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Inverse Problems 3 for Selected Topics in Mathematics I | MATH 491, Assignments of Optimization Techniques in Engineering

Material Type: Assignment; Class: Sptp:Optimization; Subject: Mathematics; University: Duquesne University; Term: Fall 2008;

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

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Math 491/CPMA 591
Inverse Problems
Fall, 2008
Problems 3
Problem 1 The following bounds are not necessarily best possible; your answers may be different.
a) khk12π,khk22π,khk1.
b) khk1= 22π,khk2=,khk=.
c) khk1=1
6[(3/4)31], khk2<p47/38,khk= 27/34.
d) khk11/e,khk2= 1
sqrte,khk=e1.
Problem 2 For one norms, if kKk<, the basic calculation is the following:
khk1=Zb
a
Zb
a
K(x, y)h(y)
dx
Zb
aZb
a|K(x, y)h(y)|dydx
khkkKkZb
aZb
a
dydx
=khkkKk(ba)2.
If kKkis not finite but kK(·, y)k1is, the basic calculation for one-norms becomes:
khk1=Zb
a
Zb
a
K(x, y)h(y)
dx
Zb
aZb
a|K(x, y)h(y)|dydx
khkZb
akKk1dx
=khkkKk1(ba).
The calculations for two norms and sup norms are similar. Using these techniques, you can show:
a) kqk1 khk,kqk2 khk,kqk khk.
b) Similar. (More generally: whenever Kis bounded and the region is finite, qwill have 1,2, and
sup norms.
c) Here the kernel is not bounded. You can still that show that kqk1,kqk2, and kqkexist, and are
bounded by some multiple of khk.
Problem 3 Calculations similar to the above yield:
a) kqk1,kqk2, and kqkexist and are bounded by some multiple of khk1.
b) Ditto.
c) Here, neither kqk1,kqk2, nor kqkneed exist.
Problem 4 Recall that a norm is a positive function which satisfies the triangle inequality and is equal
to zero if and only if its argument is zero. Thus:
a) not a norm (has non-zero kernel)
b) norm (verify)
c) not a norm (could be negative)
pf2

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Math 491/CPMA 591 Inverse Problems Fall, 2008

Problems 3

Problem 1 The following bounds are not necessarily best possible; your answers may be different.

a) ‖h‖ 1 ≤ 2 π, ‖h‖ 2 ≤

2 π, ‖h‖∞ ≤ 1. b) ‖h‖ 1 = 2

2 π, ‖h‖ 2 = ∞, ‖h‖∞ = ∞. c) ‖h‖ 1 = 16 [(3/4)− 3 − 1], ‖h‖ 2 <

47 / 38 , ‖h‖∞ = 2^7 / 34. d) ‖h‖ 11 /e, ‖h‖ 2 = 1 sqrte, ‖h‖∞ = e−^1.

Problem 2 For one norms, if ‖K‖∞ < ∞, the basic calculation is the following:

‖h‖ 1 =

∫ (^) b

a

∫ (^) b

a

K(x, y)h(y)

dx

∫ (^) b

a

∫ (^) b

a

|K(x, y)h(y)| dydx

≤ ‖h‖∞‖K‖∞

∫ (^) b

a

∫ (^) b

a

dydx

= ‖h‖∞‖K‖∞(b − a)^2.

If ‖K‖∞ is not finite but ‖K(·, y)‖ 1 is, the basic calculation for one-norms becomes:

‖h‖ 1 =

∫ (^) b

a

∫ (^) b

a

K(x, y)h(y)

dx

∫ (^) b

a

∫ (^) b

a

|K(x, y)h(y)| dydx

≤ ‖h‖∞

∫ (^) b

a

‖K‖ 1 dx

= ‖h‖∞‖K‖ 1 (b − a).

The calculations for two norms and sup norms are similar. Using these techniques, you can show:

a) ‖q‖ 1 ≤ ‖h‖∞, ‖q‖ 2 ≤ ‖h‖∞, ‖q‖∞ ≤ ‖h‖∞. b) Similar. (More generally: whenever K is bounded and the region is finite, q will have 1,2, and sup norms. c) Here the kernel is not bounded. You can still that show that ‖q‖ 1 , ‖q‖ 2 , and ‖q‖∞ exist, and are bounded by some multiple of ‖h‖∞.

Problem 3 Calculations similar to the above yield:

a) ‖q‖ 1 , ‖q‖ 2 , and ‖q‖∞ exist and are bounded by some multiple of ‖h‖ 1. b) Ditto. c) Here, neither ‖q‖ 1 , ‖q‖ 2 , nor ‖q‖∞ need exist.

Problem 4 Recall that a norm is a positive function which satisfies the triangle inequality and is equal to zero if and only if its argument is zero. Thus:

a) not a norm (has non-zero kernel) b) norm (verify) c) not a norm (could be negative)

Problem 5 Note the kernel is separable, and thus any solution must lie in the span of the αi(x)’s. In this case, α 1 (x) = x and α 2 (x) = 1. Since x^2 + 1 does not lie in the span of the 1 and x, we conclude there is no solution.

Problem 6 Set ψˆi := y100+1^ for i = 1, 2 , · · ·. Note that ψˆi does not lie in the span of the βi (since it is impossible to write a high order polynomial as a sum of low order polynomials.) If we let P denote projection on the span of the βi and define

ψi(x) = ψ(ˆx) − P ( ψ(ˆx)),

then ψi is orthogonal to the βi for all i. Concretely, we have:

ψi(x) = ψˆ(x) −

∑^100

j=

ψ(ˆx), βi(x)

βi(x)/‖βi(x)‖^2.

Problem 7 The kernel is separable, with α 1 (x) = sin x, α 2 (x) = sin 2x, and α 3 (x) = cos 2x. The left hand side of the equation is a linear combination of these elements, with coefficients 1,3, and 1, respectively. Thus the equation might have a solution. To find out, suppose f is a solution. Note that f must be a linear combination of the βi’s, where in this case β 1 = cos y, β 2 = cos 2y, and β 3 = sin 2y. Let Γ be the matrix whose (i, j)’th entry is 〈βi, βj 〉. Then

π/ 2 0 0 0 π/ 4 0 0 0 π/ 4

and the equation has a solution if and only if the matrix equation  

f 1 f 2 f 3

has a solution. Clearly, the solution is f 1 = 2/π, f 2 = 12/π, and f 3 = 4/π.

Problem 8 As above, form the matrix Γ, this time with the (i, j)th entry equal to 〈βi, αj 〉. We are seeking a solution for some scalar λ (the eigenvalue) and vector y = [y 1 , y 2 ] (the eigenvector). The system has a solution if and only if the matrix equation

Γy = λy.

Since Γ =

we see from linear algebra that the eigenvalues will be ±8. The eigenvectors are v 1 = [−. 71 , .71] and v 2 = [. 71 , .71].