**To obtain an estimate for the the lowest eigenvalue
k1 of the standard Sturm-Liouville**
**eqution (py')'+[-q+kry]=0
subjected to homogeneous boundary conditions at x=a and x=b, one**
**multiplies this equation by y, and the integrates
the product over the given range of x. Doing so**
**leads, after integration by parts, to-**

**
<p(y'^2)+q(y^2)>=k<r(y^2)**

**where the bra-ket notation indicates integration
over the indicated range of x. Thus, if the functions p, q and r remain
positive definite in a<x<b, one can conclude that the lowest eigenvalue
k1 will satisfy the inequality-**

**
k1<{<p(f')^2+q(f)^2>}/{<r(f^2)>}**

**where f is any trial function satisfying
the boundary conditions on y. Of course, the closer the trail function
matches the functional form of the lowest eigenmode y1 the better will
be the indicated upper bound for k1. Although one generally does
not know beforehand the exact shape of y1, one can approximate its shape
reasonably well by allowing the trail function to contain an adjustable
parameter.**

**Let us demonstrate. Take the boundary value
problem-**

**
y"-kxy=0 subject to y(0)=y(1)=0**

**which yields the variational upper bound-**

**
k1<{<f')^2>}/{x(f)^2>}**

**Choosing a trial function of the form
f=x^a*(1-x) , where 'a'
is an undetermined parameter, one finds the inequality-**

**
k1<[a^2/(2a-1)-(a+1)+(a+1)^2/(2a+1)]/[1/(2a+1)-2/(2a+3)+1/(2a+4)]**

**An evaluation of the right side of this inequality
shows its minimum value to be 19.15
occuring at a=1.24.
This estimate is quite good and within 1.6% of the exact lowest eigenvalue
of k1=18.9563 obtainable
for the present equation by an exact analytic solution involving Airy functions(or
Bessel functions of 1/3 order).**