CALCULUS OF VARIATIONS
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xy(x)Figure 22.10 Trial solutions usedto estimate the lowest eigenvalueλof
−y′′=λywithy(0) =y′(1) = 0. They are: (a)y= sin(πx/2), the exact result;
(b)y=2x−x^2 ;(c)y=x^3 − 3 x^2 +3x;(d)y=sin^2 (πx/2).Estimate the lowest eigenvalue of the equation−
d^2 y
dx^2=λy, 0 ≤x≤ 1 , (22.29)with boundary conditions
y(0) = 0,y′(1) = 0. (22.30)We need to find the lowest valueλ 0 ofλfor which (22.29) has a solutiony(x) that satisfies
(22.30). The exact answer is of coursey=Asin(xπ/2) andλ 0 =π^2 / 4 ≈ 2 .47.
Firstly we note that the Sturm–Liouville equation reduces to (22.29) if we takep(x)=1,
q(x)=0andρ(x) = 1 and that the boundary conditions satisfy (22.26). Thus we are able
to apply the previous theory.
We will use three trial functions so that the effect on the estimate ofλ 0 of making better
or worse ‘guesses’ can be seen. One further preliminary remark is relevant, namely that the
estimate is independent of any constant multiplicative factor in the function used. This
is easily verified by looking at the form ofI/J. We normalise each trial function so that
y(1) = 1, purely in order to facilitate comparison of the various function shapes.
Figure 22.10 illustrates the trial functions used, curve (a) being the exact solution
y=sin(πx/2). The other curves are (b)y(x)=2x−x^2 ,(c)y(x)=x^3 − 3 x^2 +3x,and(d)
y(x)=sin^2 (πx/2). The choice of trial function is governed by the following considerations:
(i) the boundary conditions (22.30)mustbe satisfied.
(ii) a ‘good’ trial function ought to mimic the correct solution as far as possible, but
it may not be easy to guess even the general shape of the correct solution in some
cases.
(iii) the evaluation ofI/Jshould be as simple as possible.