Green's function for an impulsive force on a string

[xago]

Homework Statement

[PLAIN]http://img836.imageshack.us/img836/2479/stepvt.png

Homework Equations

H'(t) = $\delta$(t)

The Attempt at a Solution

So far I've taken the derivatives of G(x,t) with respect to xx and tt and gotten
$G_{xx}$(x,t) = -$\frac{θ^{2}}{c}$ and
$G_{tt}$(x,t) = $θ^{2}$c

which gives $θ^{2}$c - $c^{2}$(-$\frac{θ^{2}}{c}$) = $\delta$(x)$\delta$(t)

= 2$θ^{2}$c = $\frac{dH(x)}{dx}$$\frac{dH(t)}{dt}$
where H(x), H(t) are the heaviside step functions for x and t.

I'm not sure how these are related or if I've gone about this in a completely wrong way. (Dirac functions are not my strong suit )

 

[xago]

Is it possible that θ(ct-x) does not equal (θct - θx) which I assumed when I did the derivatives, but that theta is a function of ct -x ? In which case I'm really lost

 

[I like Serena]

Hi xago!

No, θ(ct-x) does not equal (θct - θx).
The first is zero for any x and t for which ct-x=0.
The second is only zero if x=0 and t=0.

I recommend working out the derivatives using product rule and chain rule.
You can use that dθ/dx=δ(x) and dδ/dx=δ'(x).

For the final expression you'll need some properties of δ in 2 dimensions to simplify it.

 

[xago]

I don't understand how dθ/dx=δ(x)

 

[I like Serena]

I don't understand how dθ/dx=δ(x)

It's a property of the Heaviside step function.
See for instance: http://en.wikipedia.org/wiki/Heaviside_step_function

Edit: I meant what you already had as a relevant equation: H'(t) = δ(t).

 

[xago]

For $G_{tt}$(x,t) I'm getting $\frac{1}{2}$$\frac{d^{2}}{dt^{2}}$(θ)(ct-x)cθ(x+ct) + $\frac{d}{dt}$(θ)(ct-x)$\frac{d}{dt}$(θ)(x+ct)c + $\frac{1}{2}$θ(ct-x)$\frac{d^{2}}{dt^{2}}$(θ)(x+ct)c

For $G_{xx}$(x,t) I get pretty much the same thign except divied by c, $\frac{1}{c}($$\frac{1}{2}$$\frac{d^{2}}{dx^{2}}$(θ)(ct-x)cθ(x+ct) + $\frac{d}{dx}$(θ)(ct-x)$\frac{d}{dx}$(θ)(x+ct)c + $\frac{1}{2}$θ(ct-x)$\frac{d^{2}}{dx^{2}}$(θ)(x+ct)c)

When I plug them into $G_{tt}$(x,t) - $c^{2}$$G_{xx}$(x,t) I get 2c$\frac{d}{dt}$(θ)(ct-x)$\frac{d}{dx}$(θ)(x+ct)

 

[I like Serena]

Looks good!
That's also what I have.

Here's a few more properties of delta (from: http://en.wikipedia.org/wiki/Dirac_delta_function" ).

$\delta(x,y)=\delta(x)\delta(y)$

$\delta(ax)={\delta(x) \over |a|}$

$\delta(R(x,y))=\delta(x,y)$ for any rotation or reflection R.

 

[xago]

Well θ is a function of x and t right? So is take $\frac{d}{dx}$(θ)(x+ct) the same as δ(θ(x,t)) = δ(x,t) ?
or $\frac{d}{dx}$(θ)(x+ct) = δ(x)?

 

[I like Serena]

Well θ is a function of x and t right? So is take $\frac{d}{dx}$(θ)(x+ct) the same as δ(θ(x,t)) = δ(x,t) ?
or $\frac{d}{dx}$(θ)(x+ct) = δ(x)?

Not quite.
What you have is that θ'(y)=δ(y).
Applying the chain rule, you get:
$${d \over dt}(\Theta(x+ct)) = \delta(x+ct) \cdot c$$
Apparently you have already applied the chain rule, so you should just replace d/dt(θ)(x+ct) by δ(x+ct).
Edit: Note that δ(x+ct) is a one-dimensional dirac delta, while δ(x, ct) is a two-dimensional dirac delta.

 

[xago]

Right so what I'm trying to do get rid of the 2c term somehow, what I'm getting is

2cδ(x+ct)δ(ct-x) which I need to make equal to δ(x)δ(t)

 

[I like Serena]

Right so what I'm trying to do get rid of the 2c term somehow, what I'm getting is

2cδ(x+ct)δ(ct-x) which I need to make equal to δ(x)δ(t)

Yep.

 

[I like Serena]

Here's an alternative method to find it:

Let f(u,v) be an unspecified function.
Then:
$$\iint f(ct+x, ct-x) \delta(ct+x) \delta(ct-x) dxdt = \iint f(u,v) \delta(u) \delta(v) |\det(J)| dudv$$
where J is the Jacobian matrix identified by the transformation from (u,v) to (x,t).

Since this equation holds for any function f, it follows that:
$$\delta(ct+x) \delta(ct-x) = \delta(x) \delta(t) |\det(J)|$$
which leaves only the Jacobian matrix to be identified.

 

[xago]

I understand it pretty good now, thanks for your help!

 

[I like Serena]

You're welcome!