Mathematical Methods for Physics and Engineering - Matematica.NET

CALCULUS OF VARIATIONS◮Show that∫ b(y′j py ′ i − y j qy i)dx = λi δ ij . (22.27)aLet y i be an eigenfunction of (22.24), corresponding to a particular eigenvalue λ i ,sothat( )py′ ′i +(q + λi ρ)y i =0.Multiplying this through by y j and integrating from a to b (the first term by parts) weobtain( )[y ] b ∫ b∫ bj py′i y j(py ′ i) ′ dx + y j (q + λ i ρ)y i dx =0. (22.28)−aaThe first term vanishes by virtue of (22.26), and on rearranging the other terms and using(22.25), we find the result (22.27). ◭We see at once that, if the function y(x) minimises I/J, i.e. satisfies the Sturm–Liouville equation, then putting y i = y j = y in (22.25) and (22.27) yields J andI respectively on the left-hand sides; thus, as mentioned above, the minimisedvalue of I/J is just the eigenvalue λ, introduced originally as the undeterminedmultiplier.a◮For a function y satisfying the Sturm–Liouville equation verify that, provided (22.26) issatisfied, λ = I/J.Firstly, we multiply (22.24) through by y to givey(py ′ ) ′ + qy 2 + λρy 2 =0.Now integrating this expression by parts we have[ ] b ∫ b) ∫ bypy ′ −(py ′2 − qy 2 dx + λ ρy 2 dx =0.aaThe first term on the LHS is zero, the second is simply −I and the third is λJ. Thusλ = I/J. ◭a22.7 Estimation of eigenvalues and eigenfunctionsSince the eigenvalues λ i of the Sturm–Liouville equation are the stationary valuesof I/J (see above), it follows that any evaluation of I/J must yield a value that liesbetween the lowest and highest eigenvalues of the corresponding Sturm–Liouvilleequation, i.e.λ min ≤ I J ≤ λ max,where, depending on the equation under consideration, either λ min = −∞ and792