3. Separation of Variables

Transcription

1 3. Seprtion of Vriles 3.. Bsics of the Method. In this lecture we review the very sics of the method of seprtion of vriles in 1D The method. The ide is to write the solution s u(x, t) = n X n (x)t n (t). (3.1) where X n (x) T n (t) solves the eqution nd stisfies the oundry conditions (ut not the initil condition(s)). We give summry using het eqution here. Given eqution u t = β 2 u + P(x, t), <x <L; u(x, ) = f(x), + oundry conditions (3.2) x2 1. Require X(x) T(t) to solve the homogeneous eqution which leds to eigenvlue prolem for X: u t = β 2 u x 2 (3.3) X K X = + oundry conditions. (3.4) Solve it to get X n nd K n. Note tht the nturl rnge of n is not lwys 1, 2, 3, 2. Expnd Expnd 3. Solve to otin T n. 4. Write down the solution f(x)= n P(x, t)= n f n X n. (3.5) p n (t)x n. (3.6) T n β K n T n = p n (t), T n ()= f n (3.7) u(x, t) = n T n (t)x n (x). (3.8) We understnd tht chnges should e mde when the eqution is different Exmples. Exmple 3.1. (Simplest cse) Solve u t =3 2 u x2, <x <; u(, t) =u(, t)=, u(x, )=sin x 4 sin 3 x. (3.9) Solution. We follow the procedure: 1. Otin X n ;. Seprte vriles. Recll tht X(x) T(t) must solve the eqution. This gives X(x)T (t) =3X (x)t(t) So we hve the equtions for T nd X: T (t) T(t) = 3 X (x) X(x). (3.1) T (t) 3KT(t) =; X (x) KX(x)=. (3.11)

2 . Boundry conditions for the X eqution: u(, t) = X()=; u(, t)= X()=. (3.12) c. Solve for X n : (It s importnt to e le to solve the eigenvlue prolem from scrtch. You my e required to write down every detil insted of just write X n = from memory in the exm.) The eigenvlue prolem is We discuss the three cses: X K X =, X()=X()=. (3.13) i. K >. Generl solution is X = C 1 e K x + C 2 e K x. (3.14) Applying the initil conditions we conclude C 1 = C 2 =. ii. K =. Generl solution is Applying the initil conditions we conclude C 1 = C 2 =. iii. K <. Generl solution is X = C 1 cos ( K X = C 1 + C 2 x. (3.15) x ) +C 2 sin ( K x ). (3.16) Applying initil conditions we conclude tht C 1 = C 2 = unless K = n 2. So the eigenvlues re with eigenfunctions K n = n 2, n = 1, 2, 3, (3.17) C sin (n x). (3.18) Thus we tke 3.1 X n = sin (n x). (3.19) 2. Expnd f(x); Thus we hve f(x)=sin x 4 sin 3x=X 1 4 X 3. (3.2) 3. Solve T n ; Now we hve oth eqution nd initil condition for T n : This gives f 1 =1, f 3 = 4, ll other f n =. (3.21) T n 3K n T n =, T n ()= f n. (3.22) Clerly if f n = then T n =. So the only nonzero ones re 4. Write down solution. We hve T n (t) =T n ()e 3Knt = f n e 3Knt. (3.23) T 1 (t)= f 1 e 3K1t =1 e 3( 1)t = e 3t ; (3.24) u(x, t)= T 3 (t)= f 3 e 3K3t = 4 e 27t. (3.25) T n (t) X n (x)=e 3t sin x 4e 27t sin 3 x. (3.26) 3.1. This is just for convenience. There is some freedom in choosing X n here. For exmple, tking X n =2 sinn x is lso OK.

3 Exmple 3.2. (Eqution with Source) Solve u t =3 2 u x 2 + et sin x, < x < ; u(, t)=u(, t) =, u(x, )=sin x 4 sin 3x. (3.27) Solution. We try to solve this prolem using the sme X n s. Recll tht we ssume u(x, t) = T n (t)x n (x) (3.28) with X n 3 K n X n =. Sustitute this u(x, t) into the eqution we rech [T n 3K n T n ] X n = t sin x =e t X 1. (3.29) Asoring the t X 1 term into the left hnd side we rech This leds to Solve T 1 from As K 1 = 1 the ove is [T 1 3K 1 T 1 e t ] X 1 + n=2 [T n 3K n T n ] X n = (3.3) T 1 3K 1 T 1 e t =; T n 3K n T n =, n > 1. (3.31) The eqution is liner with generl solution Using the initil condition we rech T n, n > 1 remins the sme s in the previous exmple: T 1 3 K 1 T 1 e t =, T 1 ()=1. (3.32) T 1 + 3T = e t, T 1 ()=1 (3.33) T 1 (t)= 1 4 et + Ce 3t. (3.34) C = 3 4 T 1(t)= 1 4 et e 3t. (3.35) So finlly u(x, t)= T 3 (t)= 4e 27t, T n (t)= for n 3. (3.36) ( 1 T n (t) X n (x)= 4 et + 3 ) 4 e 3t sin x 4 e 27t sin 3x. (3.37) Exmple 3.3. (Dirichlet oundry condition) Find the solution to the het flow prolem Solution. We use seprtion of vriles. u = 5 2 u t x2, <x<, t > (3.38) u(, t)=u(, t) =, t > (3.39) 1. Seprte the vriles. Write u =X(x)T(t), the eqution ecomes The equtions for X is u(x, ) = f(x)=1 cos2x, <x <. (3.4) T X =5X T The left hnd side of the eqution for T is T 5T = X = K. (3.41) X X K X =, X()=X()=; (3.42) T 5KT =. (3.43)

4 2. Solve the eigenvlue prolem X K X =, X()=X()=; (3.44) As we hve solved it efore, we omit the detils (You need to include the detils in the exm though). The eigenvlues re n 2, n =1, 2, 3,, nd the corresponding eigenfunctions re So K n = n 2, X n = sin (n x). 3. Expnd the initil condition. We hve n sin (n x), n =1, 2, 3, (3.45) 1 cos 2x=u(x, )= n sin (n x). (3.46) Thus ll we need to do is to find the Fourier sine series for 1 cos2 x. As the intervl is [,] we hve T =. We compute for n =1, 2, 3, n = 2 (1 cos2x) sin (n x)dx = 2 sin (n x) dx 2 sin (n x) cos (2 x)dx = 2 n cos(n x)n 1 [sin (n + 2)x+sin (n 2)x] dx = 2 n [1 ( 1)n ] 1 sin (n +2)xdx 1 sin (n 2)xdx. (3.47) We evlute For the lst term, there re two cses. sin (n+2)xdx= 1 n + 2 cos (n+2)xn = 1 ( 1)n+2. (3.48) n + 2 If n = 2, then sin (n 2)x= nd sin (n 2)xdx=. (3.49) If n 2, we compute sin (n 2)xdx= 1 n 2 cos (n 2)xN = 1 ( 1)n 2. (3.5) n 2 Putting everything together, we hve ( 2 n 1 ) 1 ( 1) n n +2 n = ( 2 n 1 n 2 1 ) 1 ( 1) n n + 2 n = 2. (3.51) n 2 Noticing tht, when n is even, we hve 1 ( 1) n =. Thus the ove formul cn e simplified y setting n = 2k 1 to ( 2 2k 1 = 2k k 3 1 ) 2, k =1, 2, 3, (3.52) 2 k +1 As P(x, t)= here, the expnsion is trivil: = sin n x. (3.53) 4. Solve the eqution for T. As p n (t) = for ll n, T n stisfies T n + 5n 2 T n = T n =c n e 5n2t. (3.54)

5 5. Putting everything together, the solution is given y u(x, t)= ( 2 2 k k 3 1 ) 2 2k + 1 e 5(2k 1)2t sin ((2 k 1)x). (3.55) k=1 Exmple 3.4. (Neumnn oundry condition) Solve u u (, t) = x x u t Solution. We use seprtion of vriles. 1. Seprte the vriles. Write u(x, t)=x(x)t(t). Then the eqution leds to which gives nd = 3 2 u x2, <x <, t > (3.56) (, t) =, t >, (3.57) u(x, ) = x, < x <. (3.58) T X = 3X T T 3 T = X X = K (3.59) X KX =, X ()=X ()=, (3.6) T 3KT =. (3.61) Note tht comes from the expnsion of P(x, t) (= for this prticulr prolem). Now Solve the eigenvlue prolem We discuss the three cses.. K <. The generl solution is X = C 1 cos ( K We compute X KX =, X ()=X ()=, (3.62) X = K C 1 sin ( K x ) +C 2 sin ( K x ) + K x ). (3.63) C 2 cos ( K x ). (3.64) The oundry conditions then led to K C 2 =, (3.65) K C 1 sin ( K ) + K C 2 cos ( K ) =. (3.66) Thus C 2 =, nd K = n. Consequently the eigenvlues re with corresponding. K =. The generl solution is K n = n 2, n = 1, 2, 3, (3.67) X n = cos(n x). (3.68) X =C 1 + C 2 x (3.69) the oundry conditions then gives C 2 = which mens is n eigenvlue nd the corresponding eigenfunctions re X = 1. c. K >. The generl solution is X = C 1 e K x + C 2 e K x. (3.7) The oundry conditions leds to K C1 K C 2 = (3.71) K e K C 1 K e K C 2 =. (3.72)

