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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 wave equation
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(1) |
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 boundary 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 position of the string and the initial velocity 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 partial differential equation (1) may be written
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(2) |
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 differential equations of second order:
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(3) |
The solutions of these are, as is well known,
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(4) |
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 functions satisfy the boundary conditions. Because of the linearity of (1), also their sum series
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(5) |
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: $$A_n = \frac{2}{p}\int_{0}^p\!f(x)\sin\frac{n\pi x}{p}\,dx,$$ $$B_n = \frac{2}{n\pi c}\int_{0}^p\!g(x)\sin\frac{n\pi x}{p}\,dx.$$
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- K. V¨AISÄLÄ: Matematiikka IV. Hand-out Nr. 141. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).
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