Solving Laplace Transforms w/ Green Functions

[mateomy]

Solve...
<br /> \frac{d^2}{dt^2}G(t,t&#039;) + \omega^2G(t,t&#039;) = \delta(t-t&#039;)<br />
Solve (above) if G=0 and \frac{dG}{dt}=0 at t=0 to obtain:
<br /> G(t,t&#039;)=\begin{cases}<br /> 0 &amp; 0&lt;t&lt;t&#039; \\<br /> \frac{1}{\omega}\sin\omega(t-t&#039;) &amp; 0&lt;t&#039;&lt;t<br /> <br /> \end{cases}<br />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.

 

[Mute]

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').

 

[mateomy]

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.

 

[HallsofIvy]

Personally, I have never liked "Laplace Transform methods"- they hide too much in the machinery. Here's how I would approach this problem:

Write G_1(t, t&#039;) for G(t, t') for 0<t< t' and G_2(t, t&#039;) for G(t, t') for t'< t< 1.

For 0< t< t', G_1(t, t&#039;) must satisfy d^2/dt^2G_1+ \omega^2G_1= 0 so G_1(t, t&#039;)= A cos(\omega t)+ B sin(\omega t). Since the initial conditions are G(0, t')= 0, dG(0, t')/dt= 0, A= B= 0 so G_1(t, t&#039;)= 0 for all 0&lt; t&lt; t&#039;.<br /> <br /> For t&#039;&lt; t&lt; 1, G_2(t, t&amp;#039;) must satisfy the same differential equation so that G_2(t, t&amp;#039;)= A cos(\omega t)+ B sin(\omega t). To find the &quot;initial conditions&quot; we use the facts that the Green&#039;s function is continuous at t= t&#039; and has a unit jump discontinuity in the first derivative at t= t&#039;.<br /> <br /> So G_2(t&amp;#039;, t&amp;#039;)= Acos(\omega t&amp;#039;)+ B sin(\omega t&amp;#039;)= 0 and dG(t&amp;#039;,t&amp;#039;)/dt= -\omega A sin(\omega t&amp;#039;)+ \omega B cos(\omega t&amp;#039;)= 0. That gives two equations to solve for A and B in terms of t&#039;.