How do i deal with integral boundary conditions?

[nocloud]

I am trying to solve an ode of the form u"(x)=a(x) where a(x) is some known function and the domain is from -inf to +inf. I am required to use Green's function.

The boundary condition is u(0) = Integral[u(x),{x,0,1}] = 0

My Green's function has the form
G(x,y) = A(y)*x+B(y) x<y
G(x,y) = C(y)*x+D(y) x>y

How the heck do I use that integral to find the coefficients A and B?

 

[HallsofIvy]

I am trying to solve an ode of the form u"(x)=a(x) where a(x) is some known function and the domain is from -inf to +inf. I am required to use Green's function.

The boundary condition is u(0) = Integral[u(x),{x,0,1}] = 0

My Green's function has the form
G(x,y) = A(y)*x+B(y) x<y
G(x,y) = C(y)*x+D(y) x>y

How the heck do I use that integral to find the coefficients A and B?

Since G(x,y) must satisfy the same boundary conditions as u, for all y, you must have
\int_0^1 G(x, y)dx= 0
for all y.
\int_0^1 G(x,y)dx= \int_0^y (Ax+ B) dx+ \int_y^1 (Cx+ D) dx
= \frac{1}{2}Ay^2+ By+ C+ D - \frac{1}{2}Cy^2- Dy= 0

You will also need to use, of course, the conditions at x= y: Ay+ B= Cy+ D and A- C= 1. That is still only three conditions to determine four functions of y. The difficulty is not that the boundary condition is integral but that there is only one boundary condition. In spite of the fact that the single boundary condition involves both endpoints, you still only have one boundary condition and you need two.