Exercises and Solutions to the Concentrated Advanced Course on Stochastic Partial Differential Equations

Fred Espen Benth

1 Introduction

In September 1998 professor Helge Holden gave an advanced course on stochastic partial differential equations at the Centre for Mathematical Physics and Stochastics (MaPhySto). The course consisted of lectures and tutorial classes.

This note contains the exercises with suggested solutions discussed in the classes. Notation and necessary theoretical background to understand the exercises can be found in the monograph, [HØUZ]. All references to exercises, equations, theorems etc below are from [HØUZ].

We have included an annotated list of references which is recommended for the ones interested in White Noise Analysis and its applications.

2 Exercises

Day 1

Exercise 1. Exercise 2.6

Exercise 2. Consider w(ϕ, ω) = ⟨ω, ϕ⟩, ϕ ∈ L^{2}(IR^{d}), ω ∈ S′(IR^{d}). Show that w(ϕ) is normally distributed with expectation zero and variance ∥ϕ∥^{2}.

Exercise 3. Consider W_{ϕ}(x, ω) = w(ϕ_{x}, ω), for ϕ ∈ L^{2}(IR^{d}), x ∈ IR^{d}, where ϕ_{x}(y) := ϕ(y − x). Prove that W_{ϕ}(x, ω) has properties (2.1.20)-(2.1.22).

Exercise 4. Show that E [H_{α}] = 0 for α ≠ 0, where H_{α} is defined in Def. 2.2.1-2, pp.19-20.

Exercise 5. Let f ∈ L^{2}(μ_{m}) with chaos expansion f = ∑_{α∈J} c_{α}H_{α}. Find E [f].

Exercise 6. Consider the 1-dimensional, 1-parameter smoothed white noise

w(ϕ, ω) = ∑_{j=1}^{∞}(ϕ, ξ_{j})H_{ϵ_{j}}(ω)

where ξ_{j} is the j’th Hermite function. Find the variance of w(ϕ) by using its chaos expansion.

Exercise 7. Let {ξ_{j}}_{j∈IN} be the Hermite functions and define T = ∑_{j=1}^{∞} ξ_{j}(t)ξ_{j}(·). Show, by using Th. 2.3.1, p. 28, that T ∈ S′(IR^{d}). Moreover, prove that T = δ_{t} in the sense of distributions, where δ_{t} is the Dirac δ-function at t.

Exercise 8. Let N = m = d = 1. Find the (formal) chaos expansion of W_{t}(ω) = ⟨ω, δ_{t}⟩. Show that W_{t} ∈ (S)_{−ρ} for all ρ ∈ [0, 1].

Exercise 9. Show that W_{ϕ}(x, ω) ∈ (S)_{1} for ϕ ∈ S(IR). (Let for simplicity N = m = d = 1).

Day 2

Exercise 1. Exercises 2.9 and 2.8 a-c

Exercise 2. Show that w(ϕ) ⋄ w(ψ) = w(ϕ) · w(ψ) if supp ϕ ∩ supp ψ = ∅.

Exercise 3. Show that

w^{⋄2}(ϕ) ⋄ w(ψ) = w^{2}(ϕ)w(ψ) − ∥ϕ∥^{2}w(ψ) − 2(ϕ, ψ)w(ϕ) (2.1)

Moreover, if supp ϕ ∩ supp ψ = ∅ then

w^{⋄2}(ϕ) ⋄ w(ψ) = w^{⋄2}(ϕ)w(ψ) (2.2)

Exercise 4. Variation of exercise 2.24: Use Th. 2.5.9, p. 52, to find the Skorohod integrals in exercise 2.24 a-c.

Day 3

Exercise 1. Exercises 2.8 e+f, 2.11, 2.13, 2.15 and 2.24 d

Exercise 2. Exercises 3.4 a-c and 3.5 a+d (In exercise a, use X(0) = G instead of X(T) = G).

Day 4

Exercise 1. Exercise 4.4

Exercise 2. Use Cor. 4.3.2, p. 150, to show that the solution of

\{ \begin{matrix}\frac{∂U}{∂t} = \frac{1}{2}△U + U ⋄ W_{t}, & (t, x) ∈ IR_{+} × IR^{d} \\ U(0, x) = f(x), & x ∈ IR^{d} \end{matrix}

is

U(t, x) = Ê [f(b(t))] ⋄ exp(B(t) − \frac{1}{2}t)

Exercise 3. Assume that f(x) ∈ (S)_{−1} for every x and that f satisfies all the conditions in Th. 4.3.1, p. 147. Prove that

U(t, x) = u(t, x) ⋄ X_{t}

is the solution of

\{ \begin{matrix}\frac{∂U}{∂t} = \frac{1}{2}△U + U ⋄ W_{t}, & (t, x) ∈ IR_{+} × IR^{d} \\ U(0, x) = f(x), & x ∈ IR^{d} \end{matrix}

where u(t, x) solves

\{ \begin{matrix}\frac{∂u}{∂t} = \frac{1}{2}△u, & (t, x) ∈ IR_{+} × IR^{d} \\ u(0, x) = f(x), & x ∈ IR^{d} \end{matrix}

and

dX_{t} = X_{t} dB_{t}, X_{0} = 1

Day 5

Exercise 4. Exercises 4.5 and 4.8

3 Solutions to exercises

Day 1

Exercise 1:

Let ϕ_{n} → ϕ in L^{2}(IR^{d}) when n → ∞, where {ϕ_{n}}_{n∈IN} is a sequence in S(IR^{d}). Then {⟨ω, ϕ_{n}⟩}_{n∈IN} is a Cauchy sequence in L^{2}(μ):

E [(⟨ω, ϕ_{n}⟩ − ⟨ω, ϕ_{m}⟩)^{2}] = E [⟨ω, ϕ_{n} − ϕ_{m}⟩^{2}]
= ∥ϕ_{n} − ϕ_{m}∥^{2} → 0

whenever n, m → ∞. Hence, {⟨ω, ϕ_{n}⟩}_{n∈IN} is convergent in L^{2}(μ) and we denote its limit ⟨ω, ϕ⟩. It is easily shown that

E [⟨ω, ϕ⟩^{2}] = ∥ϕ∥^{2}

and

E [⟨ω, ϕ⟩⟨ω, ψ⟩] = (ϕ, ψ)

for any ψ ∈ S(IR^{d}).

Let {ψ_{n}}_{n∈IN} be another sequence in S(IR^{d}) converging to ϕ in L^{2}(IR^{d}). Then

E [(⟨ω, ψ_{n}⟩ − ⟨ω, ϕ⟩)^{2}] = ∥ψ_{n}∥^{2} − 2(ψ_{n}, ϕ) + ∥ϕ∥^{2}

But ∥ψ_{n}∥ → ∥ϕ∥ and (ψ_{n}, ϕ) → ∥ϕ∥^{2} when n → ∞. Therefore, ⟨ω, ψ_{n}⟩ converges to ⟨ω, ϕ⟩. The definition of ⟨ω, ϕ⟩ is thus independent of the choice of the approximating sequence.

Exercise 2:

From the Bochner-Minlos Theorem, p. 12, the characteristic function of ⟨ω, ϕ⟩ is

E [exp(i⟨ω, ϕ⟩)] = exp(−\frac{1}{2}∥ϕ∥^{2})

Thus, ⟨ω, ϕ⟩ is normally distributed with zero expectation and variance ∥ϕ∥^{2}.