6 Solving it gives C 1 = C 2 =. Therefore there is no positive eigenvlues. Summrizing, the eigenvlues re 2. Expnd f(x). We hve K n = n 2, n =, 1, 2, 3, (3.73) X n = cos(n x), n =, 1, 2, 3, x = 2 + n cos(n x). (3.74) All we need to do is to find the cosine series for x: < x <. We compute = 2 x dx=, (3.75) n = 2 x cos(n x) dx= 2 [ ] x sin (n x)n sin (n x)dx = 2[( 1)n 1] n n 2. (3.76) Note tht ( 1) n 1= for ll n even. Thus we hve 2k =, 3. Solve the eqution for T n : 2 T n + 3n 2 T n =, T n ()= 4. Summrizing, we hve u(x, t)= 2 k=1 4 2k 1 = (2 k 1) 2. (3.77) n= n even 4 n 2 n odd Exmple 3.5. (Other oundry conditions) Solve n = 2 T n = n even. (3.78) 4 t n 2 n odd e 3n2 4 (2k 1) 2 e 3(2k 1)2t cos((2 k 1)x). (3.79) u t = κu xx <x<l, t > (3.8) u(x, ) = f(x) x l, (3.81) u(, t) = t, (3.82) u x (l, t) = h u(l, t). t. (3.83) Here h >. Solution. Applying the method of seprtion of vriles, we rech We discuss the cses: i. λ >. The generl solution is Now X λx =, X()=, h X(l)+X (l) =. (3.84) Ae λ x +B e λ x. (3.85) X()= A +B = (3.86) h X(l)+X (l) = ( ) h+ λ e λ l A+ ( ) h λ e λ l B =. (3.87) The two equtions cn e written ( 1 1 ( ) h + λ e λ l ( ) h λ e λ l ) ( A B ) ( ) =. (3.88)

7 For the solution to e non-zero, we hve to hve ( 1 1 =det ( ) h + λ e λ l ( ) h λ e λ l )= ( ) h λ e λ l ( ) h+ λ e λ l. (3.89) As h > nd λ >, this is not possile. ii. λ =. The generl solution is The oundry conditions led to iii. λ <. The generl solution is A cos ( λ Now A+Bx (3.9) A =, h A+(h l +1)B = A=B =. (3.91) ) ( ) x + B sin λ x. (3.92) X()= A=, (3.93) h X(l)+X (l)= h sin ( ) ( ) λ l + λ cos λ l =. (3.94) Therefore the solution is of the form X = sin (px) with p stisfying It is esy to see tht the solutions form n infinite series Therefore our solution to the PDE cn e written s where n is determined y tn (p l)= p/h. (3.95) < p 1 < p 2 < < (3.96) n e κp n 2 t sin (p n x) (3.97) 1 f(x)= 1 n sin (p n x). (3.98) Now how should we determine n? And furthermore how cn we know whether the infinite sum gives the solution or equivlently whether similr properties s those hold for the Fourier series hold for our series with sin (p n x)? Keep in mind tht it is not possile to otin formul for the p n s. We compute, for n m, l sin (p n x) sin (p m x)dx = 1 l [cos(p n p m ) x cos (p n + p m )x] dx 2 = 1 [ sin (pn p m ) l sin (p ] n + p m )l 2 p n p m p n + p [ m sin (pn l) cos (p m l) sin (p m l) cos (p n l) Now using the fct tht we hve = 1 2 h sin (p n l)+ p n cos(p n l) = l p n p m sin (p n l) cos (p m l)+sin (p m l) cos (p n l) p n + p m ]. (3.99) sin(p n l)= p n h cos (p n l) (3.1) sin (p n x) sin (p m x) dx =. (3.11)

8 Therefore we cn determine n y l f(x) sin (p n x)dx n = l. (3.12) sin 2 (p n x)dx Exmple 3.6. (Wve eqution) 2 u t 2 = 4 2 u + xt, <x <, t > (3.13) x2 u(, t)=, u(, t)=1 t > (3.14) u(x, ) = x, <x<, (3.15) u (x, ) = 1, t <x<. (3.16) Remrk. Note tht trying to find stedy-stte solution first would fil this time: No w(x) cn possily stisfy Furthermore, ny effort of finding w(x, t) such tht in hope of v =u w stisfying the simple system 4w xx + xt=, w() =, w()=1. (3.17) w tt = 4w xx + xt, w(, t)=, w(, t)=1 (3.18) 2 v t 2 = 4 2 v x2, <x<, t > (3.19) v(, t)=, v(, t)= t > (3.11) u(x, ) = x w(x, ), < x <, (3.111) u w (x, ) = 1 (x, ), t t <x <. (3.112) does not mke much sense s solving the w eqution is not relly esier thn solving the originl u eqution. In summry, in such generl cse, it is not possile to tke cre of the oundry conditions nd the source term in one single step. They hve to e delt with seprtely. Solution. Step. Tke cre of oundry conditions. We find w(x) stisfying w xx =, w()=, w()=1. This is esy: w(x)=x/. Now set v =u w, we rech 2 v t 2 = 4 2 v +xt, <x<, t > (3.113) x2 v(, t)=, v(, t)= t > (3.114) u(x, ) = x x/, <x<, (3.115) u (x, ) = 1, <x<. (3.116) t Step 1. Find the eigenvlue prolem nd solve it. Applying seprtion of vriles, we found out tht the eigenvlue prolem is X λx =, X() =X()= (3.117) which leds to (detils omitted due to hving een done severl times efore) Step 2. Find out equtions for T n. We write λ n = n 2, X n = A n sin (n x), n =1, 2, 3, (3.118) v(x, t)= T n X n = T n (t) sin (n x) (3.119)

9 Note tht the ritrry constnt A n hs een sored into T n (t). Sustitute into the eqution: Therefore the eqution ecomes 2 v t 2 = T n (t) sin (n x); (3.12) 2 v x 2 = T n (t) [sin (n x)] = n 2 T n (t) sin (n x); (3.121) xt= 2 v t v x 2 = [T n +4n 2 T n ] sin (n x). (3.122) On the other hnd, t t = we hve ( 1 1 ) x=v(x, ) = T n () sin (n x); (3.123) nd 1 = v t (x, ) = T n () sin (n x). (3.124) Therefore, if we expnd xt, ( 1 1 ) x, 1 into their Fourier Sine series: then T n stisfies the initil vlue prolem Now we compute: h n (t): n : x t = h n (t) sin (n x); (3.125) ( 1 1 ) x = n sin (n x); (3.126) 1 = n sin (n x); (3.127) T n + 4n 2 T n = h n (t); T n () = n, T n ()= n. (3.128) h n (t) = 2 (x t) sin (n x) dx = 2 t x sin (n x) dx = 2 t ( 1 ) xdcos (n x) n = 2t [ ] x cos (n x) n cos(n x) dx = 2t n [ ( 1)n ] = ( 1) n+1 2 t. (3.129) n n = 2 ( 1 1 ) x sin (n x)dx ( = 1 1 )[ ] 2 x sin (n x)dx ( = 1 1 ) ( 1) n+1 2 n = ( 1) n+1 2 ( 1). (3.13) n

