The general relationship has the formwheren∑∫ bφ n (x) = f(x) + λ m K m (x, y)f(y)dy, (84)m=1aK 1 (x, y) = K(x, y), K m (x, y) =∫ baK(x, t)K m−1 (t, y)dt. (85)and K m (x, y) is called the mth iterated kernel. One can easliy prove that the iterated kernelssatisfy a more general relationshipK m (x, y) =∫ baK r (x, t)K m−r (t, y)dt, r = 1, 2, . . . , m − 1 (m = 2, 3, . . .). (86)Assuming that successive approximations (84) converge we obtain the (unique) solution to IE(70) by passing to the limit in (84)∞∑∫ bφ(x) = f(x) + λ m K m (x, y)f(y)dy. (87)m=1aIn order to prove the convergence of this series, write∫ band estimate C m . Setting r = m − 1 in (86) we haveaK m (x, y) =|K m (x, y)| 2 dy ≤ C m ∀x ∈ [a, b], (88)∫ bApplying to (89) the Schwartz inequality we obtaina|K m (x, y)| 2 =Integrating (90) with respect to y yields∫ bawhich gives∫ b|K m (x, y)| 2 dy ≤ B 2 |K m−1 (x, t)| 2 ≤ B 2 C m−1 (B =and finally the required estimateaK m−1 (x, t)K(t, y)dt, (m = 2, 3, . . .). (89)∫ ba∫ b|K m−1 (x, t)| 2 |K(t, y)| 2 dt. (90)a∫ b ∫ baa|K(x, y)| 2 dxdy) (91)C m ≤ B 2 C m−1 , (92)C m ≤ B 2m−2 C 1 . (93)Denoting√ ∫ bD = |f(y)| 2 dy (94)aand applying the Schwartz inequality we obtain∫ b2 ∫ bK∣ m (x, y)f(y)dy≤a∣ a∫ b|K m (x, y)| 2 dy |f(y)| 2 dy ≤ D 2 C 1 B 2m−2 . (95)a16
Thus the common term of the series (87) is estimated by the quantity√D C 1 |λ| m B m−1 , (96)so that the series converges faster than the progression with the denominator |λ|B, which provesthe theorem.If we take the first n terms in the series (87) then the resulting error will be not greaterthan|λ|D√C n+1 B n11 − |λ|B . (97)5.5 ResolventIntroduce the notation for the resolvent∞∑∫ bΓ(x, y, λ) = λ m−1 K m (x, y) (98)m=1aChanging the order of summation and integration in (87) we obtain the solution in the compactform∫ bφ(x) = f(x) + λ Γ(x, y, λ)f(y)dy. (99)aOne can show that the resolvent satisfies the IE∫ bΓ(x, y, λ) = K(x, y) + λ K(x, t)Γ(t, y, λ)dt. (100)a6 IEs as linear operator equations. Fundamental propertiesof completely continuous operators6.1 Banach spacesDefinition 3 A complete normed space is said to be a Banach space, B.Example 9 The space C[a, b] of continuous functions defined in a closed interval a ≤ x ≤ bwith the (uniform) norm||f|| C = maxa≤x≤b |f(x)|[and the metricρ(φ 1 , φ 2 ) = ||φ 1 − φ 2 || C = maxa≤x≤b |φ 1(x)| − φ 2 (x)|]is a normed and complete space. Indeed, every fundamental (convergent) functional sequence{φ n (x)} of continuous functions converges to a continuous function in the C[a, b]-metric.Definition 4 A set M in the metric space R is said to be compact if every sequence of elementsin M contains a subsequence that converges to some x ∈ R.17
- 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
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- Page 11 and 12: Example 7 The space C[a, b] of cont
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- Page 15: and so on, obtaining for the (n + 1
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- 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 and 34: ∫ bc n = . . .a∫ ba∣K(y 1 , y
- Page 35 and 36: is called the Euclidian space.In a
- 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
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- 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
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Coefficient a 0 vanishes because (s
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Expand the right-hand side in the C
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We see that the equation has the un
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or, in other notation,andorL = {l n
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where T n (x) = cos(n arccos x) are
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and the integral equationwhereΦ N
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Let us expand the functiong(s) = 1l
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Assume that X n and Y n are the spa