Exercise 3:

Property (2.1.22): Since

∥ϕ_{x}∥^{2} = ∫_{IR_{d}} ϕ^{2}(y − x) dy = ∥ϕ∥^{2}

we get from exercise 2 above that W_{ϕ}(x) ∼ N (0, ∥ϕ∥^{2})

Property (2.1.21): We consider the case (W_{ϕ}(x), W_{ϕ}(y)) for x, y ∈ IR^{d}. The general case will follow similarly. Our argument will in fact prove that (W_{ϕ}(x), W_{ϕ}(y)) is a Gaussian vector. Let c_{1}, c_{2} ∈ IR and h > 0. By the Bochner-Minlos Theorem on p. 12,

E [exp(ic_{1}W_{ϕ}(x + h) + ic_{2}W_{ϕ}(y + h)] = E [exp(i⟨ω, c_{1}ϕ_{x+h} + c_{2}ϕ_{y+h}⟩)]

= exp(−\frac{1}{2}∥c_{1}ϕ_{x+h} + c_{2}ϕ_{y+h}∥^{2})

But

∥c_{1}ϕ_{x+h} + c_{2}ϕ_{y+h}∥^{2} = c_{1}^{2}∥ϕ_{x+h}∥^{2} + 2c_{1}c_{2}(ϕ_{x+h}, ϕ_{y+h}) + c_{2}^{2}∥ϕ_{y+h}∥^{2}

= c_{1}^{2}∥ϕ∥^{2} + 2c_{1}c_{2}(ϕ_{x}, ϕ_{y}) + c_{2}^{2}∥ϕ∥^{2}

= [c_{1}, c_{2}] [ \begin{matrix}∥ϕ∥^{2} & (ϕ_{x}, ϕ_{y}) \\ (ϕ_{x}, ϕ_{y}) & ∥ϕ∥^{2} \end{matrix} ] [ \begin{matrix}c_{1} \\ c_{2} \end{matrix} ]

Hence, (W_{ϕ}(x + h), W_{ϕ}(y + h)) is Gaussian with zero expectation and covariance matrix

[ \begin{matrix}∥ϕ∥^{2} & (ϕ_{x}, ϕ_{y}) \\ (ϕ_{x}, ϕ_{y}) & ∥ϕ∥^{2} \end{matrix} ]

which we see is independent of h. We have proved stationarity (and Gaussianity). Property (2.1.20): If suppϕ_{x_{1}} ∩ suppϕ_{x_{2}} = ∅, then

E [W_{ϕ}(x_{1})W_{ϕ}(x_{2})] = (ϕ_{x_{1}}, ϕ_{x_{2}}) = 0

Hence, since (W_{ϕ}(x_{1}), W_{ϕ}(x_{2})) is Gaussian and W_{ϕ}(x_{1}) and W_{ϕ}(x_{2}) are independent.

Exercise 4:

From p. 22 (proof of Th. 2.2.3),

E [H_{α}] = E [H_{α}H_{0}] = \{ \begin{matrix}1, & α = 0 \\ 0, & α ≠ 0 \end{matrix}

Exercise 5:

From the exercise above,

E [f] = ∑_{α} c_{α}E [H_{α}] = c_{0}

Exercise 6:

By Th. 2.2.4, p. 23:

E [w^{2}(ϕ)] = ∑_{j=1}^{∞} ϵ_{j}!(ϕ, ξ_{j})^{2}

= ∑_{j=1}^{∞}(ϕ, ξ_{j})^{2}

= ∥ϕ∥^{2}

where we used Parseval’s identity in the last equality. Exercise 7: Note that with d = 1 in Th. 2.3.1, p. 28, we have the multi-index ordering

{δ^{(j)}}_{j=1}^{∞} = {(1), (2), (3), … } = {j}_{j=1}^{∞}

Recall (2.2.5), p. 19:

sup_{x∈IR} |ξ_{j}(s)| = O(j^{−1/12})

We must find a θ > 0 such that (2.4.6), p. 29, holds, i.e.

∑_{j=1}^{∞} ξ_{j}^{2}(t)(δ^{(j)})^{−θ} ≤ K ∑_{j=1}^{∞} j^{−1/6} · j^{−θ}

= K ∑_{j=1}^{∞} j^{−(1/6+θ)}

This sum is finite if θ > 5/6. Hence, T ∈ S′(IR).

Let ϕ ∈ S(IR). By definition of δ_{t}:

⟨δ_{t}, ϕ⟩ = ϕ(t)

Write ϕ = ∑_{j=1}^{∞}(ϕ, ξ_{j})ξ_{j}. Then

⟨T, ϕ⟩ = ∑_{j=1}^{∞} ξ_{j}(t)(ϕ, ξ_{j})

= ∑_{j=1}^{∞}(ϕ, ξ_{j})ξ_{j}(t)

= ϕ(t)

Hence, T = δ_{t}. Exercise 8: From exercise 7 above, δ_{t} = ∑_{j=1}^{∞} ξ_{j}(t)ξ_{j}. We calculate (formally)

⟨ω, δ_{t}⟩ = ∑_{j=1}^{∞} ξ_{j}(t)⟨ω, ξ_{j}⟩

= ∑_{j=1}^{∞} ξ_{j}(t)H_{ϵ_{j}}(ω)

Note that ⟨ω, δ_{t}⟩ is only a suggestive notation (we are not allowed to write the dual pairing of two distributions). But its formal chaos expansion will define an element in (S)_{−ρ}, as we now prove:

From Def. 2.3.2, pp. 29–30, we must find a q > 0 for every ρ ∈ [0, 1] such that

∑_{α} b_{α}^{2}(α!)^{1−ρ}(2IN)^{−αq} < ∞

We have b_{α} = ξ_{j}(t) when α = ϵ_{j}, and zero otherwise. When α = ϵ_{j}, −qϵ_{j} is a vector with zeros everywhere except at coordinate j, where it has value −q. Formula (2.3.8), p. 29, becomes

(2IN)^{−qϵ_{j}} = Π_{i}(2i)^{(−qϵ_{j})_{i}} = (2j)^{−q}

where we have used the notation (qα)_{i} to indicate the i’th coordinate. Hence,

∑_{j=1}^{∞} ξ_{j}^{2}(t)(1!)^{1−ρ}(2j)^{−q} = ∑_{j=1}^{∞} ξ_{j}^{2}(t)(2j)^{−q}

≤ K2^{−q} ∑_{j=1}^{∞} j^{−1/6}j^{−q}

This sum is finite whenever q > 5/6, and we can conclude that W_{t} ∈ (S)_{−ρ}. Note that formally we have

\frac{d}{dt}B(t) = ∑_{j=1}^{∞} \frac{d}{dt} ∫_{0}^{t} ξ_{j}(s) dsH_{ϵ_{j}} = W_{t}

Exercise 9:

The chaos expansion of W_{ϕ}(x) is

W_{ϕ}(x) = ∑_{j=1}^{∞}(ϕ_{x}, ξ_{j})H_{ϵ_{j}}

We must show ((2.3.9), p. 29)

∑_{j=1}^{∞}(ϕ_{x}, ξ_{j})^{2}(ϵ_{j}!)^{2}(2IN)^{kϵ_{j}} < ∞

for all k ∈ IN. But

∑_{j=1}^{∞}(ϕ_{x}, ξ_{j})^{2}(ϵ_{j}!)^{2}(2IN)^{kϵ_{j}} = 2^{k} ∑_{j=1}^{∞}(ϕ_{x}, ξ_{j})^{2}j^{k}

Recall that ϕ_{x} ∈ S(IR), and by Th. 2.3.1 on p. 28,

∑_{j=1}^{∞}(ϕ_{x}, ξ_{j})^{2}j^{q} < ∞

for all q ≥ 0. Hence, we can conclude that W_{ϕ}(x) ∈ (S)_{1}. Note that this will also hold for d > 1.

