Author Topic: HA6-P2  (Read 859 times)

Yeming Wen

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HA6-P2
« on: October 28, 2015, 05:17:27 PM »
« Last Edit: October 29, 2015, 12:00:39 PM by Yeming Wen »

Yeming Wen

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Re: HW6-P2
« Reply #1 on: October 28, 2015, 05:19:08 PM »
Just wondering, the boundary condition in this question is $X(0)=0$ and $X'(l)+\beta X(l)=0$?

Victor Ivrii

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Re: HW6-P2
« Reply #2 on: October 29, 2015, 05:35:39 AM »
Just wondering, the boundary condition in this question is $X(0)=0$ and $X'(l)+\beta X(l)=0$?
Indeed: the same as before on the right

Emily Deibert

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Re: HW6-P2
« Reply #3 on: October 29, 2015, 11:40:47 AM »
This problem follows similar logic to the first problem.

(a) We let $\lambda = \omega^2$. We can then proceed as in the first problem (check that post if you would like to see the steps) to get that:
X(x) = A\cos(\omega{}x) + B\sin(\omega{}x)
With the derivative being:
X'(x) = -A\omega{}\sin(\omega{}x) + B\omega{}\cos(\omega{}x)

This time we have a Dirichlet condition that $X(0) = 0$. So plugging this in:
X(0) = A = 0

Therefore $A=0$. Thus the solution is:
X_n = \sin(\omega{}_nx)

We have chosen $B=1$ for convenience.

Edit: the similar tangent relation is, as you can easily show using the boundary condition on the right end, is:
\tan(\omega{}l) = -\frac{\omega{}}{\beta{}}
« Last Edit: October 29, 2015, 11:46:20 AM by Emily Deibert »

Emily Deibert

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Re: HW6-P2
« Reply #4 on: October 29, 2015, 11:55:21 AM »
(b) We let $\lambda = - \gamma^2$. We can again proceed as in the first problem to get that:
X(x) = A\cosh(\gamma{}x) + B\sinh(\gamma{}x)
With the derivative being:
X'(x) = A\gamma{}\sinh(\gamma{}x) + B\gamma{}\cosh(\gamma{}x)

Plugging in the Dirichlet condition as in part (a):
X(0) = A = 0

Therefore $A=0$. Thus the solution is:
X_n = \sinh(\gamma{}_nx)

We have chosen $B=1$ for convenience.

The relation is:
\tanh(\gamma{}l) = -\frac{\gamma{}}{\beta{}}