# vibrating string with variable density

The unidimensional wave’s problem may be stated as

 $\frac{\partial^{2}u}{\partial t^{2}}-c^{2}(x)\frac{\partial^{2}u}{\partial x^{% 2}}=f(x,t),\qquad x\in(0,l),\quad t>0,$
 $\begin{cases}\,\,u(x,0)=f(x),\\ \,\,\frac{\partial u}{\partial t}(x,0)=g(x),\end{cases}$
 $\begin{cases}\,\,u(0,t)=0,\\ \,\,u(l,t)=0,\end{cases}$

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

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

initial conditions

 $\begin{cases}\,\,u(x,0)=f(x),&\textrm{initial string's form},\\ \,\,\frac{\partial u}{\partial t}(x,0)=0,&\textrm{string starts from the rest}% ,\end{cases}$

and boundary conditions

 $\begin{cases}\,\,u(0,t)=0,&\textrm{string's left end fixed},\\ \,\,u(1,t)=0,&\textrm{string's right end fixed}.\end{cases}$

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

 $a(a-1)+\lambda=0.\qquad\textrm{That is},\qquad a=\frac{1}{2}(1\pm\sqrt{1-4% \lambda}).$

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

 $2^{\frac{1}{2}(1+\sqrt{1-4\lambda})}-2^{\frac{1}{2}(1-\sqrt{1-4\lambda})}=0,% \quad\textrm{or}\quad 2^{\sqrt{1-4\lambda}}=1.$

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

1. 1.

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

2. 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. 3.

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

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

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

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

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

 $f(x)\,\,\sim\,\,\sum_{n=1}^{\infty}c_{n}X_{n}(x),$

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 $f(x)\in\mathcal{PC}^{1}(0,1)$, $f(0)=f(1)=0$ and $\int_{0}^{1}f^{\prime}\,{}^{2}(x)dx$ is finite. 11A result due to Green-Parseval-Schwarz (GPS) and Bessel’s inequality.
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

 $u(x,t)\,\,\sim\,\,\sum_{n=1}^{\infty}c_{n}u_{n}(x,t)=\sum_{n=1}^{\infty}c_{n}X% _{n}(x)T_{n}(t).$

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

 $u(x,t)=\sum_{n=1}^{\infty}c_{n}(1+x)^{\frac{1}{2}}\sin\left(n\pi\frac{\log(1+x% )}{\log 2}\right)\cos\left(\sqrt{\frac{n^{2}\pi^{2}}{\log^{2}2}+\frac{1}{4}}\;% t\right).$ (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 22i.e. $f(x)\in\mathcal{C}^{2}(0,1)$., 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. 33By GPS, again.

Title vibrating string with variable density VibratingStringWithVariableDensity 2013-03-22 17:26:42 2013-03-22 17:26:42 perucho (2192) perucho (2192) 9 perucho (2192) Example msc 35L05