Outline. Math Partial Differential Equations. Rayleigh Quotient. Rayleigh Quotient. Sturm-Liouville Problems Part C

Transcription

1 Mth 53 - Prtil Differentil Equtions Sturm-Liouville Problems Prt C Outline Tril Functions Joseph M. Mhffy, jmhffy@mil.sdsu.edu Deprtment of Mthemtics nd Sttistics Dynmicl Systems Group Computtionl Sciences Reserch Center Sn Diego Stte University Sn Diego, CA Eigenvlue Asymptotic Behvior Approximtion Properties Spring 9 Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45) Tril Functions Tril Functions The Sturm-Liouville Differentil Eqution problem: ( d p(x) dφ ) + q(x)φ + λσ(x)φ =. Multiply by φ nd integrte: φ d ( p(x) dφ ) ] + q(x)φ + λ The eigenvlue stisfies: λ = φ σ(x) =. φ d ( p(x) dφ ) ] + q(x)φ. φ σ(x) Integrte the eigenvlue eqution by prts: pφ dφ b b ( ) ] dφ + p q(x)φ λ =, φ σ(x) which is the. The eigenvlues re nonnegtive (λ ), if pφ dφ b, q. These conditions commonly hold for Physicl problems, where q or energy-bsorbing. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (3/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (4/45)

2 Minimiztion Principle Tril Functions Tril functions Tril Functions The eigenvlue stisfies: Theorem (Minimiztion Principle) The minimum vlue of the Ryleigh quotient for ll continuous functions stisfying the BCs (not necessrily the differentil eqution) is the lowest eigenvlue: b ( ) ] pu du b du + p q(x)u λ = min u u σ(x) This minimum occurs t u = φ, the lowest eigenfunction., Tril functions: Cnnot test ll continuous functions stisfying the BCs, but select tril functions, u T, b ( ) ] du pu T T b + dut p q(x)u T λ RQu T ] = u T σ(x), This provides n upper bound for λ. Exmple: Consider the Sturm-Liouville problem: φ + λφ =, φ() = nd φ() =. This exmple hs n eigenvlue, λ = π, with n ssocited eigenfunction, φ = sin(πx). Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (5/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (6/45) Tril functions Tril Functions Tril functions Tril Functions Exmple: We compute the Ryleigh quotient with 3 test functions, u (x), u (x), nd u 3 (x): Tent function: { x, x < u (x) =, x, x. Qudrtic function: u (x) = x x. Eigenfunction: u 3 (x) = sin(πx). ut u 3(x) u (x) u (x) x We insert ech of these functions into the Ryleigh quotient. Exmple: The Ryleigh quotient with { x, x < u (x) =, x, x, stisfies: λ RQu ] = = = ( ) du u + du u, / + / / x + / ( x), =. 4 Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (7/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (8/45)

3 Tril functions Tril Functions Ryleigh quotient Tril Functions Exmple: The Ryleigh quotient with u (x) = x x stisfies: ( ) du u + du λ RQu ] = u, = ( x) (x x ) = The Ryleigh quotient with u 3 (x) = sin(πx) stisfies: ( ) du u du3 λ RQu 3 ] = u 3, = π cos (πx) sin (πx), = π =. = π Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (9/45) : The proof of the Ryleigh quotient generlly uses the Clculus of Vritions, which cnnot be developed here. Our proof is bsed on eigenfunction expnsion. We ssume u is continuous function stisfying homogeneous BCs Assuming homogeneous BCs gives the equivlent form for the Ryleigh quotient: RQu] = ul(u) u σ, where L is the Sturm-Liouville opertor. We tke u expnded by the eigenfunctions u(x) = n φ n (x). Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45) Ryleigh quotient Tril Functions Ryleigh quotient Tril Functions (cont): Since L is liner opertor, we expect L(u) = n L(φ n (x)) = n λ n σφ n (x), where lter we show the interchnge of the summtion nd opertor when u is continuous nd stisfies homogeneous BCs of the eigenfunctions. With different dummy summtions, the Ryleigh quotient becomes RQu] = ( m= m n λ n φ m φ n σ) ( m= m n φ m φ n σ). We interchnge the summtion nd integrtion nd use orthogonlity to give RQu] = nλ n φ nσ n φ nσ. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45) : The previous eqution gives the exct expression for the Ryleigh quotient in terms of the generlized Fourier coefficients n of u. If λ is the lowest eigenvlue, then we obtin: RQu] λ n φ nσ n φ nσ = λ. Note tht equlity holds only if n = for n >, which gives the minimiztion result tht RQu] = λ for u = φ. The proof is esily extended to show tht if = for the eigenfunction expnsion of u, then RQu] = λ when n = for n > nd u = φ. Thus, the minimum vlue for ll continuous functions u tht re orthogonl to the lowest eigenfunction nd stisfy the homogeneous BCs is the next-to-lowest eigenvlue. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45)

