J. Math. Kyoto Univ. (JMKYAZ) 00-0 (0000), 000–000

An approach to the Biharmonic pseudo process by a Random walk

By

Sadao Sato^{∗}

1. Introduction

We consider the partial differential equation

\frac{∂u}{∂t} = −\frac{1}{8} \frac{∂^{4}u}{∂x^{4}}  (1.1)

Here Δ^{2} = ∂^{4}/∂x^{4} is called the biharmonic operator. The fundamental solution of this equation is

q(t, x) = \frac{1}{2π} ∫_{−∞}^{∞} dλe^{−ixλ− \frac{1}{8} λ^{4}t}  (1.2)

As shown by Hochberg[H], this function is not positive valued. Hence the usual probabilistic method is not applicable. However, there are several probabilistic approaches to this equation.

Krylov[K],later Hochberg[H] and Nishioka[N1] considered a signed finitely additive measure on a path space, which is essentially the limit of a Markov chain with the signed distribution q(t, x). Krylov obtained a path continuity of the biharmonic pseudo process. Note that there does not exist the σ-additive measure on a path space realizing a stochastic process related to the biharmonic operator. We can consider the biharmonic pseudo process only through a limit procedure from a finite additive measure. Our aim of this paper is to propose a new limit procedure for the biharmonic pseudo process.

Nishioka considered the first hitting time to (−∞, 0) and obtained the joint distribution of the first hitting time and the first hitting place. His remarkable result is that there appears the differentiation at 0 for the distribution of the first hitting place(see section 6). However, since the jump distribution q(t, x) is spread on the entire of R, it is difficult to understand this fact intuitively.

1991 Mathematics Subject Classification(s). N/A Additional infomation Communicated by N/A Additional infomation(s)

Revised Date Revised Dates Received Date

∗Tokyo Denki University

In this paper, we define a random walk , which has the only finite jumps with signed measure. And we will show Nishioka’s result by a scaling limit. We can get short and elementary proofs for his result. And we can understand that the differentiation at 0 is caused by the hitting measures on the two boundary points, which are infinite as a limit and have opposite signs (see Proposition 5.1). We can also get the distribution of the first exit time and place for finite intervals(section 8).

2. Definition of a biharmonic random walk

Let (Y_{i}) be the i.i.d. random variables with the signed distibution defined by

P (Y = 0) = p, P (Y = ±1) = q, P (Y = ±2) = r,  (2.1)

where each of probabilities should be prescribed later. Consider the random walk

X_{n} = X_{0} + Y_{1} + ⋯ + Y_{n}.

Let f be a measurable function and we shall write

E[f(X_{n})] = ∑_{k} f(k)P [X_{n} = k]

and

E_{x}[f(X_{n})] = E[f(X_{n})|X_{0} = x].

Also we shall write

P_{x}[X_{n} = k] = P [X_{n} = k|X_{0} = x].

By the Taylor expansion, we have

E_{x}[f(X_{1})] − f(x) = (p + 2q + 2r − 1)f(x) + (q + 4r)f′′(x) + (\frac{q}{12} + \frac{4r}{3} )f′′′′(x) + · · · .

If we set

p + 2q + 2r = 1, q + 4r = 0,

that is

p = 1 + 6r, q = −4r,

then we have

E_{x}[f(X_{1})] − f(x) ≃ rΔ^{2}f(x).  (2.2)

Here, number r must be negative, since we match (2.2) to (1.1) . Now we compute the characteristic function of Y.

M(μ) =^{def} E(e^{iμY}) = p + 2q cos μ + 2r cos 2μ
= 4r(1 − cos μ)^{2} + 1.

Since r is negative, we get max_{μ} M(μ) = 1 and min_{μ} M(μ) = 16r + 1. Let r = −\frac{1}{8}, and we have −1 ≤ M(μ) ≤ 1. Since

E_{x}(e^{iμX_{n}}) = e^{iμx}M(μ)^{n},

this quantity converges to zero as n → ∞ if not | cos μ| = 1. Now we have

p = \frac{1}{4}, q = \frac{1}{2}, M(μ) = 1 − \frac{1}{2}(1 − cos μ)^{2}.

Let {F_{n}}_{n≥0} be the filtration generated by {X_{n}}_{n≥0}. Since

p + 2q + 2|r| = \frac{3}{2},

the total variation of the measure restricted on F_{n} is equal to (\frac{3}{2})^{n}. Note that F_{n} is a finite set, which consists of 5^{n} atoms, and the mean E_{x}[·] is defined for any event of F_{n}. Set

p(n, k) = P_{0}[X_{n} = k].

Then we have

∑_{k} p(n, k)e^{ikμ} = M(μ)^{n}.

Thus p(n, k) is the Fourier cofficient of the right-hand . Therefore

p(n, k) = \frac{1}{2π} ∫_{−π}^{π} M(μ)^{n}e^{−ikμ}dμ  (2.3)

and we have

lim_{n→∞} p(n, k) = 0.  (2.4)

Remark 1. We can take any r in[−\frac{1}{8}, 0)for our purpose. For example, if we take r = −\frac{1}{16}, then we have 0 ≤ M ≤ 1 and this situation looks like better for some cases. However, we point out that the value −\frac{1}{8} seems to be best for the numerical simulation to be rapid.

