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# set5 - Lecture 13 Qualitative Behaviour of Solutions to...

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Lecture 13 Qualitative Behaviour of Solutions to ODEs (cont’d) Applications of the Sturm Comparison Theorem The most natural class of DEs with which we can perform a Sturm-type comparison are those with constant coefficients, i.e., y ′′ + k 2 y = 0 , k R , (1) with solutions of the form y ( x ) = A cos( kx φ ) . (2) The period of these solutions is T = 2 π/k . Consequently, the distance between consecutive zeros of these solutions is π/k . For example, suppose that for the following DE in normal form, u ′′ + q ( x ) u = 0 , (3) we have been able to establish that q ( x ) > k 2 (4) on some sufficiently large domain, e.g, [ x 0 , ). Now let u ( x ) be a solution to (3). Then by the Sturm Comparison Theorem (SCT), for any solution y in (2), a zero of u ( x ) must lie between two consecutive zeros of y . At first, one might think that it is sufficient that two consecutive zeros of u can be no further apart that 2 π/k , as shown in the left figure below. But the SCT must hold for all solutions of the form (2) which are obtained by continuously varying the phase φ . (Note that we do not pay attention to the amplitudes since they do not affect the positioning of the zeros.) If we slide the y solution in the above figure enough, then we produce the unacceptable situation on the right. We cannot have two consecutive zeros of y enclosing a region that is devoid of a zero of u . 1

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x π/k y ( x ) = A cos k ( x φ 1 ) π/k u ( x ) u ( x ) y ( x ) = A cos k ( x φ 2 ) x Now imagine sliding the y wave continuously to the right: Between any two consecutive zeros of the moving y wave - a kind of moving grating - we must always be able to find a zero of v ( x ). This leads to the following consequence: Every interval [ a, b ] of length π/k must contain at least one zero of u . And the consequence of this consequence is that the distances between consecutive zeros of v must be less than or equal to π/k . Application to Bessel’s Equation Recall Bessel’s equation of order p 0: x 2 y ′′ + xy + ( x 2 p 2 ) y = 0 (5) The substitution y ( x ) = 1 x u ( x ) (6) yields the following normal or reduced form equation for u(x): u ′′ + parenleftBigg 1 + 1 4 p 2 4 x 2 parenrightBigg u = 0 . (7) Recall that all zeros of u are zeros of y and vice versa – it is much easier to study properties of zeros, etc., from the u equation. There are three cases to consider: Case No. 1: p = 1 / 2. Then q ( x ) = 1 so that the general solution of (7), written in phase- shifted form is u ( x ) = A cos( x φ ) . (8) 2
Thus the general solution of Bessel’s DE for p = 1 / 2 is y ( x ) = A x cos( x φ ) . (9) The distance between successive zeros of y ( x ) is π . Case No. 2: 0 p < 1 / 2. Then q ( x ) > 1. From the SCT, every interval of length π must contain a zero of u or y . However, we can say even more: As x → ∞ , we see that q ( x ) 1, so we expect that the distances between consecutive zeros of u ( x ), hence y ( x ), will approach π . In fact, let’s assume that x > 0 is sufficiently large so that, for a small ǫ > 0, 1 < 1 + 1 4 p 2 4 x 2 < 1 + ǫ. (10) The leftmost term, “1”, corresponds to the DE y ′′ + y = 0 , whose solutions have consecutive zeros that are π apart.

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