Sturm's Comparison Theorem

When confronted with the regular boundary value problem

$$\left[ \frac{d}{dx} p(x)\frac{d}{dx} +q(x)+\lambda \rho (x)\right] u(x) =$$ 0 (318)

$$\alpha u(a)+\alpha 'u'(a) =$$ 0 (319)

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

we must ask

1. How does the oscillatory nature of $u(x;\lambda )$ , a solution to Eq. 3.18, depend on $\lambda$ ?
2. Why do the values of $\lambda$ permitted by (3.18) and (3.19) form a discrete and semi-infinite sequence
   $$\lambda_0 <\lambda_1 <\cdots <\lambda_n <\cdots$$
   with a smallest eigenvalue $\lambda_0$ and with $\lambda_n\to\infty$ as $n\to\infty$ ?

The ``oscillatory nature'' of a solution $u(x,\lambda)$ is expressed qualitatively by the location of the zeroes of its graph. One could also inquire about its behaviour between successive zeroes. However, we shall see that such an inquiry always leads to the same answer: Provided $q(x)+\lambda \rho (x)$ is positive between a pair of successive zeroes, the graph of $u(x;\lambda )$ has only a single maximum (or minimum). This means that $u(x,\lambda)$ can not oscillate between two of its successive zeroes.

Thus the most important issue is the existence and location of the zeroes, which are controlled entirely by the phase of a given solution $u(x,\lambda)$ . This phase is a scalar function from which one directly constructs the solution. It is preferrable to discuss the behavior of the solution $u(x;\lambda )$ in terms of its phase because the key qualitative properties of the latter are very easy to come by. As we shall see, one only needs to solve a first order differential equation, not the second order S-L equation.

However, before establishing and solving this differential equation, let us use the second order S-L differential equation directly to determine how the zeroes of $u(x\lambda )$ are affected if the parameter $\lambda$ is changed. We express this behaviour in terms of the

Sturm Comparison Theorem:

Whenever $\lambda_1 <\lambda_2$ , then between two zeroes of the nontrivial solution $u(x,\lambda _1)$ , there lies a zero of $u(x,\lambda_2)$ .

This theorem demands that one compare the two different solutions

$$u(x,\lambda_1 )\equiv u_1(x)~~\quad~~\textrm{and}~~\quad~~ u(x,\lambda_2) \equiv u_2(x)$$

of Eq. 3.18 corresponding two different constants $\lambda_1$ and $\lambda_2$ . The conclusion is obtained in three steps:

Figure 3.4: Graph of a solution $u(x,\lambda _1)$ which satisfies the mixed D-N boundary condition at $x=a$ .

Step 1: Multiply these two equations respectively by $u_2$ and $u_1$ , and then form their difference. The result, after cancelling out the $q(x)u_1u_2$ term, is

$$\frac{d}{dx} \left[ p(x)\left( u_2\frac{du_1}{dx} -u_1\frac{du_2}{dx}\right) \right] = (\lambda_2-\lambda_1)u_1u_2 \rho (x)\,,$$

which is the familiar Lagrange identity, Eq.(3.15), in disguise. Upon integration one obtains

$$p(x) (u_2u'_1-u_1u'_2)\bigg\vert^x_a = (\lambda_2-\lambda_1)\int^x_a u_1u_2 \rho \,dx\,.$$

If both $u_1$ and $u_2$ satisfy the mixed Dirichlet-Neumann boundary conditions $\alpha u(a) + \alpha ' u'(a)=0$ at $x=a$ , then

$$(u_2u'_1-u_1u'_2)_{x=a}=-\frac{\alpha }{\alpha '} [u_2(a)u_1(a)-u_1(a)u_2(a)]=0\,.$$

If $x=a$ is a singular point of the differential Eq.(3.18), $p(a)$ is zero. Thus, if $u$ and $u'$ are finite at $x=a$ , then the left hand side vanishes again at the lower limit $x=a$ . Thus both for a regular and for this singular Sturm-Liouville problem we have

$$p(x)\left( u_2(x)\frac{du_1(x)}{dx} - u_1(x) \frac{du_2(x)}{dx} \right) = (\lambda_2-\lambda_1)\int^x_a u_1u_2 \rho \,dx\,.$$

Step 2: Now assume that, for some range of values $\lambda$ , each of the corresponding solutions $u(x,\lambda)$ satisfying the boundary condition at $x=a$ , oscillates. In other words, as $x$ increases beyond $x=a$ , $u(x,\lambda)$ reaches a maximum, then decreases, passes through zero, reaches a minimum, increases and so on. That such a range of $\lambda$ -values exists, we shall see later.

Let $\lambda_1$ lie in this range, and let $x=\zeta$ be the first zero of $u_1(x)$ as in Figure 3.4. Consequently,

$$\left( p(x) u_2 \frac{du_1}{dx}\right)_\zeta = (\lambda_2-\lambda_1) \int^\zeta_a u_1u_2\rho \,dx\,.$$

Step 3. One must now conclude that if $\lambda _2 > \lambda _1$ , then the zeroes of $u_2(x)$ are more closely spaced than those of $u_1(x)$ . Why must this statement, the Sturm Comparison Theorem, be true?

Figure 3.5: Graph of a solution $u(x,\lambda _1)$ which is zero at $\zeta$ and $\xi$ .

a) Assume the contrary, i.e., assume that $u_2(x)$ has no zero in $a<x<\zeta$ . See Figure 3.4. In that case $u_2(x)>0$ for all $a<x<\zeta$ . This implies

$$\left( p(x)u_2\frac{du_1}{dx}\right)_\zeta = (\lambda_2-\lambda_1) \int^\zeta_au_1u_2 \rho \,dx >0\,.$$

But $p(\zeta )>0$ , $u_2(\zeta )>0$ and $\frac{du_1(\zeta )}{dx} <0$ , so that $\left( p(x) u_2 \frac{du_1}{dx}\right)_\zeta <0$ . This is a contradiction. Hence our assumption was wrong; the function $u_2$ does have a zero in $a<x<\zeta$ .

b) Now consider the circumstance where $u_1(x)$ has two successive zeroes at $\zeta$ and $\xi$ : $u_1(\zeta )=u_1(\xi )=0$ . In that case one obtains

$$p(\xi ) u_2(\xi )u'_1(\xi )-p(\zeta )u_2(\zeta )u'_1(\zeta )=(\lambda_2- \lambda_1)\int^\xi_\zeta u_2u_1 \rho\,dx\,.$$

If $u_2(x)$ does not change sign in $\zeta <x<\xi$ , as in Figure 3.6, then we again have a contradiction because if $u_1(x)<0$ , then $u'_1(\zeta )<0$ , $u'_1 (\xi ) >0\Rightarrow (r.h.s.)\times (l.h.s.)<0$ . In other words, the picture in Figure 3.6 is impossible. We conclude, therefore, that $u_2(x)$ must have a zero inside the open interval $(\zeta ,\xi )$ .

Figure 3.6: If $\lambda _2 > \lambda _1$ it is impossible that $u(x,\lambda _2) u(x,\lambda _1) >0$ for all $x$ in the interval $[\zeta ,\xi ]$ .