Let us prove that this is a contraction mapping of the complete set C ∗ = C[x 0 − d, x 0 + d] ofcontinuous functions defined in a closed interval x 0 − d ≤ x ≤ x 0 + d (see Example 7). We have∫ x∫ x∫ x|ψ(x) − y 0 | =∣ f(t, φ(t))dt∣ ≤ |f(t, φ(t))| dt ≤ k dt = k(x − x 0 ) ≤ kd. (54)x 0 x 0 x 0∫ x|ψ 1 (x) − ψ 2 (x)| ≤ |f(t, φ 1 (t)) − f(t, φ 2 (t))| dt ≤ Md max |φ 1(x)| − φ 2 (x). (55)x 0 x 0 −d≤x≤x 0 +dSince Md < 1, the mapping ψ = Aφ is a contraction.From this it follows that the operator equation φ = Aφ and consequently the integralequation (33) has one and only one solution.This result can be generalized to systems of ODEs of the first order.To this end, denote by ¯f = (f 1 , . . . , f n ) an n-dimensional vector-function and write theinitial value problem for a system of ODEs of the first order using this vector notationdȳdx = ¯f(x, ȳ), ȳ(x 0 ) = ȳ 0 . (56)where the vector-function ¯f is defined and continuous in a region G of the n + 1-dimensionalspace R n+1 such that G contains the ’n+1-dimensional’ point x 0 , ȳ 0 . We assume that f satisfiesthe the Lipschitz condition (51) termwise in variables ȳ 0 . the initial value problem (56) isequivalent to a system of IEs¯φ(x) = ȳ 0 +∫ xx 0¯f(t, ¯φ(t))dt. (57)Since function ¯f(t, ȳ) is continuous we have ||f(x, y)|| ≤ K in a region G ′ ⊂ G which contains(x 0 , ȳ 0 ). Now choose a number d such that(x, ȳ) ∈ G ′ if |x 0 − x| ≤ d, ||ȳ 0 − ȳ|| ≤ KdMd < 1.Then we consider the mapping ¯ψ = A ¯φ defined by¯ψ(x) = ȳ 0 +∫ xx 0¯f(t, ¯φ(t))dt, |x0 − x| ≤ d. (58)and prove that this is a contraction mapping of the complete space ¯C ∗ = ¯C[x 0 − d, x 0 + d] ofcontinuous (componentwise) vector-functions into itself. The proof is a termwise repetitiion ofthe reasoning in the scalar case above.5 Unique solvability of the Fredholm IE of the 2nd kindusing the contraction mapping. Neumann series5.1 Linear Fredholm IE of the 2nd kindConsider a Fredholm IE of the 2nd kindf(x) = φ(x) + λ∫ ba12K(x, y)φ(y)dy, (59)
where the kernel K(x, y) is a continuous function in the squareΠ = {(x, y) : a ≤ x ≤ b, a ≤ y ≤ b},so that |K(x, y)| ≤ M for (x, y) ∈ Π. Consider the mapping g = Af defined byof the complete space C[a, b] into itself. We have∫ bg(x) = φ(x) + λ K(x, y)φ(y)dy, (60)a∫ bg 1 (x) = λg 2 (x) = λa∫ bρ(g 1 , g 2 ) = max |g 1(x) − g 2 (x)| ≤a≤x≤b∫ baK(x, y)f 1 (y)dy + φ(x),K(x, y)f 2 (y)dy + φ(x),≤ |λ| |K(x, y)||f 1 (y) − f 2 (y)|dy ≤ |λ|M(b − a) max |f 1(y)| − f 2 (y)| (61)aa≤y≤b1< ρ(f 1 , f 2 ) if |λ|
- Page 1 and 2: Karlstad UniversityDivision for Eng
- Page 3 and 4: 10.5 The Hilbert-Schmidt theorem .
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- Page 29 and 30: so thatanda 11 =a 12 =a 21 =a 22 =f
- Page 31 and 32: Therefore the solution isφ(x) = f(
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- Page 35 and 36: is called the Euclidian space.In a
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A similar statement is valid for th
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In some cases one can determine the
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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
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