The lecture 12 proof of Comparison Theorem (Sturm and Picone) by contradiction is below over 6 pages of notes for v>0 and u>0. To my understanding, to examine consequences of i), ii) and iii) we need to look at the top of page 3 of notes and consider the signs of terms of the identity marked with a star at the top of page 3. Could you please provide detailed explanations for i), ii) and iii) so I can learn from it. Thank you.

Reply to comment: What other information do you need about this sum?

Question 1. During lecture 12 we examined a proof by contradiction of the Comparison Theorem based on the assumptions that v(x) > 0 on the closed interval [z1,z2), where zi and z2 were consecutive zeros of u(x), and u(x) > 0 on the open interval (z1,z2). Modify and build on the proof presented during lectures to examine the consequences of the assumptions: (i) v(x) 0 on (21, z2); (ii) v(x) > 0 on [21, 22] with u(x) 0 and b(x) > a(x) Vx. Further, suppose two successive zeros of u are z, and zz with 0) o that eiro ( > Z, Z V and agerton 2 by 1 Multidy a by M gives: which Vill" 0 3 4 2 t da) u v euil + blauw then -4 gues: Vei" - uv": + (aca) - b) Delv = 0 Now I (ver - elvi) = Ver tu' e - evlevo Vel" - uuh Pessor vt'-ww') +(91)-ko) love Integrating from 2 to Z urt a giver: ( . vel-ew) dr + cain) - bla) uvan (ala)-b(x)) evde =0 da 0 & C zo then z, Uz, ZI 20 122 nZz vel - un' + UZI V(22) ll'(22) 41(22)V'(22) - V(2,)"(21) -e (2)v'(27) + 24400) -Wn) ev dr = 0 UZ, 21 And since el 70 Na el 2) = 1(22) = 0, therefor: V(2)u' (t2) - V12 Du! (21) + 3, 2(6) bauda -bo) = 0 We now consider the sins of the terms on the LHS of of the abre ichulity. Firal consider V (22) u (22). By asampon V (2) >0. And on (21, 22, li (22) to. If u (22) = 0, then M"talne both es (22) = 0 and e(22) and since the odution to the OPE unique this forsial solution would be the only solution. This for on-trinial odution, (22) 2 mon- Since llo X (242 o Z / Z2 But if ll (7) =0 1 C olnce Nent, consider - V(2,4 (21). Again, by assumption v(2) >0. 122) > ll (2) 70. then u (2) = el' (2) = 0 = "=0, so to avoid trivial solution l (2) >0. This - (2) e' (2) 0 on the closed interval [z1,z2), where zi and z2 were consecutive zeros of u(x), and u(x) > 0 on the open interval (z1,z2). Modify and build on the proof presented during lectures to examine the consequences of the assumptions: (i) v(x) 0 on (21, z2); (ii) v(x) > 0 on [21, 22] with u(x) 0 and b(x) > a(x) Vx. Further, suppose two successive zeros of u are z, and zz with 0) o that eiro ( > Z, Z V and agerton 2 by 1 Multidy a by M gives: which Vill" 0 3 4 2 t da) u v euil + blauw then -4 gues: Vei" - uv": + (aca) - b) Delv = 0 Now I (ver - elvi) = Ver tu' e - evlevo Vel" - uuh Pessor vt'-ww') +(91)-ko) love Integrating from 2 to Z urt a giver: ( . vel-ew) dr + cain) - bla) uvan (ala)-b(x)) evde =0 da 0 & C zo then z, Uz, ZI 20 122 nZz vel - un' + UZI V(22) ll'(22) 41(22)V'(22) - V(2,)"(21) -e (2)v'(27) + 24400) -Wn) ev dr = 0 UZ, 21 And since el 70 Na el 2) = 1(22) = 0, therefor: V(2)u' (t2) - V12 Du! (21) + 3, 2(6) bauda -bo) = 0 We now consider the sins of the terms on the LHS of of the abre ichulity. Firal consider V (22) u (22). By asampon V (2) >0. And on (21, 22, li (22) to. If u (22) = 0, then M"talne both es (22) = 0 and e(22) and since the odution to the OPE unique this forsial solution would be the only solution. This for on-trinial odution, (22) 2 mon- Since llo X (242 o Z / Z2 But if ll (7) =0 1 C olnce Nent, consider - V(2,4 (21). Again, by assumption v(2) >0. 122) > ll (2) 70. then u (2) = el' (2) = 0 = "=0, so to avoid trivial solution l (2) >0. This - (2) e' (2)