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).