Green's method- linear differential operator

[binbagsss]

Homework Statement

2. Homework Equations

3. The Attempt at a Solution [/B]- So with the (from what i interpret of the notes this is needed) the same boundary conditions when time is fixed, we can relate the 'fundamental problem'- the initial condition $t=0$ given by a delta function$c_f(c,t=0)=\delta(x)$, to a more general problem with any initial condition here, for a linear operator $L$ via:
$\int c_f (x-s)g(s) ds = c(x)$ (*)where $c_f(x)$ is the solution to the fundamental problem and $g(x)=c(x,t=0)$
part a is fine but I will give my working if needed - basically just Fourier transform to turn a PDE into an ODE and Fourier transform the intial conditions also.

part b seems to be going very wrong:
- I think I am going wrong with the following conclusion from the quoted result:
Since $\int cos(nx) sin (mx) =0$ for any n, m, the non zero contribiution from the quoted result:
$F[cos(ax)](k)=\sqrt{\frac{\pi}{2}} (\delta(k-a)+\delta(k+a))$
$= \int (cos(-kx)-i sin (kx)) cos (ax) dx= \int cos (ax) (cos (kx)) dx$
$= \int e^{ikx} cos (ax) dx$
using that $\int cos(nx) sin (mx) =0$

comes from the cos.cos term only, and so since cos is even function this result is the same for the inverse Fourier transform: $F^{-1}[cos(ak)](x)=\sqrt{\frac{\pi}{2}} (\delta(x-a)+\delta(x+a))$
With this I am getting...
$c_f(x,t)=F^{-1}[\frac{1}{\sqrt{2\pi}}cos(\sqrt{D}kt)](x)=\frac{1}{2}(\delta(x-\sqrt{D}t)+\delta(x+\sqrt{D}t)$
And therefore using (*) and the intial condition for $c(x,t=0)$ I have:
$c(x,t)= \frac{1}{2} \int\limits^{a}_{-a} \delta (x-s-\sqrt{D}t)+\delta(x-s+\sqrt{D}t) ds$

which doesn't feel like it's on the right tracks...

Am I incorrect that $F^{-1}[cos(ak)](x)=\sqrt{\frac{\pi}{2}} (\delta(x-a)+\delta(x+a))$ and if so where did my reasoning go wrong?

Many thanks in advance.

 

[Orodruin]

the same boundary conditions when time is fixed

Word of warning. You need to have the same homogeneous boundary condition. Otherwise you are going to need additional terms in the solution.

Am I incorrect that F−1[cos(ak)](x)=√π2(δ(x−a)+δ(x+a))F−1[cos(ak)](x)=π2(δ(x−a)+δ(x+a)) F^{-1}[cos(ak)](x)=\sqrt{\frac{\pi}{2}} (\delta(x-a)+\delta(x+a)) and if so where did my reasoning go wrong?

This was given in the problem.

Why do you think your result is wrong? Did you try to see if it satisfies the correct PDE and initial conditions?

Also, you may want to perform the integrations.

 

[binbagsss]

Why do you think your result is wrong? Did you try to see if it satisfies the correct PDE and initial conditions?

Also, you may want to perform the integrations.

Because I thought we don't know enough to simplify this:

$c(x,t)= \frac{1}{2} \int\limits^{a}_{-a} \delta (x-s-\sqrt{D}t)+\delta(x-s+\sqrt{D}t) ds$

I know this is either equal to $1,0,$ or $2$ depending upon whether $s=x-\sqrt{D}t$ or $s=x+\sqrt{D}t$ are both in the range, not in the range, only one is within the range $s \in [-a,a]$ but I'm pretty confused, how do we know this? what is $x$ here?
Thanks

 

[Orodruin]

You have put the Heaviside functions into the integral boundaries instead of keeping them in the integrand. I think this was a mistake in terms of simplifying your expression. I suggest putting them back and performing the integral using the delta functions.