3. Scaling limit to continuous time and space

For ϵ > 0, set

x = kϵ, t = nϵ^{4}.

4 Sadao Sato

Then we have

p_{ϵ}(t, x) =^{def} p(n, k)
= \frac{1}{2π} ∫_{−π}^{π} M(μ)^{t/ϵ^{4}}e^{−ixμ/ϵ}dμ
= \frac{1}{2π} ∫_{−π/ϵ}^{π/ϵ} M(ϵy)^{t/ϵ^{4}}e^{−ixy}dy · ϵ.

We will compute the following limit

q(t, x) =^{def} lim_{ϵ→0} \frac{1}{ϵ}p_{ϵ}(t, x).

Let δ(x) be Dirac’s delta function and we clearly have

q(0, x) = δ(x).

Since M(ϵy) is an even function, we consider

\frac{1}{ϵ}p_{ϵ}(t, x) = \frac{1}{π} (∫_{π/ϵ−π/\sqrt{ϵ}}^{π/ϵ} + ∫_{π/ϵ^{1/4}}^{π/ϵ−π/\sqrt{ϵ}} + ∫_{0}^{π/ϵ^{1/4}})M(ϵy)^{t/ϵ^{4}} cos(xy)dy
= I_{1} + I_{2} + I_{3} (say).

First, we get

|I_{1}| ≤ \frac{1}{π} ∫_{π/ϵ−π/\sqrt{ϵ}}^{π/ϵ} |1 − \frac{1}{2}(1 − cos ϵy)^{2}|^{t/ϵ^{4}}dy
= \frac{1}{π} ∫_{0}^{π/\sqrt{ϵ}}(−1 + \frac{1}{2}(1 + cos ϵu)^{2})^{t/ϵ^{4}}du
≈ \frac{1}{π} ∫_{0}^{π/\sqrt{ϵ}}(1 − ϵ^{2}u^{2})^{t/ϵ^{4}}du
≈ \frac{1}{π} ∫_{0}^{π/\sqrt{ϵ}} e^{−u^{2}t/ϵ^{2}}du → 0 (ϵ → 0).

Second we estimate I_{2}. Note that M(μ) is monotone in the interval [0, π] and

M(π − π\sqrt{ϵ}) ≈ −1 + ϵπ^{2}, M(πϵ^{3/4}) ≈ 1 − \frac{1}{8}ϵ^{3/2}π^{2}.

Therefore

|I_{2}| ≤ \frac{π}{ϵ} (1 − \frac{1}{8}ϵ^{3/2}π^{2})^{t/ϵ^{4}} ≈ \frac{π}{ϵ} e^{−π^{2}t/8ϵ^{5/2}} → 0.

Finally, in the interval of the integral I_{3}, we have

M(ϵy) ≈ 1 − \frac{1}{8}ϵ^{4}y^{4}.

Now we obtain the following conclusion:

An approach to the Biharmonic pseudo process by a Random walk 5

Theorem 3.1. By the scaling x = kϵ, t = nϵ^{4}, we have

q(t, x) =^{def} lim_{n→∞} \frac{1}{ϵ}p(n, k)
= \frac{1}{2π} ∫_{−∞}^{∞} e^{− \frac{1}{8} y^{4}t−ixy}dy,

which is the fundamental solution of the equation

\frac{∂u}{∂t} = −\frac{1}{8} \frac{∂^{4}u}{∂x^{4}}.

Next we consider the initial value problem for (1.1).

Theorem 3.2. Let f be a continuous function in L^{2}(R). Suppose that f is the Fourier transform of a function f̂ in L^{1}(R) , that is

f(x) = ∫_{−∞}^{∞} e^{−iλx}f̂(λ)dλ.  (3.1)

Set

u_{ϵ}(t, x) =^{def} E_{[ \frac{x}{ϵ}]}(f(ϵX_{[ \frac{t}{ϵ^{4}} ]})),  (3.2)

where [·] denotes its integer part. Then there exists the limit

u(t, x) = lim_{ϵ→0} u_{ϵ}(t, x)

which satisfies (1.1) and

u(0, x) = f(x).

Proof. Let

l = [x/ϵ], n = [t/ϵ^{4}].

Noting that E_{l}[·] is the measure on a finite set, we have

u_{ϵ}(t, x) = E_{l}[∫_{−∞}^{∞} e^{−iλϵX_{n}} f̂(λ)dλ]
= ∫_{−∞}^{∞} dλf̂(λ) ∑_{k} e^{−iλϵ(l+k)}p(n, k)
= ∫_{−∞}^{∞} dλf̂(λ)e^{−iλϵl}M(ϵλ)^{n}.

