Spreading of a pulse as it propagates in a dispersive medium

[Aguss]

Hello everyone! I am sorry but i couldn't put this into the template.
I am studying the spreading of a pulse as it propagates in a dispersive medium, from a well known book. My problem arise when i have to solve an expression.

Firstly i begin considering that a 1-dim pulse can be written as:


u(x,t) = 1/2*1/√2∏* ∫A(k)*exp(ikx-iw(k)t) dk + cc (complex conjugate)

where: k: wave number
w: angular frecuency



and then i showed that A(k) can be express in terms of the initial values of the problem, taking into account that w(k)=w(-k) (isotropic medium):

A(k) = 1/√2∏ ∫ exp(-ikx) * (u(x,0) + i/w(k) * du/dt (x,0)) dx

I considered du/dt(x,0)=0 which means that the problems involves 2 pulses with the same amplitud and velocity but oposite directions.
So A(k) takes the form:

A(k) = 1/√2∏ ∫ exp(-ikx) * u(x,0)

Now i take a Gaussian modulated oscilattion as the initial shape of the pulse:

u(x,0) = exp(-x^2/2L^2) cos(ko x)


Then the book says that we can easily reach to the expression:

A(k) = 1/√2∏ ∫ exp(-ikx) exp(-x^2/2L^2) cos (ko x) dx



= L/2 (exp(-(L^2/2) (k-ko)^2) + exp(-(L^2/2) (k+ko)^2)

How did he reach to this?? How can i solve this last integral?


Then, with the expression of A(k) into u(x,t) arise other problem. How can i solve this other integral.


Thank you very much for helping me!

 

[mark726]

Hello, thank you for sharing your problem with us. It seems like you are studying the propagation of a pulse in a dispersive medium, which can be a challenging topic. Let me try to help you with your questions.

Firstly, to reach the expression for A(k), we can use the initial shape of the pulse, u(x,0), and substitute it into the expression for A(k). This will give us:

A(k) = 1/√2∏ ∫ exp(-ikx) * (exp(-x^2/2L^2) cos(ko x)) dx

Using the trigonometric identity cos(a+b) = cos(a)cos(b) - sin(a)sin(b), we can rewrite the expression as:

A(k) = 1/√2∏ ∫ exp(-ikx) * (exp(-x^2/2L^2) * (cos(ko x - x^2/2L^2) + sin(ko x - x^2/2L^2))) dx

Since we are only interested in the real part of A(k), we can ignore the imaginary part (sin component) and focus on the real part (cos component). This will give us:

A(k) = 1/√2∏ ∫ exp(-ikx) * (exp(-x^2/2L^2) * cos(ko x - x^2/2L^2)) dx

Using the trigonometric identity cos(a-b) = cos(a)cos(b) + sin(a)sin(b), we can further rewrite the expression as:

A(k) = 1/√2∏ ∫ exp(-ikx) * (exp(-x^2/2L^2) * cos(ko x) * cos(x^2/2L^2) + exp(-x^2/2L^2) * sin(ko x) * sin(x^2/2L^2)) dx

Since the integral of sin(x^2/2L^2) and cos(x^2/2L^2) are known, we can solve this integral and get:

A(k) = 1/√2∏ ∫ (1/2 * exp(-(k-ko)^2/2) + 1/2 * exp(-(k