Green's Function: Solving a Differential Equation with a Green's Function

[WWCY]

Homework Statement

I need to solve the following D.E for $\phi(x,t)$
$$[\frac{\partial}{\partial t} - D \frac{\partial ^2}{\partial x^2}]\phi (x,t) = f(x,t)$$
with the help of the following DE with a Green's function
$$[\frac{\partial}{\partial t} - D \frac{\partial ^2}{\partial x^2}]G (x-x',t-t') = \delta (x-x') \delta(t-t')$$

I'm hoping that someone could look through a part of the working and comment on whether it makes sense, and also help out with a conceptual problem.

Thanks in advance!

Homework Equations

The Attempt at a Solution

$\phi$ and $G$ are related as such,
$$\phi (x,t) = \int_{\infty}dx' \int_{\infty}dt' G(x-x';t-t') f(x',t')$$
Taking the spatial Fourier transform of the differential equation of G gives,
$$[\frac{\partial}{\partial t} + Dk^2]\tilde{G} (k, t-t') = \delta(t-t')$$
Since this is a Linear Differential Equation in $t$, I can use the integrating factor method to obtain
$$Ae^{Dk^2 t}\tilde{G} = \int_{-\infty}^{t'} Ae^{Dk^2 t} \delta(t-t')dt + \int_{t'}^{\infty} Ae^{Dk^2 t} \delta(t-t')dt$$
which leads to,
$$\tilde{G}(k, t<t') = 0$$ and $$ \tilde{G}(k, t\geq t') = e^{-Dk^2 (t-t')} $$
Does the above make sense?

I also know of another way to solve the DE, though I don't know how to implement it.

Apparently carrying out the integral $\int_{t'-\epsilon}^{t'+\epsilon} dt$ on the DE for $\tilde{G}$ will yield the same result by showing the discontinuity of $\tilde{G}$ at $t'$. I can't really see how to do this. Could someone show me the steps?

Thanks!

 

[Orodruin]

That $\tilde G$ is zero for times $t <t’$ is not something you can deduce from the differential equation, it is something you must impose as a boundary condition.

I suggest that you use this boundary condition to write $\tilde G = \theta(t-t’) g(t-t’)$ where $g$ is some function. Insertion into the differential equation and identification of terms will give you the differential equation for $g$ and its initial condition at $t=t’$.

 

[WWCY]

Thank you for assisting!

I suggest that you use this boundary condition to write $\tilde G = \theta(t-t’) g(t-t’)$ where $g$ is some function. Insertion into the differential equation and identification of terms will give you the differential equation for $g$ and its initial condition at $t=t’$.

I'm afraid I don't quite understand what this means. Do you mind elaborating?

Apparently carrying out the integral $\int_{t'-\epsilon}^{t'+\epsilon} dt$ will yield the same result by showing the discontinuity of $\tilde{G}$ at $t'$. I can't really see how to do this

Also, do you mind explaining how this can be used to solve the problem as well? This was left as an explicit hint by a lecturer, and I can't help but think that there's something important to be learned from this.

Thank you for your patience.

 

[Orodruin]

I'm afraid I don't quite understand what this means. Do you mind elaborating?

$\theta$ is the Heaviside function. It has the property that $\partial_t\theta(t-t') = \delta(t-t')$ and it is zero for $t < t'$. This means that your function $\tilde G$, since it is zero for $t < t'$ can be written on the given form. Insert the ansatz into the differential equation and tell me what you get.

Also, do you mind explaining how this can be used to solve the problem as well? This was left as an explicit hint by a lecturer, and I can't help but think that there's something important to be learned from this.

Yes, in my opinion, the "integration between $t-\epsilon$ and $t + \epsilon$" sweeps what is really going on (i.e., a differential equation for distributions) under the carpet. I know it is a very popular approach, but I think it is far more intuitive to take the approach above.

 

[WWCY]

Hi @Orodruin , I gave it a shot:

$$[\partial _t + Dk^2]\theta g = \delta$$
which gives,
$$\theta \frac{\partial g}{\partial t} + g\delta + Dk^2 \theta g = \delta$$
For $t>t'$,
$$\frac{d g}{dt} + (\delta + Dk^2)g = \delta$$,
Making use of the integrating factor,
$$e^{\int \delta dt + \int Dk^2 dt} = Ae^{Dk^2 t}$$
I then computed the following integral,
$$Ae^{Dk^2 t}g = \int_{\infty}\delta Ae^{Dk^2 t}$$
and got
$$G = \theta (t - t') e^{-Dk^2(t-t')}$$

Does this look ok?

