HW7, Math 322, Fall 2016

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1 HW7, Math 322, Fall 216 Nasser M. Abbasi November 4, 216 Contents 1 HW Problem a part a Problem Problem Part a Part b Problem b Problem

2 461, since a1 f, uola d/ fa la d. We now proceed in a similar way. Since 1\2 < An for n > 2, it follows that RQ[u] > A2, and furthermore the equality holds only if an for n > 2 [i.e., u a22j since al already. We have just proved the following theorem: The minimum 1 HWvalue 7for all continuous functions u that are orthogonal to the lowest eigenfunction and satisfy the boundary conditions is the net-to-lowest eigenvalue. Further 1.1 Problem generalizations also a follow directly from EXERCISES Use the Rayleigh quotient to obtain a reasonably accurate upper bound for the lowest eigenvalue of a with 2 and 1 b d +a4with d and *c St +A. with - and!ttl See Eercise a Consider the eigenvalue problem part a subject to and d 2 21 φ. Show that A > be sure to show that A. d 2 + λ 2 φ Prove that is valid in the following φ way. Assume Lu/a is piecewise smooth so that φ 1 Lu _ E bnn Putting the equation in the form d 2 φ Determine bn. [Hint: Using dgreen's formula 5.5.5, show that bn -anan if u and du/d are continuous 2 2 φ λφ and if u satisfies the same homogeneous And comparing it to the standard Sturm-Liouville form boundary conditions as the eigenfunctions On.] Shows that Now the Rayleigh quotient is p d2 φ dφ + p + qφ λσφ d2 d p 1 q 2 σ 1 λ pφφ 1 + p φ 2 qφ 2 d σφ2 d Substituting known values, and since φ, φ 1 the above simplifies to Now we can say that λ λ min λ 1 φ φ 2 d φ2 d φ φ 2 d φ2 d 2 1

3 We now need 194 a trial solution φ trial to use Chapter in the 5. above, Sturm-Liouville which needs Eigenvalue only toproblems satisfy boundary conditions to use to estimate lowest λ min. The simplest such function will do. The boundary conditions are We φ thus have,φ a 1 minimization. We theorem see for eample for the lowest that φeigenvalue trial A1. 2 We 1 can works, since φ trial ask 2, if and there φ are trial corresponding and φ trial theorems 1 1for 1 the higher. So will eigenvalues. use this in Interesting 1 generalizations immediately follow from λ min λ If we 2 1 insist that al, then 2 d F O O_ 2 anan 1 f n 2 a d RQ[u] d b En2 an fn 'Vna d d This means that in addition we are 1 restricting our + 1 function d u to be orthogonal to 461, since a1 f, uola d/ fa 1 la d We now 2 4 proceed + 2 in a similar way. Since d 1\2 < An for n > 2, it follows that d RQ[u] > A2, d and furthermore the equality 1 holds 4 2 only 2 + if 1d an for n > 2 [i.e., u a22j since al already. We have just proved the following theorem: The minimum value for all continuous functions u that are orthogonal 7to the lowest eigenfunction and satisfy the boundary conditions is the net-to-lowest eigenvalue. Further generalizations also follow directly 15 from EXERCISES Use the Rayleigh quotient to obtain a reasonably accurate upper bound Hence for the lowest eigenvalue of λ a with 2 and Problem d b +a4with d and *c St +A. with - and!ttl See Eercise a Consider the eigenvalue problem subject to A. and 21. Show that A > be sure to show that Prove that is valid in the following way. Assume Lu/a is piecewise smooth so that Lu _ E bnn d 2 φ Determine bn. [Hint: Using d 2 + λ 2 φ Green's formula 5.5.5, show that bn -anan if u and du/d are continuous and φ if u satisfies the same homogeneous boundary conditions as the eigenfunctions φ 1 On.] Putting the equation in the form d 2 φ d 2 2 φ λφ 3