$x$ is the position at which you want to know $c(x,t)$, $t$ is the time.

 

[binbagsss]

You have put the Heaviside functions into the integral boundaries instead of keeping them in the integrand. I think this was a mistake in terms of simplifying your expression. I suggest putting them back and performing the integral using the delta functions.

.

Thank you for your reply.
Sorry I do not understand though.
So we multiply the 'fundamental solution' by $g(s)$ the initial condiiton and intergrate over V, where V is the domain the IVP is specified on - $x \in (-\infty, \infty)$ right?
So the integral over $-\infty, +\infty$ has reduced to integral over $[-a,a]$ since $g(x)$ is 1 here and 0 everywhere else?

 

[Orodruin]

Thank you for your reply.
Sorry I do not understand though.
So we multiply the 'fundamental solution' by $g(s)$ the initial condiiton and intergrate over V, where V is the domain the IVP is specified on - $x \in (-\infty, \infty)$ right?
So the integral over $-\infty, +\infty$ has reduced to integral over $[-a,a]$ since $g(x)$ is 1 here and 0 everywhere else?

Yes, but, as I said, it is much clearer if you do not use $g(x)$ to reduce the integration boundary and instead use the delta functions to actually perform the integral.

 

[binbagsss]

Yes, but, as I said, it is much clearer if you do not use $g(x)$ to reduce the integration boundary and instead use the delta functions to actually perform the integral.

So instead I am looking at: $c(x,t)= \frac{1}{2} \int\limits^{a}_{-a} \delta (x-s-\sqrt{D}t)+\delta(x-s+\sqrt{D}t) ds$ so since the range is $\infinity$ these $s=x-\sqrt{D}t$ and $s=x+\sqrt{D}t$ must both be within this range, which would give $2$ and then we have some info on the values of $x$ relating to $a$ I'm guesing...but that's not right since $g(x)$ is zero for part of this volume so I'm not using the relation between $C_f(x,t)$ and $c(x,t)$ correctly..

sorry i don't understand..

 

[Orodruin]

No, you need to keep $g$ when you extend the integration to the entire real line.

 

[binbagsss]

No, you need to keep $g$ when you extend the integration to the entire real line.

i don't understand

 

[Orodruin]

Why don't you write down the integral without using $g$ to reduce the integration domain?

 

[binbagsss]

Why don't you write down the integral without using $g$ to reduce the integration domain?

was that not my post 7 reply? this was my attempt as this, but i didn't really understand

 

[Orodruin]

No, the integral you have there still has $-a$ and $a$ as limits. I want you to write out the integral without using the function $g$ to change the limits.

 

[binbagsss]

No, the integral you have there still has $-a$ and $a$ as limits. I want you to write out the integral without using the function $g$ to change the limits.

oh that was a typo, the limits should read $\pm \infty$ or the comment I wrote after wouldn't have followed...

 

[Orodruin]

Then it is wrong since it is missing a factor $g(s)$.

 

[binbagsss]

Then it is wrong since it is missing a factor $g(s)$.

but $g(s)$ is 1?

Then it is wrong since it is missing a factor $g(s)$.

ok so i have $\int \limits^{\infty}_{\infty} (\delta (x-s-\sqrt{D}t) + \delta (x-s+\sqrt{D}t) ) g(s) ds = g(x-\sqrt{D}t)+g(x+\sqrt{D}t)$

And now $g(x)$ is only $1$ for the range $x \in [-a,a]$ and 0 everywhere else, but there's nowhere to include this information in $g(x-\sqrt{D}t)+g(x+\sqrt{D}t)$ that I can see, so I guess this is wrong...

(missed a factor of 1/2 also)

ta

 

[Orodruin]

but g(s) is 1?

