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Numerical solution of an ODE be Singular at endpoint

  1. May 20, 2010 #1
    Hi all,
    I have an ODE with the form
    [tex]\frac{\mathrm{d}^{2}\phi}{\mathrm{d}x^{2}}+a\exp(b\phi)=0[/tex]

    B.C. [tex]\frac{\mathrm{d}\phi}{\mathrm{d}{x}}\right|_{x=0}=-c,\frac{\mathrm{d}\phi}{\mathrm{d}{x}}\right|_{x=L}=0[/tex]

    where [tex]a,b,c,L[/tex] are all positive.

    I solved the problem with Maple using the dsolve routine:
    dsolve({bc, equ}, type = numeric, method = bvp[midrich], range = 0 .. L)
    but it says "matrix is singular". My Fortran program using the FVM method also indicate the coefficients matrix is singular at [tex]x=L[/tex]

    But if i change the B.C. at [tex]x=L[/tex] to [tex]\phi\right|_{x=L}=0[/tex] the solution can be got.

    So i wonder
    1 why the previous Newmann B.C. cause the singularity at [tex]x=L[/tex] but the latter Dirichlet one not.
    2 how to evaluate the influence of such a boundary to the nature of the solution?

    any suggestions will be appreciated
    hui
     
  2. jcsd
  3. May 20, 2010 #2
    Unless I am mistaken, if you have Neumann conditions on both sides then there is nothing to "fix" the integration constant. So your solution will be unique up to a constant offset.
     
  4. May 20, 2010 #3
    My fault. i checked the simplified form of the eqution
    [tex]\frac{\mathrm{d}^{2}\phi}{\mathrm{d}x^{2}}+ab\phi=0[/tex]
    to get guess values of [tex]\phi[/tex] and it works with two Newmann B.C.s so i think it is the same with the original equation...

    if we substitute [tex]\phi(x)=f(x)+C1[/tex] into the original equation it can be found [tex]C1[/tex] may not be zero.

    Lord Crc, thanks for your reply.

     
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