to the library and find "Basic Partial Differential Equations" by Bleecker and Csordas. It's there I bet. You study the first couple chapters which deal with the intro and first-order PDEs and do the work then I guarantee you will never, ever have a problem with (solvable) first-order PDEs again. The text is an intro and reads nicely.
     
  7. Sep 21, 2013 #6
    I'll have to check out that text too. I'm just beginning my foray into PDEs myself. :)
     
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