Day 2

Exercise 1: Answer to 2.9: Let m=N=1. The chaos expansions are

w(ϕ) = ∑_{j=1}^{∞}(ϕ, η_{j})H_{ϵ_{j}}
w(ψ) = ∑_{j=1}^{∞}(ψ, η_{j})H_{ϵ_{j}}

From Def. 2.4.1, p.39,

w(ϕ) ⋄ w(ψ) = ∑_{j,k=1}^{∞}(ϕ, η_{j}) · (ψ, η_{k})H_{ϵ_{j}+ϵ_{k}}

Def. 2.2.1, p. 19, says:

j ≠ k : H_{ϵ_{j}+ϵ_{k}} = ⟨ω, η_{j}⟩ · ⟨ω, η_{k}⟩ = H_{ϵ_{j}}H_{ϵ_{k}}
j = k : H_{2ϵ_{k}} = h_{2}(⟨ω, η_{j}⟩) = ⟨ω, η_{j}⟩^{2} − 1 = H_{ϵ_{j}}^{2} − 1

Thus

w(ϕ) ⋄ w(ψ) = ∑_{j,k}^{∞}(ϕ, η_{j})(ψ, η_{k})H_{ϵ_{j}}H_{ϵ_{k}} − ∑_{j,k}(ϕ, η_{j})(ψ, η_{k})

= w(ϕ) · w(ψ) − ∑_{j=1}^{∞}(ϕ, η_{j})(ψ, η_{k})

But if ϕ = ∑_{j}(ϕ, η_{j})η_{j}, ψ = ∑_{k}(ψ, η_{k})η_{k}, then

(ϕ, ψ) = ∑_{j=1}^{∞}(ϕ, η_{j})(ψ, η_{j})

Thus, we have proved

w(ϕ) ⋄ w(ψ) = w(ϕ) · w(ψ) − (ϕ, η)

Answer to 2.8 a: Follows by the exercise 2.9 above with ψ = ϕ. Answer to 2.8 b: We find the chaos expansion for f(ω) = B^{⋄2}(x, ω). By (2.2.24), p. 24,

B(x, ω) = ∑_{j=1}^{∞} ∫_{0}^{x} η_{j}(u) du · H_{ϵ_{j}}

By def. 2.4.1, p. 39,

B^{⋄2}(x, ω) = ∑_{j,k=1}^{∞} ∫_{0}^{x} η_{j}(u) du · ∫_{0}^{x} η_{k}(u) duH_{ϵ_{j}+ϵ_{k}}

= ∑_{j,k=1}^{∞} ∫_{0}^{x} η_{j}(u) du · ∫_{0}^{x} η_{k}(u) duH_{ϵ_{j}}H_{ϵ_{k}}

− ∑_{j=1}^{∞} (∫_{0}^{x} η_{j}(u) du)^{2}

= B^{2}(x, ω) − x_{1}x_{2} ⋯ x_{d}

Answer to 2.8 c: From b above,

B^{2}(x, ω) = B^{⋄2}(x, ω) + x_{1}x_{2} ⋯ x_{d}

= x_{1}x_{2} ⋯ x_{d} + ∑_{j,k=1}^{∞} ∫_{0}^{x} η_{j}(u) du · ∫_{0}^{x} η_{k}(u) du H_{ϵ_{jk}}

where ϵ_{jk} = ϵ_{j} + ϵ_{k}. Exercise 2: From exercise 1 (2.9) above:

w(ϕ) ⋄ w(ψ) = w(ϕ) · w(ψ) − (ϕ, ψ)

Note that (ϕ, ψ) = 0 when supp ϕ ∩ supp ψ = ∅. Hence,

w(ϕ) ⋄ w(ψ) = w(ϕ) · w(ψ)

See (2.4.3), p. 40, for a general statement. Exercise 3: Assume first that

w^{⋄2}(ϕ) ⋄ w(ψ) = w^{2}(ϕ)w(ψ) − ∥ϕ∥^{2}w(ψ) − 2(ϕ, ψ)w(ϕ)

holds. If supp ϕ ∩ supp ψ = ∅, then

w^{⋄2}(ϕ) ⋄ w(ψ) = w^{2}(ϕ)w(ψ) − ∥ϕ∥^{2}w(ψ)

= (w^{2}(ϕ) − ∥ϕ∥^{2}) w(ψ)

= w^{⋄2}(ϕ)w(ψ)

Thus, we have shown the second part of the exercise. Now, let’s prove the first statement: From the definition of Wick products,

w^{⋄2}(ϕ) ⋄ w(ψ) = ( ∑_{i,j=1}^{∞}(ϕ, ξ_{i})(ϕ, ξ_{j})H_{ϵ_{ij}} ) ⋄ (∑_{k=1}^{∞}(ψ, ξ_{k})H_{ϵ_{k}})

= ∑_{i,j,k}(ϕ, ξ_{i})(ϕ, ξ_{j})(ϕ, ξ_{k})H_{ϵ_{ijk}}

Def. 2.2.1, p. 19, yields,

H_{ϵ_{ijk}} = \{ \begin{matrix}⟨ω, ξ_{i}⟩⟨ω, ξ_{j}⟩⟨ω, ξ_{k}⟩, & i ≠ j ≠ k \\ (⟨ω, ξ_{i}⟩^{2} − 1)⟨ω, ξ_{k}⟩, & i = j ≠ k \\ ⟨ω, ξ_{i}⟩(⟨ω, ξ_{j}⟩^{2} − 1), & i ≠ j = k \\ (⟨ω, ξ_{i}⟩^{2} − 1)⟨ω, ξ_{j}⟩, & i = k ≠ j \\ ⟨ω, ξ_{i}⟩^{3} − 3⟨ω, ξ_{i}⟩, & i = j = k \end{matrix}

Thus,

w^{⋄2}(ϕ) ⋄ w(ψ) = ∑_{i≠j≠k}(ϕ, ξ_{i})(ϕ, ξ_{j})(ψ, ξ_{k})H_{ϵ_{i}}H_{ϵ_{j}}H_{ϵ_{k}} + ∑_{i≠k}(ϕ, ξ_{i})^{2}(ψ, ξ_{k})H_{ϵ_{i}}^{2}H_{ϵ_{k}}

+ 2 ∑_{i≠j}(ϕ, ξ_{i})(ϕ, ξ_{j})(ψ, ξ_{j})H_{ϵ_{i}}H_{ϵ_{j}}^{2} + ∑_{i}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}}^{3}

− ∑_{i≠k}(ϕ, ξ_{i})^{2}(ψ, ξ_{k})H_{ϵ_{k}} − 2 ∑_{i≠j}(ϕ, ξ_{i})(ϕ, ξ_{j})(ψ, ξ_{j})H_{ϵ_{i}}

− 3 ∑_{i}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}}

= ⟨ω, ϕ⟩^{2}⟨ω, ψ⟩ − ∑_{i≠j}(ϕ, ξ_{i})^{2}(ψ, ξ_{j})H_{ϵ_{j}} − 2 ∑_{i≠j}(ϕ, ξ_{i})(ϕ, ξ_{j})(ψ, ξ_{j})H_{ϵ_{i}}

− 3 ∑_{i}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}}

We now investigate the second and the third term:

