FREDHOLM ALTERNATIVE. y Y. We want to know when the

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1 FREDHOLM ALTERNATIVE KELLY MCQUIGHAN (I) Statement and Theory of Fredholm Alternative Let X, Y be Banach spaces, A L(X, Y), equation Ax = y is solvable for x X. y Y. We want to know when the (a) Intuition Let T L(X, Y), closed. Then T L(Y, X ) and N(T ) R(T ) when interpreted as the Y Y inner product. We wish to know for which values of b Y there exists a solution to T x = b; in other words, we wish to know which values of b are in the range of T. Let y N(T ). Then, b R(T ) b, y =. (b) Fredholm Alternative Let Let X, Y be Banach spaces, A be a Fredholm Operator, b Y. Ax = bsolvableforx B R(A) b, y y N(A ). [2] Then Def: We say A L(X, Y) is a Fredholm operator if R(A) is closed and if both dim N(A) < and codim R(A) <. Then we define the Fredholm index ind A = dim N(A) codim R(A). [3] ex: Let T : R n R m for n > m. Then let dim R(T ) = l, l m so that codim R(T ) = m l. Then N(T ) = n l and ind T = n m. Thus we can interpret the Fredholm index as the number of variables minus the number of equations. (II) Counter-example if T is Not Closed [4] Consider the operator T : L 2 (R, R) L 2 ((R), (R), u u x, with domain H 1 (R, R). Then the L 2 adjoint is T u = u x by integration by parts, defined again on L 2. First calculate N(T ) = {v L 2 : v x = }, which is clearly given by the span{1}. However, 1 / L 2 (R), hence N(T ) = {}. Then, if the Fredholm Alternative applies, we would have R(T ) = L 2. We construct a counter example to show that this is not true. Take any odd, continuous, bounded function h(x) with 1

2 2 KELLY MCQUIGHAN (1) h(x) = 1 x 3 2 x 1 Then R h(x)2 dx = 1 1 h(x)2 dx + 2 dx 1 < 2 [sup(h(x)) 2x ] = 2[sup(h(x) + x ], and thus h(x) L 2. However, we will show that h(x) / R(T ). If it were, then T u = h would be solved by u x = h(x) so that u(x) is given by (2) u(x) = a + x 1 h(y)dy For x 1 we have u(x) = a 2 y x 1 = a x so that a R, u(x) / L 2. Therefore, h / R(T ). So what went wrong with the hypotheses of the Fredholm Alternative? We do not have R(T ) closed in L 2 since h(x) can be approximated by the L 2 functions (3) h n (x) = (III) Application to Wave Trains [1] (a) Review of Setup { 1 x n x 1 h(x) x 1 Recall that we have been studying wave train solutions u(x, t) = u (ω t k x) to (4) t u = D xx u + f(u) where u (θ) is 2π-periodic in its argument. Substituting u into (4) we see that u must satisfy (5) k 2 D θθ u ω θ u + f(u) = for k = k and ω = ω, with linearization (6) L = k 2 D θθ ω θ + f (u (θ)) We assumed that λ = is a simple eigenvalue of L so that

3 (b) Calculation FREDHOLM ALTERNATIVE 3 its null space is one dimensional, spanned by the derivative θ u of the wave train λ = simple also implies that θ u is not in the range of L. If it were, then u 1 s.t. L u 1 = u, implying that λ = is part of a Jordan block of dimension at least 2, which violates the simple hypothesis. We have shown that there exists a neighborhood V of (k, ω ) such that wave trains given by u(x, t) = u (ω nl (k)t kx; k) are still solutions to (4). The purpose of this lecture is to derive an expansion for ω nl near k. We assume that the hypotheses of the Fredholm Alternative hold; this will be verified later. The adjoint of L is (7) L = k2 D θθ u + ω θ u + f (u (θ)) T u The null space of L is spanned by u ad, normalized so that u ad, θ u L 2 = 1. Note that this normalization is possible by the Fredholm Alternative since θ u is not in the range of L. We then substitute u = u and ω = ω nl (k) into (5) and take the derivative with respect to k (8) (k 2 D θθ ω θ + f (u (θ))) k u + 2kD θθ u dw nl(k ) θ u = Evaluated at u = u and k = k we have the first two derivates of (5) are given by (9) L k u + 2k D θθ u dw nl(k ) θ u = L kk u + 2k D θθk u dw nl(k ) θk u + f (u )[ k u (θ), k u (θ)] + 2D θθ u + 2k D θθk u dw nl(k ) θk u d2 w nl (k ) 2 θ u = Rearranging we find

