vibrating string with variable density

The unidimensional wave’s problem may be stated as

$$\frac{{\partial }^{2}u}{\partial t^2}-c^2(x)\frac{{\partial }^{2}u}{\partial x^2}=f(x,t),x\in (0,l),t>0,$$

with initial conditions

$$\{\begin{array}{cc}u(x,0)=f(x),\hfill & \\ \frac{\partial u}{\partial t}(x,0)=g(x),\hfill & \end{array}$$

and boundary conditions

$$\{\begin{array}{cc}u(0,t)=0,\hfill & \\ u(l,t)=0,\hfill & \end{array}$$

which may be specialized to a string’s motion if we physically interpret $c^2(x)=T_0/\rho (x)$ as the ratio between the string’s initial tension and its linear density. We will discuss the free string’s vibrations (i.e. $f(x,t)\equiv 0$) given by the string’s problem

$$\frac{{\partial }^{2}u}{\partial t^2}-{(1+x)}^2\frac{{\partial }^{2}u}{\partial x^2}=0,x\in (0,1),t>0,$$ (1)

initial conditions

and boundary conditions

Without loss of generality, we assume unitary the natural undeformed string’s length. The solution of this problem approaches to a string’s motion whose linear density is proportional to ${(1+x)}^{-2}$. The method of separation of variables (i.e. $u(x,t)=X(x)T(t)$) gives the equations

$$X^{\prime \prime }+\frac{\lambda }{{(1+x)}^2}X=0,$$ (2)

with boundary conditions $X(0)=X(1)=0$, and

$$T^{\prime \prime }+\lambda T=0,$$ (3)

with initial conditions $T(0)=1$, $T^{\prime }(0)=0$. In these equations, $\lambda$ is a constant parameter.
In (2), we are dealing with a Sturm-Liouville problem. To find out the eigenvalues, one searches the solution of (2) on the form $X(x)={(1+x)}^a$, as we realize that (2) outcomes the associated characteristic equation

$$a(a-1)+\lambda =0.\mathit{\hspace{1em}\hspace{1em}}\text{That is},a=\frac{1}{2}(1\pm \sqrt{1-4\lambda }).$$

In order to satisfying $X(0)=0$, let us choose

$$X(x)={(1+x)}^{\frac{1}{2}(1+\sqrt{1-4\lambda })}-{(1+x)}^{\frac{1}{2}(1-\sqrt{1-4\lambda })}.$$

Thus, the boundary condition $X(1)=0$ becomes

$$2^{\frac{1}{2}(1+\sqrt{1-4\lambda })}-2^{\frac{1}{2}(1-\sqrt{1-4\lambda })}=0,\text{or}\mathit{\hspace{1em}}{2}^{\sqrt{1-4\lambda }}=1.$$

We next study all the possible cases for the eigenvalue $\lambda$ in the last above equation.

1. . Then $\sqrt{1-4\lambda }$ is real, and the equation does not have solution.

2. $\lambda =1/4$. Then the pair of solutions, above indicated, will not be independent. Indeed the functions ${(1+x)}^{1/2}$ and ${(1+x)}^{1/2}\log (1+x)$ are linearly independent solutions of (2), in $(0,1)$. Nevertheless, although the last one satisfies the boundary condition at $x=0$, it does not vanish at $x=1$. Hence, $\lambda =1/4$ is not an eigenvalue.

3. $\lambda >1/4$. Then $\sqrt{1-4\lambda }$ is imaginary. We may even get two solutions by setting ($i=\sqrt{-1}$)

   $$Z(x)=X_1(x)+i{X}_2(x)={(1+x)}^{\frac{1}{2}(1+i\sqrt{4\lambda -1})}={(1+x)}^{\frac{1}{2}}{e}^{\frac{1}{2}i\sqrt{4\lambda -1}\log (1+x)}$$

   $$={(1+x)}^{\frac{1}{2}}\left\{\cos \left(\sqrt{\lambda -\frac{1}{4}}\log (1+x)\right)+i\sin \left(\sqrt{\lambda -\frac{1}{4}}\log (1+x)\right)\right\},$$

   being the real and imaginary parts of $Z(x)$ two linearly independent solutions. For satisfying $X(0)=0$ one sets

