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If $y(x,t)$ satisfies the 1-dimensional wave equation

$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}\quad\text{for }0\leq x \leq l$$

with boundary conditions

\begin{align} y(0,t)&=0,\\ \frac{\partial y}{\partial x}(l,t)&=0 \end{align}

and initial conditions

\begin{align} y(x,0)&=x(2l-x),\\ \frac{\partial y}{\partial t}(x,0)&=0, \end{align}

is it true that

$$y(l,t)=l^2-c^2t^2\quad\text{for }0\leq ct\leq l\text{ ?}$$

Given that the Fourier series of $x(2l-x)$ is

$$\frac{32l^2}{\pi^3}\sum_{m=0}^{\infty}\frac{\sin((2m+1)\pi x/2l)}{(2m+1)^3},$$

I've got the full solution to be

$$y(x,t)=\sum_{m=0}^{\infty}\frac{32l^2}{(2m+1)^3\pi^3}\cos\left(\frac{(2m+1)\pi ct}{2l}\right)\sin\left(\frac{(2m+1)\pi x}{2l}\right)$$

(which seems to satisfy the equation and all 4 conditions), but when I set $x=l$, the resulting series doesn't come to $l^2-c^2t^2$.

Is there a mistake in the question asking me to show $y(l,t)=l^2-c^2t^2$, or have I made a slip in the calculations?

Many thanks for any help with this!

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It is not cleat to me why do you expect $y(l,t)=l^2-c^2t^2$. – enzotib Aug 20 '12 at 18:01
That's what the question asks to be shown. – Harry Macpherson Aug 21 '12 at 7:48
Maybe you should calculate from the general solution $y(x,t)=C_1xt+C_2x+C_3t+C_4+\sum_{n=1}^\infty C_5(n)\sin\dfrac{\pi xn}{l}\sin\dfrac{\pi ctn}{l}+\sum_{n=1}^\infty C_6(n)\sin\dfrac{\pi xn}{l}\cos\dfrac{\pi ctn}{l}+\sum_{n=1}^\infty C_7(n)\cos\dfrac{\pi xn}{l}\sin\dfrac{\pi ctn}{l}+\sum_{n=1}^\infty C_8(n)\cos\dfrac{\pi xn}{l}\cos\dfrac{\pi ctn}{l}~.$ – doraemonpaul Aug 23 '12 at 0:46
I used this general solution, along with the given boundary and initial conditions, to get the series for $y(x,t)$. The only problem is that this series doesn't give the answer it's supposed to when $x=l$, although it seems to satisfy all the conditions! – Harry Macpherson Aug 26 '12 at 16:01
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Apr 21 '13 at 2:36
up vote 2 down vote accepted

Your approach is that first handle these three B.C.s and I.C.s $y(0,t)=0$ , $\dfrac{\partial y}{\partial x}(l,t)=0$ and $\dfrac{\partial y}{\partial t}(x,0)=0$ . You find that the separation constant is best to take $-\dfrac{(2m+1)^2\pi^2c^2}{4l^2}$ and you get $y(x,t)=\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}\cos\dfrac{(2m+1)\pi ct}{2l}$ .

Now you handle the I.C. $y(x,0)=x(2l-x)$ , so you have to face $\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}=x(2l-x)$ . But the following should be extremely carefull!

You are facing to find an unusual kernel inversion, so all the calculations should be start from first principle!

Luckily the method in still hold in this case.

$\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}=x(2l-x)$

$\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}=x(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}$

$\int_0^l\sum\limits_{m=0}^\infty C(m)\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}dx=\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$\sum\limits_{m=0}^\infty C(m)\int_0^l\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}dx=\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$\because\int_0^l\sin\dfrac{(2m+1)\pi x}{2l}\sin\dfrac{(2n+1)\pi x}{2l}dx$

$=\int_0^l\dfrac{1}{2}\biggl(\cos\dfrac{(m-n)\pi x}{l}-\cos\dfrac{(m+n+1)\pi x}{l}\biggr)dx$

$=\begin{cases}\biggl[\dfrac{l}{2(m-n)\pi}\sin\dfrac{(m-n)\pi x}{l}-\dfrac{l}{2(m+n+1)\pi}\sin\dfrac{(m+n+1)\pi x}{l}\biggr]_0^l&\text{when}~m\neq n~\text{and}~m+n\neq-1\\\biggl[\dfrac{x}{2}-\dfrac{l}{2(m+n+1)\pi}\sin\dfrac{(m+n+1)\pi x}{l}\biggr]_0^l&\text{when}~m=n\\\biggl[\dfrac{l}{2(m-n)\pi}\sin\dfrac{(m-n)\pi x}{l}-\dfrac{x}{2}\biggr]_0^l&\text{when}~m+n=-1\\\left[0\right]_0^l&\text{when}~m=n~\text{and}~m+n=-1\end{cases}$

$=\begin{cases}0&\text{when}~m~\text{and}~n~\text{are integers and}~m\neq n~\text{and}~m+n\neq-1\\\dfrac{l}{2}&\text{when}~m~\text{and}~n~\text{are integers and}~m=n\\-\dfrac{l}{2}&\text{when}~m~\text{and}~n~\text{are integers and}~m+n=-1\end{cases}$

