Green Function for Second Order Differential Equation with Initial Value Problem

[mwg]

Homework Statement

need to find the green function for

d^2X / dt^2 + 2 dX/dt +(1+k^2)X= f(t)

with IV X(0)=dX/dt (=0) = 0

Homework Equations

eq. attached

The Attempt at a Solution

attached

whenever I try to solve the homogonous part, I get 0 due to the IV.
guessing my strategy is wrong.

what should it be??

 

[HallsofIvy]

A "Green's function", G(t, \tau), for a problem like this (the coefficient of the second derivative is 1) must satisfy several properties:
1) It must satisfy the homogenous d.e. in t for every \tau.
2) It must satisfy the boundary conditions.
3) It must be continuous.
4) The "jump" in the first derivative at t= \tau must be one.

Here, the general solution to the homogeneous equation is
e^t(C_1cos(kt)+ C_2sin(kt))
which means that the Green's function must be of the form
G(t, \tau)= \begin{pmatrix}e^t(Acos(kt)+ Bsin(kt) & 0\le t\le \tau \\ e^t(Ccos(kt)+ Dsin(kt) & \tau\le t\le 1\end{pmatrix}

To satisfy the first boundary condition, since 0\le\tau, we must have
G(0,\tau)= A= 0
To satisfy the second boundary conditon, we must have
G_t(0,\tau)= A+ B= 0
which, since A= 0, give B= 0. That is, G(t, \tau) is identically 0 for all 0\le t\le \tau.

But that does NOT say that G(t,\tau) must be 0 for \tau\le t\le 1.

The continuity condition gives that
\lim_{t\to\tau^+}G(t,\tau)= e^{\tau}(C cos(k\tau)+ Dsin(k\tau))= \lim_{t\to\tau^-} G(t,\tau)= 0
so we must have
e^{\tau}(C cos(k\tau)+ D sin(k\tau))= 0
which leads to
D= -\frac{cos(k\tau)}{sin(k\tau)}C

The "jump" condition on the derivative says that
\lim_{t\to\tau^-}G_t(t,\tau)- \lim_{t\to\tau^+}G_t(t,\tau)= 1
which gives
e^\tau(Ck cos(kt)+ Dk sin(kt))+ e^\tau(-Ck sin(kt)+ Dk cos(kt))= 1

those two equations give non-zero values for C and D.

 

[mwg]

Thank you so much!

one more question though, do I set sum lambda G = f(x) and solve? (sorry, couldn't figure the Latex here)