MHB Solution of the Damped Wave Equation under Certain Boundary Conditions

[Dustinsfl]

$$
u_{tt} + 3u_t = u_{xx}\Rightarrow \varphi\psi'' + 3\varphi\psi' = \varphi''\psi.
$$
$$
u(0,t) = u(\pi,t) = 0
$$
$$
u(x,0) = 0\quad\text{and}\quad u_t(x,0) = 10
$$
\[\varphi(x) = A\cos kx + B\sin kx\\\]
\begin{alignat*}{3}
\psi(t) & = & C\exp\left(-\frac{3t}{2}\right)\exp\left[t\frac{\sqrt{9 - 4n^2}}{2}\right] + D\exp\left(-\frac{3t}{2}\right)\exp\left[-t\frac{\sqrt{9 - 4n^2}}{2}\right]
\end{alignat*}
The general sol would be
\begin{eqnarray}
u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\

&+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right]

\end{eqnarray}
Correct?

 

[Dustinsfl]

$$
u_{tt} + 3u_t = u_{xx}\Rightarrow \varphi\psi'' + 3\varphi\psi' = \varphi''\psi.
$$
$$
u(0,t) = u(\pi,t) = 0
$$
$$
u(x,0) = 0\quad\text{and}\quad u_t(x,0) = 10
$$
\[\varphi(x) = A\cos kx + B\sin kx\\\]
\begin{alignat*}{3}
\psi(t) & = & C\exp\left(-\frac{3t}{2}\right)\exp\left[t\frac{\sqrt{9 - 4n^2}}{2}\right] + D\exp\left(-\frac{3t}{2}\right)\exp\left[-t\frac{\sqrt{9 - 4n^2}}{2}\right]
\end{alignat*}
The general sol would be
\begin{eqnarray}
u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\

&+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right]

\end{eqnarray}
Correct?

Assuming that the gen soln is correct. Here is what I did to solve for the coefficients. Is this correct?
I haven't been able to solve for $B_1$ though. Hopefully, someone will have some insight.
Using the first boundary condition, we have
\begin{alignat*}{5}
u(x,0) & = & A_1\sin x + \sum_{n = 2}^{\infty}C_n\sin nx & = & 0\\
& \Rightarrow & \sum_{n = 2}^{\infty}C_n\sin nx & = & -A_1\sin x\\
& \Rightarrow & C_n & = & -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\\
& & & = & \frac{2A_1\sin n\pi}{\pi(n^2 - 1)}\\
& & & = & 0
\end{alignat*}
That is, $C_n = 0$.
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right] + \exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}D_n\sin nx\sin t\frac{\sqrt{4n^2 - 9}}{2}
\end{alignat*}
Again, using the first boundary condition, we have (Is it okay to use the BC twice?)
\begin{alignat*}{3}
u(x,0) & = & A_1\sin x & = & 0.
\end{alignat*}
Therefore, $A_1 = 0$ too.
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]B_1\sin x\sinh\frac{t\sqrt{5}}{2} + \exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}D_n\sin nx\sin t\frac{\sqrt{4n^2 - 9}}{2}
\end{alignat*}
Using the second boundary condition, we have
\begin{alignat*}{5}
u_t(x,0) & = & \frac{\sqrt{5}}{2}B_1\sin x + \sum_{n = 2}^{\infty}D_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10\\
& \Rightarrow & \sum_{n = 2}^{\infty}D_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10 - \frac{\sqrt{5}}{2}B_1\sin x\\
& \Rightarrow & D_n & = & \frac{4}{\pi\sqrt{4n^2 - 9}}\int_0^{\pi}\left(10 - \frac{\sqrt{5}}{2}B_1\sin x\right)\sin nxdx\\
& & & = & -\frac{40(\cos n\pi - 1)}{n\pi\sqrt{4n^2 - 9}}\\
& & & = & -\frac{40((-1)^n - 1)}{n\pi\sqrt{4n^2 - 9}}\\
& & & = & \begin{cases}
0, & \text{if n is even}\\
\frac{80}{n\pi\sqrt{4n^2 - 9}}, & \text{if n is odd}
\end{cases}
\end{alignat*}
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]B_1\sin x\sinh\frac{t\sqrt{5}}{2}\\
& + & \frac{80\exp\left[-\frac{3t}{2}\right]}{\pi}\sum_{n = 2}^{\infty}\frac{1}{(2n - 1)\sqrt{\left(n^2 -\frac{1}{2}\right)^2 - \frac{9}{16}}}\sin nx\sin t\frac{\sqrt{\left(n^2 -\frac{1}{2}\right)^2 - \frac{9}{16}}}{2}
\end{alignat*}

 

[Sudharaka]

Hi dwsmith, :)

Can you please explain how you got,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]

 

[Dustinsfl]

Hi dwsmith, :)

Can you please explain how you got,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]

Fourier coefficient

 

[Sudharaka]

Fourier coefficient

Note that when you substitute \(t=0\) in,

\begin{eqnarray} u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\ &+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right] \end{eqnarray}

you get,

\[\sum_{n = 2}^{\infty}C_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx = -A_1\sin x\]

 

[Dustinsfl]

Note that when you substitute \(t=0\) in,

\begin{eqnarray} u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\ &+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right] \end{eqnarray}

you get,

\[\sum_{n = 2}^{\infty}C_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx = -A_1\sin x\]

$0\cdot\frac{\sqrt{4n^2 - 9}}{2} =0$ and the cosine of 0 is 1

 

[Sudharaka]

$0\cdot\frac{\sqrt{4n^2 - 9}}{2} =0$ and the cosine of 0 is 1

Ah, I was confused by the lack of parenthesis. I thought that \(\cos t\) was multiplied by \(\frac{\sqrt{4n^2 - 9}}{2}\).