∑_{j=1}^{∞}(ψ, ξ_{j})H_{ϵ_{j}} ∑_{i=1,i≠j}^{∞}(ϕ, ξ_{i})^{2} = ∑_{j=1}^{∞}(ψ, ξ_{j})H_{ϵ_{j}}∥ϕ∥^{2} − ∑_{j=1}^{∞}(ψ, ξ_{j})(ϕ, ξ_{j})^{2}H_{ϵ_{j}}

= ∥ϕ∥^{2}w(ψ) − ∑_{i=1}^{∞}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}}

The third term is:

∑_{i=1}^{∞}(ϕ, ξ_{i})H_{ϵ_{i}} ∑_{j=1,j≠i}^{∞}(ϕ, ξ_{j})(ψ, ξ_{j}) = ∑_{i=1}^{∞}(ϕ, ξ_{i})H_{ϵ_{i}}(ϕ, ψ) − ∑_{i=1}^{∞}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}}

= (ϕ, ψ)w(ϕ) − ∑_{i=1}^{∞}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}}

Putting all together yields,

w^{⋄2}(ϕ) ⋄ w(ψ) = w^{2}(ϕ)w(ψ) − ∥ϕ∥^{2}w(ψ) + ∑_{i}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}} − 2(ϕ, ψ)w(ϕ)

+ 2 ∑_{i}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}} − 3 ∑_{i}(ϕ, ξ_{i})^{2}(ψ, ξ_{i})H_{ϵ_{i}}

= w^{2}(ϕ)w(ψ) − ∥ϕ∥^{2}w(ψ) − 2(ϕ, ψ)w(ϕ)

Exercise 4:

Answer to a: We apply Th. 2.5.9, p. 52, to get

∫_{0}^{T} B(t_{0})δB(t) = ∫_{0}^{T} B(t_{0}) ⋄ W (t) dt

= ∫_{0}^{T} (∑_{j} ∫_{0}^{t_{0}} ξ_{j}(u) duH_{ϵ_{j}} ) ⋄ (∑_{j} ξ_{j}(t)H_{ϵ_{j}} ) dt

= ∫_{0}^{T} ∑_{j,k} (∫_{0}^{t_{0}} ξ_{j}(u) duξ_{j}(t)) H_{ϵ_{j}k} dt

= ∑_{j,k} ∫_{0}^{t_{0}} ξ_{j}(u) du · ∫_{0}^{T} ξ_{j}(t)H_{ϵ_{jk}} dt

= (∑_{j} ∫_{0}^{t_{0}} ξ_{j}(u) duH_{ϵ_{j}} ) ⋄ (∑_{j} ∫_{0}^{T} ξ_{j}(u) duH_{ϵ_{j}} )

= B(t_{0}) ⋄ B(T)

= B(t_{0}) · B(T) − (1_{[0,t_{0})}, 1_{[0,T)})

= B(t_{0}) · B(T) − t_{0}

Note that we could have used Cor. 2.5.12, p. 54, to obtain this result immediately. Answer to b: We calculate,

∫_{0}^{T} ∫_{0}^{T} g(s) dB(s) δB(t) = ∫_{0}^{T} (∫_{0}^{T} g(s) dB(s)) ⋄ W (t) dt
= ∫_{0}^{T} g(s) dB(s) ⋄ ∫_{0}^{T} W (t) dt
= B(T) ⋄ ∫_{0}^{T} g(s) dB(s)

By theory on p. 15,

∫_{0}^{T} g(s) dB(s) = ∫_{IR} 1_{[0,T]}(s) dB(s) = ⟨ω, 1_{[0,T]}g⟩

Thus, by exercise 1 (2.9) above,

∫_{0}^{T} ∫_{0}^{T} g(s) dB(s) δB(t) = B(T) ⋄ ∫_{0}^{T} g(s) dB(s)

= B(T) ∫_{0}^{T} g(s) dB(s) − (1_{[0,T]}, 1_{[0,T]}g)

= B(T) ∫_{0}^{T} g(s) dB(s) − ∫_{0}^{T} g(s) ds Answer to c: Since B^{⋄2}(t_{0}) + t_{0} = B^{2}(t_{0}),

∫_{0}^{T} B^{2}(t_{0}) δB(t) = ∫_{0}^{T} B^{2}(t_{0}) ⋄ W (t) dt
= B^{2}(t_{0}) ⋄ B(T)
= B^{⋄2}(t_{0}) ⋄ B(T) + t_{0}B(t_{0})

From exercise 1 (2.9) above,

B^{⋄2}(t_{0}) ⋄ B(T) = B^{2}(t_{0})B(T) − t_{0}B(T) − 2t_{0}B(t_{0})

Altogether,

∫_{0}^{T} B^{2}(t_{0}) δB(t) = B^{2}(t_{0})B(T) − t_{0}B(T) − 2t_{0}B(t_{0}) + t_{0}B(T)
= B^{2}(t_{0})B(T) − 2t_{0}B(t_{0})

Day 3

Exercise 1: Answer to 2.8 e: By (2.6.48), p. 65,

exp^{⋄}(w(η_{1})) = ∑_{n=0}^{∞} \frac{1}{n!}w^{⋄n}(η_{1})

= ∑_{n=0}^{∞} \frac{1}{n!}h_{n}(w(η_{1}))

= ∑_{n=0}^{∞} \frac{1}{n!}h_{n}(⟨ω, η_{1}⟩)

where we have used (2.4.17), p.44. From def. 2.2.1, p.19,

H_{α}(ω) = Π_{i=1}^{∞}h_{α_{i}}(⟨ω, η_{1}⟩)

Thus

H_{nϵ_{1}}(ω) = h_{n}(⟨ω, η_{1}⟩)

which implies,

exp^{⋄}(w(η_{1})) = ∑_{n=0}^{∞} \frac{1}{n!}H_{nϵ_{1}}(ω)

Answer to 2.8 f: From (2.6.55), p. 66,

exp^{⋄}(w(η_{1}) = exp(w(η_{1})) exp(−1/2)

By exercise above we get,

exp(w(η_{1})) = ∑_{n=0}^{∞} \frac{\sqrt{e}}{n!}H_{nϵ_{1}}

Answer to 2.11: From (2.6.54), p. 66,

E [K_{ϕ}(x)] = E [exp^{⋄}(W_{ϕ}(x))]

= exp (E [W_{ϕ}(x)]) = exp(0) = 1

By (2.6.56), p. 67,

K_{ϕ}(x, ω) = exp (W_{ϕ}(x, ω) − \frac{1}{2}∥ϕ∥^{2})

Thus,

K_{ϕ}^{2}(x) = exp (2W_{ϕ} − ∥ϕ∥^{2})

= exp (W_{2ϕ} − \frac{1}{2}∥2ϕ∥^{2}) · exp (\frac{1}{2}∥2ϕ∥^{2} − ∥ϕ∥^{2})

= exp^{⋄}(W_{2ϕ}) · exp(∥ϕ∥^{2})

We can calculate,

Var [K_{ϕ}(x)] = E [K_{ϕ}^{2}(x)] − E [K_{ϕ}(x)]^{2}
= exp(∥ϕ∥^{2}) · exp(E[W_{2ϕ}]) − 1
= exp ∥ϕ∥^{2} − 1

Answer to 2.13: Define

cos^{⋄} X = ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n)!} X^{⋄2n}

Let X = w(ϕ) and apply the Hermite transform: From def. 2.6.14, p. 65,

(cos^{⋄} w(ϕ))^{}˜ (z) = cos(w̃(ϕ)(z)) = ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n)!} (w̃(ϕ)(z))^{2n}

