PlanetPhysics/Solving the Wave Equation Due to D Bernoulli

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 $$\begin{matrix} \frac{\partial^2u}{\partial t^2} = c^2\cdot\frac{\partial^2u}{\partial x^2} \end{matrix}$$ 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 $$\begin{matrix} c^2\cdot\frac{X}{X} = \frac{T}{T}. \end{matrix}$$ 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: $$\begin{matrix} X = -\left(\frac{k}\right)^2 X,\quad T = -k^2T \end{matrix}$$ The solutions of these are, as is well known, $$\begin{matrix} \begin{cases} X = C_1\cos\frac{kx}+C_2\sin\frac{kx}\\ T = D_1\cos{kt}+D_2\sin{kt}\\ \end{cases} \end{matrix}$$ 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 $$C_2\sin\frac{kp} = 0.$$ If one had\, $$C_2 = 0$$,\, then $$X(x)$$ were identically 0 which is naturally impossible.\, So we must have $$\sin\frac{kp} = 0,$$ which implies $$\frac{kp} = n\pi \quad (n \in \mathbb{Z}_+).$$ 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 $$\begin{matrix} u(x,\,t) := \sum_{n=1}^\infty\left(A_n\cos\frac{n\pi c}{p}t+B_n\sin\frac{n\pi c}{p}t\right)\sin\frac{n\pi}{p}x \end{matrix}$$ 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: $$A_n = \frac{2}{p}\int_{0}^p\!f(x)\sin\frac{n\pi x}{p}\,dx,$$ $$B_n = \frac{2}{n\pi c}\int_{0}^p\!g(x)\sin\frac{n\pi x}{p}\,dx.$$