4 4 KELLY MCQUIGHAN L k u = 2k D θθ u + dw nl(k ) θ u (1) L kk u = 4k D θθk u + 2 dw nl(k ) θk u f (u )[ k u (θ), k u (θ)] 2D θθ u + d2 w nl (k ) 2 θ u We can conclude from (1) that the right hand side of each equation must be in the range of L. We apply the Fredholm Alternative, which states that therefore RHS, u ad =. Using the normalization property we have = 2k D θθ u + dw nl(k ) θ u, u ad = 2k D θθ u, u ad + dw nl(k ) = 4k D θθk u + 2 dw nl(k ) θk u f (u )[ k u (θ), k u (θ)] 2D θθ u + d2 w nl (k ) 2 θ u, u ad (11) = 4k D θθk u + 2 dw nl(k ) θk u f (u )[ k u (θ), k u (θ)] 2D θθ u, u ad + d2 w nl (k ) 2 From which we can solve for the first two terms in the expansion of w nl (k ). (12) dw nl (k ). = c g = 2k D θθ u, u ad d 2 w nl (k ) 2 = 4k D θθk u + 2c g θk u f (u )[ k u (θ), k u (θ)] 2D θθ u, u ad (c) Expansion of λ lin We can similarly solve for the first two terms in the expansion of λ lin () as follows. Recall the set-up for λ lin. We linearize (4) in the comoving frame θ = ω t k x to find (13) t u = k 2 D θθ u ω θ u + f (u (θ))u and substitute in the ansatz u(θ) = e νθ/k v(θ; ν) to obtain ) 2 ) (14) L ν v = kd ( 2 θ νk v ω ( θ νk v + f (u (θ))v

5 FREDHOLM ALTERNATIVE 5 Since λ = is a simple eigenvalue, we find that there is an analytic curve of eigenvalues λ lin (ν) near zero for ν ir near zero. Therefore, we have the following equation ) 2 ) (15) kd ( 2 θ νk v ω ( θ νk v + f (u (θ))v = λ lin (ν)v We proceed as before taking the first and second derivatives of (15) with respect to ν evaluated at ν =, using the fact that v = θ u at ν =. We find (16) L ν v = (λ lin () c p) θ u + 2k D θθ u L νν v = λ lin () θu + 2λ lin () νv + 4k D θ ν v 2c p ν v 2D θ u where c p = w k above we get is the phase speed at k. Using the Fredholm Alternative as (17) λ lin () = c p c g λ lin () = 4k D kθ u + 2D θ u (d) Verification of the Hypotheses of the Fredholm Alternative We now verify that the hypotheses of the Fredholm Alternative have indeed been met. (i) L is closed By writing L as a system of first order ODEs, we can reduce the problem to showing that T. = x A is a closed operator from L 2 per(, 2π) L 2 per(, 2π) with domain H 1 per(, 2π). We take a sequence {h n } L 2 per(, 2π) such that h n n h L 2 per (, 2π) and such that n u n : T u n = h n. We need to show that u n u H 1 per(, 2π) and T u = h. Let Φ(x, ) be the fundamental matrix solution to T u =. Then we have (18) u n (x) = Φ(x, )u n () + x Φ(x, y)h n (y)dy

6 6 KELLY MCQUIGHAN We know that the integral converges because x Φ(x, y)h n(y)dy = Φ(x, y), h n (y) L 2, so by Hilbert space theory we can conclude that Φ(x, y), h n (y) L 2 Φ(x, y), h(y) L 2. Therefore, it suffices to show that u n () u(). We use the fact that u n (2π) = u n () n to get (19) u n () = u n (2π) = Φ(2π, )u n () + Φ(x, y)h n (y)dy and it suffices to solve (2) (I Φ(2π, ))u n () = Φ(x, y)h n (y)dy We decompose R n. = N(I Φ(2π, )) + X so that we can write un () = α n u + a n where u n N(I Φ(2π, )) and a n X. We can redefine α n so that they converge since they are in the null space. Then we have (21) (I Φ(2π, ))a n () = Φ(x, y)h(y)dy which we can invert on the range of (I Φ(2π, )) to get (22) a n () = (I Φ(2π, )) 1 Φ(x, y)h(y)dy The integral on the right hand side converges as noted above. It is in the range of (I Φ(2π, )) since the range is a finite dimensional subspace of R n and necessarily closed; therefore, the right hand side converges, and since X is again a finite dimensional subspace a n () a() X. Now we can write u (x) = αu +a, and u(x) = Φ(x, )u()+ Φ(x, y)h(y)dy necessarily satisfies T u = h. Moreover, u(x) H 1 per(, 2π) since Φ(x, y) is. (ii) dim N(L ) < By hypothesis that λ = is simple. (iii) codim R(L ) < As we saw in bullet (i), ind(l ) is equivalent to ind(i Φ(2π, )). We show that ind(i Φ(2π, )) =, hence codim R(L ) = 1 <. The

7 FREDHOLM ALTERNATIVE 7 claim follows from the fact that I Φ(2π, ) : R n R n, and hence as we showed previously, ind(i Φ(2π, )) = n n =. (e) Observation about Fredholm Alternative We make an observation of the formulation of L u = h in terms of (23) (I Φ(2π, )) u() = Φ(x, y)h(y)dy. We make the assumption that N(I Φ(2π, ) is 1 dimensional so that!ψ R(I Φ(2π, )). Then (23) has a solution (24) = = = Ψ, Φ(2π, y)h(y) dy Φ(2π, y) Ψ, h(y) dy Ψ(x), h(y) dy where Ψ(x) solves the adjoint problem dw dx = A (x)w. References [1] Arjen Doelman, Bjorn Sandstede, Arnd Scheel, and Guido Schneider. The dynamics of modulated wave trains, chapter [2] Grigorios A. Pavliotis and Andrew M. Stuart. Multiscale Methods. Averaging and Homogenization, pages Springer Science+Business Media, 28. [3] Michael Renardy and Robert C. Rodgers. An Introduction to Partial Differential Equations, page 279. Springer-Verlag New York, [4] Bjorn Sandstede. Closedness of the range in the definition of fredholm operators. Notes.