   $$X(x)={(1+x)}^{\frac{1}{2}}\sin \left(\sqrt{\lambda -\frac{1}{4}}\log (1+x)\right),$$

   then the boundary condition $X(1)=0$ gives

   $$2^{\frac{1}{2}}\sin \left(\sqrt{\lambda -\frac{1}{4}}\log 2\right)=0.$$

   Therefore, $\sqrt{\lambda -1/4}\log 2$ must be an integral multiple of $\pi$, i.e. $\sqrt{\lambda -1/4}\log 2=n\pi$, or

   $$\lambda \mapsto {\lambda }_n=\frac{n^2{\pi }^2}{{\log }^{2}2}+\frac{1}{4},n\in {\mathbb{Z}}^{+}.$$

   To these eigenvalues correspond the eigenfunctions

   $$X_n(x)={(1+x)}^{\frac{1}{2}}\sin \left(n\pi \frac{\log (1+x)}{\log 2}\right),$$

and these form a complete system, whenever we impose to $f(x)$ certain conditions that we shall see later. Moreover, $f(x)$ may be expanded in Fourier series

$$f(x)\sim \sum _{n=1}^{\mathrm{\infty }}{c}_n{X}_n(x),$$

where

$$c_n=\frac{{\int }_0^{1}f(x){(1+x)}^{-\frac{3}{2}}\sin \left(n\pi \frac{\log (1+x)}{\log 2}\right)𝑑x}{{\int }_0^1{(1+x)}^{-1}{\sin }^2\left(n\pi \frac{\log (1+x)}{\log 2}\right)𝑑x}=\frac{2}{\log 2}{\int }_0^{1}f(x){(1+x)}^{-\frac{3}{2}}\sin \left(n\pi \frac{\log (1+x)}{\log 2}\right)𝑑x.$$

The completeness above mentioned and the Fourier series converges absolutely and uniformly to $f(x)$ in $(0,1)$, only if $f(x)\in 𝒫{𝒞}^1(0,1)$, $f(0)=f(1)=0$ and ${\int }_0^1{f}^{\prime }{}^2(x)dx$ is finite. .
On the other hand, for satisfying (3) and its initial conditions, we choose the eigenfunction

$$T(t)\mapsto T_n(t)=\cos \sqrt{{\lambda }_n}t.$$

Thus, a solution of (1) is given by

$$u_n(x,t)=X_n(x)T_n(t)={(1+x)}^{\frac{1}{2}}\sin \left(n\pi \frac{\log (1+x)}{\log 2}\right)\cos \sqrt{{\lambda }_n}t,$$

and the general solution of (1) may be determined as a linear (infinite) combination of these eigenfunctions, that is

$$u(x,t)\sim \sum _{n=1}^{\mathrm{\infty }}{c}_n{u}_n(x,t)=\sum _{n=1}^{\mathrm{\infty }}{c}_n{X}_n(x)T_n(t).$$

So that, the string’s problem (1) has the formal solution

$$u(x,t)=\sum _{n=1}^{\mathrm{\infty }}{c}_n{(1+x)}^{\frac{1}{2}}\sin \left(n\pi \frac{\log (1+x)}{\log 2}\right)\cos \left(\sqrt{\frac{n^2{\pi }^2}{{\log }^{2}2}+\frac{1}{4}}t\right).$$ (4)

This series converges uniformly, and hence satisfies the initial and boundary conditions, as the series for $f(x)$ converges uniformly. However, in order to assure continuous derivatives and the partial differential equation (1) to be satisfied, we need suppose that the series for $f^{\prime \prime }(x)$ converges uniformly, i.e. we must suppose that $f(x)$ to be regular , that $f(0)=f(1)=f^{\prime \prime }(0)=f^{\prime \prime }(1)=0$, and that ${\int }_0^1{f}^{\prime \prime \prime }{}^2(x)dx$ to be finite. .

Title	vibrating string with variable density
Canonical name	VibratingStringWithVariableDensity
Date of creation	2013-03-22 17:26:42
Last modified on	2013-03-22 17:26:42
Owner	perucho (2192)
Last modified by	perucho (2192)
Numerical id	9
Author	perucho (2192)
Entry type	Example
Classification	msc 35L05