The Simple String and Poisson's Equation

Consider a simple string with a force applied to it. For such a string let

$F_Tdx$

$=$ net transverse force acting on the string segment between $x$ and $x+dx$ due to tension $T$ only.

 

$=$ vertical force component

 

$=~T \displaystyle \left.\frac{du}{dx}\right\vert _{x+dx} -T \displaystyle \left.\frac{du}{dx}\right\vert _{x}$ $\left(\begin{array}{l} \vspace{.1cm}\textrm{assuming } \frac{du}{dx}\ll 1, \te... ...1+\left(\frac{du}{dx}\right)^2}} \approx \frac{du}{dx}=\tan \end{array} \right)$

so that

$$\frac{d}{dx}T\frac{du}{dx} =\frac{\textrm{(Force~due~to~tension)}... ...0\\ <0\textrm{(downward)}&\textrm{if~string~curvature}<0 \end{array}\right.~.$$

Let $F(x)=$ applied force density $\left(\displaystyle \frac{\textrm{(Force)}}{\textrm{(length)}}\right)$ .

If the string is in equilibrium then there is no acceleration. Consequently, the total force density is zero:

$$\frac{d}{dx}T\frac{du}{dx}+F(x)=0$$

or

$$\frac{d}{dx} T\frac{du}{dx}=-F(x)$$

For constant tension one obtains

$$\boxed{ \frac{d^2 u}{dx^2}=-f(x) \quad \textrm{where} ~f(x)=\frac{F(x)}{T} ~~.}$$ (413)

This is the one-dimensional Poisson equation.

Example

Consider a cable whose linear mass density is $\rho$ and which is suspended between two horizontal points, $x=0$ and $x=L$ , in a uniform gravitational field whose acceleration is $g$ .

Figure 4.1: A cable of length $L$ suspended between two horizontal points. If its slope is small then its deviation away from the dotted horizontal is governed by Poisson's equation. If the slope is not small then the deviation is described by a catenary.

The force density on such a cable is $\rho g$ . If the tension in the cable is $T$ , then the equilibrium profile $u(x)$ is governed by

$$\frac{d^2 u}{dx^2}=\frac{\rho g}{T}~~.$$

The solution is evidently

$$u(x)=c_1 +c_2 x+\frac{1}{2}\frac{\rho g}{T}x^2 ~~,$$

where the integration constants are determined by $u(0)=0$ and $u(L)=0$ . It follows that the cable's profile away from the straight horizontal is

$$u(x)=x(x-L)\frac{\rho g}{2T}~~.$$

Exercise 43.1 (ADJOINT OF AN OPERATOR)
Find the adjoint differential operator $L^\ast$ and the space on which it acts if

(a)

$Lu = u'' + a(x) u' + b(x) u$ where

$$u(0) = u'(1)\quad \textrm{and} \quad u(1) = u'(0).$$

(b)

$Lu = -\left(p\left(x\right)u'\right)' + q(x) u$ where

$$u(0) = u(1) \quad \textrm{and} \quad u'(0) = u'(1).$$

Assume that the scalar product is

$$\langle u,v\rangle = \int^1_0 ~u~v~dx.$$

Exercise 43.2 (ADJOINT EIGENVALUE PROBLEM)
Let $L$ be a differential operator defined over that domain ${\mathcal S}$ of functions which satisfy the given homogeneous boundary condition $B_1 (u) = 0$ and $B_2 (u) = 0$ . Let $L^\ast$ be the corresponding adjoint operator defined on the domain ${\mathcal S}^\ast$ of functions which satisfy the corresponding adjoint boundary conditions, $B_1^\ast (v) = 0$ and $B_2^\ast (v) = 0$ .

Let $u\in {\mathcal S}$ be an eigenfunction of $L$ :

$$Lu = \lambda u$$

Similarly let $v\in {\mathcal S}^\ast$ be an eigenfunction of $L^\ast$ :

$$L^\ast v = \lambda' v.$$

(i)

Make a guess as to the relationship between the eigenvalues $\lambda$ of $L$ and the eigenvalues $\lambda '$ of $L^*$ and give a reason why.

(ii)

Prove: If $\lambda\ne \bar{\lambda'}$ then $\langle v,u\rangle = 0$ . i.e. An eigenfunction of $L$ corresponding to the eigenvalue $\lambda$ is orthogonal to every eigenfunction of $L^\ast$ which does not correspond to $\bar \lambda$ . Here the overline means complex conjugate, of course.

Exercise 43.3 (BESSEL OPERATORS)
Find the Green's function for the Bessel operators

(a)

$Lu(x) = \displaystyle {d\over {dx}} ~x~ {du(x)\over {dx}}$

(b)

$Lu(x) = \displaystyle {d\over {dx}} ~x~ {du(x)\over {dx}} - {n^2\over {x}} u(x)$ with $y(0)$ finite and $y(1) = 0$ ,

i.e. solve the equations $Lu = -\delta (x - \xi)$ with the given boundary conditions.

Exercise 43.4 (DIFFERENT ENDPOINT CONDITIONS)

1. Find the Green's function for the operator with
   $$L ={d^2\over {dx^2}} + \omega^2\quad \textrm{with} \begin{array}{c} u(a) = 0\\ u(b) = 0 \end{array}\quad a < b$$
   and $\omega^2$ a fixed constant. i.e. solve $Lu = -\delta (x - \xi)$ with the given boundary conditions.
2. Does this Green's function exist for all values of $\omega$ ? If NO, what are the exceptional values of $\omega$ ?
3. Having found the Green's function in part (1), suppose one wishes to find the Green's function for the same differential equation, but with different end point conditions, namely $u(a)=0$ and $u'(a)=0$ . How would one find this new Green's function with a minimal amount of work? Go ahead, find it.

Exercise 43.5 (ADJOINT FOR GENERIC ENDPOINT CONDITIONS)
Suppose that $Lu = u''$ where

$$a_1 u(0) + b_1 u' (0) + c_1 u(1) + d_1 u'(1) = 0$$

and

$$a_2 u(0) + b_2 u' (0) + c_2 u(1) + d_2 u'(1) = 0.$$

1. Find $L^\ast$ and the space on which it acts if one uses the scalar product $\langle u,v\rangle = \int^1_0 ~u~v~dx$ .
2. For what values of the constants $a_1, b_1, \dots, c_2, d_2$ is the operator self adjoint?

Lecture 29