10 n : Note tht we hve tken dvntge of the fct tht n = ( 1 1 ) hn (t)/t. n = 2 Summrizing, T n stisfies sin (n x)dx = 2 n [( 1)n 1] = 2 n [( 1)n+1 + 1]. (3.131) T n + 4 n 2 T n =( 1) n+1 2 n t; T n()=( 1) n+1 2( 1) ; T n n ()= 2 n [( 1)n+1 + 1]. (3.132) Solve T n. As T n stisfies nonhomogeneous 2nd order liner constnt coefficient eqution, we hve to solve the corresponding homogeneous eqution T +4n 2 T = (3.133) first to get its generl solution, nd then find prticulr solution of T n + 4 n 2 T n = ( 1) n+1 2 n t. Inspecting the right hnd side, we conclude tht the est pproch to get the prticulr solution should e undetermined coefficients. T + 4n 2 T =. The generl solution is Prticulr solution. The correct form is Sustitute into the eqution we esily otin T = C 1 cos (2 n t)+c 2 sin (2 n t). (3.134) T p = At. (3.135) Thus T n hs to e of the form T p = ( 1)n+1 2 n 3 t. (3.136) Now enforce the initil conditions: which gives Therefore T n = C 1 cos (2 n t)+c 2 sin (2 n t)+ ( 1)n+1 2n 3 t. (3.137) T n ()=( 1) n+1 2( 1) n C 1 =( 1) n+1 2 ( 1) ; (3.138) n T n ()= 2 n [( 1)n+1 + 1] 2 n C 2 + ( 1)n+1 2 n 3 = 2 n [( 1)n+1 + 1] (3.139) C 1 = ( 1) n+1 2( 1), C 2 = 1 n n 2 [( 1)n+1 + 1] ( 1)n+1 4n 4. (3.14) T n (t) = ( 1) n+1 2 ( 1) n ( 1) n+1 2n 3 t. Write down the solution. We hve v(x, t) = ( 1) n+1 2n 3 [ ( 1) n+1 2 ( 1) n ] t sin (n x), cos (2 n t) + cos (2 n t) + ( ) 1 n 2 [( 1)n+1 + 1] ( 1)n+1 4n 4 sin (2 n t) + (3.141) ( ) 1 n 2 [( 1)n+1 + 1] ( 1)n+1 4 n 4 sin (2 n t) + (3.142)

11 nd [ ( 1) n+1 2( 1) n ( 1 n 2 [( 1)n+1 + 1] u(x, t) = v(x, t) + w(x) = x + cos (2 n t) + ) ] ( 1) n+1 4n 4 sin (2nt) + ( 1)n+1 2 n 3 t sin (n x). (3.143) Discussions. The success of the ove method relies on the following: 1. The process of seprting the vriles leds to certin eigenvlue prolem; 2. This eigenvlue prolem yields sequence of eigenvlues, nd ech eigenvlue hs one-dimensionl eigenspce tht is ny two eigenfunctions of the sme eigenvlue re linerly dependent; 3. It is possile to expnd ny resonly smooth function into (infinite) liner comintion of these eigenfunctions. The first is esy to verify. The second nd third though re not ovious t ll. When the eigenfunctions re sin n x n x or cos, we hve seen in erlier PDE courses tht 2. cn e shown y directly solving the L L eigenvlue prolem, while 3. follows from the theory of Fourier sine/cosine series. In the generl cse, we need the so-clled Sturm-Liouville theory, which will e discussed in sections 3.2 nd 3.3. Exercises. Exercise 3.1. Consider the Telegrpher s eqution (recll tht λ > ) over the intervl x [, L] suject to conditions u xx = u tt + λ u t (3.144) u(, t)=u(l, t) =; u(x,) = f(x), u t (x, )=h(x). (3.145) Use the method of seprtion of vriles to study the limiting ehvior of u s t. Exercise 3.2. Consider the het eqution with Dirichlet oundry condition: Now consider the semi-discretiztion of the eqution: Replce u xx y t) where h=l/n, the eqution ecomes n ODE system: U(t) with U =, nd A = κ h Un(t) u t = κ u xx, u(, t)=u(, t)=. (3.146) u(x+ h, t) 2 u(x, t) + u(x h, t) h 2. If we set U i (t)=u(i h, U t = A U, U (t) = U n (t)= (3.147) 1 2 ) Show tht U (m) with U j (m) = sin ( m j n ( ij = 2 i = j 1 i j =1 ). i j > 1 ) is n eigenvector of the mtrix A. ) Show tht ny solution to (3.147) cn e otined s follows: Set U = v m (t) U (m) nd solve n ODE for v m (t). c) By tking limit n, formlly justify the method of seprtion of vriles for (3.146). Exercise 3.3. Consider the ritrry liner first order PDE: (x, y) u x + (x, y) u y + c(x, y) u = d(x, y). (3.148) For wht d is this eqution solvle through seprtion of vriles? Descrie the solution procedure nd use it to solve the eqution Exercise 3.4. Consider n ritrry liner second order PDE: x u x + y u y = u. (3.149) (x, y) u xx + (x, y) u xy + c(x, y) u yy + d(x, y) u x + e(x, y) u y + f(x, y) u = g(x, y). (3.15) For wht g is this eqution solvle through seprtion of vriles?

12 3.1. Higher Dimensionl Prolems nd Specil Functions Rectngulr domins. Exmple 3.7. Find forml solution to the initil-oundry vlue prolem u = 2 u t x u y2, <x <, < y <, t > (3.151) u u (, y, t) = (, y, t) =, < y <, t > (3.152) x x u(x,, t) = u(x,, t) =, <x<, t > (3.153) u(x, y, ) = y, <x <, < y <. (3.154) Solution. We follow the procedure of seprtion of vriles. 1. Seprte the vriles. Write u(x, y, t)=x(x)y (y)t(t). Sustitute into the eqution, we hve Divide oth sides y T XY : T X Y = T X Y + T X Y. (3.155) T T = X X + Y Y. (3.156) As the left hnd side only depends on t while the right hnd side is independent of t, oth sides hve to e constnt. Applying the sme rgument one more time, we conclude tht The equtions for X, Y, T re then X X = λ, Y Y = µ, T = λ+ µ. (3.157) T X λx =, X ()=X ()=; (3.158) Y µ Y =, Y ()=Y ()=; (3.159) T (λ+ µ) T =. (3.16) 2. Solve the eigenvlue prolems. Now there re two eigenvlue prolems. We solve them one y one. i. Solve X n. X λx =, X ()=X ()=; (3.161) We hve eigenvlues λ n = n 2, n =, 1, 2, nd eigenfunctions X n = n cos(n x), n = 1, 2, 3. ii. Solve Y m. Y µ Y =, Y ()=Y ()=; (3.162) We hve eigenvlues µ m = m 2, m = 1, 2, 3, nd eigenfunctions Y m = m sin (m y), m = 1, 2, 3, 3. Solve T n,m. We hve 4. Write T n,m + (n 2 + m 2 )T n,m = T n,m =T n,m ()e (n2 +m 2 )t. (3.163) u(x, y, t) = n= c nm cos(n x) sin (m y)e (n2 +m 2 )t. (3.164) m=1 5. Compute the coefficients. We hve u(x, y, )= n= c nm cos (nx) sin (m y)e (n2 +m 2 )t. (3.165) m=1

13 To determine the coefficients, we first need to understnd the integrtions cos(n x) sin (m y) cos (n x) sin (m y) dxdy. (3.166) We compute [ cos (n x) sin (m y) cos(n x) sin (m y)dx dy = [ ] cos (n x) cos (n x)dx ] sin (m y) sin (m y)dy. (3.167) Recll tht for n, n {, 1, 2, 3, } we hve n = n = cos (n x) cos (n x) dx = 2 n = n, (3.168) n n nd for m, m {1, 2, 3, } Therefore As consequence, sin (m y) sin (m y)dy = 2 m = m (3.169) m m n = n =, m =m 2 cos (n x) sin (m y) cos (n x) sin (m y)dxdy = 2 n = n, m =m. (3.17) 4 n n or m m Now we compute Thus Therefore Summrizing, we hve 2 u(x, y, ) cos (nx) sin (m y)dx dy = 2 c m n = 2 4 c nm n = 1, 2, 3, 2 y sin (m y)dxdy = y sin (m y)dy = y dcos(m y) m = [ y cos (m y)n m y cos (n x) sin (m y) dxdy = ] cos (m y)dy. (3.171) = 2 m ( 1)m. (3.172) c m = 2 m ( 1)m+1. (3.173) [ ][ ] cos (n x)dx y sin (m y) dy =. (3.174) c nm =, n = 1, 2, 3, ; m = 1, 2, 3, (3.175) u(x, y, t)= 2 ( 1) m+1 e m2t sin (m y). (3.176) m m=1

