代做MA3XJ/MA4XJ Integral Equations Problem Sheet 3: 2023-2024代写数据结构程序
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Problem Sheet 3: 2023-2024
1. Suppose that X is a Banach space, K is a bounded linear operator on X , g ∈ X, and that yN ∈ X is defined, for each N ∈ N, by
yN = g + Kg + ... + KNg.
Show that
g + KyN = yN+1 , for N ∈ N.
Hence show that if the sequence yN is convergent to some limit z, in symbols yN → z, then
z = g + Kz.
(This is the last step of the proof on page 147 of the notes. Hint: use the fact that every bounded operator is continuous – see page 141 of the notes.)
2. If A and B are two bounded linear operators on a Banach space X, the product op- erator AB is defined by (AB)ϕ := A(Bϕ), for ϕ ∈ X, i.e. AB is the composition of the mappings A and B .
(a) Show that AB is linear.
(b) Show that AB is bounded with
||AB|| ≤ ||A|| ||B||.
(The definition of a linear operator/mapping is on p. 69 of the slides, the definition of a bounded linear operator and its norm on p. 83.)
(c) Suppose that A, B, and C are bounded linear operators on X . Show that (AB)C = A(BC),
i.e. that operator products obey the associative law. (Hint: use that if A : X → X and B : X → X then saying that A = B means that Aϕ = Bϕ for all ϕ ∈ X.)
(d) Where the bounded linear operator An is defined for n ∈ N as in the lectures (or see PS2 Q4), show that, for m,n ∈ N, it holds that
AnAm = An+m ,
i.e. that the product of the operators An and Am is An+m. (Hint: fix m ∈ N and then prove that the above formula holds for all n ∈ N by induction (on n), using the result from (c).)
3. Define K : C[−1, 1] → C[−1, 1] by
where k > 0 is some positive constant, so that K is the integral operator on C[−1, 1] with kernel k(x,t) = eikxt.
(a) Show that (hint: reverse the order of integration at an appropriate point, which is justified as the integrand is continuous)
where sinc is a standard notation for the function sincx := x-1 sin x. Thus K2 is also an integral operator, with kernel 2sinc(k(x + t)).
(b) Show that ∥K∥ = 2. (Hint: just use our standard formula on p. 89.)
(c) Deduce from applying the result in question 2(b) that ∥K2 ∥ ≤ 4.
(d) Show, using the formula from (a) which makes clear that K2 is an integral operator with a continuous kernel, that
and deduce that ∥K2 ∥ < 1 if k > 0 is large enough.
4. Suppose K is a bounded linear operator on a Banach space X and that y ∈ X sat- isfies the equation
y = Ky.
(a) Show, by induction, that
y = Kny
for n = 2, 3, . . ..
(b) Hence show that y = 0 if ∥Kn ∥ < 1 for some n ∈ N.
(c) Deduce that, if ∥Kn ∥ < 1 for some n ∈ N, then, for every g ∈ X, the equation y = g + Ky. has at most one solution.
5. Let BC(R) denote the set of functions ϕ : R → C that are continuous and bounded. This is a linear space (with addition of functions and multiplication of a function by a com- plex number defined in the usual obvious way). Let ∥ · ∥∞ be the norm on BC(R) defined by
Equipped with this norm, BC(R) is a normed space, in fact a Banach space. Given a constant λ > 0, define the integral operator K : BC(R) → BC(R) by
(a) Show from first principles (i.e. starting from the definitions on p. 83 in the notes) that the linear operator K is bounded and that
(b) Suppose that g, y ∈ BC(R) and that
y = g + Ky.
Show that
y = yN + KN+1y, for N ∈ N,
where
yN := g + Kg + ... + KNg,
and that
Deduce that, if λ > 2, then y is given by the convergent Neumann series y = g + Kg + K2g + ....
6. Suppose that K : C[a,b] → C[a,b] is an integral operator with a continuous or weakly singular kernel, and consider the integral equation
y = g + Ky (1)
where g ∈ C[a,b].
(a) By applying results from question 4 and the Fredholm Alternative, deduce that (1) has exactly one solution y ∈ C[a,b] for all g ∈ C[a,b] if ∥Kn ∥ < 1 for some n ∈ N.
(b) In the case that ⅡKn Ⅱ < 1 for some n ∈ N, and by applying the results from question 2, show that ⅡKN Ⅱ → 0 as N → ∞ . (Hint: write N as N = mn + r where m is a non- negative integer, and r is the remainder after dividing N by n.)
(c) Applying the results from (a) and (b), and arguing similarly to question 5, part (b), deduce that (1) has exactly one solution for every g ∈ C[a,b], and that this solution is given by the (convergent) Neumann series
y = g + Kg + K2g + ....
(Note that, once you have done (c), you have proved the theorem on page 160 in the special case that X = C[a,b] and K is an integral operator with continuous or weakly singular kernel.)