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Solving a PDE

  1. Dec 2, 2011 #1
    Hi, I have a problem solving this PDE:

    (y^2)*u(x,y)''+2*y*u(x,y)'-2*u(x,y) = 0

    Every derivate of u is in fonction of y.

    What I tried:

    I said that (y^2)*u(x,y)''+2*y*u(x,y)' = (u(x,y)'*y^2)' and make

    v=u(x,y)'*y^2
    then I tried to isolate u(x,y) and I arrive to u(x,y)=-v/y+C(X)

    But I can't go much further... I am blocked right there. Thank You.

    PytLov
     
  2. jcsd
  3. Dec 2, 2011 #2

    HallsofIvy

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    If every derivative is with respect to y, treat x like a constant so you have just an ordinary d.e.

    [itex]y^2u''+ 2yu'- 2u= 0[/itex]. That's a "Cauchy-Euler" equation. The substitution s= ln(y) changes that equation (in the variable y) to an equation with constant coefficients (in the variable s).

    Or you can take, as a trial solution, [itex]u= y^r[/itex]. The only effect 'x' has on this is that the "undetermined constants" you would get for an o.d.e. are "undetermined functions of x".
     
    Last edited: Dec 2, 2011
  4. Dec 2, 2011 #3
    Hi, I'm not sure of how to make your substitution. I understand that is u=ln(y), then u'=1/y and u''=-1/u^2 but I don't know how to put it in the ODE...

    y2*u''+2y*u'-2u


    PytLove
     
  5. Dec 2, 2011 #4

    HallsofIvy

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    That's not what I meant. But I got the variable mixed up- I have gone back and edited my post.

    If t= ln(y) then
    [tex]\frac{du}{dy}= \frac{du}{dt}\frac{dt}{dy}= \frac{1}{y}\frac{du}{dt}[/tex]
    then
    [tex]\frac{d^2u}{dy^2}= \frac{d}{dy}\left(\frac{1}{y}\frac{du}{dt}\right)[/tex]
    [tex]= -\frac{1}{y^2}\frac{du}{dt}+ \frac{1}{y^2}\frac{d^2u}{dt^2}[/tex]

    The equation becomes
    [tex]\left(\frac{d^2u}{dt^2}- \frac{du}{dt}\right)+ 2\frac{du}{dt}- 2u= 0[/tex]
    [tex]\frac{d^2u}{dt^2}+ \frac{du}{dt}- 2u= 0[/tex].

    What are the solutions to that?

    Or, as I also said, if you look for a solution of the form [tex]u= y^r[/tex], then [tex]u'= ry^{r-1}[/tex] and [tex]u''= r(r-1)y^{r- 2}[/tex]. Putting those into the d.e., we get
    [tex]r(r-1)y^r+ 2ry^r- 2y^r= (r^2+ r- 2)y^r= 0[/tex].

    What must r equal so that is true for all y?
     
  6. Dec 2, 2011 #5
    Thank you for you help, I understand... But I'm really not sure I would think of that in exam.

    PytLov
     
  7. Dec 3, 2011 #6

    HallsofIvy

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    Most people who are taking partial differential equations have taken at least an introductory course in ordinary differential equations. And "Cauchy-Euler" equations (also called just "Cauchy type equation" or "equi-potential equations" are a standard topic in ordinary differential equations.
     
  8. Dec 3, 2011 #7
    Yes, I know that type of equation, I have seen it in my advanced mathematics class, but we didn't exactly solve them, we we're trying to find the eigenfunctions as an introduction to Fourier series. But if you tell me that Cauchy Euler equation can always be solved with the substitution up there, that would be really useful to me.

    PytLov
     
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