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1. Define K : C[0, 2] → C[0, 2] by
Kϕ(x) = (x − t)2 ϕ(t)dt, for 0 ≤ x ≤ 2 and ϕ ∈ C[0, 2].
(a) Show, by using the formula on p. 89 in the lecture notes, that
∥K∥ = 3/8. (1)
(b) Show the same result from first principles, i.e. directly from the definition of the norm of a bounded linear operator on p. 83. Specifically, directly from the definition of K above (see part (a) of the example on pages 98-103 of the notes for a model), show that
∥Kϕ∥∞ ≤ 3/8 ∥ϕ∥∞ , for every ϕ ∈ C
Also, show that, if ψ(t) = 1 for 0 ≤ t ≤ 2, then ∥ψ∥∞ = 1 and
∥Kψ∥∞ ≥ |Kψ(0)| = 3/8.
Explain why these results in part (b) imply that (1) holds.
2. Define K : C[0, 2] → C[0, 2] by
Kϕ(x) = (x − t)3 ϕ(t)dt, for 0 ≤ x ≤ 2 and ϕ ∈ C[0, 2].
Using the theorem on page 89 in the lecture notes, compute ∥K∥ .
3. Define K : C[0, 1] → C[0, 1] by
Kϕ(x) = |x − t|−1/2 ϕ(t)dt, for 0 ≤ x ≤ 1 and ϕ ∈ C[0, 1].
Using the theorem on page 97 in the lecture notes, compute ∥K∥ .
4. Let X = C[a,b] and suppose that K : X → X is a bounded linear operator.
(a) Prove that K2 : X → X defined by K2 ϕ = K(Kϕ), for all ϕ ∈ X, is also a bounded linear operator and that
∥K2 ∥ ≤ ∥K∥2 .
(b) For n = 2, 3, ... define Kn : X → X by Kn ϕ = K(Kn−1ϕ), for ϕ ∈ X, where K1 ϕ := Kϕ .
Show by induction that Kn is a bounded linear operator for n = 2, 3, ... and that ∥Kn ∥ ≤ ∥K∥n.
5. Define K : C[0,π/2] → C[0,π/2] by
dt, for 0 ≤ x ≤ and ϕ ∈ C
(a) Show that, for 0 ≤ x ≤ π/2 and ϕ ∈ C[0,π/2],
|Kϕ(x)| ≤ x∥ϕ∥∞ .
(b) Using the result from part (a), show that
|K2 ϕ tdt∥ϕ∥∞ = ∥ϕ∥∞ , for 0 ≤ x ≤ and ϕ ∈ C[0, π/2].
(c) Using the result from part (a), show by induction that, for n = 1, 2, ...,
Knϕ(x)| ≤ x n n! ∥ϕ∥∞, for 0 ≤ x ≤ π 2 and ϕ ∈ C[0, π/2].
(d) Deduce from part (c) that
∥Kn ∥ ≤ π n n!2
and that ∥Kn ∥ → 0 as n → ∞ .
(e) From part (d) it follows that ∥K∥ ≤ π/2. Using the theorem on page 105 in the lecture notes, compute the exact value of ∥K∥ .
6. Suppose K : C[−π,π] → C[−π,π] is defined by
Kϕ(x) = Z − π π sin(x) sin(t) ϕ(t)dt, −π ≤ x ≤ π.
(a) Using the formula for ∥K∥ on page 89 of your notes, show that ∥K∥ = 4.
(b) By definition,
∥K∥ = sup ϕ∈C[−π,π]\{0} ∥Kϕ∥∞ ∥ϕ∥∞.
Since ∥K∥ = 4, this implies that, for every ϵ > 0 there exists a non-zero ϕ ∈ C[−π,π] such that
∥Kϕ∥∞ ∥ϕ∥∞ ≥ 4 − ϵ.
Defining ψ ∈ C[−π,π] by
ψ := ϕ ∥ϕ∥∞ ,
it follows that ∥ψ∥∞ = 1 and that
∥Kψ∥∞ = ∥Kϕ∥∞ ∥ϕ∥∞ ≥ 4 − ϵ.
Find an explicit function ψ for which this is true! I.e., for every ϵ > 0, construct a function ψ ∈ C[−π,π] with ∥ψ∥∞ = 1 such that
∥Kψ∥∞ ≥ 4 − ϵ .
[Hint: examine the proof of the theorem on page 89 for inspiration!]
7. Consider the integral equation that was discussed on slide 114 of the lecture notes, the integral equation
λy(x) = g(x) + k(x,t)y(t)dt, a ≤ x ≤ b,
with a = 0, b = 1, λ = 1, and
g(x) = x2 , k(x,t) = x2 − t2 ,
for x,t ∈ [a,b] = [0, 1].
Copying the solution method on slides 115-120 of the lecture notes, compute the Degen- erate Kernel Method approximate solution yN in the (simplest) case N = 1. To help you visualise what is going on, draw the figure on slide 115 that applies in this case. (The blue curve, the plot of k(x,t) against t when x = √3/2 is the same, what is different is the graph of kN (x,t). Do also look at page 110/111/112 for the graph in the general case.)
8. Consider the integral equation
λy(x) = g(x) + k(x,t)y(t)dt, a ≤ x ≤ b,
in the case that a = 0, b = 2, λ = 2, and
g(x) = x, k(x,t) = x − t,
for x,t ∈ [a,b] = [0, 1].
(a) Compute the exact solution y ∈ C[0, 2] of this equation.
(b) Copying the solution method on slides 115-121 of the lecture notes, compute the Degenerate Kernel Method approximate solution yN in the case N = 2.
9. Suppose that a < b, k,ℓ ∈ C[a,b], and define the integral operator K : C[a,b] → C[a,b] by
Kϕ(x) = k(x)ℓ(t)ϕ(t)dt, for a ≤ x ≤ b and ϕ ∈ C[a,b],
and consider the integral equation
λy = Ky (2)
where λ ∈ C.
(a) Show that, if λ ≠ 0 and
λ ≠ Z a b k(t)ℓ(t) dt, (3)
then the integral equation (2) only has the trivial solution y = 0.
(b) Using the result from part (a), what does the Fredholm Alternative (slide 124) tell
us about whether or not, given g ∈ C[a,b], the inhomogeneous equation
λy = g + Ky
has any solutions y ∈ C[a,b]?
10. 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 (4)
where g ∈ C[a,b]. Show that this equation has exactly one solution if ∥Kn ∥ < 1 for some n ∈ N. [Hint: show first that, if ∥Kn ∥ < 1 for some n ∈ N, then the equation y = Ky only has the trivial solution, and then appeal to the Fredholm Alternative.]