Undergrad Confused about Finding the Green Function

[Summer95]

Suppose we have a differential equation with initial conditions $y_{0}=y^{\prime}_{0}=0$ and we need to solve it using a Green Function. Then we set up our differential equation with the right side "forcing function" as $\delta(t^{\prime}-t)$ (or with $t^{\prime}$ and $t$ switched I'm a little confused about that) and want to solve this for the Green Function.

So what I have been trying to do is solve it for $t<t^{\prime}$ in which case the boundary conditions at $0$ apply. Then solve it for $t>t^{\prime}$ (without the boundary conditions at 0). And apply a continuity condition at $t^{\prime}$ kind of like $G(t^{\prime},t^{\prime})=G(t^{\prime},t^{\prime})$. I know I am supposed to be able to find all of the constants (in terms of $t^{\prime}$), but there are not enough conditions on this solution to find all of them. What am I missing?

Thank you very much in advance! I would be happy to also ask in terms of a specific example but I'm afraid it would be removed.

 

[Orodruin]

or with t′t′t^{\prime} and ttt switched I'm a little confused about that

Don't be, the delta distribution is symmetric.

So what I have been trying to do is solve it for t<t′t<t′tt^{\prime} (without the boundary conditions at 0). And apply a continuity condition at t′t′t^{\prime} kind of like G(t′,t′)=G(t′,t′)G(t′,t′)=G(t′,t′)G(t^{\prime},t^{\prime})=G(t^{\prime},t^{\prime}). I know I am supposed to be able to find all of the constants (in terms of t′t′t^{\prime}), but there are not enough conditions on this solution to find all of them. What am I missing?

This depends on the problem you are trying to solve. It is not certain that you will get continuity. You need to take the derivative of your Green's function and identify which terms need to be zero. Hint: You already know your Green's function is zero for t<t', write the Green's function on the form G(t,t') = H(t-t') g(t,t'), insert it into the differential equation and identify the conditions g(t,t') has to fulfil at t=t'.

 

[Summer95]

Don't be, the delta distribution is symmetric.

Ok, good! That makes sense.

Unfortunately I am still not getting it, when I assume continuity at $t^{\prime}$ I seem to be able to get one of the constants pretty successfully. I get $G(t,t^{\prime})=C*sin\omega(t-t^{\prime})$ When I try to use your hint I just end up with identities that don't tell me anything about $g(t,t^{\prime})$.

 

[Orodruin]

Ok, good! That makes sense.

Unfortunately I am still not getting it, when I assume continuity at $t^{\prime}$ I seem to be able to get one of the constants pretty successfully. I get $G(t,t^{\prime})=C*sin\omega(t-t^{\prime})$ When I try to use your hint I just end up with identities that don't tell me anything about $g(t,t^{\prime})$.

You still have not told us which differential equation you have ... based on your Green's function, I would say you have the harmonic oscillator, but how can we know if you do not tell us?

Why don't you show us what you get when you make the ansatz $G(t,t') = H(t-t') g(t,t')$? Remember that $H' = \delta$ and that $\delta'$ is also a non-zero distribution.

 

[Summer95]

Here is a different example of one of the problems I can't solve: $y^{\prime \prime}+2y^{\prime}+y=f(t)$ with initial conditions $y_{0}=y^{\prime}_{0}=0$ and $f(t)=\begin{cases} 1 & \text{if } 0<t<a \\ 0 & \text{if } t>a \end{cases}$

$\frac {d^{2}} {dt^{2}} G(t,t^{\prime})+2\frac {d} {dt} G(t,t^{\prime})+G(t,t^{\prime})=\delta(t-t^{\prime})$ of course if $t\neq t^{\prime}$ then $\frac {d^{2}} {dt^{2}} G(t,t^{\prime})+2\frac {d} {dt} G(t,t^{\prime})+G(t,t^{\prime})=0$ which has the general solution $G(t,t^{\prime})=(At+B)e^{-t}$

If $t<t^{\prime}$ then from the initial conditions $0=Be^{-t} \rightarrow$ $B=0 \rightarrow$ and from the other initial condition $A=0$ so $G(t,t^{\prime})=0$.

If $t>t^{\prime}$ we still have $G(t,t^{\prime})=(At+B)e^{-t}=(At+B)e^{-t^{\prime}}e^{-(t-t^{\prime})}$

So I tried plugging this into the DE, integrating and then differentiating and then evaluating at $t=t^{\prime}$ from which I obtained $B=-e^{t^{\prime}}$ and then to find A I used $(At^{\prime}+B)e^{-t^{\prime}}=0$ and obtained $A=e^{-t^{\prime}}/t^{\prime}$ but when I use these and try to integrate

$\int_0^t G(t,t^{\prime})dt^{\prime}$ (for the $t<a$ case) this is cannot be evaluated so I presume have not found A and B correctly.

 

[Orodruin]

You are not following the procedure I described.

I also suggest always working with t' = 0 whenever the coefficients are constant. This will save you a lot of writing.

 

[Summer95]

Problem:

$y^{\prime \prime}+2y^{\prime}+y=f(t)$ with initial conditions $y_{0}=y′_{0}=0$ and $f(t) = 1 \text{ if } 0<t<a \text{ or } 0 \text{ if } t>a$

Here is my solution, in case it might help anyone, because I was able to find so few resources on solving these:

Think of the Greens function as the response of the system to a unit impulse at $t^{\prime}$. This means that to find the Greens function you must solve $\frac {d^{2}} {dt^{2}}g(t,t^{\prime})+2\frac {d} {dt}g(t,t^{\prime})+g(t,t^{\prime})=\delta(t,t^{\prime})$ (The Greens function is $g(t,t^{\prime})$ lower case $g$)

Take the Laplace Transform of both sides:

$p^{2}G+2pG+G=L(\delta(t,t^{\prime}))$ or $G=\frac {1} {(p+1)(p+1)}L(\delta(t,t^{\prime}))=L(te^{-t})L(\delta(t,t^{\prime}))$

Now use convolution to get $g$:

$g(t,t^{\prime})=\int_0^t (t-\tau)e^{-(t-\tau)}\delta(\tau-t^{\prime})d\tau=\begin{cases} 0 & \text{if } t<t^{\prime} \\ (t-t^{\prime})e^{-(t-t^{\prime})} & \text{if } t>t^{\prime} \end{cases}$

But we have the forcing function $f(t)$ so imagine you are adding up the responses to some combination of unit functions and

$y(t)=\int_0^\infty g(t,t^{\prime})f(t^{\prime})dt^{\prime}$ Compute for $0<t<a$ and $t>a$ separately and the DE is solved.