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For the following wave equation and initial condition where G is a constant due to gravity, how would I go about finding the steady state solution:

$\frac{∂^2u}{∂t^2}= c \frac{∂^2u}{∂x^2}+ G, \quad 0\le x \le L,\: t\gt 0 $

$u(0, t) = 0 ,\quad t > 0,$

$u(L, t) = H ,\quad t > 0, $

$u(x, 0) = 0 ,\quad 0 < x < L$

$\frac{∂u}{∂t}(x, 0) = 0 ,\quad 0 < x < L$

I'm assuming that as t tends to infinity, $\frac{∂^2u}{∂t^2} = 0$ but I've never dealt with wave equations with gravity before so I'm not sure what to do with the G term.

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  • $\begingroup$ Are the boundary conditions correct as currently stated? $\endgroup$
    – Leucippus
    Mar 31, 2015 at 1:11
  • $\begingroup$ I believe so, is there something wrong with it? $\endgroup$ Mar 31, 2015 at 2:26

2 Answers 2

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For the equation $u_{tt} = c^{2} u_{xx} + a$ with the conditions \begin{align} u(0, t) &= 0 \hspace{10mm} u(L, t) = H \\ u(x,0) &= 0 \hspace{10mm} u_{t}(x,0) = 0 \end{align} consider a solution of the form $u(x,t) = F(x) G(t) + f(x)$, where $f(x)$ is the steady-state solution, for which \begin{align} u_{xx} &= F'' G + f'' \\ u_{tt} &= F G'' \end{align} and the equation becomes \begin{align} F G'' = c^{2} F'' G + f'' + a. \end{align} Now, let $0 = c^{2} f'' + a$ in order to obtain the conditions $f(0) = 0$ and $f(L) = H$. The solution to this equation is \begin{align} f(x) = - \frac{a x^{2}}{2 c^{2}} + c_{1} x + c_{2}. \end{align} The condition $f(0) = 0$ yields $c_{2} = 0$. The condition $f(L) = H$ yields $c_{1} = H/L + aL/2c^{2}$ for the general form \begin{align} f(x) = \frac{H \, x}{L} - \frac{a \, x(x - L)}{2 \, c^{2}}. \end{align}

The remaining equations are derived from \begin{align} \frac{G''}{G} = - \mu^{2} = c^{2} \frac{F''}{F} \end{align} and are \begin{align} G'' + \mu^{2} G &= 0 \\ c^{2} F'' + \mu^{2} F &= 0 \end{align} which have solutions \begin{align} G(t) &= A_{1} \cos(\mu t) + B_{1} \sin(\mu t) \\ F(x) &= A_{2} \cos\left( \frac{\mu}{c} \, x \right) + B_{2} \sin\left( \frac{\mu}{c} \, x \right) \end{align} The boundary conditions for $F$ are $F(0) = 0$, $F(L) = 0$ which yields \begin{align} A_{2} &= 0 \\ 0 &= B_{2} \sin\left( \frac{\mu}{c} \, L \right) \end{align} The value for $\mu$ is obtained from $0 = \sin(\mu L/c)$ and is \begin{align} \mu = \frac{n \pi c}{L} \end{align} for $n$ and integer.

The conditions for $G$ are $G(0) = 0$, $G_{t}(0) = 0$, as stated in the problem, lead to the results \begin{align} A_{1} &= 0 \\ \mu B_{1} &= 0. \end{align} This yields the result for $G(t) = 0$ which says the only solution to this equation for these conditions is that of $f(x)$ which is independent of time and not an actual wave form solutions.

Bypassing the conditions provided the general solution of the form \begin{align} u(x,t) = \frac{H \, x}{L} - \frac{a \, x(x - L)}{2 \, c^{2}} + \sum_{n=1}^{\infty} \left( A_{n} \cos\left(\frac{n \pi c t}{L} \right) + B_{n} \sin\left( \frac{n \pi c t}{L} \right) \right) \, \sin\left( \frac{n \pi \, x}{L} \right) \end{align}

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  • $\begingroup$ for c1, shouldn't it be H/L+aL/2c2 instead of H/L−aL/2c2 ? $\endgroup$ Apr 1, 2015 at 21:31
  • $\begingroup$ @Tommy_Smith Yes, the resulting form for $f(x)$ is correct though. $\endgroup$
    – Leucippus
    Apr 1, 2015 at 23:21
  • $\begingroup$ And below is another part of the same question. "Now consider u(x, t) = φ(x, t) + w(x). Obtain a homogenous wave-equation for φ with homogenous boundary conditions derived from the problem for u. Similarly, derive initial conditions for φ: one of these should be non-zero! Use separation of variables to find the solution for φ, and hence u." $\endgroup$ Apr 8, 2015 at 19:08
  • $\begingroup$ How do i write my answer u(x,t) as in the form of φ(x, t) + w(x)? $\endgroup$ Apr 8, 2015 at 19:09
  • $\begingroup$ @Tommy_Smith The w(x) is the two term section and $\phi(x,t)$ would be the summation and enclosed terms. In general it is often easier to write $u(x,t) = w(x) + \phi(x,t)$. $\endgroup$
    – Leucippus
    Apr 8, 2015 at 19:14
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Steady-state generally means not depending on the time $t$, so you'd be looking at the ODE $c \dfrac{d^2 u}{dx^2} + G = 0$ with boundary conditions $u(0)=0$, $u(L)=H$. It's easy to solve...

However, you should be aware that in this case there is no dissipation, so the "transient" solutions don't die away as $t \to \infty$.

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  • $\begingroup$ And below is another part of the same question. "Now consider u(x, t) = φ(x, t) + w(x). Obtain a homogenous wave-equation for φ with homogenous boundary conditions derived from the problem for u. Similarly, derive initial conditions for φ: one of these should be non-zero! Use separation of variables to find the solution for φ, and hence u." $\endgroup$ Apr 9, 2015 at 17:17
  • $\begingroup$ For the first part, Obtain a homogenous wave-equation for φ with homogenous boundary conditions derived from the problem for u, do I simply start with φ (0,t)=0 and φ (L,t)=H? $\endgroup$ Apr 9, 2015 at 17:18
  • $\begingroup$ $\varphi(x,t)= u(x,t) - w(x)$, where $w(x)$ is what I called $u(x)$ above, while $u(x,t)$ is supposed to satisfy the original problem. What does this say about $\frac{\partial \varphi}{\partial t} - c\frac{\partial^2 \varphi}{\partial x^2}$, $\varphi(0,t)$, $\varphi(L,t)$, $\varphi(x,0)$ and $\frac{\partial \varphi}{\partial t}(x,0)$? $\endgroup$ Apr 9, 2015 at 18:03

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