Wave Equation when initial displacement $f(x)=\sin{(mx)}, m=1,2,...$

Question

I am studying 1D Wave Equation $u_{tt}=c^2 u_{xx}$ with B.C. $u(0,t)=0,u(L,t)=0$ and I.C. $u(x,0)=f(x), \frac{\partial u}{\partial t}|_{t=0}=g(x) $. The question is what is $u(x,t)$ when $L=\pi, c=1,g(x)=0 \ and \ f(x)=0.01\sin{3x}$. Since f(x) in this case is sine function which is an odd function of cycle $2\pi$, I can use d'Alambert's solution formula and answer is $u(x,t)=0.01(\sin{3x} \cos{3t})$. However, since,in this case, $$ u(x,t)=\frac{1}{2}\sum_{n=1}^{\infty} B_n \cos{(nt)}\sin{(nx)}$$ when I try to use the formula of Fourier coefficient $$ Bn=\frac{2}{\pi} \int_0^{\pi} f(x) \sin{(nx)} dx $$ $$ =0.01 \frac{2}{\pi} \int_0^{\pi} \sin{(3x)} \sin{(nx)} dx $$ $$ =0.01 \frac{2}{\pi} \int_0^{\pi} \frac{1}{2} [\cos{(3-n)}x+\cos{(3+n)}x ] dx $$ $$=\frac{0.01}{\pi} [\frac{\sin{(3-n)}\pi}{(3-n)}+\frac{\sin{(3+n)}\pi}{(3+n)}]=0 \ (n=1,2...)$$ $B_n$ seems to me always gives 0, which is obvious solution. My observation is that in case where f(x) is a sine function of $\sin{(mx)}, m=1,2...,$ Fourier coefficient cannot obtained. Please kindly check if my observation is correct or not. Thanks in advance.

Answer 1

I have solved this problem by d'Alambert's solution formula approach as follows. From the given condition, $$u(x,t)=\sum_{n=1}^{\infty} B_n \cos{nt} \sin{nt} $$ $$=\sum_{n=1}^{\infty} Bn \frac{1}{2} [ \sin{(n(x-t))}+\sin{(n(x+t))}] \tag{1}$$ Noting $f(x)=u(x,0)=\sum_{n=1}^{\infty} B_n \sin{nx} $, and rewite x to x-t and x+t , $$ f(x-t)=\sum_{n=1}^{\infty} B_n\sin{n(x-t)} \tag{2}$$ $$ f(x+t)=\sum_{n=1}^{\infty} B_n \sin{n(x+t)} \tag{3}$$ Inserting (2) and (3) into (1)above,$$u(x,t)=\frac{1}{2} [ f(x-t)+f(x+t)] \tag{4}$$ In case $f(x)=0.01 \sin{3x}$ $$f(x-t)=0.01[\sin{3(x-t)}] \tag{5} $$ $$f(x+t)=0.01[\sin{3(x+t)}] \tag{6} $$inserting (5)(6) into (4) above, $$ u(x,t)=\frac{0.01}{2}[\sin{3(x-t)}+\sin{3(x+t)}] $$ $$= \frac{0.01}{2}[2\sin{\frac{6x}{2}}\cos{\frac{-6t}{2}} ]$$ $$=0.01 (\sin{3x}\cos{3t})$$ This way I avoid computation of $Bn$.

Answer 2

Turns out that the solution can be expressed as a Fourier sum. Here's how.

Lets look at the problem:

$$ \bar{u} _{\bar{t}\bar{t}} = c^2 \bar{u}_{\bar{x}\bar{x}} \tag 1$$

$$ \bar{u}(0, \bar{x}) = \bar{A}\sin(\bar{w}\bar{x}) \tag 2$$

$$ \bar{u}_\bar{t}(0, \bar{x}) = 0 \tag 3$$

$$ \bar{u}(\bar{t}, 0) = 0 \tag 4$$

$$ \bar{u}(\bar{t}, L) = 0 \tag 5$$

First, we scale our problem. Let $\bar{u} = Lu$, $\bar{x} = Lx$, $\bar{t} = Lt/c$, $\bar{A}/L = A$ and $\bar{w} L = w$. Now we have:

$$ u_{tt} = u_{xx} \tag 6 $$

$$ u(0, x) = A \sin (wx) \tag 7 $$

$$ u_{t}(0, x) = 0 \tag 8 $$

$$ u(t, 0) = 0 \tag 9 $$

$$ u(t, 1) = 0 \tag {10} $$

The boundary conditions are homogeneous, so we can use the separation of variables to solve the problem. Let $u(t,x)=T(t)X(x)$. Plugging this into $(6)$ gives us:

$$ \ddot{T} X=TX'' \tag {11}$$

$$ {\ddot{T} \over T} ={X''\over X} = C \tag {12}$$

where $C$ is a constant. Note that only if $C$ is strictly negative do we get non-trivial solutions to our ODE's. Let $C=-k^2$. The solution to the ODE:

$${X''\over X}=-k^2 \tag {13}$$ $$X''+k^2 X=0 \tag {14}$$

is:

$$X = B \sin (kx) + D \cos (kx) \tag {15}$$

Plugging in our BC's gives us:

$$ X = \sum_{n=0}^\infty B_n \sin(n \pi x) \tag {16}$$

where $k=n \pi$. The solution to our temporal ODE is:

$$T = \sum_{n=0}^\infty \Bigg (E_n \sin(n \pi t) + F_n cos(n \pi t) \Bigg ) \tag {17}$$

Now our solution is in the form:

$$u= \sum_{n=0}^\infty \Bigg( E_n \sin(n \pi t) + F_n \cos(n \pi t) \Bigg )B_n \sin(n \pi x) \tag{18}$$

One coefficient can be absorbed if it goes into brackets so we get:

$$u= \sum_{n=0}^\infty \Bigg( G_n sin(n \pi t) + H_n cos(n \pi t) \Bigg )\sin(n \pi x) \tag{19}$$

Now it's time to use the IC's and the orthogonality theorem. First, we solve:

$$ u(0,x) = \sum_{n=0}^\infty H_n \sin(n \pi x) =A \sin(wx) \tag{20} $$

$$ \int_0^1 \sum_{n=0}^\infty H_n \sin(n \pi x) \sin(m \pi x)dx=\int_0^1 A \sin(wx)\sin(m \pi x)dx \tag{20} $$

$$H_n = 2A\int_0^1 \sin(wx) \sin(n \pi x) dx \tag {21}$$

$$ H_n = 2A{n \pi \sin(w) \over w^2-n^2 \pi ^2} \tag {22}$$

Note that I used wolfram to solve the integral. The second IC gives us $G_n=0$. Now all you have to do is rescale and plug in your values. Let me know if there is something I need to clarify. But the final answer to your question is that the solution can be expressed in the Fourier infinite sum form.

EDIT

I rescaled the solution and plugged in the values. It seems that the coefficient really is zero because of the sin function. So, the rescaled solution is:

$$ \bar{u} = 2 \bar{A} \sum_{n=0}^{\infty}{n \pi \sin(\bar{w}L) \over \bar{w}^2L^2-n^2 \pi ^2} \sin \Bigg({n \pi \bar{x} \over L}\Bigg) \cos \Bigg({n \pi c \bar{t} \over L} \Bigg) \tag {23}$$

In your problem, $\bar{w}=3$ and $L =\pi$ which means $\sin(\bar{w}L)= \sin(3 \pi)=0$. And there you go, your coefficient $B_n$ really is equal to zero. However, that is the correct solution to the problem! :D