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Find a Formal Solution to the Vibrating String Problem

The unidimensional wave's problem may be stated as

2 u t 2 - c 2 ( x ) 2 u x 2 = f ( x , t ) , x ( 0 , l ) , t > 0 ,

with initial conditions Mathworld Planetmath

{ u ( x , 0 ) = f ( x ) , u t ( x , 0 ) = g ( x ) ,

and boundary conditions

which may be specialized to a string's motion if we physically interpret c 2 ( x ) = T 0 / ρ ( 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 ) 0 ) given by the string's problem

2 u t 2 - ( 1 + x ) 2 2 u x 2 = 0 , x ( 0 , 1 ) , t > 0 , (1)

initial conditions

{ u ( x , 0 ) = f ( x ) , initial string's form , u t ( x , 0 ) = 0 , string starts from the rest ,

and boundary conditions

{ u ( 0 , t ) = 0 , string's left end fixed , u ( 1 , t ) = 0 , string's right end fixed .

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 Mathworld Planetmath (i.e. u ( x , t ) = X ( x ) T ( t ) ) gives the equations

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

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

a ( a - 1 ) + λ = 0 . That is , a = 1 2 ( 1 ± 1 - 4 λ ) .

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

X ( x ) = ( 1 + x ) 1 2 ( 1 + 1 - 4 λ ) - ( 1 + x ) 1 2 ( 1 - 1 - 4 λ ) .

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

2 1 2 ( 1 + 1 - 4 λ ) - 2 1 2 ( 1 - 1 - 4 λ ) = 0 , or 2 1 - 4 λ = 1 .

We next study all the possible cases for the eigenvalue λ in the last above equation.

  1. 1.

    λ < 1 / 4 . Then 1 - 4 λ is real, and the equation does not have solution.

  2. 2.

    λ = 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 Mathworld Planetmath 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, λ = 1 / 4 is not an eigenvalue.

  3. 3.

    λ > 1 / 4 . Then 1 - 4 λ is imaginary. We may even get two solutions by setting ( i = - 1 )

    Z ( x ) = X 1 ( x ) + i X 2 ( x ) = ( 1 + x ) 1 2 ( 1 + i 4 λ - 1 ) = ( 1 + x ) 1 2 e 1 2 i 4 λ - 1 log ( 1 + x )
    = ( 1 + x ) 1 2 { cos ( λ - 1 4 log ( 1 + x ) ) + i sin ( λ - 1 4 log ( 1 + x ) ) } ,

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

    X ( x ) = ( 1 + x ) 1 2 sin ( λ - 1 4 log ( 1 + x ) ) ,

    then the boundary condition X ( 1 ) = 0 gives

    Therefore, λ - 1 / 4 log 2 must be an integral multiple Mathworld Planetmath of π , i.e. λ - 1 / 4 log 2 = n π , or

    λ λ n = n 2 π 2 log 2 2 + 1 4 , n + .

    To these eigenvalues correspond the eigenfunctions

    X n ( x ) = ( 1 + x ) 1 2 sin ( n π log ( 1 + x ) log 2 ) ,

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

where

c n = 0 1 f ( x ) ( 1 + x ) - 3 2 sin ( n π log ( 1 + x ) log 2 ) 𝑑 x 0 1 ( 1 + x ) - 1 sin 2 ( n π log ( 1 + x ) log 2 ) 𝑑 x = 2 log 2 0 1 f ( x ) ( 1 + x ) - 3 2 sin ( n π log ( 1 + x ) log 2 ) 𝑑 x .

The completeness above mentioned and the Fourier series converges absolutely and uniformly to f ( x ) in ( 0 , 1 ) , only if f ( x ) 𝒫 𝒞 1 ( 0 , 1 ) , f ( 0 ) = f ( 1 ) = 0 and 0 1 f ( x ) 2 d x is finite. 1 1A result due to Green-Parseval-Schwarz (GPS) and Bessel's inequality Mathworld Planetmath .
On the other hand, for satisfying (3) and its initial conditions, we choose the eigenfunction

Thus, a solution of (1) is given by

u n ( x , t ) = X n ( x ) T n ( t ) = ( 1 + x ) 1 2 sin ( n π log ( 1 + x ) log 2 ) cos λ n t ,

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

u ( x , t ) n = 1 c n u n ( x , t ) = n = 1 c n X n ( x ) T n ( t ) .

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

u ( x , t ) = n = 1 c n ( 1 + x ) 1 2 sin ( n π log ( 1 + x ) log 2 ) cos ( n 2 π 2 log 2 2 + 1 4 t ) . (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 Mathworld Planetmath derivatives and the partial differential equation Mathworld Planetmath (1) to be satisfied, we need suppose that the series for f ′′ ( x ) converges uniformly, i.e. we must suppose that f ( x ) to be regular 2 2i.e. f ( x ) 𝒞 2 ( 0 , 1 ) . , that f ( 0 ) = f ( 1 ) = f ′′ ( 0 ) = f ′′ ( 1 ) = 0 , and that 0 1 f ′′′ ( x ) 2 d x to be finite. 3 3By GPS, again .

Find a Formal Solution to the Vibrating String Problem

Source: https://planetmath.org/vibratingstringwithvariabledensity