14 Lplce s equtions in polr nd sphericl coordintes. We consider Lplce s eqution in the unit disc nd unit ll. Exmple 3.8. (Polr) Solve Solution. We get This leds to nd The Θ eqution is esily solved, giving u xx + u yy =, x 2 + y 2 <1; u= f(θ) x 2 + y 2 =1. (3.177) r 2 R R + r R R = Θ =λ. (3.178) Θ Θ +λθ=; Θ()=Θ(2), Θ ()=Θ (2 ) (3.179) r 2 R + rr λr = ; R() ounded; R(1)= f(θ). (3.18) λ n = n 2, n =, 1, 2, 3, (3.181) Θ =1, Θ n1 = cos n θ, Θ n2 = n sin n θ. (3.182) Sustituting the λ n s into the R eqution nd expnding f(θ) into Fourier series we otin updted R eqution s f(θ)= 2 + [ n cos n θ + n sin n θ], (3.183) r 2 R +rr =, R () ounded, R (1)= 2 ; (3.184) r 2 R ni + rr ni n 2 n i=1 R =, R ni () ounded, R ni (1)={ n i=2. (3.185) Notice tht the R eqution is Cuchy-Euler, which mens it cn e solved y setting R=r α with α stisfying α (α 1)+α n 2 = α 1,2 = ±n. (3.186) When n =, we hve repeted root, therefore the generl solution for R is R (r)=c 1 +C 2 ln r. (3.187) The oundry conditions then dicttes tht C 2 =, C 1 = 2 ; Similrly, we hve R n1(r) = n r n nd R n2 (r) = n r n. Summrizing, the solution to the prolem is u(r, θ)= 2 + r n [ n cosnθ + n sin n θ] (3.188) where n, n comes from (3.183), the Fourier expnsion of f. Exmple 3.9. (Sphericl with rottionl symmetry) We consider the Lplce eqution u xx +u yy + u zz =, u= f on x 2 + y 2 + z 2 = 1. (3.189) It is cler tht we should turn to sphericl coordintes x=r cosθ sin ϕ, y =r cos θ cos ϕ, z = r cos ϕ (3.19) where θ is the ngle (on x-y plne) from the x xis (tht is tn θ = y/x), nd ϕ the verticl ngle from the z xis (tht is cos ϕ =z/r). Clerly < θ 2, ϕ. The eqution now ecomes u rr + 2 r u r + 1 [ r 2 u ϕϕ + (cot ϕ) u ϕ + 1 ] sin 2 ϕ u θθ =, u(1, θ, ϕ)= f(θ, ϕ). (3.191)

15 We first consider the cse where f hs rottionl symmetry, tht is f = f(ϕ). Then it is resonle to expect u=u(r, ϕ). The prolem now reduces to Let u(r, ϕ) =R(r)Φ(ϕ), we rech u rr + 2 r u r + 1 r 2 [u ϕϕ +(cot ϕ) u ϕ ] =, < r < 1, < ϕ < (3.192) u(1, ϕ) = f(ϕ), < ϕ < (3.193) r 2 R + 2 rr λr =, R, R ounded s r +; (3.194) The Φ eqution, under the chnge of vrile x=cos ϕ, ecomes Φ + cot ϕφ + λφ=. (3.195) (1 x 2 ) y 2xy + λy =, 1<x<1. (3.196) where y(x) 6 Φ(ϕ). The resonle oundry condition for the y equtions should e y, y remin ounded s x 1 nd x 1 +. It turns out tht such oundry condition lredy determines list of eigenvlues nd eigenfunctions: λ n =n(n +1), n =, 1, 2, ; P n = 1 d n 2 n n! dx n[(x2 1) n ]. (3.197) These P n s re clled Legendre polynomils. With λ n s known, we cn esily solve R n =r n. So the solution is where n comes from the expnsion with n = u(r, ϕ) = n= f(ϕ)= n= n r n P n (cos ϕ) (3.198) n P n (cos ϕ). (3.199) f(ϕ)pn (cos ϕ) sin ϕdϕ P 2. (3.2) n (cosϕ) sinϕ dϕ Exmple 3.1. (Sphericl, generl cse) Now we consider the generl cse u rr + 2 r u r + 1 [ r 2 u ϕϕ + (cot ϕ) u ϕ + 1 ] sin 2 ϕ u θθ =, u(1, θ, ϕ)= f(θ, ϕ). (3.21) Setting u(r, θ, ϕ)=r(r)θ(θ)φ(ϕ), we rech r 2 R +2rR λ R =, (3.22) ( Θ + µ Θ =, (3.23) Φ + cot ϕ Φ + λ µ 1 ) sin 2 Φ =. (3.24) ϕ Here R, Φ suject to similr oundry conditions s in the lst exmple, while Θ enjoys the periodic oundry condition. It is cler tht Θ should e solved first to yield µ m = m 2, m =, 1, 2, (3.25) with eigenfunctions 1 (for m = ) nd cos m θ, sin m θ for m = 1, 2, 3,. Tking the chnge of vrile x=cos ϕ nd set y(x) 6 Φ(ϕ) we rech ( (1 x 2 ) y 2 xy + λ m2 1 x 2 ) y =, 1 <x <1. (3.26)

16 This is clled ssocited Legendre s eqution. The eigenvlues re still n (n + 1), n =, 1, 2, while the (ounded) eigenfunctions re now the ssocited Legendre functions of first kind P n m (x) 6 ( 1) m (1 x 2 ) m/2 d m dx mp n(x) (3.27) where P n (x) re the Legendre polynomils in the lst exmple. Note tht the ove formul gives P m n (x)= when m > n. The solution now reds u(ρ, θ, ϕ) = n r n [ nm cos(m θ) + nm sin (m θ)] P m n (cosϕ) (3.28) n= with nm, nm given y nm = nm = m= 2 f(θ, m ϕ)pn (cos ϕ) cos (m θ) sinϕdθdϕ 2 [P m n (cos ϕ) cos (m θ)] 2 sinϕdθ dϕ ; (3.29) 2 f(θ, ϕ)p m n (cos ϕ) sin (m θ) sinϕdθdϕ 2 [P m n (cos ϕ) sin (m θ)] 2 sinϕdθ dϕ. (3.21) Remrk Note tht we cn expnd ny function defined on the sphere y cos (m θ) P n m (cos ϕ) nd sin (m θ)p n m (sin ϕ). This is clled sphericl hrmonics expnsion Het eqution in the cylinder. Exmple Consider the het eqution in 2D disc x 2 + y 2 1: u t = κ (u xx +u yy ) (3.211) u(x, y, ) = f(x, y) (3.212) u(x, y, t) = x 2 + y 2 = 1. (3.213) Solution. Due to the specil geometry of the domin, it is nturl to consider the prolem using polr coordintes (r, θ) stisfying x=r cosθ, y = r sinθ. (3.214) Now we chnge the vriles from x, y to r, θ. Differentiting the ove reltion we hve consequently Therefore (cosθ)r x r (sinθ) θ x = 1 (3.215) (cosθ)r y r (sinθ)θ y = (3.216) (sinθ)r x +r (cosθ) θ x = (3.217) (sinθ) r y +r (cosθ)θ y = 1 (3.218) r x = x r, r y = y r, r xx= 1 r x2 r 3, r yy = 1 r y2 r3; (3.219) θ x = sinθ r = y r 2, θ y = cosθ = x r r 2, θ xx= 2 xy r 4, θ yy = 2xy r 4 ( ) x u xx = u 2 rr r 2 u 2 xy rθ r 3 + u y 2 1 θθ r 4 +u r r x2 2 xy r 3 + u θ r 4, (3.22) ( ( y u yy = u 2 rr r 2 + u 2xy rθ r 3 +u x 2 1 θθ y 4 +u r )+ r y2 r 3 u θ 2 xy ) r 4. (3.221) The eqution nd the initil-oundry conditions in polr coordinte form re ( u t = κ u rr + 1 r u r + 1 ) r 2 u θθ (3.222) u(r, θ, ) = f(r, θ) (3.223) u(1, θ, t) =. (3.224)

