AMS 212A Applied Mathematical Methods I Appendices of Lecture 06 Copyright by Hongyun Wang, UCSC. ( ) cos2

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1 AMS 22A Applied Mathematical Methods I Appendices of Lecture 06 Copyright by Hongyun Wang UCSC Appendix A: Proof of Lemma Lemma : Let (x ) be the solution of x ( r( x)+ q( x) )sin 2 + ( a) 0 < cos2 where p(x) > 0 and r(x) > 0 for x in [a b] (from the Sturm-Liouville problem). Then (x ) has the 3 properties below. For x > a a) (x ) is a strictly increasing function of. b) lim ( x )+ + c) lim ( x ) 0 Proof of a): Consider 2 >. We need to show (x 2 ) (x ) > 0 for x > a. We prove it in a few steps. Step A: Suppose (x 0 2 ) (x 0 ) 0 and sin( 0 ) 0 at some x 0. Then (x 2 ) (x ) is a strictly increasing function of x in a neighborhood of x 0. Proof of Step A: ( x ) ( r( x)+ q( x) )sin > 0 x cos2 > x ( x 0 2 ) x ( x 0 ) ( 2 )r( x 0 )sin 2 Taylor expansion of (x 2 ) (x ) at x 0 : ( x 0 + s 2 ) ( x 0 + s ) ( 2 )r( x 0 )sin 2 0 s + O s 2 Step B: The conclusion of Step A is still true if sin( 0 ) 0. Suppose (x 0 2 ) (x 0 ) 0 and sin( 0 ) 0 at some x 0. Then (x 2 ) (x ) is a strictly increasing function of x in a neighborhood of x

2 AMS 22A Applied Mathematical Methods I Proof of Step B: x ( x 0 2 ) x ( x 0 ) 0 Let us calculate xx (x ). xx ( x ) ( r( x)+ q( x) )sin ( r( x)+ q( x) )2sin cos x > xx ( x 0 2 ) xx ( x 0 ) 0 cos 2 2cos sin x ext we calculate xxx (x ). We will only look at its value at x 0. At x 0 we have (x 0 2 ) (x 0 ) 0 where sin( 0 ) 0. It follows that x xx0 0 independent of sin 0 independent of xx 0 These results significantly simplify the calculation of xxx (x 0 ). xxx xx0 cos 2 + 2( r( x)+ q( x) )cos 2 x xx p( x) cos2 ( x ) 2 xx 0 2 xx 0 > xxx ( x 0 2 ) xxx ( x 0 ) 2( 2 )r( x 0 ) Taylor expansion of (x 2 ) (x ) at x 0 : 0 2 > 0 Step C: ( x 0 + s 2 ) ( x 0 + s ) ( 3 2 )r( x 0 ) 0 2 s 3 + O( s 4 ) (x 2 ) and (x ) are the same at x a : (a). Results of Steps A and B tell us that (x 2 ) (x ) > 0 for x slightly above a (it does not matter whether or not sin 0)

3 AMS 22A Applied Mathematical Methods I Suppose x 0 is the location where (x 2 ) (x ) transitions from positive to non-positive. Mathematically that is ( x 0 2 ) ( x 0 ) 0 ( x 2 ) ( x ) 0 for x slightly above x 0. But this contradicts with the results of Steps A and B. Therefore we conclude that (x 2 ) (x ) > 0 for x > a. Proof of b): We first calculate an integral. sin 2 + cos 2 d d tan 2 2 tan dx x ds 2 s as + Since r(x) and p(x) are positive on [a b] there exist r 0 q 0 p 0 and such that r( x) r 0 > 0 q( x) q 0 for x [ a b] p( x) p 0 > 0 for x [ a b] Let ( c) q 0 / r 0. For sufficiently large we have r( x)+ q( x) ( ( c) )r( x)+ ( c) r( x)+ q x ( ( c) for x [ a b] The differential equation for becomes x ( r( x)+ q( x) )sin 2 + ( c) ( sin 2 + cos 2 cos2 for x [ a b] Thus for sufficiently large (x ) is an increasing function of x. Let x n be the location where (x ) hits n. We re-write the differential equation for as d ( ( c) sin 2 + cos 2 dx - 3 -

4 AMS 22A Applied Mathematical Methods I Integrate from x n to x n+ we have ( n+) d ( ( c) ( x sin 2 + cos 2 n+ x n ) Using the analytical expression of the integral we obtain d ( x n+ x n ) ( ( c) sin 2 + cos as + ( ( c) > ( x a) x n x n n + ( x 0 a) 0 as + ( ( c) (Recall that we showed that (x ) is an increasing function of x when is large.) It follows that for any x > a and > 0 we can make (x ) > when is large enough Therefore we conclude lim ( x )+ + ote: The proof shows that when is large (b ) hits at O( 2 ) which means O( 2 ) This is consistent with the solution of u u. u( 0) 0 u( L) 0 Proof of c): We first show that (x ) 0 for x > a. At 0 the evolution equation of tells us r( x)+ q x x 0 ( ) sin 2 + p( x) cos2 > (x ) will never go below 0 for x > a. 0 p( x) > 0 Since r(x) and p(x) are positive on [a b] there exist r 0 q 0 p 0 and such that - 4 -

