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Given the general solution of the wave equation:

$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$

Find the particular solution given that:

$\displaystyle \frac{\partial}{\partial t}u(x,0) = x$ for $0\leq x \leq 1$

First Attempt:

$\displaystyle u_t(x,0) = \sum_{n=0}^\infty b_nc\left(n+\frac{1}{2}\right)\pi\cdot \sin\left[\pi\left(n+\frac{1}{2}\right) x\right] = x $

Then we find $b_n$ as shown below:

$b_n = \frac{2}{\pi} \int_0^\pi \pi x c(n+\frac{1}{2})) \cdot sin\big(\pi(n+\frac{1}{2}))\big)dx $

which simplifies to:

$2c(n+\frac{1}{2})\Big(\Big[-x\pi (n+\frac{1}{2}) \cdot cos\big(\pi x(n+\frac{1}{2})\big)\Big]_0^\pi + \pi (n+\frac{1}{2}) \int_0^\pi cos\big( x\pi (n+\frac{1}{2})\big)\Big)$

But is this the right way to go?

Second Attempt:

$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$

let $\alpha = (n+\frac{1}{2})$

$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \cdot sin\big(ct\alpha \pi\big) \cdot sin\big(\pi\alpha x\big)$

$\displaystyle u_t(x,t) = \sum_{n=0}^\infty b_n \cdot c\pi \alpha \cdot cos\big(ct\alpha \pi\big) \cdot sin\big(\pi\alpha x\big)$

How do I calculate $b_n$ from here?

Me attempting to find $b_n$:

$\displaystyle b_n = \frac{2}{\pi} \int_0^1 \pi x c\alpha \cdot sin\big(\pi\alpha\big)dx $

goes to: $b_{n} = \frac{2}{(c \pi^{2} \alpha)} \int_{0}^{1} x \sin (\alpha \pi x) dx$

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  • $\begingroup$ This problem was asked a couple of days ago (I remember because I edited the question) though I am unable to find it now. Which probably means it was deleted. $\endgroup$ – mattos Dec 7 '16 at 16:04
  • $\begingroup$ Yes my friend asked it however there was no answer, so he deleted it. $\endgroup$ – user2250537 Dec 7 '16 at 16:05
  • $\begingroup$ Any idea? Because we are struggling. $\endgroup$ – user2250537 Dec 7 '16 at 16:07
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    $\begingroup$ If anything, your orthogonality condition gives $$b_{n} = \frac{2}{c \pi^{2} (n + 1/2)} \int_{0}^{\pi} x \sin ((n+1/2)\pi x) dx$$ You multiplied the $c(n+1/2)\pi$ instead of dividing. Then integrate by parts. It won't be pretty though. $\endgroup$ – mattos Dec 7 '16 at 16:27
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    $\begingroup$ You have to differentiate $u$ first, then find $b_{n}$, much as you have tried. And no, your limits are $0 \le x \le 1$ so you should be integrating from $0$ to $1$. $\endgroup$ – mattos Dec 7 '16 at 16:59
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$$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$$ $$\displaystyle u(x,t) = \sum_{n=0}^\infty \frac{b_n}{2}\left[ \cos\left[\pi\left(n+\frac{1}{2}\right)(x-ct) \right]-\cos\left[\pi\left(n+\frac{1}{2}\right) (x+ct)\right] \right]$$ $$u(x,t) =f(x-ct)-f(x+ct)$$ Where $$f(y)=\sum_{n=0}^\infty \frac{b_n}{2} \cos\left[\pi\left(n+\frac{1}{2}\right)(y) \right]$$ Now the boundary condition($\displaystyle \frac{\partial}{\partial t}u(x,0) = x$) gives $f(x)= -\frac{x^2}{4c}+d$ which gives $$u(x,t) =\frac{(x+ct)^2}{4c} -\frac{(x-ct)^2}{4c}$$

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  • $\begingroup$ Where did u get the boundary condition from? $\endgroup$ – user2250537 Dec 7 '16 at 19:49
  • $\begingroup$ $\displaystyle \frac{\partial}{\partial t}u(x,0) = x$ $\endgroup$ – MereMortal47 Dec 7 '16 at 20:01
  • $\begingroup$ how did u get $f(x)= -\frac{x^2}{4c}+d$ tho? $\endgroup$ – user2250537 Dec 7 '16 at 20:25
  • $\begingroup$ $\frac{\partial}{\partial t}u(x,t) = -cf'(x-ct)-cf'(x+ct)$ then $\frac{\partial}{\partial t}u(x,0) = -2cf'(x) = x$ now integrate. $\endgroup$ – MereMortal47 Dec 7 '16 at 20:54

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