17 We pply seprtion of vriles to solve this eqution. First we try to find non-trivil sic solutions of the form u(r, θ, t)=r(r)θ(θ)t(t). (3.225) Sustituting this into the eqution we rech ( R(r) Θ(θ)T (t)=κ R (r)θ(θ)+ 1 r R (r)θ(θ)+ 1 ) r 2 R(r)Θ (θ) T(t). (3.226) Dividing oth sides y R(r) Θ(θ)T(t) we rech T (t) T(t) = κ ( R (r) R(r) + 1 R (r) r R(r) + 1 Θ (θ) r 2 Θ(θ) As the LHS only involves t nd the RHS only r, θ there is constnt λ such tht nd Multiply oth sides y r 2 we hve ). (3.227) T (t) = κ λ (3.228) T(t) R (r) R(r) + 1 R (r) r R(r) + 1 Θ (θ) r 2 = λ. (3.229) Θ(θ) r 2 R (r) R(r) + r R (r) R(r) +λr2 = Θ (θ) Θ(θ). (3.23) The LHS only involves r nd the RHS only θ. Thus there is constnt µ such tht Θ (θ) Θ(θ) = µ, As Θ(θ) is oviously 2 periodic, we hve nd r 2 R (r) R(r) On the other hnd, the eqution for R now ecomes with the oundry condition The generl solution is + r R (r) R(r) +λr2 = µ. (3.231) µ = n 2, n =1, 2, 3, (3.232) Θ(θ) =Acos (nθ)+b sin (n θ). (3.233) r 2 R + rr +(λr 2 n 2 )R=, (3.234) R(1) =, R() ounded. (3.235) ( R(r)=C 1 J n λ ) ( ) r + C2 Y n λ r (3.236) with J n,y n Bessel functions of the first nd second kinds. It turns out tht Y n (r) is unounded s r +, therefore we hve ( ) R n (r)=j n λ r. (3.237) Applying the oundry condition R(1) = gives Now we expnd f(r, θ) = n,k R n,k (r) =J n (α n,k r), 2 λ n,k = α n,k. (3.238) n,k R n,k (r) cos (nθ)+ n,k R n,k sin (n θ). (3.239) nd the solution is given y [ n,k R n,k (r) cos (n θ)+ n,k R n,k sin (nθ)] e λn,kt (3.24) n,k

18 Discussions. We see tht similr to the 1D sitution, the success of the ove method relies on the following: 1. The process of seprting the vriles leds to certin eigenvlue prolem; 2. This eigenvlue prolem yields sequence of eigenvlues, nd ech eigenvlue hs one-dimensionl eigenspce tht is ny two eigenfunctions of the sme eigenvlue re linerly dependent; 3. It is possile to expnd ny resonly smooth function into (infinite) liner comintion of these eigenfunctions. In the higher dimensionl cse, it is quite uncler which weight we should use for the expnsion formuls. Reference. John M. Dvis, Introduction to Applied Prtil Differentil Equtions, Chp. 5, 6. Exercises. Exercise 3.5. Consider the oundry vlue prolem for u(x, y) in the nnulr region: { u xx + u yy = ρ 2 < x 2 + y 2 f x < 1; u(x, y)= 2 + y 2 = ρ 2 g x 2 + y 2 =1. (3.241) Otin the formul for the solution using seprtion of vriles. Exercise 3.6. Consider the Lplce eqution in the disc x 2 + y 2 < ρ 2, with oundry condition u = f. Using seprtion of vriles, derive Poisson s formul: u(r, θ)= ρ2 r 2 2 f(ϕ) 2 ρ 2 dϕ. (3.242) 2 ρ r cos(θ ϕ) + r2 Cn we otin similr formul for solutions of the ove nnulus prolem? Exercise 3.7. Using Poisson s formul (3.242) to prove the following men vlue property: u(, θ) = 1 2 f(ϕ) dϕ. (3.243) 2 Then estlish the following mximum principle: If u xx + u yy = for (x, y) Ω R 2, nd u= f on Ω, then mx Ω u = mx Ω f. Cn this line of rgument e used to prove the uniqueness of the clssicl solution to u xx + u yy = g in Ω, u= f on Ω? Why? Exercise 3.8. Consider the eqution r 2 R + r R + (λ r 2 n 2 ) R =, R() ounded, R(1)=. (3.244) ) Prove tht there re no negtive eigenvlues. ) Prove tht λ = is not n eigenvlue. c) Let λ k λ l e eigenvlues, prove tht the eigenfunctions R k (r), R l (r) stisfy 1 R k (r) R l (r) r dr =. (3.245) Exercise 3.9. Let n N. Consider the eqution (1 x 2 ) y 2 x y + n(n +1) y =, 1 < x <1 (3.246) Use power series method to show tht the generl solution is y = c 1 P n + c 2 Q n where P n is polynomil of degree n, nd Q n is unounded t 1 nd 1 +. Exercise 3.1. Solve the virting drum: with (Ans: c n J (k n r) cos(k n t)). u tt = u xx + u yy = u rr + r 1 u r + r 2 u θθ. (3.247) u(r, θ, )= f(r), u t (r, θ, )=, u(1, θ, t)=. (3.248)

19 3.2. Sturm-Liouville theory. First recll how we prove convergence for the 1D cses. This relies on the explicit formul for prtil sums nd cnnot e esily generlized Sturm-Liouville prolems. The stndrd Sturm-Liouville (SL) prolem is of the form (p(x) y ) + q(x) y +λr(x) y =, < x < (3.249) α y() +α 1 y () =, (3.25) β y()+ β 1 y () =. (3.251) where ll the functions nd numers re rel. For simplicity we ssume the coefficients re s smooth s we need. The prolem is clled regulr when p, q,r re ounded on [,] (tht is the intervl x ), p,r > for ll x, nd α, α 1 rel, not oth, nd β, β 1 rel, not oth. singulr when ny one or more of the following hppens The intervl (, ) is infinite, tht is either = or =+ or oth occurs. p(x)= for some x [, ] or r(x)= for some x [, ]. One or severl coefficient function ecomes t or, or oth. Exmple We check the systems we hve delt with We hve y +λy =, y()= y(l) = (3.252) =, = l; p(x)=1, q(x)=, r(x)=1; α = 1, α 1 =, β = 1, β 1 =. (3.253) The system is regulr SL prolem. We hve y + λy =, y ()= y (l)= (3.254) =, = l; p(x)=1, q(x)=, r(x)=1; α =, α 1 =1, β =, β 1 = 1. (3.255) This is lso regulr SL prolem. We hve y + λy =, y()=, y (l)= h y(l). (3.256) =, = l; p(x)=1, q(x)=, r(x)=1; α = 1, α 1 =, β = h, β 1 =1. (3.257) x 2 y + x y +(λx 2 n 2 ) y =, y() ounded, y(1) =. (3.258) At first sight this prolem is not n SL prolem. However we cn trnsform it through the following opertions: We serch for multiplier h(x) such tht h(x)[x 2 y +xy + (λx 2 n 2 ) y] = (p y ) + q y +λry. (3.259)

20 Compring the two sides, we hve which leds to h(x)x 2 = p(x), h(x)x= p(x) (3.26) p(x) = 1 x p(x) p(x)=x h(x)= 1 x. (3.261) Thus we see tht the eqution is equivlent to which corresponds to (xy ) n2 x y + λxy = (3.262) =, =1; p(x)=x, q(x) = n2 x, r(x)=x; β =1, β 1 =. (3.263) This is singulr SL prolem. Any λ tht the prolem hs non-trivil solutions is clled n eigenvlue, the corresponding solutions re clled eigenfunctions Properties of regulr Sturm-Liouville prolems. We see from the following theorem tht the solutions to SL prolem enjoy similr properties s the functions sin ( n x ) nd cos ( n x ) in the Fourier series. More specificlly we hve the following theorem. l l Theorem A regulr SL prolem hs the following properties. 1. Aout eigenvlues: It hs nonzero solutions for countly infinite set of vlues of λ, clled eigenvlues of the prolem. These eigenvlues re ll rel. The set of eigenvlues does not hve ny limit points in R. These eigenvlues re ounded from elow if α α 1 nd β β 1. These eigenvlues re ounded from elow y if furthermore q. In summry, if we hve α α 1 nd β β 1, then then eigenvlues cn e enumerted s: λ 1 λ Aout eigenfunctions: ) For ech fixed eigenvlue λ n, the solution spce is one-dimensionl. Tht is, there is ϕ n such tht ll other solutions for the sme λ is multiple of ϕ n. ) (Orthogonlity) ϕ n (x) ϕ m (x)r(x)dx= for ny n m. c) (Bessel s inequlity) If ϕ n s re chosen such tht We hve the following Bessel s inequlity ϕ n (x) 2 r(x)dx = 1 (3.264) f(x) 2 r(x)dx c n 2. (3.265)

21 d) (Completeness) c1) The only continuous function f on [, ] with f. f(x) ϕ n (x) r(x) dx= for ll n is c2) For ny f hving two continuous derivtives on [, ] nd stisfying the oundry conditions, the infinite sum c n ϕ n (3.266) where f(x) ϕ n (x)r(x) dx c n = ϕ n (x) 2 r(x)dx (3.267) converges solutely uniformly to f(x). By solutely uniformly we men c n ϕ n < (3.268) nd the convergence to f is uniform. 1 c3) (Prsevl s equlity) If function f(x) is continuous nd ϕ n s re chosen such tht ϕ n (x) 2 r(x)dx = 1 (3.269) We hve the following f(x) 2 r(x)dx = c n 2. (3.27) Remrk We note tht the completeness prt is quite schizophreni f(x) is continuous in one line ut required to e twice continuously differentile in the next. This will e resolved in lter sections when we tke higher, functionl nlytic, point of view. Exmple We check the systems we hve delt with We hve r(x) = 1, therefore the eigenfunctions stisfy nd the expnsion reds y +λy =, y()= y(l) = (3.271) l ϕ n (x) ϕ m (x)dx = n m (3.272) f(x) = c n ϕ n, c n = l f(x) ϕn (x)dx l [ϕ n (x)] 2 dx. (3.273) y + λy =, y ()= y (l)= (3.274) We hve gin r(x) = 1, thus lthough the eigenfunctions re now different, the orthogonlity reltion nd expnsion formul remin the sme s ove.

