Solving Laplace Transforms w/ Green Functions: A Guide

In summary: If we solve for A and B, we get A=\frac{1}{\omega} sin(\omega t') and B=-\frac{1}{\omega} cos(\omega t').
  • #1
mateomy
307
0
Solve...
[tex]
\frac{d^2}{dt^2}G(t,t') + \omega^2G(t,t') = \delta(t-t')
[/tex]
Solve (above) if G=0 and [itex]\frac{dG}{dt}=0[/itex] at t=0 to obtain:
[tex]
G(t,t')=\begin{cases}
0 & 0<t<t' \\
\frac{1}{\omega}\sin\omega(t-t') & 0<t'<t

\end{cases}
[/tex]I'm supposed to use Laplace Transforms to figure this out. (I'm going out of Boas Chapter 8, Section 12 problem 1) To be honest, I'm having a really difficult time getting my head around Green Functions so this is really pushing on me.

The rest of the problem states, "Use laplace transforms to find the inverse transform".

I don't even know where to begin.
 
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  • #2
Well, you know what a Laplace transform is, I'm assuming. Do you know the rules for evaluating the Laplace transforms of derivatives,

$$\mathcal L\left[\frac{d^nf}{dt^n}\right],$$

in terms of ##\mathcal L[f]##, for any function f(t)?

What happens if you apply the Laplace transform to your differential equation? i.e., what is

$$\mathcal L\left[ \frac{d^2}{dt^2}G(t,t') + \omega^2 G(t,t')\right]$$
in terms of the Laplace transform of ##G(t,t')##? (transforming with respect to t, not t').

Once you've found that, you know it must also be equal to the Laplace transform of the delta function, so you will then need to solve for ##\mathcal L[G]## and inverse transform to get G(t,t').
 
  • #3
Hmmm, I'm rereading the Laplace section carefully. I'll reaffirm that and then I'll try your hints. I'll come back with potential/probable issues. Thank you.
 
  • #4
Personally, I have never liked "Laplace Transform methods"- they hide too much in the machinery. Here's how I would approach this problem:

Write [itex]G_1(t, t')[/itex] for G(t, t') for 0<t< t' and [itex]G_2(t, t')[/itex] for G(t, t') for t'< t< 1.

For 0< t< t', [itex]G_1(t, t')[/itex] must satisfy [itex]d^2/dt^2G_1+ \omega^2G_1= 0[/itex] so [itex]G_1(t, t')= A cos(\omega t)+ B sin(\omega t)[/itex]. Since the initial conditions are G(0, t')= 0, dG(0, t')/dt= 0, A= B= 0 so [itex]G_1(t, t')= 0 for all 0< t< t'.

For t'< t< 1, [itex]G_2(t, t')[/itex] must satisfy the same differential equation so that [itex]G_2(t, t')= A cos(\omega t)+ B sin(\omega t)[/itex]. To find the "initial conditions" we use the facts that the Green's function is continuous at t= t' and has a unit jump discontinuity in the first derivative at t= t'.

So [itex]G_2(t', t')= Acos(\omega t')+ B sin(\omega t')= 0[/itex] and [itex]dG(t',t')/dt= -\omega A sin(\omega t')+ \omega B cos(\omega t')= 0[/itex]. That gives two equations to solve for A and B in terms of t'.
 

1. What is a Laplace Transform?

A Laplace Transform is a mathematical tool used to solve differential equations. It converts a function of time into a function of frequency, making it easier to analyze and solve complex problems.

2. What is a Green Function?

A Green Function is a mathematical function that helps to solve differential equations by taking into account the boundary conditions of the problem. It represents the response of a system to a unit impulse, and can be used to find the response to any input.

3. How do you solve Laplace Transforms using Green Functions?

To solve Laplace Transforms using Green Functions, you first need to find the Green Function for the specific problem at hand. Then, you can use the Green Function to set up an integral equation, which can then be solved to find the solution to the Laplace Transform.

4. What are the benefits of using Green Functions in Laplace Transform problem solving?

Using Green Functions in Laplace Transform problem solving allows for a more systematic and efficient approach. It takes into account the boundary conditions of the problem, making it easier to find the solution compared to other methods. It also allows for the solution to be easily generalized to different problems with similar boundary conditions.

5. Are there any limitations to using Green Functions in Laplace Transform problem solving?

One limitation of using Green Functions is that they can be difficult to find for more complex problems. They also may not exist for some problems, in which case another method would need to be used. Additionally, Green Functions can only be used for linear problems, so they may not be applicable to all situations.

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