4 Het Eqution with BC of Third Kind: Consider the PDE with the BCs u(, t) = u t = k u x, nd u (L, t) = hu(l, t). x If h >, then this is physicl problem nd the right endpoint represents Newton s lw of cooling with n environmentl temperture of. Note: The problem solving below cn be done eqully well with the String Eqution, u tt = c u xx, where the right BC represents restoring force for h > nd is clled n elstic BC. If h <, either problem is not physicl, s the het eqution would be hving het constntly pumped into the rod, nd the string eqution hs destbilizing force on the right end. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (3/45) Seprtion of Vribles: Let u(x, t) = G(t)φ(x), then s before, the time dependent ODEs re Het Flow: Vibrting String: The Sturm-Liouville problem becomes: dg dt = λkg, d G dt = λc G. d φ + λφ =, φ() = nd φ (L) + hφ(l) =, where h is physicl nd h < is nonphysicl. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (4/45) Positive eigenvlues: Let λ = α >, then The BC, φ() =, implies c =. φ(x) = c cos(αx) + c sin(αx). The other BC, φ (L) + hφ(l) =, implies tht c (α cos(αl) + h sin(αl)) = or tn(αl) = α h = αl. This is trnscendentl eqution in α, which cnnot be solved exctly. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (5/45) Eigenvlue eqution is given by This eqution cn only be solved numericlly, such s Mple or MtLb This sketch is for the physicl cse, h >. Visully, cn see tht symptoticlly: ( α n L n ) π, s n tn(αl) = αl, h >. z z = tn(αl) αl 5 z = αl π/ π 3π/ π 5π/ Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (6/45)

5 Agin the eigenvlue eqution is given by This sketch is for the nonphysicl cse, < <, which is of 3 cses. There is lowest eigenvlue, λ < π. Asymptoticlly: ( α n L n ) π, s n tn(αl) = αl, < <. z 5 5 z = αl z = tn(αl) π/ π 3π/ π 5π/ Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (7/45) αl There re two dditionl cses for the nonphysicl problem, where tn(αl) = αl, = or <. In both cses, the first positive eigenvlue stisfies π < λ < 3π. z 5 z = αl z = tn(αl) π/ π 3π/ π 5π/ h = αl z 5 z = αl z = tn(αl) π/ π 3π/ π 5π/ h < Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (8/45) αl The nonphysicl problem with = hs its first positive eigenvlue, αl (λ = α ). Zero E.V.: Consider λ =, which gives the solution φ(x) = c x + c The BC φ() = c =. The other BC φ (L) + hφ(l) = c ( + ) =, so if =, then λ = is n eigenvlue with ssocited eigenfunction, φ (x) = x. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (9/45) Negtive E.V.: We don t expect negtive eigenvlues for physicl problems, s it produces n exponentilly growing t-solution. Suppose λ = α <, so φ α =, which hs the generl solution: The BC φ() = c =. The remining BC gives: which is nontrivil if φ(x) = c cosh(αx) + c sinh(αx). c (α cosh(αl) + h sinh(αl)) =, tnh(αl) = α h = αl, which is nother trnscendentl eqution. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45)

