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# vibrating string with variable density

The unidimensional wave’s problem may be stated as

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

with initial conditions

$\begin{cases}u(x,0)=f(x),\\ \frac{\partial u}{\partial t}(x,0)=g(x),\end{cases}$ |

$\begin{cases}u(0,t)=0,\\ u(l,t)=0,\end{cases}$ |

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. $\lambda<1/4$. 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 ($i=\sqrt{-1}$)

$Z(x)=X_{1}(x)+iX_{2}(x)=(1+x)^{{\frac{1}{2}(1+i\sqrt{4\lambda-1})}}=(1+x)^{{% \frac{1}{2}}}e^{{\frac{1}{2}i\sqrt{4\lambda-1}\log(1+x)}}$ $=(1+x)^{{\frac{1}{2}}}\left\{\cos\left(\sqrt{\lambda-\frac{1}{4}}\log(1+x)% \right)+i\sin\left(\sqrt{\lambda-\frac{1}{4}}\log(1+x)\right)\right\},$ 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*. ^{1}^{1}A 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* ^{2}^{2}i.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. ^{3}^{3}By GPS, again.

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## Comments

## Equations

I am fairly sure your equation is wrong. It should be \partial_x c^2 \partial_x w not \partial_x^2 w.

Energy form etc. should give you this

## Re: Equations

>I am *fairly sure* your equation is wrong. It should be \partial_x c^2 \partial_x w not \partial_x^2 w.

Fairly sure? Are you sure? As I see there neither any explanation to do so contundent claim nor I agree with your strange notation. On the contrary my notation is crystal. We're talking about the linear differential operator

L[u] = (\partial^2/\partial{t^2} - c^2(x)\partial^2/\partial{x^2})[u],

where L is applied to the transverse response u = u(x,t) which give us the *place* occuped for any particule of the vibrating string at any time.

As far as c^2(x) = T_0/\rho(x), it is interpreted as the ratio between the initial string tension and the linear variable density of the string, which is assumed an unidimensional, flexible continuum media as well as small strains, i.e. e<<1. I don't want to do my explanation so large, as this PDE appears in any book about elementary continuum mechanics. (as well as the inherent assumed assumptions)

>Energy form etc. should give you this

Energy form? What is that? Energy equation? I don't like it! I prefer use Euler-Poisson equation(a.ka. specialized Newton's Eq. for a continuum) since the little string cross section, (l>>d) enables us to disregard any shear force against the longitudinal tension of the string.

So that the problem about initial and boundary values

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

u(x,0) = f(x),

\partial{u}/\partial{t}(x,0) = g(x),

and

u(0,t) = u(l,t) = 0,

1.- It is well-posed.

2.- Enjoys existence and uniqueness.

So my dear meylan: I am fairly sure my equation *is not wrong*.

Regards,

perucho

## Re: Equations

I now understand that your problem has variable density but constant elasticity. I missed this point and was thinking of the more general string. It might be worthwhile adding this equation in as I may not be the only one who makes this mistake. Also, it might be good to point out that you have divided by density so the forcing function is dependent on this variable density.

## Re: Equations

Hi meylan,

>I now understand that your problem has variable density but constant elasticity.

-Yes, in my entry I'm discussing a perfectly alastic string. But also it is possible to study another mathematical models like for instance a Maxwellian (viscoelastic) model (I wrote a paper about this matter). Indeed all depends about the assumed *constitutive equation* which is an experimental-empirical relation or an equation developed by the rational observation about the behavior of the string's material properties. You may assume constitutive functionals so general as

T(s,t) = L(e(s,t),s),

where T(s,t) is the generic string's tension, e(s,t) the strain (deformation) function and "s" is a parameter (usually the arc length). However, such general cases imply advanced numerical methods, a lot of programmation and the adoption of finite elements, perturbations, potential boundary value or discretisation.

In my paper I could get an exact solution for a non-Hookean model under certain restrictive conditions, obtaining a PDE with variable coefficients which was reduced to two ODEs once made suitable transformations; one of which was solved by Fourier-Bessel technique, and the another one was reduced to a Sturm-Liouville problem . General solution was represented by a series given by a linear (infinite) combination of the eigenfunctions and its correspondent eigenvalues.

>Also, it might be good to point out that you have divided by density so the forcing function is dependent on this variable density.

-Well, after of some working one arrives to the PDE

\rho(x).\partial^2{u}/\partial{x^2} - T_0.\partial^2{u}/\partial{t^2} = \rho(x).f(x,t). (1)

So, what you call "forcing function" is simply dividing (1) by \rho(x), and renaming T_0/\rho(x) = c^2(x).

perucho

## Re: Equations

Thanks for this. My area is water waves and we obtain a similar equation for shallow water waves in variable depth, except that the depth term appears exactly in the way I suggested the density should for a string. This is what made me think your equations were wrong.