Thread: Finding the Green's Function for Simple ODE with Initial Conditions

1. Finding the Green's Function for Simple ODE with Initial Conditions

I need to find the Green’s Function for the problem
u'' = f(t); u(0) = 0; u(1) = 1

I know that the general solution of the homogenious equation is
u=at+b
where a and b are constants, and the Green's function is
G(t,z)=(u1(z)*u2(t)-u2(z)*u1(t))/W(z)
where u1 and u2 are independent particular solutions of the homogenious equation, but I'm not sure how to apply the initial values to find the right u1 and u2. Can someone please point me in the right direction?

2. Saying that the solution to the homogeneous equation is of the form at+ b means, since G(t, z) must satisfy the equation for all t except possibly t= z, that G(t, z) is of the form At+ B for t< z and Ct+ D where t> z. At t= z, of course, the two must match so you must have G(z, z)= Az+ B= Cz+ D. G(0, z)= A(0)+ B= B= 0 (0< z, of course). G(1, z)= C(1)+ D= C+ D= 1, D= 1- C. Putting those into Az+ B= Cz+ D, we have Az= Cz+ 1- C= C(z- 1)+ 1.

You will need one more condition to find C and so A and D: the difference between the two derivatives, taken from the left and right, at t= z, must be 1:
for t< z, dG/dt= A and for t> z, dG/dt= C: A- C= 1 so that A= C+ 1. Putting that int Az= C(z- 1)+ 1 gives (C+1)z= C(z- 1)+ 1 so Cz+ z= Cz- C+ 1. The "Cz" terms cancel: z= 1- C so C= 1-z. then A= C+ 1= 1- z+ 1= -z, D= 1- C= 1- (1-z)= z, and we already had B= 0:
$\displaystyle G(t, z)= \{\begin{array}{cc}At+ B= -zt & t< z \\ Ct+ D= (1- z)t+ z & t\ge z\end{array}$

3. Thank you, but I don't really understand how you got that. The only mention of Green's function in my textbook is that it's G(t,z)=(u1(z)*u2(t)-u2(z)*u1(t))/W(z) where u1 and u2 are independent solutions of the homogenious equation, and that the integral from t0 to t of G(t,z)f(t)dz is a particular solution of the inhomogenious equation. Why does G have to satisfy the same initial values as u? And how do you know the difference of the derivatives?