6 There re 4 cses to consider solving Physicl cse ( > ) hs negtive slope, so only intersects origin. When < <, only intersects origin. When =, line is tngent to origin. When <, there is unique positive eigenvlue tnh(αl) = αl. z < < < = z = tnh(αl) αl Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45) - Physicl Problem Het Eqution: Consider the PDE with the BCs nd ICs u(, t) = nd u t = k u x, u (L, t) = hu(l, t), h >, x u(x, ) = f(x). The Sturm-Liouville problem hd eigenvlues, λ n = αn, where α n, n =,,... solves tn(α n L) = α nl, nd corresponding eigenfunctions φ n = sin(α n x). Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (/45) - Physicl Problem Het Eqution (cont): The time dependent solution is G n (t) = e kλ nt = e kα n t. With the product solution, u n (x, t) = G n (t)φ n (x), the superposition principle gives: u(x, t) = A n e kα n t sin(α n x), where α n stisfies tn(α n L) = α nl. The generlized Fourier coefficients stisfy: A n = L f(x) sin(α nx) L sin (α n x). - Physicl Problem Het Eqution (cont): However, with sin(α n L) = α n h cos(α n L) L sin (α n x) = α nl sin(α n L) = Lh + cos (α n L). 4α n h Thus, the generlized Fourier coefficients stisfy: L h f(x) sin(α n x) A n = Lh + cos, (α n L) nd the temperture in the rod is given by u(x, t) = A n e kα n t sin(α n x). Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (3/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (4/45)

7 - Physicl Problem Tke L =, k =, nd h =.5 nd suppose f(x) = for x. The Fourier coefficients re redily found: Solution with terms. A n = h ( cos(α nl)) α n (Lh + cos (α n L)). - Physicl Problem % Solutions to the het flow eqution % on one-dimensionl rod length L 3 % Right end with Robin Condition 4 formt compct; 5 L = ; % width of plte 6 Temp = ; % Constnt temperture of... rod, initilly 7 tfin = ; % finl time 8 k = ; % het coef of the medium 9 h =.5; % Newton cooling constnt NptsX=5; % number of x pts NptsT=5; % number of t pts Nf=; % number of Fourier terms 3 x=linspce(,l,nptsx); 4 t=linspce(,tfin,nptst); 5 X,T]=meshgrid(x,t); Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (5/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (6/45) - Physicl Problem - Physicl Problem 7 figure() 8 clf 9 = zeros(,nf); b = zeros(,nf); U = zeros(nptst,nptsx); z =.7; 3 for :Nf 4 z = fsolve(@(x) h*l*sin(x)+x*cos(x),z); 5 (n) = z/l; 6 b(n)=(*temp*h/((n)*(l*h+(cos((n)*l))ˆ)))... 7 *(-cos((n)*l)); % Fourier coefficients 8 Un=b(n)*exp(-k*((n))ˆ*T).*sin((n)*X); %... Temperture(n) 9 U=U+Un; 3 z = z + pi; 3 end 3 set(gc,'fontsize',]); 33 surf(x,t,u); 34 shding interp 35 colormp(jet) 36 xlbel('$x$','fontsize',,'interpreter','ltex'); 37 ylbel('$t$','fontsize',,'interpreter','ltex'); 38 zlbel('$u(x,t)$','fontsize',,'interpreter','ltex'); 39 xis tight 4 view(4 ]) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (7/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (8/45)

