代写MA3XJ/MA4XJ Integral Equations Problem Sheet 4: 2023-2024代写C/C++语言

2024.08.14 - 首页 >> Java编程

MA3XJ/MA4XJ Integral Equations

Problem Sheet 4: 2023-2024

1. Let K denote the integral operator on C[a, b] which occurs in the integral equation on page 195 of the lecture notes, i.e.

where k0 > 0,

and the function k : [a, b] → R is positive and piecewise continuous.

(a) Show that

where

(b) The solution to the 1D scattering problem discussed in the lectures satisfies the integral equation

where ui (x) := exp(ik0x), for a ≤ x ≤ b. The Born approximation u b to the scattering problem is given by the same formula, but with u replaced by u i on the right hand side, i.e. ub is given by

Write both these equations in operator form. using the notation from part (a).

u − ub = K2u

and that, if ∥K∥ < 1, it holds that

∥u∥∞ ≤ (1 − ∥K∥)−1

and that

(d) Combining the results of parts (a) and (c), work out a condition on L, k0 and ∥k 2 − k20∥∞ which will ensure that u b is in error by no more than 0.01 on [a, b], i.e. which guarantees that

|ub (x) − u(x)| ≤ 0.01,

for a ≤ x ≤ b.

2. Suppose that k : R → (0,∞) is continuous and that, for some a < b and some k0 > 0,

k(x) = k0 for x ≥ b and x ≤ a.

Where BC(R) is the Banach space as defined in question 5 on Problem Sheet 3, define the integral operator K : BC(R) → BC(R) by

where the function Φ is as defined in question 1.

(a) Show that

where L = b − a. (This looks as if it is the same result as in question 1 (and it is almost the same!), but in this question K is defined so as to be an integral operator on BC(R), so that

with ∥ · ∥∞ as defined in question 5 on Problem Sheet 3.)

(b) Suppose that u ∈ C 2 (R) and that:

(i) u ′′(x) + k 2 (x)u(x) = 0, for x ∈ R;

(ii) for some T ∈ C, u(x) = T exp(ik0x), for x ≥ b;

(iii) for some R ∈ C, u(x) = exp(ik0x) + R exp(−ik0x), for x ≤ a.

Show that u satisfies the integral equation

u = ui + Ku, (1)

where u i ∈ BC(R) is defined by u i (x) = exp(ik0x), for x ∈ R. (Hint: this is just the argu-ment that we went through in the lectures (and see slides 199-207). You basically just need to repeat this, but it all simplifies a little bit as we are assuming here that k is continuous – in the lectures and on the slides we did the slightly more complicated case where k is piecewise continuous.)

Show that if u ∈ BC(R) satisfies this equation then u satisfies conditions (ii) and (iii) in part (b), and get explicit formulae for R and T as integrals involving the values of k and u on [a, b]. (Hint: note that exp(ik0|x − y|) in the definition of Φ(x, y) can be written without the modulus signs if a ≤ y ≤ b and x ≥ b, or if a ≤ y ≤ b and x ≤ a.)

(d) Let u b ∈ BC(R) be the Born approximation to u, given by

(This is the same definition as in question 1, but now we are thinking of u b as defined on the whole real line, not just on [a, b].) Adapting the arguments from question 1, show that

3. Consider the following problems:

Problem 1 (Volterra integral equation). Find y ∈ C[0, 1] such that

Problem 2 (IVP for an ODE). Find y ∈ C 2 [0, 1] such that

y'' (x) = (1 + x − x 2 )y' (x) + (2 − 2x)y(x), 0 ≤ x ≤ 1,

and y(0) = 1, y' (0) = 2.

Show that if y satisfies Problem 1, then it satisfies Problem 2.

4. Suppose that y ∈ C[0, π] and that

Show that y ∈ C 2 [0, π], that

y'' (x) = 0, for 0 ≤ x ≤ π,

and that y(0) = 1, y ′ (0) = 0. Hence deduce that y(x) = 1, for 0 ≤ x ≤ π.

5. Consider the following problems:

Problem 1 (Volterra integral equation). Find y ∈ C[0, 1] such that

Problem 2 (IVP for an ODE). Find y ∈ C 1 [0, 1] such that

y' (x) = 1 + (1 − x 2 )y(x), 0 ≤ x ≤ 1,

and y(0) = 1.

(a) Show that y satisfies Problem 1 if and only if it satisfies Problem 2.

(b) Do Problems 1 and 2 have solutions? If so how many?

|y(x)| ≤ 2ex , 0 ≤ x ≤ 1.

(d) Assuming that y satisfies Problem 1, compute approximations to y(0.5) and y(1) by applying the trapezium rule with step size h = 0.5.

(e) The integral equation in Problem 1 can be written in operator form. as y = g + Ky, where g(x) = 1 + x, for 0 ≤ x ≤ 1, and K : C[0, 1] → C[0, 1] is defined by

Compute the first three terms of the Neumann series, i.e. compute

y2 := g + Kg + K2g.

(f) Where K, g, and y2 are as defined in part (e), compute ∥K∥ and show that, if y ∈ C[0, 1] satisfies Problem 1, then

6. Let y ∈ C[0, 2] be the unique solution of the integral equation

(a) By applying the Gronwall inequality, show that |y(x)| ≤ exp(ex), for 0 ≤ x ≤ 2.

(b) Compute approximations to y(0.5) and y(1) using the trapezium rule with step size h = 0.5.

2y' (x) = 3y(x) − 1, 0 ≤ x ≤ 2.

(d) Find the exact solution of the integral equation by solving the initial value problem in (c). In particular compute the exact values of y(0.5) and y(1).

7. Suppose that u ∈ C(R) and k ∈ C(R) and that, for some k0 > 0 and a < b, k(x) = k0 for x > b and for x < a. Suppose further that

where

and u i (x) = exp(ik0x), for x ∈ R. Show that u ∈ C 2 (R) and that

u'' (x) + k 2 (x)u(x) = 0

for x ∈ R.

Hint. First observe that we will have shown that u ∈ C2 (R) and that the above equa-tion holds for all x ∈ R if, for all sufficiently large A > 0, we show that u ∈ C 2 [−A, A] and that the above differential equation holds in [−A, A]. Then observe that, if A is large enough so that [a, b] ⊂ [−A, A], then, for −A ≤ x ≤ A,

Now proceed by applying the Leibniz rule to the integrals in the above expression to show that u ∈ C 2 [−A, A] and get an expression for u'' (x).

8. Show that the trapezium rule approximation with step-length h = 1/N applied to the equation

leads to the equations defining the approximation yn to y(tn) (where tn = hn):

Using h = 0.5 compute approximations to y(0), y(0.5) and y(1). Compare these approxi-mations to the exact solution to the integral equation obtained via converting the integral equation to an initial value problem.