22 y + λy =, y()=, y (l)= h y(l). (3.275) We hve r(x) = 1. Thus the orthogonlity reltion nd expnsion formul remin the sme lthough we cnnot write the explicit formuls for the eigenfunctions nymore! Bessel functions. x 2 y + x y +(λx 2 n 2 ) y =, y() ounded, y(1) =. (3.276) Recll tht y multipliction of h(x) = 1/x the prolem is turned into the Sturm-Liouville form (xy ) n2 x y + λxy = (3.277) Although this prolem is singulr, it turns out tht its eigenvlues/eigenfunctions enjoy similr properties s those in the regulr cse. In prticur, we hve the orthogonlity reltion nd expnsion formuls: 1 1 ϕ n (x) ϕ m (x)xdx=, c n = f(x) ϕn (x) xdx 1 [ϕ n (x)] 2 x dx. (3.278) Legendre functions. [(1 x 2 ) y ] +λy =, 1 <x <1. (3.279) As p(x)=1 x 2 vnishes t oth ends, the oundry conditions should e tken s y, y remin ounded s x ±1. (3.28) The eigenvlues re λ n = n (n + 1). Here r(x)=1, so the corresponding eigenfunctions stisfy Hermite functions. 1 1 P m (x)p n (x)dx=, n m. (3.281) u 2xu + λu=, < x < (3.282) To write this prolem into SL prolem, we multiply the eqution y e x2 to otin ( e x 2 u ) +λe x 2 u =, < x <. (3.283) This is regulr S-L prolem. Now tht we hve p(x) = e x2 which tends to s x ±, the oundry conditions should e u, u remin ounded s x ±. (3.284) The eigenvlues re λ n =2 n for nonnegtive integers n. Since r(x)=e x2, the orthogonlity property reds H n (x) H m (x)e x2 dx =, n m. (3.285) Proof. (Of the esy prts of theorem 3.14) The proofs for some of the ove clims re more technicl. We postpone them to the next lecture nd only prove the esy ones (those in lue) here. 1. Properties of the eigenvlues. These eigenvlues re ll rel.

23 Let λ e n eigenvlue nd let ϕ e corresponding eigenfunction. We compute = [(p y ) + q y + λ ry] ȳ dx = (p y ) ȳ + q y 2 + λ r y 2 = (py ) ȳ N py ȳ + q y 2 +λ r y. (3.286) On the other hnd, tking the complex conjugte of we otin (p y ) + q y + λry = (3.287) (p ȳ ) + q ȳ + λ r ȳ =. (3.288) In other words, λ is lso n eigenvlue with eigenfunction ȳ. Multiplying this eqution y y nd integrte, we hve = [(p ȳ ) + q ȳ + λ rȳ ] y = (p ȳ ) y + q y 2 +λ r y 2 = (p ȳ ) yn p y ȳ + q y 2 + λ r y 2. (3.289) Comining the ove, we rech (λ λ ) r y 2 = p() [y () ȳ () ȳ () y()] p()[y () ȳ () ȳ () y()]. (3.29) Using the oundry conditions we see tht Therefore α y() +α 1 y () =, (3.291) β y()+ β 1 y () =. (3.292) y () ȳ () ȳ () y()=, y () ȳ () ȳ () y() =. (3.293) which leds to λ = λ, or λ is rel. (λ λ ) r y 2 dx = (3.294) These eigenvlues re ounded from elow if α α 1 nd β β 1. These eigenvlues re ounded from elow y if furthermore q. We hve = [(p y ) + q y + λry] ȳ dx = (p y ) ȳ + q y 2 + λ r y 2 = (p y ) ȳ N p y ȳ + q y 2 +λ r y 2. = p() y () ȳ () p() y () ȳ () [p y 2 q y 2 ] +λ r y 2. (3.295)

24 Thus { p() y () ȳ ()+ p() y () ȳ ()+ } [p y 2 q y 2 ] λ=. (3.296) r y 2 Using the oundry conditions we hve p() y () ȳ () = p() β β 1 y() 2, (3.297) p()y () ȳ ()= p() α α 1 y() 2. (3.298) When α α 1 nd β β 1, oth terms re non-negtive which mens If furthermore q, we see tht λ too. 2. Properties of eigenfunctions. λ q y 2 r y. (3.299) 2 For ech fixed eigenvlue λ, the solution spce is one-dimensionl. Tht is, there is y λ such tht ll other solutions for the sme λ is multiple of y λ. Fix λ. Let y(x) nd z(x) e two eigenfunctions. Tht is nd (p(x) y ) + q(x) y +λr(x) y =, < x < (3.3) α y() +α 1 y () =, (3.31) β y()+ β 1 y () =. (3.32) (p(x)z ) + q(x)z + λr(x) z =, < x < (3.33) α z() +α 1 z () =, (3.34) β z()+ β 1 z () =. (3.35) Multiplying the y eqution y z nd z eqution y y, nd sutrct, we hve We conclude tht = (p y ) z (p z ) y = (p (y z z y)). (3.36) p(x)(y z z y)(x) = p()(y z z y)(). (3.37) As y, z oth stisfy the oundry conditions, we hve which leds to p()(y ()z() z () y())= (3.38) p (y z z y) = y z z y = (3.39) for ll x s p(x) >. Finlly, y z z y = (Orthogonlity) y y = z z ln y ln z = constnt y/z = constnt. (3.31) ϕ n (x) ϕ m (x)r(x)dx= for ny n m. It suffices to show tht if λ, µ re two distinct eigenvlues, nd y, z the corresponding eigenfunctions, then yzr dx =.

25 Exercises. Using the equtions we hve [(p y ) + q y + λry] z [(p z ) + q z + µ r z] y dx =. (3.311) After using the oundry conditions, we cn show tht LHS = (λ µ) yzr dx. (3.312) Therefore As λ µ, we hve (Bessel s inequlity) From orthogonlity we hve [ f(x) N (λ µ) y(x)z(x)r(x)dx=. (3.313) Tking limit n we otin Bessel s inequlity. y(x)z(x)r(x)dx=. (3.314) ] 2 c n ϕ n r(x)dx = f(x) 2 r(x)dx N c n 2. (3.315) Exercise Write the following equtions into S-L form nd discuss whether they re regulr or singulr. Determine wht is the orthogonlity reltion their eigenfunctions should stisfy. ) Legendre s eqution: ) Cheyshev s eqution c) Lguerre s eqution d) Hermite s eqution e) Bessel s eqution of order n (1 x 2 ) y 2 x y + λ y =, 1 < x < 1 (3.316) (1 x 2 ) y x y + λ y =, 1 < x <1 (3.317) x y + (1 x) y + λ y =, < x < (3.318) y 2 x y + λ y =, < x < (3.319) x 2 y + x y + (λ x 2 n 2 ) y =, < x < 1. (3.32) Exercise Give ny second order eqution (x) y + (x) y + c(x) y =. (3.321) Prove tht there exists multiplier h(x) such tht h(x) [(x) y + (x) y + c (x) y] =(p(x) y ) + q(x) y. (3.322) Note tht the term of first order derivtive disppers. Exercise (Dvis) Consider the S-L prolem (p y ) + q y + λ y =, < x <, y()=, y()=. (3.323) Show tht if p(x), q(x) M, then ny eigenvlue λ M. References. Anthony W. Knpp, Advnced Rel Anlysis, 1.3. John M. Dvis, Introduction to Applied Prtil Differentil Equtions, Chp. 4.

