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Differential Equations

  • #1
1,444
2
Of the partial kind

We have just finished solving the one dimensional wave equation for a vibrating string of length L using d'Alembert's solution which is

[tex] u(x,t) = \frac{1}{2} \displaystyle{[} f(x+ct) + f(x-ct) \displaystyle{]} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds [/tex]

the question is
If g(x) satisifes the conditions taht g(x) = -g(-x) and g(x) = -g(2l -x) then
[tex] \int_{x}^{x+2l} g(s) ds = 0 [/tex]
Well according to a theorem when a periodic function is integrated over a symmetric interval is zero. Also since the velocity of the wave is ct then ct = 2L

[tex] \int_{x-ct}^{x+ct} g(s) ds = 0 [/tex]
[tex] \int_{x-ct}^{x} g(s) ds + \int_{x}^{x+ct} g(s) ds = 0 [/tex]
[tex] \int_{x-2L}^{x} g(s) ds = -\int_{x}^{x+2L} g(s) ds [/tex]
but here is where i am stuck. How do i change the limit of integration over the left hand side

ALso im not even sure is if this is the correct way to go? Is it the right way? Your help and advice would be greatly appreciated!
Thank you!
 

Answers and Replies

  • #2
Physics Monkey
Science Advisor
Homework Helper
1,363
34
To convert the integral on the left hand side of your final equaiton into something more useful, try the change of variable [tex] s = s' - 2L [/tex].
 

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