a review of the generic regular BC problem

A review of the generic regular BC problem PhL 8.10.11 Lemma 0: 1 Unmixed BC's and Full Self-adjointness of L (L is symmetric) 2 Lemma 1: 2 Corollary 1: 2 Lemma 2: 2 Lemma 3: 3 The Three Basic Theorems for Symmetric L's: real, real and orthogonal. 3 Why g is symmetric 4 Solution u in terms of g. 4 The Eigenfunction expansion for g 5 Location of the eigenvalues. 6 Completeness of the eigenfunctions 6 The integral of g formula 7 Why the spectrum is discrete 7 Corollary 4: 8 This subject is treated in Stak Chap 4. He does a long example, then addresses the general case on page 268. This is the 1D "Sturm-Liouville" problem in which various assumptions are made. My Chap 4 meta meta notes are not too bad and have all the major results, but they are not systematically derived there, something I will try to do here. Our starting point is having some formally self-adjoint 2nd order differential operator L constructed from functions p and q, and then we can at least talk about this grid of equations: operator: Lλ = L - λs L = -∂x(p∂x) + q Fred general: Lλu = f Lu - λsu = f // inhomo 2 Set f = 0: Lλφλ=0 Lφλ = λsφλ // homo 2 = the EV problem Set λ = 0: L0u = f Lu = f // inhomo1 Set f = δ: Lλg = δ Lg - λsg = δ // Green's function problem with λ It is required that functions s(x)>0 and p(x)>0 on our interval (a,b), q(x) is also real. Lemma 0: An L of this form is formally self adjoint. Proof: !Syntax Error, Idx v*Lu = !Syntax Error, Idx v*[(pu')' +qu] = !Syntax Error, Idx [v*(pu')' +qv*u] = !Syntax Error, Idx (-v*'(pu') +qv*u ) = !Syntax Error, Idx (-v*'pu' +quv* ) = !Syntax Error, Idx (+(v*'p)'u +quv* ) = !Syntax Error, Idx u[ (v*'p)' + qv*] = !Syntax Error, Idx u (Lv)* => <v,Lu> = <Lv,u> if we ignore the parts In the above we did two parts integrations and "threw out" the parts. The meaning of formally self adjoint is that <v,Lu> = <Lv,u> if you ignore the parts. Unmixed BC's and Full Self-adjointness of L (L is symmetric) At each endpoint we have a standard unmixed homogeneous boundary condition meaning Au(a) + Bu'(a) = 0 where A and B must be real. The following Lemma will be useful at once, and is never obvious to me: Lemma 1: If two functions u and v satisfy the BC at x=a, then W[u,v; a] = W[u,; a] = 0. Proof: Au(a)+Bu'(a) = 0 => = 0 Av(a)+Bv'(a) = 0 => AB= 0 => W[u,v;a] = 0 Since A and B are real, we can CC the second BC to get A(a)+B'(a) = 0, so W[u,; a] = 0 as well. Since there is some similar BC at endpoint b, we know also that W[u,v; b] = W[u,; b] = 0. Corollary 1: If functions u satisfies an arbitrary unmixed BC at x=a, then W[u,u; a] = W[u,; a] = 0. The first result is a 2=2 statement, but the second result is more interesting. Lemma 2: Another less than obvious fact. For ANY functions u and v, we have Lu - u= ∂x { p(x) W[u,; x] } Proof: L = -∂x(p∂x) + q => Lu = -(pu')'+qu v*Lu - u(Lv)* = v*[ -(pu')'+qu] - u[-(pv*')'+qv*] = - v*(pu')' + u(pv*')' = - v*[p'u'+pu"]+ u[p'v*'+pv*"] = p {-v*u" + uv*"} + p' {- v*u' + uv*'} Now note that W[u,v*] = uv*' - u'v* => (W[u,v*])' = u'v*' + uv*" - u"v* - u'v* = uv*" - u"v* Thus we have shown above that v*Lu - u(Lv)* = p {-v*u" + uv*"} + p' {- v*u' + uv*'} = p(W[u,v*])' + p' W[u,v*] = { p W[u,v*]} ' QED. If we now integrate Lemma2 from a to b, we get !Syntax Error, Idx { Lu - u} = p(b) W[u,; b] - p(a) W[u,; a] = < v,Lu> - <Lv,u> Now the pW stuff here is "the parts" and if we ignore the parts, we get <v,Lu> = <Lv,u>, so this is a second proof of Lemma 0 above which said that an L of our form was formally self adjoint. But now we apply Lemma 1 at the two end points and we find then that !Syntax Error, Idx { Lu - u} = p(b) 0 - p(a) 0 = 0 = < v,Lu> - <Lv,u> and now we see that <v,Lu> = <Lv,u> even if we keep the parts! So this is