Numerical solution of an ODE be Singular at endpoint

[SHmech]

Hi all,
I have an ODE with the form
$$\frac{\mathrm{d}^{2}\phi}{\mathrm{d}x^{2}}+a\exp(b\phi)=0$$

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

where $$a,b,c,L$$ 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 $$x=L$$

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

So i wonder
1 why the previous Newmann B.C. cause the singularity at $$x=L$$ 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

 

[Lord Crc]

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.

 

[SHmech]

My fault. i checked the simplified form of the eqution
$$\frac{\mathrm{d}^{2}\phi}{\mathrm{d}x^{2}}+ab\phi=0$$
to get guess values of $$\phi$$ and it works with two Newmann B.C.s so i think it is the same with the original equation...

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

Lord Crc, thanks for your reply.

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.