We know from calculus that

cos(w̃(ϕ)(z)) = \frac{1}{2} (exp(iw̃(ϕ)(z)) + exp(−iw̃(ϕ)(z)))

By inverting the Hermite transform,

cos^{⋄}(w(ϕ)) = \frac{1}{2} (exp^{⋄}(iw(ϕ)) + exp(−iw(ϕ)))
= \frac{1}{2} (exp^{⋄}(w(iϕ)) + exp^{⋄}(w(−iϕ)))

From (2.6.55), p. 66,

exp^{⋄}(w(iϕ)) = exp (w(iϕ) − \frac{1}{2}∥iϕ∥^{2})

By definition of the real L^{2}(IR)-norm:

∥iϕ∥^{2} = ∫_{IR}(iϕ(x))^{2} dx = i^{2} ∫_{IR} ϕ(x)^{2} dx = −∥ϕ∥^{2}

Hence,

exp(\frac{1}{2}∥iϕ∥^{2}) = exp(\frac{1}{2}∥ϕ∥^{2})

Similarly, ∥ − iϕ∥^{2} = −∥ϕ∥^{2} and

exp(−\frac{1}{2}∥ − iϕ∥^{2}) = exp(\frac{1}{2}∥ϕ∥^{2})

We get,

cos^{⋄}(w(ϕ)) = exp(\frac{1}{2}∥ϕ∥^{2})\frac{1}{2} (exp(iw(ϕ)) + exp(−iw(ϕ)))

= exp(\frac{1}{2}∥ϕ∥^{2}) cos w(ϕ)

Proof of b is similar. Answer to 2.15 a: Since,

\frac{1}{2}(x − t)^{2} = \frac{1}{2}x^{2} − xt + \frac{1}{2}t^{2}

and

\frac{1}{2}x^{2} − \frac{1}{2}(x − t)^{2} = xt − \frac{1}{2}t^{2}

we get

exp(tx − \frac{1}{2}t^{2}) = exp(\frac{1}{2}x^{2}) exp(−\frac{1}{2}(x − t)^{2})

Taylor expansion of the last term around t = 0 gives,

exp(−\frac{1}{2}(x − t)^{2}) = ∑_{n=0}^{∞} \frac{1}{n!} \frac{d^{n}}{dt^{n}} exp(−\frac{1}{2}(x − t)^{2}) |_{t=0} · t^{n}

= ∑_{n=0}^{∞} \frac{1}{n!} \frac{d^{n}}{du^{n}} exp(−\frac{1}{2}u^{2}) |_{u=x} · (−1)^{n} · t^{n}

= ∑_{n=0}^{∞} \frac{(−1)^{n}}{n!} (\frac{d^{n}}{du^{n}} exp(−\frac{1}{2}u^{2})) · t^{n}

Invoking def. (C.1.), p. 207, we get

exp(−\frac{1}{2}(x − t)^{2}) = exp(−\frac{1}{2}x^{2}) ∑_{n=0}^{∞} \frac{t^{n}}{n!} h_{n}(x)

Thus,

exp(tx − t^{2}/2) = ∑_{n=0}^{∞} \frac{t^{n}}{n!} h_{n}(x)

Answer to 2.15 b: Put t = ∥ϕ∥ and x = w(ϕ)/∥ϕ∥ in the formula from exercise a above:

exp(w(ϕ) − \frac{1}{2}∥ϕ∥^{2}) = ∑_{n=0}^{∞} \frac{∥ϕ∥^{2}}{n!} h_{n}(\frac{w(ϕ)}{∥ϕ∥})

Answer to 2.15 c: The result follows immediately with ϕ = 1_{[0,t]} in exercise b above.

Answer to 2.15 d: By definition of the Wick exponential,

exp^{⋄} w(ϕ) = ∑_{n=0}^{∞} \frac{1}{n!}w^{⋄n}(ϕ)

Lemma 2.6.16, p. 66, yields,

exp^{⋄} w(ϕ) = exp(w(ϕ) − \frac{1}{2}∥ϕ∥^{2})

Hence,

∑_{n=0}^{∞} \frac{1}{n!}w^{⋄n}(ϕ) = ∑_{n=0}^{∞} \frac{1}{n!} (∥ϕ∥^{n}h_{n}(\frac{w(ϕ)}{∥ϕ∥}))

By uniqueness of the chaos expansion,

w^{⋄n}(ϕ) = ∥ϕ∥^{n}h_{n}(\frac{w(ϕ)}{∥ϕ∥})

Answer to 2.24 d: By cor. 2.5.12, p. 54,

∫_{0}^{T} exp(B(T)) δB(t) = exp(B(T)) ⋄ ∫_{0}^{T} W (t) dt
= exp^{⋄}(B(T)) exp(T/2) ⋄ B(T)
= exp(T/2) (B(T) ⋄ exp^{⋄}(B(T)))

By def. og the Wick exponential and exercise 2.15 d above (with ϕ = 1_{[0,T]}):

B(T) ⋄ exp^{⋄}(B(T)) = ∑_{n=0}^{∞} \frac{1}{n!}B^{⋄(n+1)}(T)

= ∑_{n=0}^{∞} \frac{1}{n!}∥1_{[0,T]}∥^{n+1}h_{n+1}(\frac{B(T)}{∥1_{[0,T]}∥})

Hence, since ∥1_{[0,T]}∥ = T^{1/2},

∫_{0}^{T} exp(B(T)) δB(t) = exp(T/2) ∑_{n=0}^{∞} \frac{T^{n+1/2}}{n!} h_{n+1}(\frac{B(T)}{\sqrt{T}})

Exercise 2:

Answer to 3.4 a: Rewrite the equation for X_{t} in integral form,

X_{t} = x + ∫_{0}^{t} rX_{s} ds + ∫_{0}^{t} αX_{s} ⋄ W_{s} ds

We apply the Hermite transform to this integral equation. Denote HX_{t}(z) = X̃_{t}(z), and note that H(X_{s} ⋄ W_{s})(z) = HX_{s}(z) · HW_{s}(z):

X̃_{t}(z) = x + ∫_{0}^{t}(r + αW̃_{s}(z))X̃_{s}(z) ds

Observe that this is an ordinary deterministic (complex-valued) differential equation for each z.

\frac{d}{dt}X̃_{t}(z) = (r + αW̃_{t}(z))X̃_{t}(z), X̃_{0}(z) = x

The solution is known to be

X̃_{t}(z) = x exp(rt) exp(α ∫_{0}^{t} W̃_{s}(z) ds)

Note that

H (exp^{⋄}(αB(t))) (z) = exp(α ∫_{0}^{t} W̃_{s}(z) ds)

Hence,

X_{t} = x exp(rt) exp^{⋄}(αB(t))
= x exp(rt) exp(αB(t) − α^{2}t/2)
= x exp((r − α^{2}/2)t + αB(t))

Answer to 3.4 b: The integral form of the equation is

X_{t} = x + ∫_{0}^{t} rX_{s} ds + ∫_{0}^{t} α ⋄ W_{s} ds

Apply the Hermite transform to get the differential equation

\frac{d}{dt}X̃_{t}(z) − rX̃_{t}(z) = αW̃_{t}(z)

