Looks good!
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 by imposing a jump condition on at and solving the differential equation for and . Then identify the Green's function for the IVP ODE
I was wondering whether someone could read through my working and see whether I have the correct Green's function.
Lets intergrate from times to
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 approaches 0. This makes the intergal of equal to zero. So we are left with evaluated from to . This equals 1.
To find the general solution, I need to solve the IVP for and . Start with .
The solution to the ODE is . Since the IVP states , This means that A=0. Therefore, for all .
For , we need a new IVP. It is found from [y] = 1 from evaluating [y] from to . Expanding this, . We found that = 0 from above, therefore . This is the new IVP for .
The ODE is . From the IVP, . This means .
Therefore, .
Finally, the solution to the ODE:
for all .
for all
This means my Green's function is
Am I right?
Thank you for your time.