# Solving third order nonlinear boundary value problem

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parham kianian on 21 Dec 2020
Edited: David Goodmanson on 22 Dec 2020
I have a boundary value problem in this form: where alpha is a nonzero constant and boudary conditions are:
f(-1) = 0
f(0) = 1
f(1) = 0
How can I solve this problem using bvp5c?
David Goodmanson on 22 Dec 2020
Edited: David Goodmanson on 22 Dec 2020
Hi parham,
This looks doable if you constuct an f vector of length 6, where the first three elements are f,f',f'' in the left hand region -1<=x<=0 and the second three elements are f,f',f'' in the right hand region 0<=x<=1. Then the differential equation part will be
dfdt(1) = f(2)
dfdt(2) = f(3)
dfdt(3) = a*f(1)*f(2) + 4*a^2*f(2)
and the same for f variables (4,5,6) since the differential equation is the same. At the boundary x = 0, the first and second derivatives of f have to be continuous across the boundary. So for the boundary part in terms of fa and fb, remembering that the a,b boundaries are -1,0 for f (1,2,3) and 0,1 for f(4,5,6) ,the follwing values are zero:
fa(1)
fb(1)-1
fa(4)-1
fb(4)
fb(2)-fa(5) % continuous 1st derivative
fb(3)-fa(6) % continuous 2nd derivative