Since f̂ is in L^{1}(R) and M(ϵλ)^{n} is bounded, Lebesgue’s dominated convergence theorem implies

u(t, x) = ∫_{−∞}^{∞} dλf̂(λ)e^{−iλx}e^{− \frac{1}{8} λ^{4}t}
= \frac{1}{2π} ∫_{−∞}^{∞} dyf(y) ∫_{−∞}^{∞} dλe^{−iλ(x−y)− \frac{1}{8} λ^{4}t},

which completes the proof.

4. Hitting measure to (−∞, 0)

For each |s| < 1, the Green operator of {X_{n}}_{n≥0} is defined by

G_{s}f(l) = ∑_{n=0}^{∞} s^{n}E_{l}[f(X_{n})].

By the Parseval’s equality, we have

∑_{k} p(n, k)^{2} = \frac{1}{2π} ∫_{−π}^{π} M(μ)^{2n}dμ = p(2n, 0) ≤ 1

which is monotone decreasing to zero as n goes to ∞. Schwarz’s inequality implies

|E_{l}[f(X_{n})]| = | ∑_{k} p(n, k − l)f(k)| ≤ \sqrt{p(2n, 0)} ∥ f ∥_{ℓ^{2}},

and we obtain the following:

Proposition 4.1. For every f of ℓ_{2}, G_{s}f is analytic in |s| < 1, and it holds that

|G_{s}f(l)| ≤ \frac{1}{1 − |s|} ∥ f ∥_{ℓ^{2}} .

We define the first hitting time to the set Z ∩ (−∞, 0) as

σ = min{n; X_{n} < 0}

and we shall write

p̃(n, l, k) =^{def} P_{l}[X_{n} = k; n < σ],

since F_{n} is a finite set, here the right-hand quantity is well-defined. Then we have

p̃(n, l, k) = P_{l}[X_{n} = k] − P_{l}[X_{n} = k; σ ≤ n]
= P_{l}[X_{n} = k] − ∑_{i=−1,−2} ∑_{m=0}^{n} P_{l}[σ = m, X_{σ} = i]P_{i}[X_{n−m} = k],

where the last equality is implied from the strong Markov property of {X_{n}}. For l and k in Z, define

g_{s}(l, k) = G_{s}1_{k}(l),

g̃_{s}(l, k) = ∑_{n=0}^{∞} s^{n}p̃(n, l, k),

H_{i}(s, l) = ∑_{n=0}^{∞} s^{n}P_{l}[σ = n, X_{σ} = i].

Note that g̃_{s}(l, k) and H_{i}(s, l) are analytic in |s| < \frac{2}{3}, since the total variation of E_{x}[·] restricted on F_{n} is equal to (\frac{3}{2})^{n}. From the definition, we get

g̃_{s}(l, k) = g_{s}(l, k) − ∑_{i=−1,−2} H_{i}(s, l)g_{s}(i, k).  (4.1)

Fix |s| < 1, and we shall use the notation

c_{n} = \frac{1}{2π} ∫_{−π}^{π} \frac{e^{inμ}}{1 − sM(μ)}dμ (= c_{−n}).  (4.2)

Since (2.1) and (4.2) imply that

g_{s}(l, k) = c_{k−l},  (4.3)

we have

g̃_{s}(l, k) = c_{k−l} − ∑_{i=−1,−2} H_{i}(s, l)c_{k−i}.  (4.4)

Especiallly, when k = −1 and −2, we have

0 = c_{l+1} − c_{0}H_{−1}(s, l) − c_{1}H_{−2}(s, l),  (4.5)
0 = c_{l+2} − c_{1}H_{−1}(s, l) − c_{0}H_{−2}(s, l).  (4.6)
Proposition 4.2. For |s| < 1 and l ∈ Z ∩ [0, ∞),
H_{−1}(s, l) = \frac{c_{l+1}c_{0} − c_{l+2}c_{1}}{c_{0}^{2} − c_{1}^{2}} ,  (4.7)
H_{−2}(s, l) = \frac{c_{l+2}c_{0} − c_{l+1}c_{1}}{c_{0}^{2} − c_{1}^{2}} .  (4.8)

Moreover H_{i}(s, l) are analytic in |s| < 1.

Proof. We know that c_{l}’s are analytic in |s| < 1. The first part of the proposition is clear from (4.5) and (4.6), if we can show that the denominator c_{0}^{2} − c_{1}^{2} has no zero in |s| < 1. We have

c_{0} − c_{1} = \frac{1}{2π} ∫_{−π}^{π} \frac{1 − cos μ}{1 − sM(μ)}dμ
= \frac{1}{2π} ∫_{−π}^{π} \frac{(1 − cos μ)(1 − s̄M(μ))}{|1 − sM(μ)|^{2}}dμ.

So, we get

Re(c_{0} − c_{1}) = \frac{1}{2π} ∫_{−π}^{π} \frac{(1 − cos μ)(1 − Re(s)M(μ))}{|1 − sM(μ)|^{2}}dμ > 0.

Similarly, we obtain Re(c_{0} + c_{1}) > 0. Now the second part is immediate and the proof is complete.

