Differential Equations

  • #1
stunner5000pt
1,455
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 I am 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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