If we multiply both sides with the integrating factor exp(−rt) and integrate, we get

exp(−rt)X̃_{t}(z) − X̃_{0}(z) = ∫_{0}^{t} α exp(−rs)W̃_{s}(z) ds

Or, by rewriting and inverting the Hermite transform,

X̃_{t} = x exp(rt) + exp(rt) ∫_{0}^{t} exp(−rs)W̃_{s}(z) ds
= x exp(rt) + ∫_{0}^{t} exp(r(t − s))W̃_{s}(z) ds
= x exp(rt) + H (∫_{0}^{t} exp(r(t − s))W_{s} ds) (z)
= x exp(rt) + H (∫_{0}^{t} exp(r(t − s) dB(s)) (z)

We can conclude,

X_{t} = x exp(rt) + ∫_{0}^{t} exp(r(t − s)) dB(s)

Answer to 3.4 c: The integral form of the problem is,

X_{t} = x + ∫_{0}^{t} r ds + ∫_{0}^{t} αX_{s} ⋄ W (s) ds

Hermite transformation,

X̃_{t}(z) = x + rt + ∫_{0}^{t} αX̃_{s}(z)W̃_{s}(z) ds

which gives the differential equation,

\frac{d}{dt}X̃_{t}(z) − αW̃_{t}(z)X̃_{t}(z) = r

Multiplying both sides with exp(−α ∫_{0}^{t} W̃_{s}(z) ds) and then integrating, yields,

X̃_{t}(z) exp(−α ∫_{0}^{t} W̃_{s}(z) ds) − X̃_{0}(z) = ∫_{0}^{t} r exp(−α ∫_{0}^{s} W̃_{u}(z) du) ds

The Hermite transformed solution is:

X̃_{t}(z) = x exp(α ∫_{0}^{t} W̃_{s}(z) ds) + ∫_{0}^{t} r exp(∫_{s}^{t} W̃_{u}(z) du) ds

Observe that

H (exp^{⋄}(α(B(t) − B(s))) (z) = exp(α ∫_{0}^{t} W̃_{u}(z) du)

Inversion of the Hermite transform provides us with the solution:

X_{t} = x exp^{⋄}(αB(t)) + ∫_{0}^{t} r exp^{⋄}(α(B(t) − B(s))) ds

= x exp(αB(t) − α^{2}t/2) + ∫_{0}^{t} r exp(α(B(t) − B(s)) + α^{2}(t − s)/2) ds

Answer to 3.5 a: Rewriting the problem on integral form:

X_{t} = G + ∫_{0}^{t} rX_{s} ds + ∫_{0}^{t} αX_{s} ⋄ W_{s} ds

Hermite transform,

X̃_{t}(z) = G̃(z) + ∫_{0}^{t} rX̃_{s}(z) ds + ∫_{0}^{t} αX̃_{s}(z)W̃_{s}(z) ds

or,

\frac{d}{dt}X̃_{t}(z) = (r + αW̃_{t}(z))X̃_{t}(z)

Solving as in the exercises above,

X̃_{t}(z) = G̃(z) exp(rt + α ∫_{0}^{t} W̃_{s}(z) ds)

Inverting the Hermite transform, using the fact H(X ⋄ Y) = HXHY,

X_{t} = G ⋄ exp^{⋄}(rt + αB(t))
= G ⋄ exp((r − α^{2}/2)t + αB(t))

Answer to 3.5 d: Rewriting,

X_{t} = G + ∫_{0}^{t} B(T) dt + ∫_{0}^{t} X_{s} ⋄ W_{s} ds

Hermite transformation,

X̃_{t}(z) = G̃(z) + B̃_{T} (z)t + ∫_{0}^{t} X̃_{s}(z)W̃_{s}(z) ds

or,

\frac{d}{dt}X̃_{t}(z) − W̃_{t}(z)X̃_{t}(z) = B̃_{T} (z)

Multiplying with the integrating factor exp(− ∫_{0}^{t} W̃_{s}(z) ds) and integrating,

X̃_{t}(z) exp(− ∫_{0}^{t} W̃_{s}(z) ds) − X̃_{0}(z) = ∫_{0}^{t} exp(− ∫_{0}^{s} W̃_{u}(z) du)B̃_{T} (z) ds

Inversion of the Hermite transform yields the solution,

X_{t} = G ⋄ exp^{⋄}(B(t)) + B(T) ⋄ ∫_{0}^{t} exp^{⋄}(B(t) − B(s)) ds

Day 4

Exercise 1:

Answer to 4.4 a: Consider the problem

\{ \begin{matrix}\frac{∂U}{∂t} = LU + U ⋄ W_{ϕ}(x, ω), & (t, x) ∈ IR_{+} × IR^{d} \\ U(0, x) = f(x) & x ∈ IR^{d} \end{matrix}

Denote the Hermite transform of U(t, x) by Ũ(t, x; z). By formally applying the Hermite transform to the problem above, we get a deterministic partial differential equation of parabolic type (with complex values):

\{ \begin{matrix}\frac{∂Ũ}{∂t} = LŨ + Ũ ⋄ W̃_{ϕ}(z), & (t, x) ∈ IR_{+} × IR^{d} \\ Ũ(0, x; z) = f(x) & x ∈ IR^{d} \end{matrix} (3.1)

By considering the real and imaginary part of Ũ, we have a Feynman-Kac representation of the solution (see e.g. [KS]),

Ũ(t, x; z) = Ê^{x} [f(X_{t}) exp(∫_{0}^{t} W̃_{ϕ}(X_{s}; z) ds)]

where Ê^{x} is the expectation with respect to the probability P̂ on the space Ω̂ (where the process X_{t} lives), conditioned on X_{0} = x. For each z, Ũ(t, x; z) solves (3.1). We must prove the following,

1 U(t, x) = H^{−1}(Ũ(t, x; z)) for a U(t, x) ∈ (S)_{−1}.

2 that we can invert the Hermite transform through (3.1), i.e.

H\frac{∂U}{∂t} (z) = \frac{∂Ũ(z)}{∂t}

HLU(z) = LŨ(z)

We consider 1, which we divide into three subparts: (i): We prove that s → W̃_{ϕ}(X_{s}(ω̂); z) is continuous on [0, t] for each z ∈ K_{q}(R) and bounded in (s, z) on [0, t] × K_{q}(R) (almost surely ω̂).

Since (ϕ_{x}, η_{j}) is the convolution between ϕ ∈ L^{2}(IR^{d}) and η_{j} ∈ S(IR^{d}), the function x → (ϕ_{x}, η_{j}) is bounded and uniformly continuous. Consider x → W̃_{ϕ}(x; z) for z ∈ K_{q}(R) (for some q, R):

W̃_{ϕ}(x; z) = lim_{N→∞} ∑_{j=1}^{N}(ϕ_{x}, η_{j})z_{j}

and thus W̃_{ϕ}(x; z) is the limit of a uniformly convergent sequence of uniformly continuous functions. Hence, W̃_{ϕ}(x; z) is continuous in x for z ∈ K_{q}(R). From stochastic analysis we know that s → X_{s}(ω̂) is continuous for almost all paths. Thus, s → W̃_{ϕ}(X_{s}(ω̂); z) is continuous for z ∈ K_{q}(R)

(ii): We show that W̃_{ϕ}(X_{s}(ω̂); z) is integrable (in s) in (S)_{−1} and

H (∫_{0}^{t} W_{ϕ}(X_{s}(ω̂), ω) ds) (z) = ∫_{0}^{t} W̃_{ϕ}(X_{s}(ω̂); z) ds

By (i) above, all assumptions in lemma 2.8.5, p. 78, are satisfied for

H(W_{ϕ}(X_{s}(ω̂), ·))(z) = W̃_{ϕ}(X_{s}(ω̂); z)

