Wave equation with initial and boundary conditions - is this function right?

Question

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!

Answer 1

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 http://tutorial.math.lamar.edu/Classes/DE/FourierSineSeries.aspx 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 http://integrals.wolfram.com/index.jsp?expr=x%282l-x%29sin%28%28%282m%2B1%29pi+x%29%2F%282l%29%29&random=false,

$\int_0^lx(2l-x)\sin\dfrac{(2m+1)\pi x}{2l}dx=\dfrac{16l^3}{(2m+1)^3\pi^3}$ 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 http://tutorial.math.lamar.edu/Classes/DE/FourierCosineSeries.aspx 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 http://integrals.wolfram.com/index.jsp?expr=%28l%5E2-c%5E2x%5E2%29cos%28%28%282m%2B1%29pi+cx%29%2F%282l%29%29&random=false,

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

Answer 2

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.