the first very major result! It says that in our BV problem situation, L is not only formally self adjoint, it is fully self adjoint (aka "symmetric") because "the parts" vanish. Lemma 3: [ W(u,v)] ' = W(u,v') Proof: We already derived this above in effect, but let's do it directly [ W(u,v)] ' = [uv'-vu']' = u'v'+uv" - v'u'-vu" = uv" -vu" = W[u,v'] QED The Three Basic Theorems for Symmetric L's: real, real and orthogonal. The following is shown on page 270 of Stak. I use the QM convention on inner products here. Theorem 1: <Lu,u> is real. Proof: <u,Lu> = <Lu,u> = <u,Lu>* Theorem 2: Eigenvalues of L are real Proof: Lφλ = λsφλ, => <φλ,Lφλ> = λ <φλ, sφλ> (Lφλ)* = λ*sφλ* => <φλ*,(Lφλ)*> = λ* <φλ*, sφλ*> = λ*<sφλ, φλ> = λ*<φλ, sφλ> where in 2nd last step we use <x*,y*>= <x,y>* = <y,x> and s=s*. Thus we have shown λ <φλ, sφλ> = <φλ,Lφλ> λ*<φλ, sφλ> = <φλ*,(Lφλ)*> = < Lφλ, φλ>* = < Lφλ, φλ> = < φλ, Lφλ> where we used Theorem 1 and then the fact that L is fully self-adjoint. Subtracting we find (λ -λ*)<φλ, sφλ> = 0 => (λ -λ*) = 0 => λ=λ* => λ = real Theorem 3: Eigenfunctions of different λ are orthogonal with weight s <Lφλ', φλ> = < λ'sφλ', φλ> = λ'* <sφλ', φλ> = λ' <sφλ', φλ> by Theorem 2 <φλ', Lφλ> = < φλ', λsφλ> = λ <sφλ', φλ> So we have shown λ' <sφλ', φλ> = <Lφλ', φλ> = <φλ', Lφλ> // since L is symmetric λ <sφλ', φλ> = <φλ', Lφλ> Therefore (λ'-λ) <φλ', Lφλ> = 0 so of λ→λ', then <φλ', Lφλ> = 0 QED. Why g is symmetric When we solve for g using the "usual method", we always write g(x|ξ) = A w(x<) z(x>) x< = min(x,ξ) x> = max(x,ξ) and we compute A from the jump condition, and we thereby construct g! This form works because we make sure function w satisfies the left BC and z the right BC. This is one reason why we wanted to have only unmixed BC's in our problem specification. Thus we can write g(x|ξ) = A w(min(x,ξ) z(max(x,ξ)) Since the min and max functions are symmetric under x↔ξ, g(x|ξ) must also be symmetric. Solution u in terms of g. Consider these two systems Lu-λsu = f Lλu = f Lg-λsg = δ Lλg = δ Let's conjecture that u(x) = !Syntax Error, Ig(x|ξ;λ)f(ξ)dξ Try this in the first equation above LHS : Lx {!Syntax Error, Ig(x|ξ;λ)f(ξ)dξ } - λs(x) {!Syntax Error, Ig(x|ξ;λ)f(ξ)dξ } = {!Syntax Error, I Lx g(x|ξ;λ)f(ξ)dξ } - λs(x) {!Syntax Error, Ig(x|ξ;λ)f(ξ)dξ } = {!Syntax Error, I [λs(x) g(x|ξ;λ) + δ(x-ξ} ] f(ξ)dξ } - λs(x) {!Syntax Error, Ig(x|ξ;λ)f(ξ)dξ } = !Syntax Error, I δ(x-ξ) f(ξ)dξ = f(x) Therefore our conjectured u(x) form solves our ODE AND since g meets the BC's, so does u, so u must be the solution. You might wonder about the possibility of having extra homo solutions floating around. They would be solutions to Lu-λsu = 0 or Lu = λsu. But if this is true, λ must be an eigenvalues. Thus, if we assume λ is NOT an EV, then there is no extra homo solution to add, and our conjectured form is the complete answer. The Eigenfunction expansion for g Consider Lu-λsu = f = sF where we have defined F ≡ f/s as shown. Then make these two expansions u = Σn unφn F = Σn Fnφn Then we have (L-λs) (Σn unφn) = s (Σn Fnφn) => Σn un(L-λs)φn = s(Σn Fnφn) => Σn un(λns-λs)φn = sΣn Fnφn => Σn un(λn-λ)φn = Σn Fnφn we then appeal to the completeness of the φn to get un(λn-λ)= Fn => un= Fn/(λn-λ) As long as λ differs from all eigenvalues, we then have a solution to our f-driven problem u(x) = Σn unφn(x) = Σn φn(x) Fn/(λn-λ) Now consider the special case where f = δ. We then have F(x) = δ(x-ξ)/s(x) Fn = <φn,sF> = <φn,sδ/s> = <φn,δ> = φn(ξ)* We have then shown that g(x) = g(x|ξ;λ) = Σn φn(x) Fn/(λn-λ) = Σn φn(ξ)*φn(x) /(λn-λ) or if we throw in the symmetry of g, we get g(x|ξ;λ) = Σn φn(x)*φn(ξ) /(λn-λ) This is a very famous and important result, and I have shown exactly how you obtain it! Location of the eigenvalues. Stak claims