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An inhomogenous equation-help !

  1. Dec 7, 2006 #1
    poisson equation with non homogeneous B.C.

    hello,
    I am having some trouble to solve a problem and I was wondering if someone could give me a clue. The problem is the following:

    u,xx+u,yy=xy , 0<x<1, 0<y<1
    BC:
    u,x(0,y)=0 & u(1,y)=y for 0<y<1
    u(x,0)=0 & u(x,1)=-x for 0<x<1

    the equation is non homogeneous and so are the B.C.'s

    thanks in advance

    -marcel
     
  2. jcsd
  3. Feb 23, 2007 #2
    Can anybody help me with this inhomogenous equation : y''=A-B(y')^2
    A and B are constants.
    Thanks for reading.
     
  4. Feb 23, 2007 #3

    arildno

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    Let z=y', and solve for z first.
     
  5. Feb 23, 2007 #4
    Thanks for fast answer.
    I have tried like that, but it seems...not finished. Could you write down some more details?
     
  6. Feb 23, 2007 #5

    arildno

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    What did you get for z?
    Or did you, perchance, not finish even that part of your task?
     
  7. Feb 23, 2007 #6
    arildno is perfectly correct in his suggestion. If, however, you are not fond of it, you can try a similar one, i.e.,

    [tex]y^{\prime} = \frac{1}{B} \frac{u^{\prime}}{u} [/tex]

    this will give you

    [tex]\frac{1}{B} \frac{u^{\prime \prime}}{u} - \frac{1}{B} \frac{u^{\prime 2}}{u^2} = A - \frac{1}{B} \frac{u^{\prime 2}}{u^2}[/tex]

    leaving you with

    [tex]u^{\prime \prime} = AB u [/tex]

    which should seem familiar to you.

    Incidentally, to further arildno's point -- if you show us your working-out when you get stuck, we will have more chance of helping you.
     
    Last edited: Feb 23, 2007
  8. Feb 23, 2007 #7
    Thank you very much Mathrew and Arildno. I do not know if this type of equation is easy for people in this forum, but for me, at least by now, it is not. Hope after some time , I will get improved.
     
  9. Feb 24, 2007 #8

    HallsofIvy

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    Just out of curiosity, why are you refusing to post what you have done?
     
  10. Feb 24, 2007 #9
    OK, It’s nice if you can have a look and check for me.

    From the hint of Matthew, I can rewrite the following:

    u''= ABu with y'=u'/(Bu)

    ==> u''-ABu=0. Let u=exp(kt)
    then we have: k^2exp(kt)-ABexp(kt)=0
    ==> k^2-AB=0.==> k=+-sqrt(AB) I just investigate A>0 and B>0

    So the general solutions of u is :

    u=C1.exp(sqrt(AB).t) + C2.exp(-sqrt(AB).t) with C1 and C1 are arbitrary constants.
    Let k=sqrt(AB)
    Then y'=u'/(Bu)={C1.k.exp(kt) - C2.k.exp(-kt)}/{B*(C1.exp(kt) + C2.exp(-kt))}
    Integrate 2 sides :
    y=(1/k)*{ln[C1.exp(2kt)-C2]-kt} +C3

    Hope that I did not make any mistake. I am not sure adding of the arbitrary C3 is correct. This is a second order equation.
     
  11. Feb 24, 2007 #10
    How about doing homogeneous solution plus particular solution.

    Then, do Fourier series to solve the boundary value problem. You have to compensate for the added particular solution when doing the FS.

    Finding a part. sol'n. is sort of easy by the symmetry of the PDE. In fact there are lots of choices for the part. sol'n. Some more may be more convenient than others.
     
  12. Feb 24, 2007 #11
    Err...looks like you may have got something slightly wrong there. If

    [tex]u(t) = C_{1}e^{kt} + C_{2} e^{-kt} [/tex]

    where [tex]k=\sqrt{AB}[/tex]

    then, as

    [tex]y^{\prime} = \frac{1}{B} \frac{u^{\prime}}{u} [/tex]

    this means that you can also write y as

    [tex]y(t) = \frac{1}{B} \ln{|u(t)|} + C_{3}[/tex]

    where [tex]C_{3}[/tex] is another constant (just integrate the [tex]y^{\prime}[/tex] equation, and you'll see).

    So, your final solution is

    [tex]y(t) = \frac{1}{B} \ln{|C_{1} e^{kt} + C_{2}e^{-kt}|} + C_{3} [/tex]
     
    Last edited: Feb 24, 2007
  13. Feb 25, 2007 #12
    Oh, it's just simple. It took me quite a time to integrate the funciton y', and there's mistakes ! Thanks again.
     
  14. Feb 27, 2007 #13
    Hi,

    Can anyone explain for me why this solution has 3 arbitrary constants. Is there anything special here? I have tried to solve in Matlab, the result has only 2 constants.
     
  15. Feb 27, 2007 #14
    Whose thread is this anyway?


    Wasn't this Marcel's thread? That's whose Q I was addressing.

    :rofl:
     
  16. Feb 27, 2007 #15
    There must be something wrong here. This thread named "An inhomogenous equation-help !" surely was posted by me. I do not understand why it is mixed with the thread by mremilli, who posted long before me. I might have made a mistake when posting this one mightn't I?
     
  17. Feb 27, 2007 #16

    HallsofIvy

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    Whatever happened to mremilli's question?

    The fact that the equation is not homogeneous is no big deal but it will simplify a lot if you find a function that satisfies just the boundary conditions. Subtract that from u to get a differential equation that is still non-homogeneous but does have homogeneous boundary conditions. You can then write the solution as a Fourier sine series.
     
    Last edited: Feb 27, 2007
  18. Feb 27, 2007 #17
    That's what I was answering back in 2/24

    :uhh:
     
    Last edited: Feb 27, 2007
  19. Feb 27, 2007 #18
    More

    Try Pinsky's book PDE's and BV Prob's w/ App's
     
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