5 AMS 22A Applied Mathematical Methods I r( x) r 0 > 0 q( x) q 0 for x [ a b] p( x) p 0 > 0 for x [ a b] Let ( c) q 0 / r 0. For negative and large we have r( x)+ q( x) ( + ( c) )r( x) ( c) r( x)+ q x ( + ( c) for x [ a b] > x ( r( x)+ q( x) )sin 2 + cos2 ( + ( c) sin 2 + for x [ a b] p 0 For any small > 0 we select M r 0 sin 2 + p 0. When ( + ( c) )< M the two statements below are true. As long as (x ) satisfies sin 2 sin 2 we have x That is (x ) decreases with large negative slope -/ until sin 2 sin 2. Once (x ) it will never go above. That is (x ) will stay below. The condition for (x ) at x a is ( a) 0 < In the case of 0 (x ) will stay below when is negative and large. In the case of > > 0 we pick small > 0 satisfying sin 2 < sin 2. When is negative and large (x ) decreases with large negative slope -/ until and stays below thereafter. Therefore we conclude lim ( x ) 0 Appendix B: Proof of Rayleigh principle Rayleigh principle: - 5 -

6 Let W span{ v 0 v } W 2 uuc BC AMS 22A Applied Mathematical Methods I { and u W } Consider the minimization of R(u) over W. Let u + arg min Ru uw u0 Claim 2: u + is an eigenfunction corresponding to eigenvalue + and + Proof of Claim 2: Step : We first prove that min Ru uw u0 r( x) Lu W This is verified directly. for u W Let v W. By definition v is a linear combination of {v 0 v v }. v c n r( x) Lu v u r x L( v) u r( x) L c n u ( n )c n 0 Ste We consider a function of a scalar variable s gs Ru ( + + su) where u W By definition of u + we have g( 0) 0 for any u W The derivative of R(u + + s u) has the expression d ds Ru ( + su) + s0 2 u u + u + r( x) Lu ( +) + Ru ( + ) u u

7 AMS 22A Applied Mathematical Methods I 2 u + u + u r( x) Lu ( +)+ Ru ( + )u + The derivative of R(u + + s u) is zero for any u W. d ds Ru ( + + su) 0 for any u W s0 > u r( x) Lu ( +)+ Ru ( + )u + 0 for any u W From Step we know r( x) Lu ( +)+ Ru + r( x) Lu ( +)+ Ru ( + )u + 0 > Lu ( + ) R( u + )r( x)u + u + W. We conclude > R(u + ) is an eigenvalue and u + is a corresponding eigenfunction. By definition of u + we have Ru ( + ) min Ru uw > Ru ( + ) Rv ( + ) + (v + is an eigenfunction corresponding to + ). On the other hand + is the smallest eigenvalue whose eigenfunctions are orthogonal to W. R(u + ) is an eigenvalue whose eigenfunctions are orthogonal to W. > + Ru ( + ) Combining these two results we conclude + Ru ( + ) min Ru uw Appendix C: Proof of the completeness of eigenfunctions For f( x)c 2 BC we need to show that for any > 0 there exist and { c n } such that f c n < where the norm is based on the inner product - 7 -

8 AMS 22A Applied Mathematical Methods I Proof: Let u u u u x b a ( ) 2 r x dx b n f v n and h f b n. We only need to show that there exits such that h h < 2 We prove it through a sequence of steps. Step: h W Ste: It is straightforward to verify that h 0 0 n. Rayleigh principle tells us + min Ru. uw u0 > + Rh ( ) > + h r( x) L ( h ) h h > h h + h r( x) L ( h ) Step 3: Using the fact that is the eigenfunction corresponding to n we have f r( x) L ( ) f n n b n r( x) L ( ) n n Step 4: We use the result of Step 3 to calculate h r( x) L ( h ). h r( x) L ( h ) f b n r x L f b n - 8 -

9 AMS 22A Applied Mathematical Methods I f r( x) L( f) 2 f r( x) L b n + b n r x L b n Step 5: f r( x) L( f) + 2 n b n Combining the results of Ste and Step 4 we obtain h h f + r( x) L( f) + b n n 2 f + r( x) L( f) K 0 2 n b n where (K 0 + ) is the number of negative eigenvalues. Since K 0 is fixed and lim + + we conclude + h h 0 as