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Some Differential Equation help needed |
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| Sep20-05, 09:49 PM | #1 |
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Some Differential Equation help needed
Of the Partial Kind
Using d'Alemberts soltuion for the vibrating string in one dimension Find u(1/2,3/2), when l-=1, c=1, f(x) = 0, g(x) = x(1-x) Now i tried simply substituting this into the solution that is (since f(x)=0) [tex] u(x,t) = \frac{1}{2} \int_{x-t}^{x+t} g(x) dx [/tex] but it yields the wrong answer. Does the length of the string have anything to do with the answer? Thank you in advance for your help! |
| Sep21-05, 05:46 AM | #2 |
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Recognitions:
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[tex]g_1[x]=x(1+x)[/tex] The equation for the interval [0,1] is just g(x): [tex]g_2(x)=x(1-x)[/tex] Now I wish to do that again for the interval [1,2], that is an odd extension of g(x) which would just be flipping it over into the interval [1,2]. The equation for that one would be: [tex]g_3(x)=-(x-1)+(x-1)^2[/tex] The third plot is all three. So: [tex] \begin{align*} u(1/2,3/2) &=\frac{1}{2} \int_{-1}^{2}\tilde{g_0}(\tau)d\tau \\ &= \frac{1}{2}\left(\int_{-1}^0 g_1(\tau)d\tau+\int_0^1 g_2(\tau)d\tau+\int_1^2 g_3(\tau)\tau \right) \end{align} [/tex] I get -1/12. Is that what you get? Edit: Stunner, I initially made a typo on g3 but corrected it above. Edit2: Forgot the 1/2 in front of the integral sign. Suppose that's -1/12 now. Sorry. |
| Sep21-05, 05:37 PM | #3 |
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it is quite clear that we are not being taught (or the material's presentation) correctly. I did not know how to extend the functions. I understand now... thank you very much!
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