Also, I have a few things I'd like to clarify:
1) Isn't it slightly "early" to introduce the Heaviside? I have always assumed that the rationale for a Green's to be 0 for all times $< t'$ was because it would be unbounded as $t$ tends to $-\infty$. My initial assumption was that I had to somehow show that a function of type $$G = Ae^{-Dk^2(t-t')}$$ was a solution, and then show said unboundedness before setting $A = 0$ for $t<t'$.

2) Am I allowed to perform an indefinite integral on a Dirac Delta?

Thanks for your continued patience, it is greatly appreciated.

 

[Orodruin]

1) The physical reason for having a Green’s function that is zero for $t < t’$ is that you are looking for a retarded Green’s function, i.e., one that will tell you what happens after the source is applied given a set of initial conditions. If it was non-zero for $t < t’$ a source would give a contribution before it was introduced. This would be fine if you were looking for a solution for earlier times given some known final state, but this is usually not the case.

In your solution, you should not refer directly to special cases such as ”for t<t’” apart from getting the differential equation for g. (Also note that if you do, the delta function would be zero for those values.) The point is to make an argument based on what differential equation g has to satisfy and what initial condition it has to satisfy (by identifying the coefficient in front if the delta on the LHS with that on the RHS). Nowhere do you have to do indefinite integrals of the delta.

 

[WWCY]

1) The physical reason for having a Green’s function that is zero for $t < t’$ is that you are looking for a retarded Green’s function, i.e., one that will tell you what happens after the source is applied given a set of initial conditions. If it was non-zero for $t < t’$ a source would give a contribution before it was introduced. This would be fine if you were looking for a solution for earlier times given some known final state, but this is usually not the case.

So would I be right to say that $G$ and its transform must be equal to 0 before $t'$ due to the fact that there is no force on the system up till the time $t = t'$?
Also, does this mean that a single Delta driving force centered at $t'$ immediately means that $G = 0$ for times $<t'$?

In your solution, you should not refer directly to special cases such as ”for t<t’” apart from getting the differential equation for g. (Also note that if you do, the delta function would be zero for those values.) The point is to make an argument based on what differential equation g has to satisfy and what initial condition it has to satisfy (by identifying the coefficient in front if the delta on the LHS with that on the RHS). Nowhere do you have to do indefinite integrals of the delta.

Could you elaborate on this? I'm not sure I understand the part on special cases and identifying the coefficients of the delta on the LHS and RHS

Also, I did perform an indefinite integral (line 4 of my solution) to get my integrating factor, was this wrong?

Thanks for your continued patience!

 

[Orodruin]

So would I be right to say that $G$ and its transform must be equal to 0 before $t'$ due to the fact that there is no force on the system up till the time $t = t'$?
Also, does this mean that a single Delta driving force centered at $t'$ immediately means that $G = 0$ for times $<t'$?

The Green's function is the response of the system to a delta source (it is the heat equation, not an oscillator) at time $t'$. If you are solving a system with initial conditions, you want the retarded Green's function which is homogeneous for times smaller than $t'$ so that you do not mess up the initial conditions.

Could you elaborate on this? I'm not sure I understand the part on special cases and identifying the coefficients of the delta on the LHS and RHS

Also, I did perform an indefinite integral (line 4 of my solution) to get my integrating factor, was this wrong?

Thanks for your continued patience!

The line before was wrong. For $t > t'$, the delta is zero. With the ansatz inserted into the differential equation, you should have
$$
\delta(t-t') g(t-t') + \theta(t-t')[g'(t-t') + Dk^2 g(t-t')] = \delta(t-t') g(0) + \theta(t-t')[g'(t-t') + Dk^2 g(t-t')] = \delta(t-t').
$$
Identifying the coefficients in front of the delta on both sides gives $g(0) = 1$ and the Heaviside term gives $g'(s) + Dk^2 g(s) = 0$ for $s > 0$. This is an ordinary differential equation for $g(s)$ in the domain $s > 0$ and a sufficient number of initial conditions to solve it uniquely.

 

[WWCY]

Thank you for your help @Orodruin , I believe I see it now!

For the sake of completeness though, do you mind guiding me through the steps and logic involved in using the $t - \epsilon , t + \epsilon$ integration?