Solving third order nonlinear boundary value problem

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]

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