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An interesting PDE perhaps?

  1. Oct 4, 2005 #1


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    In the HW section, someone proposed:

    [tex]u^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0;\quad u(x,0)=x[/tex]

    As per "Basic PDEs" by Bleecker and Csordas", treating this as:

    [tex]F(x,y,u,p,q)=0\quad\text{with}\quad \frac{\partial u}{\partial x}=p\quad\text{and}\quad\frac{\partial u}{\partial y}=q[/tex]

    and solving for parametric functions as a function of two variables t and s such that:

    [tex]\frac{dx(s,t)}{dt}=\frac{\partial F}{\partial p}[/tex]

    [tex]\frac{dy(s,t)}{dt}=\frac{\partial F}{\partial q}[/tex]

    [tex]\frac{du(s,t)}{dt}=p\frac{\partial F}{\partial p}+q\frac{\partial F}{\partial q}[/tex]

    Solving the ODEs subject to boundary conditions:

    [tex]u(x,0)=u(f(s),g(s))=s [/tex]





    Solving for u:


    To what extent can I rely upon this method to solve more complex non-linear PDEs? Guess that would involve some analysis of sorts limited by the ability to affect some integration or another. Some limitations I can think of include:

    1. I just glossed-over the need sometimes to solve a set of 5 equations instead of the simpler 3 sets like above. Seems though should be able to numerically integrate the 5 equations (or 3) no matter how complex and thus arrive at least in principle to a parametric solution described above.

    2. u(x,y) cannot allways be determined explicitly in terms of x and y although the parametric forms are equally valid albeit a bit more difficult to manipulate. Suppose I can work on them a bit but I mean, that could take a whole semester to fully analyze:


    Probably been done already anyway. :confused:
  2. jcsd
  3. Oct 5, 2005 #2
    [tex] u = \pm \sqrt{ \frac{x}{y}} [/tex]
    isn't a solution to
    [tex] u^2 \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} =0, \:
    u(x,0)=x. [/tex]

    the problem is in the u(x,0)=x part.
    how's about:
    [tex] u = \frac{-1 + \sqrt{1+4xy}}{2y} [/tex]
  4. Oct 5, 2005 #3


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    Good for you Qbert. Yea, I thought about not meeting the initial conditions last night. Problem lies in:

    [tex]\frac{dx(s,t)}{dt}=u^2=s^2;\quad x(s,0)=s[/tex]



    and solving for u(x,y):


    However, the negative one does not converge to u(x,0)=x but the positive one does.

    Dang it! I hate when that happens. :yuck: And I'm not surprised you detected it. Thanks! Guess I gotta' go and update the one in the HW section now. :rolleyes:
  5. Oct 5, 2005 #4


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    Alright, suppose I glossed-over something again. This:

    [tex]\mathop\lim\limits_{y\to 0}\left(\frac{-1+\sqrt{1+4xy}}{2y}\right)=x[/tex]

    Well, it's not obvious to me anyway. :surprised .

    Anyway, since they both go to zero, I can justify using . . . wait, let me look up the spelling . . . L'Hopital's rule. That give me:

    [tex]\frac{x}{\sqrt{1+4xy}}[/tex] which then goes to x as y goes to zero.
  6. Oct 5, 2005 #5


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    I have a question:

    So for:

    [tex]u^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0;\quad u(x,0)=x[/tex]

    we got the solution in parametric form:




    Now suppose I couldn't solve for u explicitly in terms of x and y and just had the parametric equations above.

    I don't know how to algebraically verify the solution using only the parametric equations. Can someone show me or should I know this one too?

  7. Oct 6, 2005 #6


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    Alright I think I have it thanks to . . . well something else I worked on in the forum:

    So we have:

    [tex]u^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0[/tex]





    so, lets start calculating partials:

    [tex]\frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+
    \frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial u}{\partial x}(2st+1)[/tex]

    [tex]\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+
    \frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=s^2\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}[/tex]


    [tex]\frac{\partial u}{\partial x}=\frac{1}{2st+1}\frac{\partial u}{\partial s}[/tex]

    [tex]\frac{\partial u}{\partial y}=\frac{\partial u}{\partial t}-\frac{s^2}{2st+1}\frac{\partial u}{\partial s}[/tex]

    Substituting these into the original PDE:

    [tex]\frac{u^2}{2st+1}\frac{\partial u}{\partial s}+\frac{\partial u}{\partial t}-\frac{s^2}{2st+1}\frac{\partial u}{\partial s}=0[/tex]

    Now, since u=s, the two partials with respect to s cancel leaving us with:

    [tex]\frac{\partial u}{\partial t}=0[/tex]

    but since u=s, that partial is zero, thus satisfying the relation.
    Last edited: Oct 6, 2005
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