# Wave Equation (Fourier Coefficients)

1. Nov 23, 2012

### zack7

For the wave equation I managed to get
the coefficient of f:

a1=2

and

the coefficient of g:

$\frac{12pi}{2pi*2}$=B2

Is these answers right, since my B2 does not match the answer I was given.

Thank you

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2. Nov 23, 2012

### haruspex

Was that the right attachment? I don't see any reference to a1, B2, f or g there.

3. Nov 23, 2012

### zack7

Basically I was taught the the first U(x,0) is f and the second is g, so I go on to name the Fourier coefficient of f coefficient as a and the second as b.

4. Nov 23, 2012

### haruspex

Ok. In future, please don't assume others use the same notation.

5. Nov 23, 2012

### zack7

Using the formula ($\frac{n*pi*a}{l}$)*b2=12

where a=2, ;L=pi, and n=2,

Therefore I get three, I think I am making a mistake with my a.

6. Nov 24, 2012

### haruspex

But I don't know where you get that formula from. What equations do you have for u(x,t) and ut(x,t) before determining b2?

7. Nov 24, 2012

### zack7

That formula was given by my teacher after forming the formula for u(x,t)

An= Fourier Coefficient of f

($\frac{n*pi*a}{l}$)*b2=Fourier Coefficient of g

The u(x,t) formula is

$\sum$(An*cos$\frac{npiat}{l}$+Bn*sin$\frac{npiat}{l}$)*sin$\frac{npix}{L}$

where a infinite series is not needed for this question

8. Nov 24, 2012

### haruspex

I understand that you get it down to a1 cos(3t) sin (x) + b2 sin(6t) sin(2x), right? And we agree that the initial condition on u gives a1=2. But show me how you use the condition on ut to get b2 =3.

9. Nov 24, 2012

### zack7

Looking at the function 12sin(2x)

I get n= 2
L= pi (initial condition)
Fourier coefficient = 12
a= 2 ( previous answer (a1)

Using the formula I get
$\frac{12pi}{2pi*2}$=B2
which gives me three which is wrong since the answer should be two

10. Nov 24, 2012

### haruspex

You have u(x, t) = a1 cos(3t) sin (x) + b2 sin(6t) sin(2x)
You are given u(x, 0) = 2 sin(x), from which you deduce a1=2.
You are given ut(x, 0) = 12 sin(2x).
Show me what equation you get for ut(x, 0), and how setting that equal to 12 sin(2x) gives you b2=3.
You refer to some other equation you have been given, but I don't know exactly what that equation is or in what context you are supposed to apply it. My guess is you are misusing it.

11. Nov 25, 2012

### zack7

This was the equation that my teacher used when he found Bn , I do not know any other way to find it.

($\frac{n*pi*a}{l}$)*b2=Fourier Coefficient of g

12. Nov 25, 2012

### haruspex

Fourier coefficients are determined by plugging in the initial condition information. There is no generic formula. I would guess the formula your teacher gave you was for a certain kind of initial condition and does not apply here. Please try my way. If you can't follow it all at first, post what you can.
- Write out the equation for ut(x, t)
(You understand that this means the partial derivative of u(x,t) wrt t, right?)
- Plug in the given fact that ut(x, 0) = 12 sin(2x).
- From that, determine b2.
In view of your misunderstanding on this point, it might also be useful for you to redo the step where you determine a1. This time, instead of using any formula your teacher gave you, work it from the other initial condition you were given.

13. Nov 25, 2012

### zack7

Using this formula $\frac{2}{L}$$\int(f(x)*sin(\frac{npix}{L}$)

I obtain An= 2

and

bn= 12

what should I do next ?

Is this the way you were talking about, now I would need to find Bn

Last edited: Nov 25, 2012
14. Nov 25, 2012

### haruspex

zack, we're never going to get through this if you keep writing things like "I obtain". Show me precisely how you obtain it. Leave nothing out. If you quote a formula, explain your understanding of what all the terms in the formula mean.

15. Nov 25, 2012

### zack7

Using this formula $\frac{2}{L}$$\int(f(x)*sin(\frac{npix}{L}$)

for f(x) = 2sin(x)
n=1 since cx=x which gives me one
L=pi from the conditions of the wave equation where x goes from 0 to pi

This gives me
$\frac{2}{pi}$$\int(2sin(x)*sin(\frac{1pix}{pi}$)

$\frac{2}{pi}$$\int(2sin(x)*sin(x)$)

$\frac{2}{pi}*2$$\int(\frac{1}{2}-\frac{1}{2}cos(2x))$)

This gives me
x-sin(x)cos(x)
Inserting the limits gives me pi*$\frac{2}{pi}$
gives me 2

Therefore A1=2

I did the same way for bn by solving the integral which gives me 12

16. Nov 25, 2012

### haruspex

Let me try a different tack. You understand that the secondary boundary condition refers to ut(), not u()? How do you use that when you apply your formula? Are you using a different formula for getting b2 from that you used to get a1?

17. Nov 25, 2012

### zack7

I used the same boundary condition for the integral below (0<x<pi)
$\frac{2}{pi}$$\int(12sin(2x)*sin(\frac{2pix}{pi}$)

This gave me 12

Is this wrong ?

18. Nov 26, 2012

### haruspex

I would think so. It is not going to be exactly the same procedure for u() and for ut(). I don't know what formula to give you instead - not an expert on Fourier analysis - but I can solve it very easily from first principles. You have a generic equation for u, so you can differentiate wrt t (partially) to get ut. Then substitute in the initial condition info.

19. Nov 26, 2012

### zack7

Thank you for all the help, I got it solved, got my teacher to explain.