5. Existence of the Hitting measure

In this section, we will compute the limit

lim_{s→1} H_{i}(s, l)

which is the interpretation of the hitting measure

∑_{m=0}^{∞} P_{l}[σ = m, X_{σ} = i] = P_{l}[X_{σ} = i, σ < ∞].

However, we note that the latter quantity may not be convergent, because these terms are signed.

At first, we study the c_{n}’s in (4.2). Set z = e^{iμ}, and change a variable. The residue theorem implies

c_{n} = \frac{1}{2π} ∫_{−π}^{π} \frac{e^{inμ}}{1 − s\{1 − \frac{1}{2}(1 − cos μ)^{2}\}}dμ  (5.1)
= \frac{1}{2πi} ∫_{C} \frac{dz}{z} \frac{z^{n}}{1 − s\{1 − \frac{1}{2}(1 − \frac{z^{2}+1}{2z})^{2}\}}  (5.2)

= \frac{1}{2πis} ∫_{C} dz \frac{8z^{n+1}}{8(\frac{1}{s} − 1)z^{2} + (z − 1)^{4}}  (5.3)
= \frac{8}{s} ∑_{|z|<1} Res(\frac{z^{n+1}}{4v^{4}z^{2} + (z − 1)^{4}} )  (5.4)

where C is the unit circle in the complex plain and

v = (2(\frac{1}{s} − 1))^{1/4}.  (5.5)

In the remainder of this section, we restrict s as a real number. So, we assume that 0 < s < 1 and v is positive. We consider zeros of the denominator f(z) = 4v^{4}z^{2} + (z − 1)^{4}. By an elementary calculation, we obtain that

α = 1 + iv^{2} − v\sqrt{\frac{\sqrt{v^{4} + 4} − v^{2}}{2}} − iv\sqrt{\frac{\sqrt{v^{4} + 4} + v^{2}}{2}}  (5.6)

is a zero point, which satisfies |α| < 1. We note that α satisfies

(1 − α)^{2} = 2iv^{2}α.  (5.7)

And ᾱ is another zero point inside of C. Other two zeros are outside of C. Therefore

\begin{aligned}c_{n} & = \frac{8}{s}(\frac{α^{n+1}}{f′(α)} + \frac{ᾱ^{n+1}}{f′(ᾱ)}) \\ & = \frac{16}{s}Re(\frac{α^{n+1}(α − 1)}{8v^{4}α(α − 1) + 4(α − 1)^{4}} ) \\ & = \frac{2}{sv^{4}} Re(\frac{α^{n}(1 − α)}{1 + α}). \end{aligned}  (5.8)

In the following, we shall use

c̃_{n} = Re(\frac{α^{n}(1 − α)}{1 + α}) = \frac{sv^{4}c_{n}}{2}  (5.9)

instead of c_{n} for simple notations. Note that v tends to zero as s tends to 1. Let v be small and Taylor expansion derives that

α ≈ 1 − (1 + i)v + iv^{2} + \frac{1}{4}(1 − i)v^{3} − \frac{1}{32}(1 + i)v^{5},  (5.10)
log(α) ≈ −(1 + i)v − \frac{1}{12}(1 − i)v^{3} − \frac{3}{160}(1 + i)v^{5},  (5.11)
\frac{1}{1 + α} ≈ \frac{1}{2} + \frac{(1 + i)}{4}v + \frac{1}{16}(1 − i)v^{3} − \frac{3}{128}(1 + i)v^{5},  (5.12)
\frac{α}{1 + α} ≈ \frac{1}{2} − \frac{(1 + i)}{4}v − \frac{1}{16}(1 − i)v^{3} + \frac{3}{128}(1 + i)v^{5},  (5.13)
α^{n} ≈ 1 − n(1 + i)v + in^{2}v^{2} + \frac{n(4n^{2} − 1)}{12}(1 − i)v^{3},  (5.14)
c̃_{n} ≈ v(\frac{1}{2} + \frac{1 − 4n^{2}}{8} v^{2} + \frac{n(n^{2} − 1)}{3}v^{3}).  (5.15)

From (5.7) and (5.9), we obtain the following simple equalities

c̃_{n} + c̃_{n+1} = Re(α^{n}(1 − α)),  (5.16)
c̃_{n} − c̃_{n+1} = Re(\frac{α^{n}(1 − α)^{2}}{1 + α})  (5.17)
= −2v^{2}Im(\frac{α^{n+1}}{1 + α})  (5.18)

which are often used in our computations. By Proposition 4.2 and the aboves, we get

H_{−1}(s, l) + H_{−2}(s, l) = \frac{c_{l+1} + c_{l+2}}{c_{0} + c_{1}} = \frac{c̃_{l+1} + c̃_{l+2}}{c̃_{0} + c̃_{1}}
= \frac{Re(α^{l+1}(1 − α))}{Re(1 − α)}
≈ \frac{v + (−l^{2} − 3l − 9/4)v^{3}}{v − 1/4v^{3}} → 1 as s → 1,

10 Sadao Sato

and

H_{−1}(s, l) − H_{−2}(s, l) = \frac{c_{l+1} − c_{l+2}}{c_{0} − c_{1}} = \frac{c̃_{l+1} − c̃_{l+2}}{c̃_{0} − c̃_{1}}
= Im(\frac{α^{l+2}}{1 + α})/Im(\frac{α}{1 + α})
≈ \frac{(l + \frac{3}{2})v^{3} + (−l^{2} − 3l − 2)v^{4}}{\frac{1}{2}v^{3} + \frac{1}{8}v^{5}} → 2l + 3 as s → 1.

