Since B 2 (x, y) ≡ 0, all subsequent c 3 , c 4 , . . ., and B 3 , B 4 , . . ., vanish (see (173) and (174)),and we obtainandD(x, y, λ) =1∑ (−1) n=B n (x, y)λ n = B 0 (x, y) − λB 1 (x, y) =n=0n!( 1= x + y − λ2 (x + y) − xy − 1 , (176)3)D(λ) ==2∑ (−1) nc n λ n = c 0 − c 1 λ + 1n=0n!2 c 2λ 2 == 1 − λ − 1 12 λ2 = − 1 12 (λ2 + 12λ − 12)Γ(x, y, λ) =D(x, y, λ)D(λ)= −12 x + y − λ ( 1(x + y) − xy − ) 12 3 ). (177)λ 2 + 12λ − 12The solution to (175) is given by formula (99)so that we obtain, using (176),φ(x) = f(x) +∫ 1φ(x) = f(x) + λ Γ(x, y, λ)f(y)dy,0∫λ1[6(λ − 2)(x + y) − 12xy − 4λ]f(y)dy, (178)λ 2 + 12λ − 12 0which coincides with formula (150).The numbers λ = λ ∗ k (k = 1, 2) determined in (144), λ ∗ 1 = −6 + √ 4 and λ ∗ 2 = −6 − √ 4, are(simple, i.e., of multiplicity one) poles of the resolvent because thay are (simple) zeros of theFredholm determinant D(λ). When λ = λ ∗ k (k = 1, 2), then IE (175) is not solvable.10 Hilbert spaces. Self-adjoint operators. Linear operatorequations with completely continuous operatorsin Hilbert spaces10.1 Hilbert space. Selfadjoint operatorsA real linear space R with the inner product satisfying(x, y) = (y, x) ∀x, y ∈ R,(x 1 + x 2 , y) = (x 1 , y) + (x 2 , y), ∀x 1 , x 2 , y ∈ R,(λx, y) = λ(x, y), ∀x, y ∈ R,(x, x) ≥ 0, ∀x ∈ R, (x, x) = 0 if and only if x = 0.34
is called the Euclidian space.In a complex Euclidian space the inner product satisfies(x, y) = (y, x),(x 1 + x 2 , y) = (x 1 , y) + (x 2 , y),(λx, y) = λ(x, y),Note that in a complex linear space(x, x) ≥ 0, (x, x) = 0 if and only if x = 0.(x, λy) = ¯λ(x, y).The norm in the Euclidian space is introduced by||x| = √ x, x.The Hilbert space H is a complete (in the metric ρ(x, y) = ||x − y||) infinite-dimensionalEuclidian space.In the Hilbert space H, the operator A ∗ adjoint to an operator A is defined by(Ax, y) = (x, A ∗ y), ∀x, y ∈ H.The selfadjoint operator A = A ∗ is defined from(Ax, y) = (x, Ay), ∀x, y ∈ H.10.2 Completely continuous integral operators in Hilbert spaceConsider an IE of the second kindφ(x) = f(x) +∫ baK(x, y)φ(y)dy. (179)Assume that K(x, y) is a Hilbert–Schmidt kernel, i.e., a square-integrable function in the squareΠ = {(x, y) : a ≤ x ≤ b, a ≤ y ≤ b}, so thatand f(x) ∈ L 2 [a, b], i.e.,∫ b ∫ baa∫ ba|K(x, y)| 2 dxdy ≤ ∞, (180)|f(x)| 2 dx ≤ ∞.Define a linear Fredholm (integral) operator corresponding to IE (179)Aφ(x) =∫ baK(x, y)φ(y)dy. (181)If K(x, y) is a Hilbert–Schmidt kernel, then operator (181) will be called a Hilbert–Schmidtoperator.Rewrite IE (179) as a linear operator equationφ = Aφ(x) + f, f, φ ∈ L 2 [a, b]. (182)35
- Page 1 and 2: Karlstad UniversityDivision for Eng
- Page 3 and 4: 10.5 The Hilbert-Schmidt theorem .
- Page 5 and 6: 2 Notion and examples of integral e
- Page 7 and 8: Subtracting termwise we obtain an o
- Page 9 and 10: with the initial conditiony(x 0 ) =
- Page 11 and 12: Example 7 The space C[a, b] of cont
- Page 13 and 14: where the kernel K(x, y) is a conti
- Page 15 and 16: and so on, obtaining for the (n + 1
- Page 17 and 18: Thus the common term of the series
- Page 19 and 20: Proof. To prove the theorem, it is
- Page 21 and 22: Theorem 9 (Superposition principle)
- Page 23 and 24: A, being completely continuous in t
- Page 25 and 26: These conditions are equivalent to
- Page 27 and 28: In particular, if λ is a regular v
- Page 29 and 30: so thatanda 11 =a 12 =a 21 =a 22 =f
- Page 31 and 32: Therefore the solution isφ(x) = f(
- Page 33: ∫ bc n = . . .a∫ ba∣K(y 1 , y
- Page 37 and 38: Now note that K N (x, y) in (189) i
- Page 39 and 40: Proof. Assume that y n ∈ Im T and
- Page 41 and 42: We see that, on the one hand, vecto
- Page 43 and 44: Multiply equality (213) by ¯φ 0 a
- Page 45 and 46: Replacing x by y abd vice versa, we
- Page 47 and 48: 12 Harmonic functions and Green’s
- Page 49 and 50: Since on Ω(x, r) we havegrad y Φ(
- Page 51 and 52: From Green’s first formula applie
- Page 53 and 54: which proves the continuity of the
- Page 55 and 56: where B 1 is a constant taken accor
- Page 57 and 58: Theorem 36 The operators I − K 0
- Page 59 and 60: Theorem 41 Let D ∈ R 2 be a domai
- Page 61 and 62: n = 2 π∫1−1f(x)U n (x) √ 1
- Page 63 and 64: A similar statement is valid for th
- Page 65 and 66: In some cases one can determine the
- Page 67 and 68: Coefficient a 0 vanishes because (s
- Page 69 and 70: Expand the right-hand side in the C
- Page 71 and 72: We see that the equation has the un
- Page 73 and 74: or, in other notation,andorL = {l n
- Page 75 and 76: where T n (x) = cos(n arccos x) are
- Page 77 and 78: and the integral equationwhereΦ N
- Page 79 and 80: Let us expand the functiong(s) = 1l
- Page 81 and 82: Assume that X n and Y n are the spa