on page 272 that there can be at most a finite number of negative λn but on the right they will march off to infinity and ∞ will be the only accumulation point. He proves this in Exercise 4.7 which includes a whole page of waypoints and which I never did, and I won't do it right here. Completeness of the eigenfunctions As Stak shows on page 272, you have to appeal to the theory of integral equations (after you convert the eigenvalue ODE Lu=λsu to an integral equation) to conclude that the functions φn form a complete orthonormal set, where one ingredient is that the kernel now g(x|ξ) is symmetric. In our work above, we always include s in the inner product as <f,sg> and thus treat s as a weight function in the integral. If we say f(x) = Σn an [φn] = (Σn an φn) => f/ = Σn an φn If we define a new function F = f/then we have F = Σn an φn . We assumed f was an arbitrary function for our expansion, but then of course F can be considered an arbitrary function as well, since for any F we have f = F which is then an arbitrary function. The point is that if φn forms a complete set, then the functions φn also form a complete set, and we can use either complete set. Now if we select the φn to be orthonormal, we then have <φm,sφn> = δn,m This tells us that the completeness relation must have this form δ(x-ξ)/s(x) = Σn φn*(x)φn(ξ) To verify this result, we take the weighed inner product against each eigenfunction of our complete basis <φm, s(x)δ(x-ξ)/s(x)> = <φm, s(x), Σn φn*(x)φn(ξ)> φm(ξ) = Σn φn(ξ) <φm, s(x), φn*(x) > = Σn φn(ξ) δm,n = φm(ξ) Since the projections of our two sides are the same for every element of a complete basis, the two sides must be equal. The integral of g formula We have already shown that g(x|ξ;λ) = Σn φn(x)*φn(ξ) /(λn-λ) where the λn lie on the real axis. If we use a CCW great circle contour, it certainly will include that real axis, and we can then say (1/2πi) ∫C dλ g(x|ξ;λ) = Σn φn(x)*φn(ξ) (1/2πi) ∫C dλ/(λn-λ) = - Σn φn(x)*φn(ξ) The contour is in the correct direction around each pole. The minus sign arise because the residue formula requires 1/(λ-λn) on the bottom! We can now combine our integral of g formula with our completeness formula to say - (1/2πi) ∫C dλ g(x|ξ;λ) = Σn φn(x)*φn(ξ) = δ(x-ξ)/s(x) <φm,sφn> = δn,m We have now proven this to be true for any regular BV problem! A common method we use to find the eigenfunctions is to compute g using standard methods, and try to get it into a Σn form with a symmetric summand, and then we can just "read off" the φn(x) normalized eigenfunctions! Why the spectrum is discrete Lemma 4: Abel's formula W[u,v; x] = C e-m(x). We go way back to Abel's formula for the Wronskian on page 60, which applies to any second order L. I will prove this result right here. Consider L = a0∂x2 + a1∂x + a2 Lu = a0u" + a1u' + a2u Assume that Lu=0 and Lv = 0 so that u and v are both homo solutions. Then consider vLu-uLv = 0-0 = 0 v(a0u" + a1u' + a2u) - u(a0v" + a1v' + a2v) = 0 v(a0u" + a1u') - u(a0v" + a1v') = 0 a0(v u" - u v") + a1(v u' - u v') = 0 a0{W[v,u]}' + a1 W[v,u] = 0 where we have used Lemma 3 above. Since we could have done u↔v at the start, we must also have a0{W[u,v]}' + a1 W[u,v] = 0 This last first order ODE has the form a0W'+a1W = 0 => dW/dx = -(a1/a0)W => dW/W = -(a1/a0)dx => lnW = -!Syntax Error, Idx' (a1(x')/a0(x')) + c => W = C exp [-!Syntax Error, Idx' (a1(x')/a0(x')) ] So we have shown that W[u,v; x] = W = C exp [-!Syntax Error, Idx' (a1(x')/a0(x')) ] = C e-m(x) and this is Abel's baby, where C is some constant one could determine from solutions u and v. Corollary 4: The function m(x) is finite for all x since we assume a0 ≠ 0. Thus, the only way for the Wronskian to vanish is if C = 0. Thus, if W[u,v; x0] ≠ 0, then W[u,v; x] for all x !