Thus we get the following proposition: Proposition 5.1. For l ∈ Z ∩ [0, ∞),

H_{−1}(1, l) = 2 + l,  (5.19)
H_{−2}(1, l) = −1 − l.  (5.20)

Corollary 5.1. As v → 0, the followings hold:

g_{s}(l, k) ≈ \frac{1}{sv^{3}} ,  (5.21)
g̃_{1}(l, k) = \{ \begin{matrix}8 + \frac{28}{3}l + 8k + 12kl + 4kl^{2} − \frac{4}{3}l^{3} & (k ≥ l) \\ 8 + \frac{28}{3}k + 8l + 12kl + 4lk^{2} − \frac{4}{3}k^{3} & (k ≤ l) \end{matrix} .  (5.22)

Proof. (5.21) is easily obtained from (4.3) and (5.8). The latter is obtained by an elementary but long calculation. We will omit the details.

Now we can explain why the differentiation appears at 0. Let f(x) be a differentiable function and consider

E_{l}[s^{σ}f(ϵX_{σ})] = f(−2ϵ)H_{−2}(s, l) + f(−ϵ)H_{−1}(s, l)

= \frac{f(−ϵ) + f(−2ϵ)}{2}(H_{−1}(s, l) + H_{−2}(s, l)) + \frac{f(−ϵ) − f(−2ϵ)}{2}(H_{−1}(s, l) − H_{−2}(s, l)).

By letting s → 1, we get

lim_{s→1} E_{l}[s^{σ}f(ϵX_{σ})] = \frac{f(−ϵ) + f(−2ϵ)}{2} + \frac{f(−ϵ) − f(−2ϵ)}{2}(2l + 3).

We take the same scaling as in Theorem 3.1, that is, set x = lϵ and fix it. As ϵ goes to zero, the right hand side of this equation goes to

f(0) + f′(0)x.

Therefore we obtain

Q_{x}[f(Z_{σ})] = f(0) + f′(0)x

for the biharmonic pseudo process {Z_{t}}, that is the ”stochastic process” obtained through the scaling limit in Theorem 3.1. Here, Q_{x}[·] denotes the expectation for {Z_{t}} starting from x. However, there does not exist such measure in the usual sense. We will consider Q_{x}[·] only through a limit procedure for our random walk {X_{n}}. In this discussion, we used double limit procedure. We will give more direct proof by using the scaling limit in the next section.

6. Scaling limit of the first hitting time and place

We will compute the Fourier-Laplace transform of the first hitting place and time:

Q_{x}[e^{iμZ_{σ}−λσ}],

which was first obtained by Nishioka[N1]. We use the same scaling as in section 3. Let f(x) be a bounded C^{1}-function and x = lϵ. We consider

Q_{x}[e^{−λσ}f(Z_{σ})] =^{def} lim_{ϵ→0} E_{l}[e^{−ϵ^{4}λσ}f(ϵX_{σ})],  (6.1)

where x = lϵ is fixed. The left-hand side of this equation has only symbolical meaning. Here

σ = inf{t > 0; Z_{t} < 0}

is the first hitting time to the set (−∞, 0) of the biharmonic pseudo process {Z_{t}}. We consider the mean Q_{x}[·] for {Z_{t}} through the scaling limit of {X_{n}}. Now we compute the right-hand side of (6.1):

E_{l}[e^{−ϵ^{4}λσ}f(ϵX_{σ})]
= f(−ϵ)E_{l}[e^{−ϵ^{4}λσ}; X_{σ} = −1] + f(−2ϵ)E_{l}[e^{−ϵ^{4}λσ}; X_{σ} = −2]
= f(−ϵ)H_{−1}(e^{−ϵ^{4}λ}, l) + f(−2ϵ)H_{−2}(e^{−ϵ^{4}λ}, l)
= \frac{1}{2}(f(−ϵ) + f(−2ϵ))(H_{−1} + H_{−2}) + \frac{1}{2}(f(−ϵ) − f(−2ϵ))(H_{−1} − H_{−2})
= I_{1} + I_{2} (say).

Since s = e^{−ϵ^{4}λ} , we have

v = (2(e^{ϵ^{4}λ} − 1))^{1/4} ≈ ϵ(2λ)^{1/4}.

