Interchanging between real part of cmplx exp and trig functions

[N00813]

Homework Statement

Given u(x,t) = sum( e^(-at/2)*cos(n*pi*x/2L) * Re[A_n*e^(i*w_n*t)+B_n*e^(-i*w_n*t)], and the boundary conditions u(-L)=u(L)=0 for all t;
du/dt = 0 for all x at t = 0;
u(x,t=0) = e^(-|x|/l)

Find A_n and B_n.

Homework Equations

N/A

The Attempt at a Solution

I have attempted to turn the Real part into coefficients of cos and sin, i.e.:
Re[A_n*e^(i*w_n*t)+B_n*e^(-i*w_n*t)] = C_n cos(w_n*t) + D_n sin (w_n*t)
then taking advantage of cos orthogonality in x to get C and D.

I can't think of how to turn C and D into A and B. So far I figured out that C is the real part of A+B, but I can't figure out how to get the imaginary parts of A and B from C and D.

I'm sure it's simple, but I just can't seem to get it.

Any help is appreciated.

 

[Dick]

Homework Statement

Given u(x,t) = sum( e^(-at/2)*cos(n*pi*x/2L) * Re[A_n*e^(i*w_n*t)+B_n*e^(-i*w_n*t)], and the boundary conditions u(-L)=u(L)=0 for all t;
du/dt = 0 for all x at t = 0;
u(x,t=0) = e^(-|x|/l)

Find A_n and B_n.

Homework Equations

N/A

The Attempt at a Solution

I have attempted to turn the Real part into coefficients of cos and sin, i.e.:
Re[A_n*e^(i*w_n*t)+B_n*e^(-i*w_n*t)] = C_n cos(w_n*t) + D_n sin (w_n*t)
then taking advantage of cos orthogonality in x to get C and D.

I can't think of how to turn C and D into A and B. So far I figured out that C is the real part of A+B, but I can't figure out how to get the imaginary part.

I'm sure it's simple, but I just can't seem to get it.

Any help is appreciated.

I don't think you need orthogonality or anything special. Just use Euler's formula. e^(i*w_n*t)=cos(w_n*t)+i*sin(w_n*t) and e^(-i*w_n*t)=cos(w_n*t)-i*sin(w_n*t). It's just D is the real part of i*A-i*B, isn't it?

 

[N00813]

I don't think you need orthogonality or anything special. Just use Euler's formula. e^(i*w_n*t)=cos(w_n*t)+i*sin(w_n*t) and e^(-i*w_n*t)=cos(w_n*t)-i*sin(w_n*t). It's just D is the real part of i*A-i*B, isn't it?

Yes, it is.
Maybe I've gone down the wrong way.

The question tells me to find A and B.

Using the non-zero initial condition given, I can find Re [A_n + B_n]. Using the zero initial condition (du/dt = 0 @ t = 0) I can find (a/2)*Re[A_n + B_n] = Re [iw_n (A_n - B_n)].

But I can't think of how to turn that into A_n and B_n by themselves.

 

[N00813]

I don't think you need orthogonality or anything special. Just use Euler's formula. e^(i*w_n*t)=cos(w_n*t)+i*sin(w_n*t) and e^(-i*w_n*t)=cos(w_n*t)-i*sin(w_n*t). It's just D is the real part of i*A-i*B, isn't it?

Hey, just had a thought:
If C = Re[A+B]
and D = Re[i(A-B)]

Then is D/i = Re[A-B]

and C + D/i Re [2A]?

This way I can find the real parts of A and B.

For the imaginary, I was thinking iC = I am [ -B -A] , D = I am [ -A + B].

Is this correct?

 

[abitslow]

You lost me as soon as you failed to close the parentheses of sum(. Then I lost interest when you failed to show ANY indication of what is being summed over.

 

[Dick]

Hey, just had a thought:
If C = Re[A+B]
and D = Re[i(A-B)]

Then is D/i = Re[A-B]

and C + D/i Re [2A]?

This way I can find the real parts of A and B.

For the imaginary, I was thinking iC = I am [ -B -A] , D = I am [ -A + B].

Is this correct?

You are thinking about it too hard. You can't really 'solve' for A and B given C and D. There are lots of choices that work. Like A=C/2+iD/2 and B=C/2-iD/2. Or A=(C/2+1)+iD/2 and B=(C/2-1)+iD/2. And many others.

 

[N00813]

You are thinking about it too hard. You can't really 'solve' for A and B given C and D. There are lots of choices that work. Like A=C/2+iD/2 and B=C/2-iD/2. Or A=(C/2+1)+iD/2 and B=(C/2-1)+iD/2. And many others.

OK.

Let's say I start again with the question. I find the expression for Re[A + B] and the expression for Re[iw_n*A-iw_n*B]. From this, you're saying it's impossible to get A and B uniquely?

 

[Dick]

OK.

Let's say I start again with the question. I find the expression for Re[A + B] and the expression for Re[iw_n*A-iw_n*B]. From this, you're saying it's impossible to get A and B uniquely?

Of course it is. I already gave you two choices that work. Taking the real part of something cancels some components. Your right side has two real parameters. The left side has two complex parameters which is the same as four real parameters. You can't determine them all. Write out A=Re(A)+i*Im(A) and B=Re(B)+iIm(b) and see how many you can determine.

 

[N00813]

Of course it is. I already gave you two choices that work. Taking the real part of something cancels some components. Your right side has two real parameters. The left side has two complex parameters which is the same as four real parameters. You can't determine them all. Write out A=Re(A)+i*Im(A) and B=Re(B)+iIm(b) and see how many you can determine.

Thanks. I suppose the question itself has holes in it.