$\therefore C(n)\dfrac{l}{2}=\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$C(n)=\dfrac{2}{l}\int_0^lx(2l-x)\sin\dfrac{(2n+1)\pi x}{2l}dx$

$C(m)=\dfrac{2}{l}\int_0^lx(2l-x)\sin\dfrac{(2m+1)\pi x}{2l}dx$

From the result of,

$\int_0^lx(2l-x)\sin\dfrac{(2m+1)\pi x}{2l}dx=\dfrac{16l^3}{(2m+1)^3\pi^3}$ when $m$ is integer

$\therefore y(x,t)=\sum\limits_{m=0}^\infty\dfrac{32l^2}{(2m+1)^3\pi^3}\sin\dfrac{(2m+1)\pi x}{2l}\cos\dfrac{(2m+1)\pi ct}{2l}$ for $0\leq x\leq l$

To check whether $y(l,t)=l^2-c^2t^2$ for $0\leq ct\leq l$ correct or not, note that you are difficult to simplify $\sum\limits_{m=0}^\infty\dfrac{32l^2}{(2m+1)^3\pi^3}\cos\dfrac{(2m+1)\pi ct}{2l}$ , so you should take the inverse process that consider $\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}=l^2-c^2t^2$ , determine whether $f(m)$ equals to $\dfrac{32l^2}{(2m+1)^3\pi^3}$ or not.

Note that you are facing to find an unusual kernel inversion again, so all the calculations should be again to start from first principle!

Luckily the method in still hold in this case.

$\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}=l^2-c^2t^2$

$\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}=(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}$

$\int_0^\frac{l}{c}\sum\limits_{m=0}^\infty f(m)\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}dt=\int_0^\frac{l}{c}(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$\sum\limits_{m=0}^\infty f(m)\int_0^\frac{l}{c}\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}dt=\int_0^\frac{l}{c}(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$\because\int_0^\frac{l}{c}\cos\dfrac{(2m+1)\pi ct}{2l}\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$=\int_0^\frac{l}{c}\dfrac{1}{2}\biggl(\cos\dfrac{(m-n)\pi ct}{l}+\cos\dfrac{(m+n+1)\pi ct}{l}\biggr)dt$

$=\begin{cases}\biggl[\dfrac{l}{2(m-n)\pi c}\sin\dfrac{(m-n)\pi ct}{l}+\dfrac{l}{2(m+n+1)\pi c}\sin\dfrac{(m+n+1)\pi ct}{l}\biggr]_0^\frac{l}{c}&\text{when}~m\neq n~\text{and}~m+n\neq-1\\\biggl[\dfrac{t}{2}+\dfrac{l}{2(m+n+1)\pi c}\sin\dfrac{(m+n+1)\pi ct}{l}\biggr]_0^\frac{l}{c}&\text{when}~m=n\\\biggl[\dfrac{l}{2(m-n)\pi c}\sin\dfrac{(m-n)\pi ct}{l}+\dfrac{t}{2}\biggr]_0^\frac{l}{c}&\text{when}~m+n=-1\\\left[t\right]_0^\frac{l}{c}&\text{when}~m=n~\text{and}~m+n=-1\end{cases}$

$=\begin{cases}0&\text{when}~m~\text{and}~n~\text{are integers and}~m\neq n~\text{and}~m+n\neq-1\\\dfrac{l}{2c}&\text{when}~m~\text{and}~n~\text{are integers and}~m=n\\\dfrac{l}{2c}&\text{when}~m~\text{and}~n~\text{are integers and}~m+n=-1\end{cases}$

$\therefore f(n)\dfrac{l}{2c}=\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$f(n)=\dfrac{2c}{l}\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2n+1)\pi ct}{2l}dt$

$f(m)=\dfrac{2c}{l}\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2m+1)\pi ct}{2l}dt$

From the result of,

$\int_0^\frac{l}{c}~(l^2-c^2t^2)\cos\dfrac{(2m+1)\pi ct}{2l}dt=\dfrac{16l^3}{(2m+1)^3\pi^3c}$ when $m$ is integer

$\therefore f(m)=\dfrac{32l^2}{(2m+1)^3\pi^3}=C(m)$

Hence $y(l,t)=l^2-c^2t^2$ for $0\leq ct\leq l$ is correct

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Thank you for taking the time to write such a long answer! Sorry I haven't accepted it before; I'd forgotten about this question before your last comment above! – Harry Macpherson Apr 22 '13 at 19:35

You just have to find the Fourier series of $l^2 - c^2t^2$ for $t \in [0, \frac lc)$, then compare the series with the solution as $x = l$. The actual period is $\frac{4l}c$, but you can do two extensions of $l^2 - c^2t^2$ to cover the period. Equivalently, you can just use the cosine terms that are present in the solution as the basis.

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Please let me know what is wrong with this if you downvote it. – Tunococ Sep 19 '12 at 15:28

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