Hello everbody,

I want to ask a question about the folowing problem:
the equation is : d(d(y)/dt)/dt=G(t,d(y)/dt,y)
this equation reduces to
the system
d(y)/dt=z
d(z)/dt=G(t,z,y)
and the boundary conditions are:
t=0, y=0
t=1, y=1
My question is if this problem is solvable with realaxation method or not? Because I tried to do but I get the message 'singular matrix, row all 0 in pinvs'!!
Thank you

I've been trying to use solvde to solve a system of equations which have the exact same form as the equation you've posted here .... not surprisingly then I'm having the exact same problem!

It appears to me to be due to the fact that the block matrix routines (red, pinvs, and bksub) have been written with a very specific problem forumlation in mind. As stressed in the text, the authors have assumed that each boundary condition applys to a separate independant variable. However, in our problem the boundary conditions are both applied to the same variable and so this assumption breaks down.

Essentially the block matrix solving functions are creating a row equivalent to 0 = 0, and ignoring the boundary condition at one end. Sadly (assuming this is indeed what is happenining) its not clear to me that modifyint these routines to account for this is as trivial as it sounds, as the variable NB is hidden in many variables such as ic1, im, nbf ... etc.

I'm going to attempt this, but if I'm wrong about the above or some one has already attempted this I'd really appricate any help.

Dear dmo,

I have avoided this problem since two or three months, may be your case is exactely the same as mine. So let me show where was my mistake?

My mistake was very simple, I corrected it by just renaming my functions in a way that at least one of the matrix element s(i,j) in each row at the first boundary (dont care about the last boundary) is not zero.

Or let me show it explictely, if you have system of 4 functions L,R,F and G, and you you at the first boundary x0, the conditions: L(x0)=a, G(x0)=b. of course if name your functions like: y(1)=L,y(2)=R,y(3)=F and y(4)=G, you will find this problem!!!
So you must put y(1) and y(2) to be L and G or G and L respectively.
I hope this will be your problem.