Homogeneous Boundary Conditions

We can now state the S-L problem. If the endpoint conditions are of the mixed Dirichlet-Neumann type,

$$\alpha u(a)+\alpha 'u'(a) =$$ 0 (34)

$$\beta u(b)+\beta 'u'(b) = 0 ~,$$

with the $\alpha$ 's and $\beta$ 's independent of $\lambda$ , then the boxed Eq. (3.3) together with Eq.(3.4) constitute a regular Sturm-Liouville system.

If, by contrast,

$$u(a)-u(b) = 0~\quad~\textrm{and}~\quad~p(a)=p(b)$$ (35)

$$u'(a)-u'(b) =$$ 0

then Eqs.(3.3) and (3.5) constitute a periodic Sturm-Liouville system.

If $p(a)=0$ and the $1^{\textrm{st}}$ b.c. in Eq.(3.4) is dropped, then we have a singular S-L system. We shall consider the properties of these S-L systems in a subsequent section.

It is difficult to overstate the pervasiveness of these S-L systems in nature. Indeed, natural phenomena subsumed under the regular S-L problem, for example, are quite diverse. Heat conduction and the vibration of bodies illustrate the point.

A. Heat conduction in one dimension.

Consider the temperature profile $u(x)$ of a conducting bar of unit length which is

1. insulated at $x=0$ (no temperature gradient), and satisfies
2. radiative boundary condition at $x=1$

Separation of variables applied to the heat equation yields the following S-L system:

$$u'' +\lambda u =0$$ (36)

with

$$u'(0) =$$ 0 (37)

$$-u'(1) = hu(1).$$ (38)

Here $h$ is a non-negative constant. Note that at $x=1$

$$h =0 \Rightarrow~~\textrm{no~radiation}$$

$$h >0 \Rightarrow~~\textrm{finite~heat~loss~due~to~radiation ~(Newton's~law~of~ cooling)}.$$

B. Vibrations in one dimension.

Alternatively, consider a vibrating string whose transverse amplitude $u(x)$ satisfies the following homogeneous endpoint conditions:

1. At $x=0$ there is no transverse force to influence the string's motion. The tension produces only a longitudinal force component. In this circumstance the string is said to be free at $x=0$ . This free boundary condition is expressed by the statement
   $$u'(0)=0 ~~.$$

2. At $x=1$ the string is tied to a spring so that the vertical spring displacement coincides with the displacement of the string away from equilibrium. Even though the tail end of the string gets accelerated up and down, the total transverse force on it vanishes because it has no mass. Consequently,
   $$-u'(1)T - ku(1) =0 ~~,$$
   or
   $$-u'(1)=hu(1) ~~,$$
   where
   $$h=\frac{k}{T} = \frac{\textrm{spring~constant}}{\textrm{string~tension}}$$

The transverse amplitude profile of the string is governed by Eq.(3.2). For constant tension and uniform mass density this equation becomes

$$u'' +\lambda u =0$$

We see that the S-L system for the heat conduction problem, Eqs.(3.6)- (3.8), coincides with that for the vibration problem.

The task of solving this regular S-L system consists of finding all possible values of $\lambda$ for which the solution $u(x,\lambda)$ is non-trivial. Consequently, there are four distinct cases to consider:

1. $\lambda =0$ ,
2. $\lambda >0$ ,
3. $\lambda <0$ , and
4. $\lambda =$ complex.

We shall have to consider cases 1.-3. only. This is because the next subsection (3.3.3) will furnish us with some very powerful theorems about the nature of the allowed values of $\lambda$ and the corresponding non-trivial solutions $u(x,\lambda)$ .

1. $\lambda =0$ leads to $u=c_1+c_2x$
   

   $$h>0 \Rightarrow c_1=c_2=0~~\textrm{i.e.,}~u(x)=0~~~\hbox{for ~all}~~~0<x<1$$

   $$h=0 \Rightarrow u(x)=c_1$$

   
   constant solution. (What physical circumstance is expressed by $u(x)=c_1$ ?)
2. $\lambda =\alpha^2 >0$ , $\alpha >0$
   The general solution to the differential equation is
   $$u(x)=c_1\cos\alpha x+c_2\sin\alpha x ~~.$$
   Now consider the boundary conditions.
   1. Eq.(3.7) $\Rightarrow c_2=0$ . Consequently, $u(x)=c_1 \cos \alpha x ~.$
      
   2. Eq.(3.8) $\Rightarrow -\alpha c_1\sin\alpha +hc_1\cos\alpha =0$ . Consequently,

      $$\tan\alpha = \frac{h}{\alpha}\,.$$ (39)

      
      

   This transcendental equation determines the allowed values of $\alpha$ and hence of $\lambda$ . How do we find them? A very illuminating way is based on graphs. Draw the two graphs (Figure 3.3)
   $$y=\tan\alpha$$
   and
   $$y=\frac{h}{\alpha}\,.$$
   Where they intersect gives the allowed values of $\alpha$ , and hence $\lambda =\alpha ^2$ , the eigenvalues of the S-L problem.

   Figure: There are two graphs in this figure: that of $tan~\alpha$ and that of the two hyperbolas $h/\alpha$ . The intersection of these two graphs is the solution set to the transcendental eigenvalue Eq.(3.9). The $\alpha$ -values of the heavy dots are the desired solutions. Note that if $\alpha$ is a solution, then $-\alpha$ is another solution, but it yields the same eigenvalue $\lambda =\alpha ^2$ .

   
   We see that there are an infinite number of intersection points
   $$\alpha_1,~\alpha_2,~\alpha_3,\dots ,\alpha_n,\dots\,.$$
   For large $n$ we have $\alpha_n\simeq (n-1)\pi$ . The corresponding eigenvalues are
   $$\lambda_n=\alpha^2_n~~\qquad~~n=1,2,3, \cdots\,.$$
   Comment: One important question is this: how do the allowed eigenvalues and eigenfunctions depend on boundary conditions? More on that later.
3. $\lambda = -\beta^2 <0\,.$ This leads to the general solution
   $$u(x)=c_1\cosh\beta x+c_2\sinh \beta x\,~.$$
   The boundary conditions yield
   $$\tanh\beta = -\frac{h}{\beta}~~.$$
   The graph of the hyperbolic tangent does not intersect the graph of the two equilateral hyberbolas. Consequently, the set of solutions for $\beta$ is the empty set. Thus the S-L system has no solution, except the trivial one $u(x)=0$ .
4. What about complex $\lambda$ ?
   We shall see in the next section that the eigenvalues of a S-L problem are necessarily real.

Lecture 20