8 Fourier Series - BC 3 rd Kind Fourier Series - BC 3 rd Kind The solution of the Het Eqution with Robin BCs used the Fourier expnsion of f(x) = with the eigenfunctions, φ n = sin(α n x). Below re grphs showing the eigenfunction expnsion. f(x) n = 5 n = n = n = x f(x) n = n = n = x % Fourier series formt compct; 3 L = ; % width of plte 4 Temp = ; % Constnt temperture of... rod, initilly 5 h =.5; % Newton cooling constnt 6 NptsX=5; % number of x pts 7 Nf=; % number of Fourier terms 8 X=linspce(,L,NptsX); 9 = zeros(,nf); b = zeros(,nf); U = zeros(,nptsx); U = zeros(,nptsx); 3 U = zeros(,nptsx); 4 U3 = zeros(,nptsx); 5 z =.7; Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (9/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (3/45) Fourier Series - BC 3 rd Kind Fourier Series - BC 3 rd Kind 6 for :Nf 7 z = fsolve(@(x) h*l*sin(x)+x*cos(x),z); 8 (n) = z/l; 9 b(n)=(*temp*h/((n)*(l*h+(cos((n)*l))ˆ)))... *(-cos((n)*l)); % Fourier coefficients Un = b(n)*sin((n)*x); % Temperture(n) U = U+Un; 3 if (n 5) 4 U = U+Un; 5 end 6 if (n ) 7 U = U+Un; 8 end 9 if (n ) 3 U3 = U3+Un; 3 end 3 z = z + pi; 33 end Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (3/45) 34 plot(x,u,'m-','linewidth',.5); 35 hold on 36 plot(x,u,'r-','linewidth',.5); 37 plot(x,u3,'-','color',.5 ],'LineWidth',.5); 38 plot(x,u,'b-','linewidth',.5); 39 plot( ], ],'k-','linewidth',.5); 4 grid; 4 legend('n = 5','n = ','n = ','n = ',... 4 'loction','southest'); 43 xlim( ]); 44 ylim( ]); 45 xlbel('$x$','fontsize',,'interpreter','ltex'); 46 ylbel('$f(x)$','fontsize',,'interpreter','ltex'); 47 set(gc,'fontsize',]); Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (3/45)

9 - Non-Physicl Problem Het Eqution with Non-Physicl BCs stisfies: PDE: u t = ku xx, BC: u(, t) =, IC: u(x, ) = f(x), u x (L, t) = hu(l, t) with h <. For < h <, the Sturm-Liouville problem is the sme s the physicl problem with eigenvlues, λ n = αn, where α n, n =,,... solves tn(α n L) = α nl, nd corresponding eigenfunctions re φ n = sin(α n x). The solution stisfies: u(x, t) = A n e kα n t sin(α n x), with the sme generlized Fourier coefficients s for the physicl problem. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (33/45) - Non-Physicl Problem Het Eqution with Non-Physicl BCs nd h = hs λ = with the eigenfunction φ (x) = x, so the solution becomes: u(x, t) = A x + with A n s before for n =,,... nd A = A n e kα n t sin(α n x), 3 L L 3 xf(x). If h < nd β solves tnh(β L) = β h, then there is the dditionl eigenfunction φ (x) = sinh(β x), so the solution becomes: u(x, t) = A e kβ t sinh(β x) + with A n s before for n =,,... nd L A n e kα n t sin(α n x), A = β f(x) sinh(β x) cosh(β L) sinh(β L) β L. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (34/45) - Physicl Problem Het Eqution with h = (insulted right end) stisfies: PDE: u t = ku xx, BC: u(, t) =, IC: u(x, ) = f(x), u x (L, t) =. This problem is solved in the norml mnner s before, nd it is esy to see tht the eigenvlues, λ n = (n ) π L, with corresponding eigenfunctions re (( ) n φ n = sin ) πx. L The solution stisfies: u(x, t) = (( ) n A n e kλnt sin ) πx, L with similr Fourier coefficients to our originl Het problem. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (35/45) Eigenvlue Asymptotic Behvior Eigenvlue Asymptotic Behvior Approximtion Properties Exmine the Sturm-Liouville eigenvlue problem in the form d p(x) dφ ] + λσ(x) + q(x)]φ =. The eigenvlues generlly must be computed numericlly. There is number of people working on detils of these problems, so the scope of this problem is beyond this course. (See Mrk Dunster) Interpret this problem like spring-mss problem for lrge λ, where x is time nd φ is position. p(x) cts like the mss. For λ lrge, λσ(x)φ cts like restoring force This solution rpidly oscilltes Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (36/45)