4 And comparing it to the standard Sturm-Liouville form Shows that p d2 φ dφ + p + qφ λσφ d2 d p 1 q 2 σ 1 Now the Rayleigh quotient is λ pφφ 1 + p φ 2 qφ 2 d σφ2 d Substituting known values, and since φ, φ 1 the above simplifies to λ φ φ 2 d φ2 d Since eigenfunction φ can not be identically zero, the denominator in the above epression can only be positive, since the integrand is positive. So we need now to consider the numerator term only: φ 2 d + 2 φ 2 d For the second term, again, this can only be positive since φ can not be zero. For the first term, there are two cases. If φ zero or not. If it is not zero, then the term is positive and we are done. This means λ >. if φ then φ 2 d and also conclude λ > thanks to the second term 2 φ 2 being positive. So we conclude that λ can only be positive. 1.3 Problem Problem Consider eigenvalue problem d dr also assume dφ dφ bounded. And r dφ dr λrφ, < r < 1 subject to B.C. φ < you may dr dr 1. a prove that λ. b Solve the boundary value problem. You may assume eigenfunctions are known. Derive coefficients using orthogonality. Notice: Correction was made to problem per class . Book said to show that λ > which is error changed to λ Part a From the problem we see that p r, q, σ r. The Rayleigh quotient is λ pφφ 1 + p φ 2 qφ 2 dr σφ2 dr rφφ 1 + r φ 2 dr rφ2 dr 1 4

5 The term rφφ 1 epands to 1 φ 1 φ 1 φ φ Since φ 1, the above is zero and Equation1 reduces to λ r φ 2 dr rφ2 dr The denominator above can only be positive, as an eigenfunction φ can not be identically zero. For the numerator, we have to consider two cases. case 1 If φ then we are done. The numerator is positive and we conclude that λ >. case 2 If φ then φ is constant and this means λ is possible hence λ. Now we need to show φ being constant is also possible. Since φ 1, then for φ to be true everywhere, it should also be φ which means φ is some constant. We are told that φ <. Hence means φ is constant is possible value since bounded. Hence φ is possible. Therefore λ. QED Part b The ODE is rφ + φ + λrφ < r < 1 1 φ < φ 1 In standard form the ODE is φ + 1 r φ + λφ. This shows that r is a regular singular point. Therefore we try φ r a n r n+α Hence n φ r φ r n n + α a n r n+α 1 n n + α n + α 1 a n r n+α 2 n Substituting back into the ODE gives r n + α n + α 1 a n r n+α 2 + n + α a n r n+α 1 + λr a n r n+α n n n + α n + α 1 a n r n+α 1 + n + α a n r n+α 1 + λ a n r n+α+1 n To make all powers of r the same, we subtract 2 from the power of last term, and add 2 to the inde, resulting in n + α n + α 1 a n r n+α 1 + n + α a n r n+α 1 + λ a n 2 r n+α 1 n n 5 n n2 n

6 For n we obtain α α 1 a r α 1 + α a r α 1 α α 1 a + α a a α 2 α + α a α 2 But a we always enforce this condition in power series solution, which implies Now we look at n 1, which gives α a 1 r n+α a 1 r n+α 1 For n 2, now all terms join in, and we get a recursive relation a 1 n n 1 a n r n 1 + n a n r n 1 + λa n 2 r n 1 For eample, for n 2, we get a 2 λ 2 2 a All odd powers of n result in a n. For n 4 And for n 6 And so on. The series is φ r n n 1 a n + n a n + λa n 2 λa n 2 a n n n 1 + n λ n 2 a n 2 a 4 λ 4 2 a 2 λ 4 2 λ 2 2 a a 6 λ 6 2 a 4 λ λ a a n r n n λ a λ a a + + a 2 r a 4 r a 6 r 6 + a λr2 2 2 a + λ2 r a λ 3 r a + λr 2 λr 4 λr 6 a

7 From tables, Bessel function of first kind of order zero, has series epansion given by 1 n z 2n J o z n! 2 2 n z 2 1 z 4 1 z z z z6 + 1 z z z By comparing 2,3 we see a match between J o z and φ r, if we let z λr we conclude that φ 1 r a J λr We can now normalized the above eigenfunction so that a 1 as mentioned in class. But it is not needed. The above is the first solution. We now need second solution. For repeated roots, the second solution will be But α, hence Hence the solution is φ 2 r φ 1 r ln r + r α φ 2 r φ 1 r ln r + n b n r n b n r n n φ r c 1 φ 1 r + c 2 φ 2 r Since φ is bounded, then c 2 since ln not bounded at zero, and the solution becomes where a is now absorbed with the constant c 1 φ r c 1 φ 1 r cj λr The boundedness condition has eliminated the second solution altogether. Now we apply the second boundary conditions φ 1 to find allowed eigenvalues. Since φ r cj 1 λr Then φ 1 implies cj 1 λ The zeros of this are the values of λ. Using the computer, these are the first few such values. λ λ λ

