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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

$$\displaystyle u(x,t) = \sum_{n=0}^\infty b_n \sin\left[ct\left(n+\frac{1}{2}\right) \pi\right]\sin\left[\pi\left(n+\frac{1}{2}\right) x\right]$$ $$\displaystyle u(x,t) = \sum_{n=0}^\infty \frac{b_n}{2}\left[ \cos\left[\pi\left(n+\frac{1}{2}\right)(x-ct) \right]-\cos\left[\pi\left(n+\frac{1}{2}\right) (x+ct)\right] \right]$$ $$u(x,t) =f(x-ct)-f(x+ct)$$ 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 $$u(x,t) =\frac{(x+ct)^2}{4c} -\frac{(x-ct)^2}{4c}$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.