Not outside of $-a \leq x \leq a$ so if you want to extend the integral to the entire real line you must include it or it will not be the same integral.

but there's nowhere to include this information in g(x−√Dt)+g(x+√Dt)g(x−Dt)+g(x+Dt)g(x-\sqrt{D}t)+g(x+\sqrt{D}t)

What do you mean there is no way to include that information? The information is right there. In fact, regardless of the shape of $g$ (you could pick any shape) the result will be two waves with exactly the same shape and half the amplitude of the initial condition - one moving to the left and one moving to the right. Of course, if you want to sketch the solution at some time, you will need to know what $g$ is. I suggest you try to figure out what the solution for $\sqrt{D} t > a$ looks like. Once you have done that you can try it for $\sqrt D t < a$.

 

[binbagsss]

Not outside of $-a \leq x \leq a$ so if you want to extend the integral to the entire real line you must include it or it will not be the same integral.

What do you mean there is no way to include that information? The information is right there. In fact, regardless of the shape of $g$ (you could pick any shape) the result will be two waves with exactly the same shape and half the amplitude of the initial condition - one moving to the left and one moving to the right. Of course, if you want to sketch the solution at some time, you will need to know what $g$ is. I suggest you try to figure out what the solution for $\sqrt{D} t > a$ looks like. Once you have done that you can try it for $\sqrt D t < a$.

I don't understand why it is $\sqrt{D} t > a$ and $\sqrt D t < a$ we want to look at.
Nor why there will always be two waves, since $g(x-\sqrt{D} t )$ is $1$ when $-a + \sqrt{D} t < x < a+ \sqrt{D} t$ and $g(x+\sqrt{D} t )$ is $1$ when $-a - \sqrt{D} t < x < a- \sqrt{D} t$ , both zero otherwise.
So only if $-a + \sqrt{D} t < x < a- \sqrt{D} t$ do both terms contribute..

 

[Orodruin]

I don't understand why it is $\sqrt{D} t > a$ and $\sqrt D t < a$ we want to look at.

Because together that will cover all of the possible cases?

Nor why there will always be two waves, since $g(x-\sqrt{D} t )$ is $1$ when $-a + \sqrt{D} t < x < a+ \sqrt{D} t$ and $g(x+\sqrt{D} t )$ is $1$ when $-a - \sqrt{D} t < x < a- \sqrt{D} t$ , both zero otherwise.
So only if $-a + \sqrt{D} t < x < a- \sqrt{D} t$ do both terms contribute..

Yes, there will always be two waves. That's physics. If you release a string with a non-trivial shape from rest, the resulting motion must be described by the superposition of one function moving left and one moving to the right. In order for the string to be at rest at the initial time, these functions must be the same.

 

[binbagsss]

Because together that will cover all of the possible cases?
Yes, there will always be two waves. That's physics. If you release a string with a non-trivial shape from rest, the resulting motion must be described by the superposition of one function moving left and one moving to the right. In order for the string to be at rest at the initial time, these functions must be the same.

Agree with this,. but both terms don't always contribute. doens't each term describe a wave?

 

[Orodruin]

Contribute to what? If you release a string from rest, then both terms must be the same or you break the initial condition that it is released from rest. Yes both terms are waves. One moving to the left and the other to the right.

 

[binbagsss]

Contribute to what? If you release a string from rest, then both terms must be the same or you break the initial condition that it is released from rest. Yes both terms are waves. One moving to the left and the other to the right.

my bottom line of post 17

 

[Orodruin]

I don't understand what the problem is. Since the waves are zero outside of a finite region, once they move away they will not contribute to the displacement. Why do you consider that a problem?

 

[vela]

I think you're not remembering that $g(x\pm vt)$ is a function of both $x$ and $t$. For a given time $t_0$, $g(x+vt_0)$ is defined for all values of $x$. In some places, it'll vanish, and in others, it'll be equal to 1. Same with $g(x-vt_0)$. If the two waves happen to be equal to 1 at the same $x$, then the waves constructively interfere at that point.