Talk:PlanetPhysics/Solving the Wave Equation Due to D Bernoulli

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: solving the wave equation due to D. Bernoulli %%% Primary Category Code: 02.30.Jr %%% Filename: SolvingTheWaveEquationDueToDBernoulli.tex %%% Version: 1 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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A string has been strained between the points\, $(0,\,0)$\, and\, $(p,\,0)$\, of the $x$-axis.\, The transversal vibration of the string in the $xy$-plane is determined by the one-dimensional \htmladdnormallink{wave equation}{http://planetphysics.us/encyclopedia/WaveEquation.html} \begin{align} \frac{\partial^2u}{\partial t^2} = c^2\cdot\frac{\partial^2u}{\partial x^2} \end{align} satisfied by the ordinates\, $u(x,\,t)$\, of the points of the string with the abscissa $x$ on the time moment\, $t\,(\geqq 0)$. The \htmladdnormallink{boundary}{http://planetphysics.us/encyclopedia/PiecewiseLinear.html} conditions are thus $$u(0,\,t) = u(p,\,t) = 0.$$ We suppose also the initial conditions $$u(x,\,0) = f(x),\quad u_t'(x,\,0) = g(x)$$ which give the initial \htmladdnormallink{position}{http://planetphysics.us/encyclopedia/Position.html} of the string and the initial \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} of the points of the string.

For trying to separate the variables, set $$u(x,\,t) := X(x)T(t).$$ The boundary conditions are then\, $X(0) = X(p) = 0$,\, and the \htmladdnormallink{partial differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} (1) may be written \begin{align} c^2\cdot\frac{X}{X} = \frac{T}{T}. \end{align} This is not possible unless both sides are equal to a same constant $-k^2$ where $k$ is positive; we soon justify why the constant must be negative.\, Thus (2) splits into two ordinary linear \htmladdnormallink{differential equations}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} of second order: \begin{align} X = -\left(\frac{k}{c}\right)^2 X,\quad T = -k^2T \end{align} The solutions of these are, as is well known, \begin{align} \begin{cases} X = C_1\cos\frac{kx}{c}+C_2\sin\frac{kx}{c}\\ T = D_1\cos{kt}+D_2\sin{kt}\\ \end{cases} \end{align} with integration constants $C_i$ and $D_i$.

But if we had set both sides of (2) equal to\, $+k^2$, we had got the solution\, $T = D_1e^{kt}+D_2e^{-kt}$\, which can not present a vibration.\, Equally impossible would be that\, $k = 0$.

Now the boundary condition for $X(0)$ shows in (4) that\, $C_1 = 0$,\, and the one for $X(p)$ that $$C_2\sin\frac{kp}{c} = 0.$$ If one had\, $C_2 = 0$,\, then $X(x)$ were identically 0 which is naturally impossible.\, So we must have $$\sin\frac{kp}{c} = 0,$$ which implies $$\frac{kp}{c} = n\pi \quad (n \in \mathbb{Z}_+).$$ This means that the only suitable values of $k$ satisfying the equations (3), the so-called eigenvalues, are $$k = \frac{n\pi c}{p} \quad (n = 1,\,2,\,3,\,\ldots).$$ So we have infinitely many solutions of (1), the eigenfunctions $$u = XT = C_2\sin\frac{n\pi}{p}x \left[D_1\cos\frac{n\pi c}{p}t+D_2\sin\frac{n\pi c}{p}t\right]$$ or $$u = \left[A_n\cos\frac{n\pi c}{p}t+B_n\sin\frac{n\pi c}{p}t\right] \sin\frac{n\pi}{p}x$$ $(n = 1,\,2,\,3,\,\ldots)$ where $A_n$'s and $B_n$'s are for the time being arbitrary constants.\, Each of these \htmladdnormallink{functions}{http://planetphysics.us/encyclopedia/Bijective.html} satisfy the boundary conditions.\, Because of the linearity of (1), also their sum series \begin{align} u(x,\,t) := \sum_{n=1}^\infty\left(A_n\cos\frac{n\pi c}{p}t+B_n\sin\frac{n\pi c}{p}t\right)\sin\frac{n\pi}{p}x \end{align} is a solution of (1), provided it converges.\, It fulfils the boundary conditions, too.\, In order to also the initial conditions would be fulfilled, one must have $$\sum_{n=1}^\infty A_n\sin\frac{n\pi}{p}x = f(x),$$ $$\sum_{n=1}^\infty B_n\frac{n\pi c}{p}\sin\frac{n\pi}{p}x = g(x)$$ on the interval\, $[0,\,p]$.\, But the left sides of these equations are the Fourier sine series of the functions $f$ and $g$, and therefore we obtain the expressions for the coefficients: $$A_n = \frac{2}{p}\int_{0}^p\!f(x)\sin\frac{n\pi x}{p}\,dx,$$ $$B_n = \frac{2}{n\pi c}\int_{0}^p\!g(x)\sin\frac{n\pi x}{p}\,dx.$$

\begin{thebibliography}{9} \bibitem{K.V.}{\sc K. V ais al a:} {\em Matematiikka IV}.\, Hand-out Nr. 141.\quad Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967). \end{thebibliography}

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