10 Eigenvlue Asymptotic Behvior Eigenvlue Asymptotic Behvior Approximtion Properties With lrge λ, the solution oscilltes rpidly over few periods, so cn pproximte the coefficients s constnts. Thus, the DE is pproximted ner ny point x by p(x ) d φ + λσ(x )φ, which is like stndrd spring-mss problem. It follows tht the frequency is pproximted by λσ(x ) ω = p(x ) Eigenvlue Asymptotic Behvior Eigenvlue Asymptotic Behvior Approximtion Properties The mplitude nd frequency re slow vrying, so With Tylor series, we write φ(x) = A(x) cos(ψ(x)). φ(x) = A(x) cosψ(x ) + ψ (x )(x x ) +...], so the locl frequency is ψ (x ), where ( ) / ψ (x ) = λ / σ(x ). p(x ) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (37/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (38/45) Eigenvlue Asymptotic Behvior Integrting ψ (x ) gives the correct phse x ( ) ψ(x) = λ / σ(x ) /. p(x ) Eigenvlue Asymptotic Behvior Approximtion Properties It cn be shown (beyond this clss) tht the independent solutions re pproximted for lrge λ by ) ] / φ(x) (σp) /4 exp ±iλ / x ( σ p If φ() =, then the eigenfunction cn be pproximted by ( ) ) / φ(x) = (σp) /4 sin If the second BC is φ(l) =, then λ / L λ / x ( σ p ( ) σ / nπ or λ p L ( σ p nπ ) /. Eigenvlue Asymptotic Behvior Exmple: Consider the eigenvlue problem d φ + λ( + x)φ =, with BCs φ() = nd φ() =. Our pproximtion gives: λ nπ ( + x ) / ] = Eigenvlue Asymptotic Behvior Approximtion Properties n π 3 ( + x ) 3/ ] = n Numericl Formul n π 4 9 (3/ ). Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (39/45) Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (4/45)

11 Approximtion Properties Eigenvlue Asymptotic Behvior Approximtion Properties Approximtion Properties Eigenvlue Asymptotic Behvior Approximtion Properties We climed tht ny piecewise smooth function, f(x), cn be represented by the generlized Fourier series of eigenfunctions: f(x) n φ n (x) By orthogonlity with weight σ(x) of the eigenfunctions n = Suppose we use finite expnsion, f(x)φ n(x)σ(x) φ n(x)σ(x). f(x) M α n φ n (x). How do we choose α n to obtin the best pproximtion? Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (4/45) How do we define the best pproximtion? Definition (Men-Squre Devition) The stndrd mesure of Error is the men-squre devition, which is given by: E = f(x) M α n φ n (x)] σ(x). This devition uses the weighting function, σ(x). It penlizes hevily for lrge devition on smll intervl. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (4/45) Approximtion Properties Eigenvlue Asymptotic Behvior Approximtion Properties Approximtion Properties Eigenvlue Asymptotic Behvior Approximtion Properties The best pproximtion solves the system: or = E α i = E α i =, i =,,..., M. f(x) ] M α n φ n (x) φ i (x)σ(x), i =,,..., M. This would be complicted, except tht we hve mutul orthogonlity of the φ i (x) s, so f(x)φ i (x)σ(x) = α i φ i (x)σ(x). Solving this system for α i gives the α i s the generlized Fourier coefficients. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (43/45) An lternte proof of this result shows tht the minimum error is: E = f σ M α n φ nσ. This eqution shows tht s M increses, the error decreses. Definition (Bessel s Inequlity) Since E, f σ M α n φ nσ. More importntly, ny Sturm-Liouville eigenvlue problem hs n eigenfunction expnsion of f(x), which converges in the men to f(x). Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (44/45)

12 Approximtion Properties Eigenvlue Asymptotic Behvior Approximtion Properties The convergence in men implies tht which gives the following: Definition (Prsevl s Equlity) Since E, f σ = lim E =, M α n φ nσ. This inequlity is generliztion of the Pythgoren theorem, which importnt in showing completeness. Joseph M. Mhffy, jmhffy@mil.sdsu.edu Sturm-Liouville Problems (45/45)