26 3.3. Proof of the Theorem (Difficult prts). Recll tht we re studying regulr Sturm-Liouville prolems: (p(x) y ) + q(x) y +λr(x) y =, < x < (3.324) α y() +α 1 y () =, (3.325) β y()+ β 1 y () =. (3.326) with p, q, r re ounded on [, ] (tht is the intervl x ), p, r > for ll x, nd α, α 1 rel, not oth, nd β, β 1 rel, not oth. We would like to prove: It hs nonzero solutions for countly infinite set of vlues of λ. These eigenvlues re ll rel. The set of eigenvlues does not hve ny limit points. For ny f hving two continuous derivtives on [, ] nd stisfying the oundry conditions, the infinite sum c n ϕ n (3.327) where f(x) ϕ n (x)r(x) dx c n = ϕ n (x) 2 r(x)dx converges solutely uniformly to f(x). By solutely uniformly we men nd the convergence to f is uniform. The only continuous function f on [, ] with If ϕ n s re chosen such tht (3.328) c n ϕ n < (3.329) 1 f(x) ϕ n (x)r(x) dx= for ll n is f. ϕ n (x) 2 r(x)dx = 1 (3.33) We hve the following Prsevl-type reltion f(x) 2 r(x)dx = c n 2. (3.331) To do these we need to first trnsform the eqution to n equivlent integrl eqution Equivlent integrl eqution. First we show tht there is function G(x; ξ) stisfying: y = G(x, ξ) f(ξ) dξ solves (p(x) y ) + q(x) y + f(x) =, < x < (3.332) α y()+α 1 y () =, (3.333) β y()+ β 1 y () =. (3.334) Such function is clled the Green s function to the prolem. We do this through explicit construction. The sic ide is s follows. We try to find solution to (p(x) y ) + q(x) y + δ(x ξ) =, <x < (3.335) α y() +α 1 y () =, (3.336) β y()+ β 1 y () =. (3.337) nd cll this solution G(x, ξ). Then it is cler tht y = G(x, ξ) f(ξ) dξ is the solution to the originl prolem. The proof of this is left s n exercise.

27 Since δ(x ξ)= for x ξ, we see tht (Here we fix ξ, so stnds for x derivtive) (p(x)g ) + q(x)g= < x < ξ, ξ < x <. (3.338) Now y the theory of 2nd order ODE, there is y 1 (x) stisfying nd y 2 (x) stisfying (p(x) y ) + q(x) y =, α y()+α 1 y ()= (3.339) (p(x) y ) + q(x) y =, β y()+ β 1 y ()= (3.34) We ssume y 1, y 2 re linerly independent. Then we know the generl solution to (p y ) + q y = is C 1 y 1 + C 2 y 2. Therefore { C1 (ξ) y G(x, ξ)= 1 (x)+c 2 (ξ) y 2 (x) <x < ξ D 1 (ξ) y 1 (x) +D 2 (ξ) y 2 (x) ξ <x <. (3.341) Since y 1, y 2 re linerly independent, y 2, y 1 cnnot stisfy the oundry condition t, respectively. Consequently C 2 = D 1 =. To ccomodte the δ(x ξ) term, we integrte, for ny ε > : ξ+ε [(p(x)g ) ξ+ε + q(x)g] dx+ δ(x ξ) dx= (3.342) ξ ε ξ ε which leds to p(ξ + ε)g (ξ + ε, ξ) p(ξ ε) G (ξ ε, ξ)+ ξ+ε ξ ε q(x)gdx+1=. (3.343) Now tke the limit ε, As q, G re ounded, the integrl term. Using (3.341) we rech Together with the continuity of G t x = ξ: we otin Thus we otin p(ξ) [C 1 y 1 D 2 y 2 ] = 1. (3.344) C 1 y 1 D 2 y 2 = (3.345) y C 1 = 2 p [y 1 y 2 y 1 y 2 ], D y 2 = 1 p [y 1 y 2 y 1 y 2 ]. (3.346) y 2(ξ) y 1 (x) p[y G(x, ξ)= 1 y 2 y 1 y < x < ξ 2 ] y. 1(ξ) y 2 (x) p[y 1 y 2 y 1 y ξ < x < 2 ] (3.347) It turns out tht G(x, ξ)=g(ξ, x). (See exercise) Now we turn ck to the Sturm-Liouville prolem We see tht the solution cn e written s y(x)=λ (p(x) y ) + q(x) y +λr(x) y =, < x < (3.348) α y() +α 1 y () =, (3.349) β y()+ β 1 y () =. (3.35) G(x, ξ) r(ξ) y(ξ) dξ (3.351) Now set z(x)=r 1/2 (x) y(x) nd k(x, ξ)=r(x) 1/2 G(x, ξ)r(ξ) 1/2, we rech z(x)=λ k(x, ξ) z(ξ) dξ. (3.352) If we denote the opertor K[z] 6 k(x, ξ) z(ξ) dξ (3.353)

28 then we hve K[z] = µ z (3.354) where µ =λ 1. It is cler tht: The integrl form prolem (3.354) nd the originl Sturm-Liouville prolem shre eigenfunctions, nd hve one-to-one correspondence etween eigenvlues. Therefore in the following we study the integrl form. Remrk Note tht there my e prolem if λ = (nd λ < during the following proofs). This is esily fixed. Since ll (possile) eigenvlues re ounded elow, we cn find numer c nd chnge the eqution to Thus the new eigenvlues re λ +c which re ll positive. (py ) + (q c r) y + (λ+c)ry =. (3.355) Remrk Also note tht the orthogonlity reltions y n (x) y m (x)r(x)dx = (3.356) ecomes since z n (x)=r(x) 1/2 y n (x), z m (x)=r(x) 1/2 y m (x). z n (x)z m (x) dx= (3.357) Some nottions. Now we introduce some nottions to simplify the following presenttion. Lter we will see tht these cn e gretly generlized. Inner product: Let z, z e two rel functions, then their inner product is defined s (z, z ) 6 z(x) z (x) dx. (3.358) Inner product is the infinite dimensionl generliztion of dot product for vectors. L 2 Norm: We define the L 2 norm s z =(z, z) 1/2. (3.359) Remrk Norms re generliztions of length of vectors, which in turn generlizes solute vlue for numers. Thus sic requirement should e z, nd z = z =. This explins the requirement r(x) >. Orthogonlity: Two functions z, z re sid to e orthogonl if (z, z )=. Liner opertors nd self-djointness. A liner opertor is mpping K: L 2 L 2 such tht The opertor is self-djoint if K( z 1 + z 2 )=K[z 1 ] + K[z 2 ]; (3.36) (Kz 1, z 2 )=(z 1, Kz 2 ). (3.361) Self-djoint opertor is generliztion of symmetric (Hermitin if complex) mtrices. Norm of opertors. The norm of n opertor is defined s K 6 Kf sup Kf = sup f =1 f f. (3.362)

29 An opertor is sid to e ounded if its norm is finite. From definition we hve, for ny f, Kf K f. (3.363) Cuchy-Schwrz inequlity. Let (, ) e n inner product, nd the ssocited norm. Then We cn show tht our opertor Kf 6 is liner, self-djoint, ounded opertor. (f, g) f g. (3.364) k(x, ξ) f(ξ) dξ (3.365) Eigenvlues re countle nd isolted. In this section we show tht the eigenvlues re countle nd hve no limit point(s), in other words, the eigenvlues (ssuming it s ounded elow) cn e listed < λ 1 < λ 2 <λ 3 <. From the duscussion in the lst section we see tht it suffices to show tht the eigenvlues of (3.354) cn e listed s µ 1 > µ 2 > >, with the only possile limit point. This is done in severl steps. 1. There is t lest one eigenvlue. Consider m 6 sup z =1 (K[z], z) where the supreme is tken over ll continuous functions with z 2 = 1. Since k(x, ξ) is ounded function, m R. We show tht m = µ 1. By definition of sup there is {z n } stisfying z n = 1 nd (K[z n ], z n ) m. All we need to do is to show tht there is susequence of z n converging uniformly. However this turns out to e not esy. So we pply the following trick: Insted of showing z n n eigenfunction, we show K[z n ] n eigenfunction. The ide is to try to pply the following Theorem 3.2. (Arzel-Ascoli) Let {f n (x)} e sequence of functions tht re Uniformly ounded: There is M > such tht f n (x) M for ll n N nd ll x [,]. Equicontinuous: For ny ε >, there is δ > such tht x 1 x 2 < δ f n (x 1 ) f n (x 2 ) < ε holds for ll n N nd ll x 1, x 2 [, ]. Then there is susequence f nk (x) converging uniformly to some continuous function f(x). Tht is for ny ε>, there is K N such tht for ll k>n, nd ll x [,], f nk (x) f(x) <ε. First we show tht K[z n ] is uniformly ounded. This follows from the inequlity ( ) 1/2 ( K[z n ] k(x, ξ) z n (ξ) dξ k(x, ξ) 2 1/2 dξ z n (ξ) dξ) 2 (3.366) nd the oundedness of k. Next we show tht K[z n ] is equicontinuous, tht is for ny ε > there is δ > such tht whenever x 1 x 2 <δ, K[z n ](x 1 ) K[z n ](x 2 ) <ε. To see this we write K[z n ](x 1 ) K[z n ](x 2 ) = [k(x 1, ξ) k(x 2, ξ)] z n (ξ)dξ k(x 1, ξ) k(x 2, ξ) z n (ξ) dξ ( ) 1/2 ( ) k(x 1, ξ) k(x 2, ξ) 2 1/2 dξ z n (ξ) 2 dξ = ( k(x 1, ξ) k(x 2, ξ) 2 dξ) 1/2. (3.367)

