Transcription of 13 Sturm{Liouville problems. Eigenvalues and eigenfunctions
1 13 Sturm{Liouville problems. Eigenvalues and eigenfunctionsIn the previous lecture I gave four examples of different boundary value problems for a second orderODE that resulted in a countable number of constants (lambdas) and a countable number of corre-sponding solutions, which were used afterwards to build the corresponding Fourier series to representsolutions for PDE. Not surprisingly, these four examples can be generalized in a relatively abstractframework, which I discuss in this lecture. In the literature this framework is calledSturm Liouvilleproblem after two mathematicians who first concentrated on this start with the definition ofSturm liouville differential operatorLon the intervalx [a;b]:Lu:= (p(x)u ) +q(x)u;( )wherep;qare continuous on [a;b] andp(x)>0 for anyx [a;b].}
2 Note that all four problems fromthe previous lecture can be written asLu= u;( )withp(x) = 1 andq(x) = 0 and some additional boundary conditions. Most of these boundaryconditions can be written in the general form 1u(a) + 2u (a) = 0; 21+ 22>0; 1u(b) + 2u (b) = 0; 21+ 22>0:( )Quite natural (recall the definition of Eigenvalues and eigenvectors for, , matrices) problem ( ) ( ) is called theSturm liouville eigenvalue problem . Sometimes instead of ( ) periodicboundary conditions are used:u(a) =u(b);p(a)u (a) =p(b)u (b):( )To study the properties of the Eigenvalues of the sturm liouville problem , I start with a derivationofLagrange s identity. Letu;vbe two arbitraryC(2)[a;b] functions, then (check the skipped steps)uLv vLu= u(pv ) +quv+v(pu ) quv=v(pu ) u(pv ) =(p(vu uv )) :From Lagrange s identity, by integrating fromatob, I getGreen s formula ba(uLv vLu) dx=p(vu uv ) ba:Both Lagrange s identity and Green s formula hold for anyuandv.
3 I claim that ifuandvare suchthat they satisfy ( ) or ( ) then Green s formula becomes ba(uLv vLu) dx= 0:Indeed, ifuandvsatisfy, , ( ) then I must have that 1u(a) + 2u (a) = 0; 21+ 22>0; 1v(a) + 2v (a) = 0;Math 483/683: Partial Differential Equations by Artem Novozhilove-mail: Spring 20201and therefore the determinant of the matrix[u(a)u (a)v(a)v (a)]must be zero, that isu(a)v (a) v(a)u (a) = 0:Similarly,u(b)v (b) v(b)u (b) = 0;which proves the stated fact for the boundary conditions ( ). I leave checking ( ) as an , using the notation for the inner product: u;v = bau(x)v(x) dx;I can rewrite Green s formula as u;Lv = Lu;v :An operator that satisfies such condition is calledself-adjoint1(think about symmetric real matrices,such thatA=A ), and hence I proved that the sturm liouville operator defined on the functionsthat satisfy ( ) or ( ) is self-adjoint (note that it is important to add the boundary conditions,without them the self-adjointness does not make any sense).
4 Now I am ready to proveLemma a self-adjoint sturm liouville operator. Then all the Eigenvalues will prove this lemma by contradiction. Let Cbe my eigenvalue andu = 0 a correspondingeigenfunction. By the properties of differential operator and linearity ofLI get that andumust beanother eigenvalue and corresponding eigenfunction. Now consider Lu;u = u;u = u;u :On the other hand, sinceLis self-adjoint, Lu;u = u;Lu = u; u = u;u :Therefore( ) u;u = ( ) ba|u|2dx;and since ba|u|2dx>0 then = ;which means that is real. Lemma a self-adjoint operator. Then if 1and 2are two different Eigenvalues andu1;u2are two corresponding eigenfunctions thenu1andu2are real life is much more complicated, I can only refer to a proper graduate course to set the matter have Lu1;u2 = 1 u1;u2 = 2 u1;u2 ;or( 1 2) u1;u2 = 0 = u1;u2 = 0: Two previous lemmas are very nice, however, they are true under the assumption that my operatorhas any Eigenvalues and eigenfunctions at all.
5 A more impressive theorem, whose proof is significantlymore involved, and hence omitted here, is as the sturm liouville eigenvalue problem , ,( )plus( )or( ).Then there exists a countable sequence of Eigenvalues 1 2 3 :::;such that k ask . The corresponding system of eigenfunctions {u1;u2;:::}is complete inL2[a;b], , any functionf L2[a;b]can be represented as a convergent generalized Fourier seriesf(x) =c1u1(x) +c2u2(x) +:::;where the coefficients are given byck= f;uk uk;uk , , sturm - liouville Theory and its Applications, 2008, by Mohammed Al-Gwaiz for allthe details. To conclude this section let me collect together all the results for the sturm liouville operatorLu= u that we got so far (in the previous lecture and in homework problems).Boundary conditionsEigenvaluesEigenfunctionsu(0) =u(1) = 0 k= 2k2;k= 1;2;:::uk(x) =Bsin kxu(0) =u(L) = 0 k= 2k2L2;k= 1;2;:::uk(x) =Bsin kLxu (0) =u (1) = 0 k= 2k2;k= 0;1;2;:::uk(x) =Acos kxu( ) =u( );u ( ) =u ( ) k=k2;k= 0;1;2;:::uk(x) =Acoskx;vk(x) =Bsinkxu(0) =u (1) +hu(1) = 0 Solutions to tan = =huk(x) =Bsin kxu(0) =u (1) = 0 k= ( (k 1=2))2;k= 1;2;:::uk(x) =Bsin (k 1=2)x;k= 1;2;:::u (0) =u(1) = 0 k= ( (k 1=2))2;k= 1;2;:::uk(x) =Bcos (k 1=2)x;k= 1;2;:::3
