1.3. Base conceptual
1.3.4. Papel de las TIC en la Formación Educativa
There are clearly a number of issues here yet to be resolved. Perhaps the biggest open question is whether it is possible to show that the solutions of the martingale problem for (AΨ, µ) have unique one-dimensional distributions.
Unfortunately we are as yet unable to answer this, although we make the fol- lowing remarks about it. If ((ξx, ηx);x∈[0, X]) is a solution of the martingale
problem for (AΨ, µ), then there is a specific noise ( ˆWx;x∈ [0, X]) such that
((ξx, ηx);x ∈ [0, X]) satisfies the system (3.4.2). We would like to think of
((ξx, ηx);x∈[0, X]) as being a strong solution for the noise ˆW and the initial
condition (ξ0, η0) = (u0, v0). One way we might make this more precise is by
thinking of ((hh1, uxi,hh2, vxi);x ∈ [0, X]) as a strong solution to (3.4.2) for
(h1, h2)∈Ψ. The problem with this is that our system is not closed- it depends
also on the termhh′
2, ξxi+hA1A2h2, ξxi+hA2h2, ηxi. In any case, our hope is that
for the given initial condition (u0, v0) and a noise ˆW there is one, and only one,
solution to (3.4.2). We denote this by (ξx, ηx) = Φ(x,(u0, v0),Wˆ). We would
thus like to show that any solution of the martingale problem for (AΨ, µ) has the
form (Φ(x,(u0, v0),Wˆ) for some noise ˆW such that hWˆ(h2),Wˆ(h4)i=hh2, h4i
for all (h1, h2),(h3, h4)∈Ψ, and hence all solutions of the martingale (AΨ, µ)
have the same one-dimensional distributions (in fact, the same law, so this is clearly more than sufficient).
The success of this approach depends largely on whether we can show that the solutions to (3.4.2) are unique given a noise ˆW. Although we are attempting to show more than we need here, this approach has the benefit that uniqueness reduces to the uniqueness of a deterministic system of equations. Indeed, if we denote the difference between two solutions of (3.4.2) by (Ξx,Θx), then for all
(h1, h2)∈Ψ we have hh1,Ξxi= Z x 0 h h1,Θyidy hh2,Θxi=− Z x 0 (hh′2,Ξyi+hA1A2h2,Ξyi+hA2h2,Θyi)dy (3.4.4)
for all x ∈ [0, X], and Ξ0 = Θ0 = 0. For any h ∈ C0∞([0,∞)) such that
(h, h)∈Ψ we can write this as d2
dx2hh,Ξxi+hh′,Ξxi+hA1A2h,Ξxi+
d
dxhA2h,Ξxi= 0. Intuitively then we would like to be able to show that the equation
∂2 ∂x2φ(x, t)− ∂ ∂tφ(x, t) +A ∗ 2A∗1φ(x, t) +A∗2 ∂φ ∂x(x, t) = 0
has only the zero solution when supplied with boundary conditions φ(0, t) =
∂
∂xφ(x, t)|x=0= 0 for allt >0. In fact, there is very little hope for this without
also knowing that φ(x,0) = 0 for x∈ [0, X]. However, we have assumed this condition for our original processu(x, t) satisfying equation (0.1.1), so we may include this condition as some further property of the spaceE.
Since this uniqueness problem is unresolved, we do not attempt to discuss it here. However, we point out that if the uniqueness of the above partial differential equation is to be enough to provide uniqueness for solutions to the weak form of our equation, we need to know that the weak form is defined for ‘sufficiently’
many test functions h, whatever is meant by that. This involves choosing Ψ as large as possible such that ((ux, vx);x≥ 0) solves the martingale problem
for (AΨ, µ). Naturally, the smaller the set Ψ, the less hope we have of finding
uniqueness to the above weak equation. Let us offer some heuristic why it seems plausible that we can take Ψ =C∞
0 ([0,∞))×C0∞([0,∞)). To begin with, it is
clear that for anyh∈C∞
0 ([0,∞)),handh′are inX1. In fact, we can easily see
thatA2h∈X1 also. Indeed, when we investigated the tail ofA2hwe saw that
|A2h(t)| ≤c 1 √ t−T − 1 √ t −√1 t− 1 √ t+T
for t≫T, where the support of his in [0, T]. Our previous estimate was rather crude, and it is in fact possible to show this bound is
cT3
√
t−T√t+T(√t−T+√t+T)(√t+√t−T)2(√t+√t+T)2
which looks liket−7 2.
ForX2, we still look to use the ideas of section 3.4.2, but a more fruitful route
could be to take
H ={h: (I−∆)−1
2m(I−∆)−12h∈L2([0,∞))}
where ∆ is the Laplacian with Dirichlet boundary conditions at zero, and m is a positive weight function. If one assumes that C12
v and (I−∆)−
1 2 have
representations as symmetric kernel operators, where the kernels have certain favourable properties, and furthermore that they commute (and we observe that ∆ andCv do commute), then it is possible to show that
X i kC12 veik2H=c Z ∞ 0 m(t) Z ∞ 0 1 p |t−t′|− 1 √ t+t′ ! k(t, t′)dt′dt
wherek(t, t′) is the kernel of (I−∆)−1, which looks likee−|t−t′| . Formally we have hh, vxi= ((I−∆) 1 2m−1(I−∆) 1 2h, v x)H.
Thus we might think of vx as a bounded linear functional on a subspace of
C0,3
4+βcontaininghsatisfyingm
−1
2(I−∆)12h∈L2([0,∞)). The benefit of this
approach then is that we reduce to checking tail properties of (I−∆)12h. It is
tempting to compare this to the tail properties ofh′. In particular, forh∈C∞ 0 ,
h′ can be integrated against any continuous weightm, and we already have an
idea of how quickly (A2h)′ decays. However, at the time of writing more work
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