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Math Help - A Challenging PDE question

  1. #1
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    A Challenging PDE question

    Dear users of MHF, I've found two PDE question very challenging, would you guys mind to give some hints?

    First question:

    1) (-x) Ux + (2y) Uy = x^2 + y^2

    a) find the characteristic curve (I solved)

    b) solve the above equation by conditions U(x,1)= x^2 , u(x,-1) = x.

    second question:

    2)Ux + Uy= x - y + U

    condition : U(0,y)= e^(-y^2).


    for the second question I tried U = x^2 - (xy) + f(x-y) e^(x)
    but I yielded a result like x - y + f(x-y)e^(x)....

    the questions just sound so impossible for me...

    and the condition in 1b was so tricky...
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  2. #2
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    Re: A Challenging PDE question

    Did you obtain tbe general solutions

    (1) u = -\frac{1}{2}x^2 + \frac{1}{4} y^2 + f(x^2y)?

    In (2) you mentioned you tried u = x^2 - x y + e^x f(x-y).  Where did this come from?
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  3. #3
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    Re: A Challenging PDE question

    Quote Originally Posted by Danny View Post
    Did you obtain tbe general solutions

    (1) u = -\frac{1}{2}x^2 + \frac{1}{4} y^2 + f(x^2y)?

    In (2) you mentioned you tried u = x^2 - x y + e^x f(x-y).  Where did this come from?
    Thanks danny, definitely I obtained (1), but when I try to use the condition given, I can't figure out how can I get u(x,-1)=x .

    for 2) I only know that  u= f(x-y)e^y or  u=f(x-y)e^x usually results in ux+uy= u, how ever the x-y parts I find it so hard...

    if i let Ux= x,  x= x^2 /2
         Uy= -y,  y= -y^2/2
    but then applying the condtion given I can't get the answer  u(x,0)= e^(-(y^2))
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  4. #4
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    Re: A Challenging PDE question

    For the first part, subs y =1 into your solution, set this equation the the IC and solve for f. For example,

    u(x,1) = - \frac{x^2}{2} + 1 + f(x^2) = x^2 so f(x^2) = \frac{3}{2}x^2- 1.

    Let  r = x^2 so f(r) = \frac{3}{2}r- 1. This then defines f. Similarly for the second IC.

    For the second question, do you know the method of characteristics?
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  5. #5
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    Re: A Challenging PDE question

    With method of characteristic, I know how to get for Ux + Uy=0 u= f(x-y)

    (at work, sorry for short reply)
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  6. #6
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    Re: A Challenging PDE question

    OK. Your answer above kinda makes sense. You can do one of two things. If you let

    u = v - x + y then your PDE

    u_x + u_y = x - y + u becomes v_x + v_y = v. Then do what you did.

    OR you can solve the charactistic equations

    \frac{dx}{1} = \frac{dy}{1} = \frac{du}{x-y+u}.
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