8 or Hence λ λ λ φ n r c n J λn r φ r φ n r c n J λn r 4 To find c n, we use orthogonality. Per class discussion, we can now assume this problem was part of initial value problem, and that at t we had initial condition of f r, therefore, we now write f r c n J λn r Multiplying both sides by J λm r σ and integrating gives where σ r f r J λm r rdr Where Ω is some constant. Therefore c m J 2 c m c m Ω c n J λm r J λn r rdr J 2 λm r rdr λm r rdr c n f r J λm r rdr Ω And φ r c n J λn r With the eigenvalues given as above, which have to be computed for each n using the computer. 8

9 *Courtesy of E. C. Gartland, Jr. 1.4 Problem b EXERCISES Estimate to leading order the large eigenvalues and corresponding eigenfunctions for pc + [,\v + qq if the boundary conditions are a 4 and L *b Consider and d L c and L + hcbl dj + A1 + o From tetbook, equation 5.9.8, we are given that for large λ large. Take into account amplitude and period variations. φ σp 1 4 ep ±i σ t λ p t dt λ σp 1 σ t λ 4 c 1 cos p t dt + c 2 sin subject to and 461. Roughly sketch the eigenfunctions for A σ t p t dt 1 Where c 1, c 2 are the two constants of integration since this is second order ODE. For φ, the integral σt pt dt σt pt dt and the above becomes Hence c 1 and 1 reduces to φ σp 1 4 c1 cos + c 2 sin c 1 σp 1 4 φ c 2 σp 1 λ σ t p t dt Hence φ c 2 σp 1 4 cos λ c 2 σp 1 4 cos λ c 2 σp 1 4 λσ p cos σ t p t dt d λ d λ p t dt σ p σ t λ σ t p t dt σ t p t dt Since φ L then c 2 σp 1 4 λσ p cos λ L σ t p t dt 9

10 n assumed accurate Error Which means, for 6 non-trivial solution, that L σ t λn p t dt n 1 *Courtesy of E. C. Gartland, Jr. π 2 Therefore, for large λ, i.e. large n the estimate is EXERCISES 5.9 n 1 λn 2 π Estimate to leading order the large L σt eigenvalues pt dt and corresponding eigenfunctions for 2 pc n + [,\v 1 λ n 2 π + qq if the boundary conditions are L σt pt dt a 4 and L 1.5 Problem *b - and d L c and L + hcbl Consider dj + A1 + o subject to and 461. Roughly sketch the eigenfunctions for A large. Take into account amplitude and period variations. φ + λ 1 + φ Comparing the above to Sturm-Liouville form pφ + qφ λσφ Shows that p 1 q σ 1 + Now, from tetbook, equation 5.9.8, we are given that for large λ φ σp 1 4 ep ±i σ t λ p t dt σp 1 4 c 1 cos λ σ t p t dt + c 2 sin λ σ t p t dt Where c 1, c 2 are the two constants of integration since this is second order ODE. For φ, the integral σt pt dt σt pt dt and the above becomes φ σp 1 4 c1 cos + c 2 sin c 1 σp

11 Hence c 1 and 1 reduces to φ c 2 σp 1 λ σ t p t dt Applying the second boundary condition φ 1 on the above gives φ 1 c 2 σp 1 λ σ t p t dt Hence for non-trivial solution we want, for large positive integer n 1 σ t λ dt nπ p t nπ λ σt pt dt nπ 1 + tdt But 1 + tdt , hence nπ λ n λ n 2 Therefore, solution for large λ is φ c 2 σp 1 λ c 2 σp n σ t c n p t dt σ t p t dt 1 + tdt c n To plot this, let us assume c 2 1 we have no information given to find c 2. What value of n to use? Will use different values of n in increasing order. So the following is plot of φ n For 1 and for n 1, 2, 3,, 8. 11

12 Plot of ϕ for n 1 Plot of ϕ for n ϕ. ϕ Plot of ϕ for n 3 Plot of ϕ for n ϕ. ϕ Plot of ϕ for n 5 Plot of ϕ for n ϕ. ϕ Plot of ϕ for n 7 Plot of ϕ for n ϕ. ϕ