Hence, (ii) follows. (iii): We show that

H (Ê^{x} [f(X_{t}) exp^{⋄}(∫_{0}^{t} W_{ϕ}(X_{s}, ω) ds)]) (z) = Ê^{x} [f(X_{t}) exp(∫_{0}^{t} W̃_{ϕ}(X_{s}; z) ds)]

Note that Ê^{x} is taken with respect to ω̂, the variable in X_{t}. What we need to prove is that the Hermite transform commutes with Ê^{x}. We haven’t specified any topology for the probability space on which X_{t} lives, and thus we cannot say anything about the continuity of the mapping ω̂ → X_{t}(ω̂). Hence, lemma 2.8.5, p. 78 cannot be used. But there exists a weaker result on integration implying that the functional in question is Pettis integrable (rather than integrable in the sense of Bochner which is the essence of lemma 2.8.5.). We don’t state the general result, but simply give the conditions for our special case: If

ω̂ → f(X_{t}(ω̂)) exp(∫_{0}^{t} W̃_{ϕ}(X_{s}(ω̂); z) ds)

is measurable and bounded on Ω̂ for every z ∈ K_{q}(R) for some positive q, K, then

f(X_{t}(ω̂)) exp^{⋄}(∫_{0}^{t} W_{ϕ}(X_{s}(ω̂), ω) ds

is Pettis integrable on Ω̂ and the Hermite transform commutes with Ê^{x}. But we can find q, K positive such that

ω̂ → f(X_{t}(ω̂)) exp(∫_{0}^{t} W̃_{ϕ}(X_{s}(ω̂); z) ds)

is bounded for every z ∈ K_{q}(R) since f ∈ L^{∞}(IR^{d}) and ϕ ∈ L^{2}(IR^{d}). Thus we have proved the result. Summing up, we can conclude

U(t, x) = Ê^{x} [f(X_{t}) exp^{⋄}(∫_{0}^{t} W_{ϕ}(X_{s}, ω) ds)]

which proves part 1. Part 2 can be proven using lemma 2.8.4, p. 77 together with regularity of Ũ(t, x; z) in (t, x), due to the fact that L is uniformly elliptic (which essentially makes Ũ(t, x; z) C^{1,2} for every z ∈ K_{q}(R).) We skip this proof here, and leave it to the interested reader.

Answer to 4.4 b: Since W_{ϕ}(x) → W_{x} in (S)_{−1} when ϕ → δ_{0}. Therefore, when ϕ → δ_{0},

exp^{⋄}(∫_{0}^{t} W_{ϕ}(X_{s}, ω) ds) → exp^{⋄}(∫_{0}^{t} W_{X_{s}} ds)

in (S)_{−1} a.s–Ω̂ (this is true since we have combined an analytic function exp() with a convergent sequence). We can conclude

Ê^{x} [f(X_{t}) exp^{⋄}(∫_{0}^{t} W_{ϕ}(X_{s}, ω) ds)] → Ê^{x} [f(X_{t}) exp^{⋄}(∫_{0}^{t} W_{X_{s}} ds)]

in (S)_{−1} when ϕ → δ_{0}.

Exercise 2:

From cor. 4.3.2, p. 150 with K(t, x) = W_{t} and σ = 1,

U(t, x) = Ê^{x} [f(b_{t}) ⋄ exp^{⋄}(∫_{0}^{t} W_{t−s} ds)]
= Ê^{x} [f(b_{t}) ⋄ exp(∫_{0}^{t} W_{s} ds)]
= Ê^{x} [f(B_{t})] ⋄ exp(B(t) − t/2)

Exercise 3:

From the assumptions on f, u(t, x) ∈ (S)_{−1} for all (t, x) ∈ IR_{+} × IR^{d}. Moreover, u(t, x) is an (S)_{−1}-solution of

\frac{∂u}{∂t} = \frac{1}{2}△u, u(0, x) = f(x)

Obviously X_{t} ∈ L^{2}(μ) since X_{t} = exp(B(t) − t/2), and thus U(t, x) = u(t, x) ⋄ X_{t} ∈ (S)_{−1}. By Wick Calculus we have the following:

\frac{∂}{∂t} (u(t, x) ⋄ X_{t}) = \frac{∂u}{∂t}(t, x) ⋄ X_{t} + u(t, x) ⋄ \frac{∂}{∂t}X_{t}

= (\frac{1}{2}△u(t, x)) ⋄ X_{t} + u(t, x) ⋄ X_{t} ⋄ W_{t}

= \frac{1}{2}△(u(t, x) ⋄ X_{t}) + (u(t, x) ⋄ X_{t}) ⋄ W_{t}

= \frac{1}{2}△U(t, x) + U(t, x) ⋄ W_{t}

Moreover, U(0, x) = u(0, x) ⋄ X_{0} = f(x) ⋄ 1 = f(x).

Day 5

Exercise 1: Answer to 4.5: Given

U(t, x) = \frac{1}{2} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s} W (s, y) dy ds

The proof that U is the unique (S)^{∗}-solution of the problem will go as follows: 1 U(t, x) ∈ (S)^{∗} 2 HU(t, x)(z) := Ũ(t, x; z) is the unique solution of

\{ \begin{matrix}\frac{∂^{2}Ũ(t,x;z)}{∂t^{2}} − \frac{∂^{2}Ũ(t,x;z)}{∂x^{2}} = W̃ (t, x; z), & (t, x) ∈ IR_{+} × IR \\ Ũ(0, x; z) = \frac{∂Ũ}{∂t} (0, x; z) = 0 \end{matrix}
for every z ∈ C_{c}^{IN}.

3 The Hermite transform commutes with differentiation, i.e

H\frac{∂^{2}U}{∂t^{2}} = \frac{∂^{2}Ũ}{∂t^{2}} and H\frac{∂^{2}U}{∂x^{2}} = \frac{∂^{2}Ũ}{∂x^{2}}

We start with 1: The chaos expansion of W (s, y) is

W (s, y) = ∑_{j=1}^{∞} η_{j}(s, y)H_{ϵ_{j}}

which implies that

U(t, x) = ∑_{j=1}^{∞} (\frac{1}{2} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s} η_{j}(s, y) dy ds) H_{ϵ_{j}}

(since W (s, y) is strongly integrable in (S)^{∗}). Since η_{j}(s, y) = (ξ_{δ_{1}^{(j)}} ˆ⊗ξ_{δ_{2}^{(j)}})(s, y)

(see p. 19, (2.2.8) for notation), we can bound the coefficients by applying prop. 2.3.7, p.36: U(t, x) ∈ (S)^{∗} if

sup_{j} (\frac{1}{2} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s} η(s, y) dy ds)^{2} (2j)^{−q} < ∞

for some q > 0. But,

sup_{(s,y)∈IR^{2}} |η_{j}(s, y)| ≤ \frac{1}{2} sup_{s} |ξ_{δ_{1}^{(j)}}(s)| sup_{y} |ξ_{δ_{2}^{(j)}}(y)|

+ \frac{1}{2} sup_{s} |ξ_{δ_{2}^{(j)}}(s)| sup_{y} |ξ_{δ_{1}^{(j)}}(y)|

≤ K(δ_{1}^{(j)})^{−1/12}(δ_{2}^{(j)})^{−1/12} ≤ K

Hence we get the bound

(∫_{0}^{t} ∫_{x+s−t}^{x+t−s} η(s, y) dy ds)^{2} ≤ K^{2}(∫_{0}^{t} 2(t − s) ds)^{2} = 4K^{2}t^{2}