For the simplicity, we set

ν = (2λ)^{1/4}.  (6.2)

By (5.11), we have

α^{l} = exp(l(−v(1 + i) + O(v^{2})))  (6.3)
≈ exp(−lϵν(1 + i) + O(ϵ))  (6.4)
→ e^{−νx(1+i)} as ϵ → 0.  (6.5)

Recall (5.10) and we have

H_{−1} + H_{−2} = \frac{Re(α^{l}(1 − α))}{Re(1 − α)}
≈ Re(α^{l}(1 + i)v)/v ≈ Re(e^{−νx(1+i)}(1 + i)) as ϵ → 0.

Thus we get

I_{1} ≈ f(0)e^{−νx}(cos(νx) + sin(νx)) as ϵ → 0.

On the other hand, we obtain

H_{−1} − H_{−2} = Im(α^{l+1} \frac{α}{1 + α})/Im(\frac{α}{1 + α})
≈ Im(α^{l+1} \frac{1}{2})/(−\frac{v}{4} ) ≈ −2Im(e^{−νx(1+i)})/v as ϵ → 0.

Thus it is implied that

I_{2} ≈ f′(0)e^{−νx} sin(νx))/ν.

Therefore we have proved the following theorem by the new method, that is the scaling limit of the Markov chain {X_{n}}.

Theorem 6.1 (K.Nishioka). For every differentiable function f, we have

Q_{x}[e^{−λσ}f(Z_{σ})] = f(0)e^{−νx}(cos(νx) + sin(νx)) + \frac{f′(0)}{ν}e^{−νx} sin(νx), (6.6)

where ν = (2λ)^{1/4}. In particular, it holds that

Q_{x}[e^{iμZ_{σ}−λσ}] = e^{−νx}(cos(νx) + sin(νx)) + \frac{iμ}{ν}e^{−νx} sin(νx).  (6.7)

Remark 2. Let λ → 0 in (6.5), and we obtain

u(x) = Q_{x}[f(Z_{σ})] = f(0) + f′(0)x

which satisfies

u′′′′ = 0, u(0) = f(0), u′(0) = f′(0).

Here we find the differentiation of f(x). This fact, which is due to K.Nishioka, seems to be remarkable for many probabilists.

As was shown in [N1], we can write the Laplace inverse of the above formula in an explicit form. From the Laplace inversion formula, a direct calculation shows that

L^{−1}(e^{−ν(a+ib)}) = \frac{i}{4π} ∫_{0}^{∞} u^{3}e^{− \frac{u^{4}t}{8}} (e^{− \frac{u}{2} (a−b+i(a+b))} − e^{− \frac{u}{2} (a+b−i(a−b))})du (6.8)

when |b| ≤ a. Then we easily obtain the following functions

K(t, x) ≡ L^{−1}(Re(e^{−ν(1+i)x}(1 + i))  (6.9)
= \frac{1}{4π} ∫_{0}^{∞} u^{3}e^{−u^{4}t/8}(sin xu − cos xu + e^{−xu})du,  (6.10)
J(t, x) ≡ L^{−1}(\frac{1}{ν} e^{−νx} sin νx)  (6.11)
= \frac{1}{4π} ∫_{0}^{∞} u^{2}e^{−u^{4}t/8}(sin xu − cos xu + e^{−xu})du,  (6.12)

which give us the density function of the first hitting time σ and the hitting place:

Q_{x}[σ ∈ dt, Z_{σ} ∈ dy] = K(t, x)dtδ(dy) − J(t, x)dtδ′(dy).  (6.13)

7. Scaling limit of the transition probability with killing

In the previous section, we obtained g̃_{1}(l, k). Here, we consider the scaling limit of g̃_{s}(l, k), which is the Laplace transform of the transition probability of the biharmonic pseudo process with the absorbing boudary . We can find the same result with that in Nishioka[N1, section 9.1].

Let lϵ = x, kϵ = y and x, y be positive . Noting that Δt = ϵ^{4}, we consider

Q_{x}[∫_{0}^{σ} e^{−λt}f(Z_{t})dt] =^{def} lim_{ϵ→0} ∑_{n=0}^{∞} E_{l}[e^{−λϵ^{4}n}f(ϵX_{n}); n ≤ σ]ϵ^{4}.  (7.1)

Hence

Q_{x}[∫_{0}^{σ} e^{−λt}δ_{y}(Z_{t})dt]
≈∑_{n=0}^{∞} E_{l}[e^{−λϵ^{4}n}1_{k}(X_{n}); n ≤ σ]ϵ^{3}
=g̃_{s}(l, k)ϵ^{3} = (c_{k−l} − H_{−1}(s, l)c_{k+1} − H_{−2}(s, l)c_{k+2})ϵ^{3},

where s = e^{−λϵ^{4}}. On the other hand, since −k − 1 < 0 < l, we have

p_{n}(l, −k − 1) = P_{l}(X_{n} = −k − 1) = P_{l}(X_{n} = −k − 1; σ < n)
= ∑_{j=−1,−2} ∑_{r=0}^{n} P_{l}[σ = r; X_{σ} = j]P_{j}[X_{n−r} = −k − 1].

Multiply the both side s^{n}, and sum in n, we obtain

c_{l+k+1} = ∑_{j=−1,−2} H_{j}(s, l)c_{k+1+j}.

