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)