Thus,

sup_{j} K^{2}t^{2}(2j)^{−q} < ∞

for every q > 0. We have proved that U(t, x) ∈ (S)^{∗}. We consider part 2: The Hermite transform of U(t, x) is seen to be

Ũ(t, x; z) = \frac{1}{2} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s} W̃ (s, y; z) dy ds

where W̃ (s, y; z) = ∑_{j=1}^{∞} η_{j}(s, y)z_{j} (we have assumed – which is also true – that the Hermite transform commutes with both integrals. Prove this!). Recall from calculus: ∂/∂t ∫_{0}^{t} f(s, t) ds = f(t, t) + ∫_{0}^{t} ∂/∂tf(s, t) ds:

\frac{∂}{∂t} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s} W̃ (s, y; z) dy ds = ∫_{x+t−t}^{x+t−t} W̃ (t, y; z) dy +

+ ∫_{0}^{t} \frac{∂}{∂t} ∫_{x+s−t}^{x+t−s} W̃ (s, y; z) dy ds

= ∫_{0}^{t} (W̃ (s, x + t − s; z) − W̃ (s, x + s − t; z)) ds

A similar calculation gives,

\frac{∂^{2}}{∂t^{2}} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s}W̃ (s, y; z) dy ds = 2W̃ (t, x; z) +

+ ∫_{0}^{t} (\frac{∂W̃}{∂x} (s, x + t − s; z) − \frac{∂W̃}{∂x} (s, x + t − s; z)) ds

On the other hand,

\frac{∂^{2}}{∂x^{2}} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s} W̃ (s, y; z) dy ds = ∫_{0}^{t} \frac{∂}{∂x} (W̃ (s, x + t − s; z) − W̃ (s, x + s − t)) ds

= ∫_{0}^{t} (\frac{∂W̃}{∂x} (s, x + t − s; z) − \frac{∂W̃}{∂x} (s, x + s − t; z)) ds

Hence,

\frac{∂^{2}Ũ}{∂t^{2}} (t, x; z) − \frac{∂^{2}Ũ}{∂x^{2}} (t, x; z) = W̃ (t, x; z)

for (t, x) ∈ IR_{+} × IR. Boundary conditions,

Ũ(0, x; z) = 0, \frac{∂Ũ}{∂x} (0, x; z) = 0

We have proved part 2. Consider part 3: We shall not investigate this point in detail, but refer to lemma 2.8.4, p. 77. This lemma however, is for processes in (S)_{−1}.The analogue for (S)^{∗} can be obtained by substituting the boundedness condition in z with an exponential growth condition. If we, however, want to show that U(t, x) is an (S)_{−1}-solution, we can apply lemma 2.8.4 as it stands. Sketch of such a proof:

(i) Show that ∂Ũ/∂t and ∂Ũ/∂x are continuous in t and x and bounded in (t, x, z) ∈ IR_{+} × IR × K_{q}(R) for some q, R.

(ii) Show the same for ∂^{2}Ũ/∂t^{2} and ∂^{2}Ũ/∂x^{2}.

From lemma 2.8.4 we can then conclude that differentiation and the Hermite transform commutes.

Answer to 4.4 b: If W (s, y) = W (y), we have

U(t, x) = \frac{1}{2} ∫_{0}^{t} ∫_{x+s−t}^{x+t−s} W (y) dy ds = \frac{1}{2} ∫_{0}^{t} (B(x + t − s) − B(x + s − t)) ds

by the relation B(t) = ∫_{0}^{t} W (x) dx.

Answer to 4.8: Notice that c_{0}(x) = E[p(x)]. Thus, using E[X ⋄ Y] = E[X]E[Y] we get

E [(K(x) ⋄ p′(x))′] = (E [K(x)] E [p′(x)])′
= (exp E[W_{ϕ}(x)] · c′_{0}(x))′
= c′′_{0}(x)
= −1

Since E[p(a)] = c_{0}(a) = c_{0}(b) = E[p(b)] = 0, we have that c_{0}(x) is the solution of the following second order differential equation:

\{ \begin{matrix}c′′_{0}(x) = −1, & x ∈ [a, b] \\ c_{0}(a) = c_{0}(b) = 0 \end{matrix}

It is straightforward to check that the solution is

c_{0}(x) = −\frac{1}{2} (x^{2} − (a + b)x + ab)

c_{0}(x) is the best ω-constant approximation to p(x, ω).

We proceed with c_{ϵ_{j}}(x): Introduce the short-hand notation c_{j} for c_{ϵ_{j}}. From (4.6.28), p. 172:

\frac{1}{2}c′′_{j}(x) + (∫ ϕ(y − x)ξ_{j}(y) dy)′ c′_{0}(x) = 0

for x ∈ [a, b]. We have used that f = −1 only gives chaos of order 0, namely −1. In addition we have the boundary conditions c_{j}(a) = c_{j}(b) = 0. Inserting the expression for c_{0}:

\frac{1}{2}c′′_{j}(x) − (ϕ′_{x}, ξ_{j}) · (−\frac{1}{2})(2x − (a + b)) = 0

or

c′′_{j}(x) + (ϕ′_{x}, ξ_{j})(2x − (a + b)) = 0

Define the function g_{j}(x) = (ϕ′_{x}, ξ_{j})(2x − (a + b)). We can write the solution c_{j}(x) on the following form,

c_{j}(x) = k_{1} + k_{2}x − ∫_{a}^{x} ∫_{a}^{y} g_{j}(z) dz dy

where the constants k_{1} and k_{2} are solutions of

k_{1} + ak_{2} = 0

and

k_{1} + bk_{2} = ∫_{a}^{b} ∫_{a}^{y} g_{j}(z) dz dy

The equations for k_{1} and k_{2} are obtained from the boundary conditions. Note that c_{ϵ_{j}}(x) is the best Gaussian approximation to p(x, ω).

References

[DPV] Th. Deck, J. Potthoff and G. Våge: A review of white noise analysis from a probabilistic standpoint. Acta Appl. Math., 48, 1997, pp. 91-112. As the title says, this is white noise analysis from the probabilistic point of view, e.g. the functional analysis part is cut down to a minimum. Strongly recommended!

[H] T. Hida: “Brownian Motion”, Springer Verlag, 1980. The foundations of the White Noise Analysis.

[HKPS] T. Hida, H.-H. Kuo, J. Potthoff and L. Streit: “White Noise Analysis”, Kluwer, 1993. A complete introduction to White Noise Analysis, with many examples from its applications in physics.

[HØUZ] H. Holden, B. Øksendal, J. Ubøe and T. Zhang: “ Stochastic Partial Differential Equations. A Modelling, White Noise Functional Approach”, Birkhäuser Verlag, 1996.

[HP] E. Hille and R. S. Phillips: “Functional Analysis and Semigroups”, Amer. Math. Soc. Colloq. Publ. 31, 1957. All about Pettis and Bochner integration, in addition to results on the Hermite functions.

[KS] I. Karatzas and S. E. Shreve: “Brownian Motion and Stochastic Calculus”, Springer Verlag, 1991. In Chapters 4.4 and 5.7 you will find the Feynman-Kac’ formuli.

[W] J. B. Walsh: An introduction to stochastic partial differential equations. In R. Carmona, H. Kesten and J. B. Walsh (editors): École d’Été de Probabilités de Saint-Flour XIV-1984. Springer LNM 1180, 1986. pp. 265-437. A good starting point for anyone interested in SPDE’s