Discretization Methods Must replace the problem of computing the unknown function f with a discrete problem that we can solve on a computer. Linear integral equation ⇒ system of linear algebraic equations. Quadrature Methods. Compute approximations ˜ f j = ˜ f (t j ) to the solution f at the abscissas t 1 , t 2 ,..., t n . Expansions Methods. Compute an approximation of the form f (n) (t )= n X j =1 ζ j φ j (t ), where φ 1 (t ),...,φ n (t ) are expansion/basis functions. Intro to Inverse Problems Chapter 3 Discretization; SVD 1 / 23
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Discretization Methods
Must replace the problem of computing the unknown function f with adiscrete problem that we can solve on a computer.
Linear integral equation ⇒ system of linear algebraic equations.Quadrature Methods.
Compute approximations f̃j = f̃ (tj) to the solution fat the abscissas t1, t2, . . . , tn.
Expansions Methods.Compute an approximation of the form
f (n)(t) =n∑
j=1
ζj φj(t),
where φ1(t), . . . , φn(t) are expansion/basis functions.
Discretization: the Galerkin MethodSelect two sets of functions φi and ψj , and write
f (t) = f (n)(t) + Ef (t), f (n)(t) ∈ span{φ1, . . . , φn}g(s) = g (n)(s) + Eg (s), g (n)(s) ∈ span{ψ1, . . . , ψn} .
Write f (n) as the expansion
f (n)(t) =n∑
j=1
ζj φj(t)
and define the function
ϑ(s) =
∫ 1
0K (s, t) f (n)(t) dt =
n∑j=1
ζj
∫ 1
0K (s, t)φj(t) dt
= ϑ(n)(s) + Eϑ(s) , ϑ(n) ∈ span{ψ1, . . . , ψn} .
Note that, in general, ϑ does not lie in the same subspace as g (n).Intro to Inverse Problems Chapter 3 Discretization; SVD 5 / 23
Computation of the Galerkin SolutionThe best we can do is to require that ϑ(n)(s) = g (n)(s) for s ∈ [0, 1].This is equivalent to requiring that the residual g(s)− ϑ(s) is orthogonalto span{ψ1, . . . , ψn}, which is enforced by
〈ψi , g〉 = 〈ψi , ϑ〉 =
⟨ψi ,
∫ 1
0K (s, t) f (n)(t) dt
⟩, i = 1, . . . , n.
Inserting the expansion for f (n), we obtain the n × n system
The singular values decay gradually to zero.No gap in the singular value spectrum.Condition number cond(A) = “∞.”Singular vectors have more oscillations as i increases.In this problem, # sign changes = i − 1.
The following pages: Picard plots with increasing noise.
The relative decay of the singular values σi and the right-hand side’s SVDcoefficients uTi b plays a major role!
The Discrete Picard Condition is satisfied if the coefficients |uTi bexact|,on the average, decay to zero faster than the corresponding singularvalues σi .
Recall that the (least squares) solution is given by
x =n∑
i=1
uTi b
σivi .
Must get rid of the “noisy” SVD components. Note that
uTi b = uTi bexact + uTi e ≈
{uTi b
exact, |uTi bexact| > |uTi e|
uTi e, |uTi bexact| < |uTi e|.
Hence, due to the DPC:“noisy” SVD components are those for which |uTi bexact| is small,and therefore they correspond to the smaller singular values σi .