Therefore

g̃_{s}(l, k) + c_{l+k+1} = c_{k−l} + R,

where

R = ∑_{j=−1,−2} H_{j}(s, l)(c_{k+1+j} − c_{k−j})

= \frac{H_{−1} + H_{−2}}{2} (c_{k−1} + c_{k} − (c_{k+1} + c_{k+2})) − \frac{H_{−1} − H_{−2}}{2} (c_{k−1} − c_{k} + c_{k+1} − c_{k+2}).

From (5.16) ,(5.17) and the equalities following them, we can easily get

sv^{4}R ≈ −4v^{2}Re(α^{l+1}(1 + i))Im(α^{k}) − 4vIm(α^{l+1})Im(α^{k})

as s → 1(v → 0). Then we obtain

ϵ^{3}R ≈ −\frac{4}{ν^{3}} Im(α^{l+1})Im(α^{k})
≈ −\frac{4}{ν^{3}} Im(e^{−(1+i)νx})Im(e^{−(1+i)νy}) = \frac{4}{ν^{3}} e^{−νx} sin(νx) sin(νy) as ϵ → 0.

By remembering (5.10) and (6.4), it is easily checked that

ϵ^{3}c_{k−l} ≈ \frac{1}{ν^{3}} Re(e^{−(1+i)ν|x−y|}(1 + i)).

We may therefore see that

∫_{0}^{∞} e^{−λt}q(t, x)dt = \frac{1}{ν^{3}} Re(e^{−(1+i)ν|x|}(1 + i)).

Thus, we conclude

q̃(t, x, y) ≡ q(t, x, y; t < σ) = q(t, x − y) − q(t, x + y) − L^{−1}(\frac{4}{ν^{3}} e^{−ν(x+y)} sin νx sin νy).

Since

L(\frac{∂}{∂x}q(t, x)) = −\frac{2}{ν^{2}} e^{−νx} sin νx

and (6.10) holds, we obtain the following result from the scaling limit of {X_{n}}.

Theorem 7.1 (K.Nishioka).

q̃(t, x, y) = q(t, x − y) − q(t, x + y) + 2J(t, y) ∗ \frac{∂}{∂x}q(t, x).

Remark 3. The third term in the right hand of this equation shows an effect from the differential term of the first hitting place. For the Brownian motion, this term does not exist and the reflection principle holds.

From the above, we also get

lim_{λ→0} ∫_{0}^{∞} e^{−λt}q̃(t, x, y)dt

= lim_{λ→0} \frac{1}{ν^{3}} (Re(e^{−(1+i)ν|x−y|}(1 + i)) − Re(e^{−(1+i)ν(x+y)}(1 + i)) − 4e^{−ν(x+y)} sin(νx) sin(νy))
= \{ \begin{matrix}4x^{2}y − \frac{4}{3}x^{3} & (y ≥ x) \\ 4xy^{2} − \frac{4}{3}y^{3} & (y ≤ x) \end{matrix} ,

An approach to the Biharmonic pseudo process by a Random walk 15

which is the scaling limit of g̃_{1}(l, k). On the other hand, for the Brownian motion, it is well known that

∫_{0}^{∞} q̃(t, x, y)dt = x + y − |x − y|
= \{ \begin{matrix}2x & (y ≥ x) \\ 2y & (y ≤ x) \end{matrix} .

8. Hitting measure to the boundary of finite intervals

In this section, we consider the first hitting time

σ = min{n; X_{n} < 0 or X_{n} > L}.

By an analogous argument as in section 4, we obtain

g̃(s, l, k) = c_{k−l} − ∑_{j=−1,−2,L+1,L+2} H_{j}(s, l)c_{k−j}  (8.1)

and

( \begin{matrix}c_{0} & c_{1} & c_{L+2} & c_{L+3} \\ c_{1} & c_{0} & c_{L+3} & c_{L+4} \\ c_{L+2} & c_{L+3} & c_{0} & c_{1} \\ c_{L+3} & c_{L+4} & c_{1} & c_{0} \end{matrix} ) ( \begin{matrix}H_{−1} \\ H_{−2} \\ H_{L+1} \\ H_{L+2} \end{matrix} ) = ( \begin{matrix}c_{l+1} \\ c_{l+2} \\ c_{L+1−l} \\ c_{L+2−l} \end{matrix} )  (8.2)

We consider the new variables

x = \frac{H_{−1} + H_{−2} + H_{L+1} + H_{L+2}}{2} , y = \frac{H_{−1} − H_{−2} + H_{L+1} − H_{L+2}}{2}
u = \frac{H_{−1} + H_{−2} − H_{L+1} − H_{L+2}}{2} , w = \frac{H_{−1} − H_{−2} − H_{L+1} + H_{L+2}}{2} .