Anyway it is incorrect that,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]

Note that the Fourier series of the sine function is zero and hence does not converge to the function itself.

 

[Dustinsfl]

Ah, I was confused by the lack of parenthesis. I thought that \(\cos t\) was multiplied by \(\frac{\sqrt{4n^2 - 9}}{2}\).

Anyway it is incorrect that,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]

Note that the Fourier series of the sine function is zero and hence does not converge to the function itself.

I have that $C_n = 0$. Can I use the same boundary condition to obtain $A_1$? How can I obtain $B_1$?

 

[Ackbach]

Ah, I was confused by the lack of parenthesis. I thought that \(\cos t\) was multiplied by \(\frac{\sqrt{4n^2 - 9}}{2}\).

This is precisely why I always insist that students put parentheses around function arguments. Don't write so that you can be understood. Write so you can't be misunderstood.

$$ \cos \left(t \cdot \frac{ \sqrt{4n^{2}-9}}{2} \right)$$
is better than
$$ \cos t \cdot \frac{ \sqrt{4n^{2}-9}}{2}\;\text{or}\;\cos t \frac{ \sqrt{4n^{2}-9}}{2},$$
because the latter is ambiguous.

It might be a tad more typing up front, but it saves typing later.

 

[Sudharaka]

I have that $C_n = 0$. Can I use the same boundary condition to obtain $A_1$? How can I obtain $B_1$?

\(C_n =A_{1}= 0\) is a trivial solution of the equation, \(\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\). You don't need to use a boundary condition for the second time to say that. However it may not be the only solution.

 

[Dustinsfl]

\(C_n =A_{1}= 0\) is a trivial solution of the equation, \(\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\). You don't need to use a boundary condition for the second time to say that. However it may not be the only solution.

How do I find $B_1$?

 

[Sudharaka]

I think the solution you have obtained for the damped wave equation is incorrect. Refer, >>this<<.

 

[Dustinsfl]

I think the solution you have obtained for the damped wave equation is incorrect. Refer, >>this<<.

There equation doesn't account for overdamped solutions. If they let $c = 5$, the first few n terms would be over damped and they would need a summation cosh and sinh + summation of cos and sine.

In my case, I have overdamping at n = 1. So I don't need a summation. I just have one term.

 

[Dustinsfl]

I think the solution you have obtained for the damped wave equation is incorrect. Refer, >>this<<.

This linked lead me to another pdf from that school that helped with another problem though :)

 

[Dustinsfl]

\(C_n =A_{1}= 0\) is a trivial solution of the equation, \(\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\). You don't need to use a boundary condition for the second time to say that. However it may not be the only solution.

If it isn't the only solution, how can we find another solution?

 

[Dustinsfl]

I have a form for $B$ but I not to sure about it
$$
B=\frac{4 \sqrt{5} \text{Csc}[x] \left(\pi -2 i \text{ArcTanh}\left[e^{-i x}\right]+2 i \text{ArcTanh}\left[e^{i x}\right]+4 \text{Sin}[x]\right)}{\pi }
$$

 

[Dustinsfl]

I have a form for $B$ but I not to sure about it
$$
B=\frac{4 \sqrt{5} \text{Csc}[x] \left(\pi -2 i \text{ArcTanh}\left[e^{-i x}\right]+2 i \text{ArcTanh}\left[e^{i x}\right]+4 \text{Sin}[x]\right)}{\pi }
$$

When I plot my soln, I don't see any damping. Is there a mistake some where?
Using the second boundary condition, we have
\begin{alignat*}{5}
u_t(x,0) & = & \frac{\sqrt{5}}{2}B_1\sin x + \sum_{n = 2}^{\infty}D_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10\\
& \Rightarrow & \sum_{n = 1}^{\infty}d_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10\\
& \Rightarrow & d_n & = & \frac{40}{\pi\sqrt{4n^2 - 9}}\int_0^{\pi}\sin nx dx\\
& & & = & -\frac{40(\cos n\pi - 1)}{n\pi\sqrt{4n^2 - 9}}\\
& & & = & \begin{cases}
0, & \text{if n is even}\\
\frac{80}{n\pi\sqrt{4n^2 - 9}}, & \text{if n is odd}
\end{cases}
\end{alignat*}
If we peel off the $n = 1$ term now, we will have the first term which is $\frac{40}{\pi}\sin x$.
So $B_1 = \frac{16\sqrt{5}}{\pi}$
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]\frac{16\sqrt{5}}{\pi}\sin x\sinh\frac{t\sqrt{5}}{2} + \frac{80\exp\left[-\frac{3t}{2}\right]}{\pi}\sum_{n = 2}^{\infty}\frac{\sin nx\sin\left(t\frac{\sqrt{4(2n - 1)^2 - 9}}{2}\right)}{(2n - 1)\sqrt{4(2n - 1)^2 - 9}}
\end{alignat*}