30 Since k is continuous over [,] [,], it is uniformly continuous, thus there is δ > such tht whenever x 1 x 2 <δ, k(x 1, ξ) k(x 2, ξ) < ε. Therefore there is susequence z nk such tht K[z nk ] converges uniformly to some function ϕ. We now show tht ϕ must e n eigenfunction with eigenvlue m. We hve K[z nk ] mz nk 2 = (K[z nk ] m z nk, K[z nk ] m z nk ) This implies ϕ 2 m 2 > ϕ. Since K is ounded, to show Kϕ=mϕ, ll we need = (K[z nk ], K[z nk ]) 2m(K[z nk ], z nk )+m 2 ϕ 2 m 2 (3.368) = Kϕ mϕ = lim k K[K[z n k ]] m K[z nk ] K[z nk ] mz nk. (3.369) We re-visit K[z nk ] m z nk 2 = (K[z nk ], K[z nk ]) 2m(K[z nk ], z nk )+m 2 m 2 (3.37) All we need to show is K[z nk ] m. Proof of Kf m f. It suffices to prove Kf m for ll f 1. Tke ny f, g with f = g = 1. We hve (K(f g), (f g)) (see exercise) which leds to (Kf, f)+(kg, g) 2(Kf, g) (Kf, g) m. (3.371) This holds for ll f, g with f = g = 1. Now tke g = Kf K[z nk ] m s desired. 2. The eigenvlues cnnot hve nonzero limit point(s). Assume the contrry. We hve Kf leds to Kf m. Thus K[z nj ] = λ j z nj (3.372) with λ j λ >. Then without loss of generlity, we cn ssume λ j λ/2 for ll j. Now since z nj re orthogonl to one nother, we hve K[z nj ] K[z nk ] =λ j 2 +λ k 2 λ 2 /2. (3.373) On the other hnd we know tht there is susequence converging uniformly. Contrdiction Generlized Fourier expnsion. In this section we show tht the eigenfunctions cn e used to expnd generl twice continuously differentile function. First we show tht if there is continuous function f such tht (f,z n )= f(x) zn (x) dx= for ll eigenfunctions z n, then f(x) =. Assume not. Then consider ll such functions. Nme this set y H. Without loss of generlity, we ssume z n s re normlized, tht is z n =1. We first show tht if f H then so does Kf: (Kf,z n )=(f,kz n )=λ n (f,z n )=. (3.374) Note tht we hve used the self-djointness of K. Next we show tht if f m H, f m f, then f H. To see this we use Cuchy-Schwrz: (f,z n ) = (f m f,z n ) f m f z n = f m f. (3.375) Therefore we cn repet the rgument of existence of eigenvlues to show tht sup f H, f =1 (Kf, f)> (ecuse (Kf, f)> for ll f ) is n eigenvlue with some eigenfunction f. But then there must e n such tht f =z n which leds to ( f, z n ), contrdiction.

31 Next we show the Prsevl reltion. By Bessel s inequlity, we hve the convergence of c n z n with (f, zn) c n =. Now f c z n 2 n z n stisfies ( ) f cn z n, z n = (3.376) thus we hve f = c n z n which leds to Prsevl s equlity. Finlly we show tht if f C 2 (nd stisfies the oundry conditions) then the expnsion converges uniformly. Notice tht every such f cn e written s K[g] with g continuous. Then Exercises. N (f,z n )z n = N n=m n=m uniformly ecuse µ n. This mens the sequence N n=m nd convergence follows. Exercise Show tht if G(x, ξ) solves then y = G(x, ξ) f(ξ)dξ solves µ n g n z n (3.377) (f, z n ) z n is Cuchy (uniformly in x) (p(x) y ) + q(x) y + δ(x ξ) =, < x < (3.378) α y() + α 1 y () =, (3.379) β y() + β 1 y () =. (3.38) (p(x) y ) + q(x) y + f(x) =, < x < (3.381) Exercise Clculte Green s functions for the following prolems. ) y =, y()= y(1) =. ) y =, y() = y (1)=. α y() + α 1 y () =, (3.382) β y()+ β 1 y () =. (3.383) Exercise In the construction of Green s function, we need the solutions to nd (p(x) y ) + q(x) y =, α y()+α 1 y () = (3.384) (p(x) y ) + q(x) y =, β y()+ β 1 y () = (3.385) to e linerly independent. Wht hppens if tht s not the cse? Exercise Prove tht the Green s function G(x, ξ) s defined in (3.347) is symmetric: G(x, ξ)=g(ξ,x). (Hint: Show tht p [y 1 y 2 y 1 y 2 ] is constnt). Exercise Show K is liner, self-djoint, ounded. Exercise Let K e defined s where with G(x; ξ) the Green s function for the opertor in the sense tht the solution to is given y Kf 6 k(x; ξ) f(ξ) dξ (3.386) k(x; ξ)=r(x) 1/2 G(x; ξ) r(ξ) 1/2 (3.387) (p y ) + q y with oundry conditions y()= y()= (3.388) (p y ) + q y = f, y()= y() = (3.389) y(x)= G(x; ξ) f(ξ) dξ. (3.39) Assume p, q. Show tht K is non-negtive opertor, tht is (Kz, z) for ll continuous functions z. References. Anthony W. Knpp, Advnced Rel Anlysis, Chp. 2. Kôsku Yosid, Lectures on Differentil nd Integrl Equtions, Chp. 2.

32 3.4. Higher Dimensionl Prolems. In contrst to Section 3.1, here we consider higher dimensionl equtions in n irregulr domin insted of regulr ones like rectngle, disc, sphere, etc. Note tht the shred property of those regulr domins is tht they re either rectngulr or cn e trnformed to rectngulr domin through chnge of vriles. In this cse one cn still formulte the eigenvlue prolems nd study their properties, ut usully one cnnot write explicit formuls for the eigenfunctions nymore Higher dimensionl seprtion of vriles. In the following we use the nottion x = (x 1,, x n ). We consider the following eqution in higher dimensions: Let Ω e domin in R n, let L e liner differentil opertor of the following form: L[u] = n (p(x)u xj ) xi + q(x)u= (p u)+ q u (3.391) i,j=1 For exmple, when p = 1, q =, L is the usul Lplce opertor We will the reson for the negtive sign lter. Then we set up the eqution s L[u] = u. (3.392) u t +L[u] = Ω R +, u(x, )= f(x) (3.393) with certin oundry conditions. Now we cn pply the ide of seprtion of vriles. Since Ω is generl domin, there is usully no chnge of vriles to mp it to rectngulr one, therefore we cn only write T(t) X(x) nd try to require it to solve the eqution. Sustituting into the eqution we rech T (t)x(x) +T(t)L[X] = L[X] λx =, T (t)+λt(t)=. (3.394) From our understnding of the method, we expect the eigenvlue prolem L[X] λ X =, oundry conditions (3.395) to hve countly mny eigenvlues which cn e listed y their sizes: λ 1 < λ 2 < nd for ech eigenvlue we expect to hve one X i (x). These X i (x) re orthogonl in the sense tht X n (x)x m (x)dx = m n (3.396) Ω nd complete in the sense tht for ny resonly smooth f(x), we cn write f(x)= f n X n (x) with f n = Ω f(x)x n(x)dx Ω X n(x) 2 dx. (3.397) Then the finl solution should e u(x, t) = f n e λnt X n (x). (3.398) Before we proceed, we hve to point out tht some of the ove expecttions re not relistic. For exmple, in higher dimensions, it my hppen tht more thn one linerly independent eigenfunctions correspond to one sme eigenvlue λ. To see this, recll the het eqution in rectngulr domin u t =u xx + u yy, < x <, < y <. (3.399) We know tht the eigenvlues re m 2 + n 2 with eigenfunctions sin n x sin m x. But clerly there is the possiility tht m n 1 2 =m 2 2 +n 2 2. Now we introduce the oundry conditions: α(x)u+ β(x) u = on Ω. (3.4) n