Hi there,

I've been completing a problem of a question based on ODE and Green's Functions. The question asks me to come up with a general solution for the IVP ODE $\displaystyle L[y] = y\prime + y\cdot sin(t)=\delta(t - \tau), y(0) = 0$ by imposing a jump condition on $\displaystyle y(t)$ at $\displaystyle t = \tau $ and solving the differential equation for $\displaystyle t < \tau$ and$\displaystyle t > \tau$. Then identify the Green's function $\displaystyle G(t,\tau)$ for the IVP ODE $\displaystyle L[y] = y\prime + y\cdot sin(t)=q(t), y(0) = 0$

I was wondering whether someone could read through my working and see whether I have the correct Green's function.

Lets intergrate $\displaystyle L[y] = y\prime + y\cdot sin(t)=\delta(t - \tau)$ from times $\displaystyle \tau + \varepsilon$ to $\displaystyle \tau - \varepsilon$

$\displaystyle \delta(t - \tau)$ will equal 1 over this interval, as it is the step function. The area under it's curve will always equal 1.

We take the limit as $\displaystyle \varepsilon$ approaches 0. This makes the intergal of $\displaystyle y\cdot sin(t)$ equal to zero. So we are left with $\displaystyle [y]$ evaluated from $\displaystyle \tau^+$ to $\displaystyle \tau^-$. This equals 1.

To find the general solution, I need to solve the IVP for $\displaystyle t < \tau$ and$\displaystyle t > \tau$. Start with $\displaystyle t < \tau$.

The solution to the ODE is $\displaystyle y(t) = A\cdot \exp^{cos(t)}$. Since the IVP states $\displaystyle y(0) = 0$, $\displaystyle 0 = A\cdot \exp^{cos(0)}$ This means that A=0. Therefore, $\displaystyle y(t) = 0$ for all $\displaystyle t < \tau$.

For $\displaystyle t > \tau$, we need a new IVP. It is found from [y] = 1 from evaluating [y] from $\displaystyle \tau^+$ to $\displaystyle \tau^-$. Expanding this, $\displaystyle y(\tau^+) - y(\tau^-) = 1 $. We found that$\displaystyle y(\tau^-)$ = 0 from above, therefore $\displaystyle y(\tau^+) = 1$. This is the new IVP for $\displaystyle t > \tau$.

The ODE is $\displaystyle y(t) = B\cdot \exp^{cos(t)} $. From the IVP, $\displaystyle 1 = B\cdot \exp^{cos(\tau)} $. This means $\displaystyle B = \exp^{-cos(\tau)}$.

Therefore, $\displaystyle y(t) = \exp^{-cos(\tau)} \cdot \exp^{cos(t)}$.

Finally, the solution to the ODE:

$\displaystyle y(t) = 0$ for all $\displaystyle t < \tau$.

$\displaystyle y(t) = \exp^{cos(t) - cos(\tau)}$ for all $\displaystyle t > \tau$

This means my Green's function is $\displaystyle \exp^{cos(t) - cos(\tau)}$

Am I right?

Thank you for your time.