Then we get a simultaneous equation

( \begin{matrix}a & c \\ c & b \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} ) = ( \begin{matrix}h_{1} \\ h_{2} \end{matrix} ) ,  (8.3)
( \begin{matrix}d & −c \\ −c & b \end{matrix} ) ( \begin{matrix}u \\ w \end{matrix} ) = ( \begin{matrix}h_{3} \\ h_{4} \end{matrix} ) ,  (8.4)

16 Sadao Sato

where

a = 2(c_{0} + c_{1}) + c_{L+2} + 2c_{L+3} + c_{L+4}
= 2(1 − Re(α)) + Re(α^{L+2}(1 − α^{2})),
b = 2(c_{0} − c_{1}) + c_{L+2} − 2c_{L+3} + c_{L+4}
= −4v^{2}Im(\frac{α}{1 + α}) − 2v^{2}Im(\frac{α^{L+3}(1 − α)}{1 + α}),
d = 2(c_{0} + c_{1}) − c_{L+2} − 2c_{L+3} − c_{L+4}
= 2(1 − Re(α)) − Re(α^{L+2}(1 − α^{2})),
c = c_{L+2} − c_{L+4}
= −2v^{2}Im(α^{L+3}),
h_{1} = c_{l+1} + c_{l+2} + c_{L+1−l} + c_{L+2−l},
h_{2} = c_{l+1} − c_{l+2} + c_{L+1−l} − c_{L+2−l},
h_{3} = c_{l+1} + c_{l+2} − c_{L+1−l} − c_{L+2−l},
h_{4} = c_{l+1} − c_{l+2} − c_{L+1−l} + c_{L+2−l}.

We set

γ = Re(α^{L+2}), ρ = Im(α^{L+2}).

Using (5.10),(5.13),(5.16) and (5.18), we obtain

a = 2v + Re(α^{L+2}2(1 + i)v) + o(v)
= 2v(1 + γ − ρ) + o(v),
b = v^{3} − 2v^{2}Im(α^{L+2} \frac{1 + i}{2}v) + o(v^{3})
= v^{3}(1 − γ − ρ) + o(v^{3}),
d = 2v(1 − γ + ρ) + o(v),
c = −2v^{2}ρ + o(v^{2}).

Thus we get

det1 =^{def} ab − c^{2} = 2v^{4}(1 − 2ρ − γ^{2} − ρ^{2}) + o(v^{4}),
det2 =^{def} db − c^{2} = 2v^{4}(1 + 2ρ − γ^{2} − ρ^{2}) + o(v^{4}).

Moreover

h_{1} = Re((1 − α)(α^{l+1} + α^{L+1−l})) ≈ vRe((1 + i)(α^{l+1} + α^{L+1−l})) + o(v),
h_{2} = −2v^{2}Im(\frac{1}{1 + α}(α^{l+2} + α^{L+2−l})) ≈ −v^{2}Im(α^{l+2} + α^{L+2−l}) + o(v^{2}),
h_{3} = vRe((1 + i)(α^{l+1} − α^{L+1−l})) + o(v),
h_{4} = −v^{2}Im(α^{l+2} − α^{L+2−l}) + o(v^{2}).

An approach to the Biharmonic pseudo process by a Random walk 17

Therefore

x = (b ∗ h_{1} − c ∗ h_{2})/det1
= \frac{Re((1 + i)(α^{l+1} + α^{L+1−l}))(1 − γ − ρ) − Im(α^{l+2} + α^{L+2−l})2ρ}{2(1 − 2ρ − γ^{2} − ρ^{2})} + o(v),
y = (−c ∗ h_{1} + a ∗ h_{2})/det1
= \frac{Re((1 + i)(α^{l+1} + α^{L+1−l}))2ρ − Im(α^{l+2} + α^{L+2−l})2(1 + γ − ρ)}{2v(1 − 2ρ − γ^{2} − ρ^{2})} + o(\frac{1}{v}),
u = (b ∗ h_{3} + c ∗ h_{4})/det2
= \frac{Re((1 + i)(α^{l+1} − α^{L+1−l}))(1 − γ − ρ) + Im(α^{l+2} − α^{L+2−l})2ρ}{2(1 − γ^{2} + 2ρ − ρ^{2})} + o(v),
w = (c ∗ h_{3} + d ∗ h_{4})/det2
= \frac{−2ρRe((1 + i)(α^{l+1} − α^{L+1−l})) + Im(α^{l+2} − α^{L+2−l})2(1 − γ + ρ)}{2v(1 − γ^{2} + 2ρ − ρ^{2})} + o(\frac{1}{v}).

Now we consider the scaling limit. We set

x = lϵ, a = Lϵ, v = ϵν

where ν = (2λ)^{1/4}. For the pseudo biharmonic process {Z_{t}}, we consider

σ = inf{t; Z_{t} < 0 or Z_{t} > a}.

Since we approximate the pseudo process by the scaling limit of {X_{n}}, we define

Q_{x}[e^{−λσ}f(Z_{σ})] =^{def} lim_{ϵ→0} E_{l}[e^{−ϵ^{4}λσ}f(ϵX_{σ})],  (8.5)

where x = lϵ and a = Lϵ are fixed and f is a bounded C^{1}-function. We shall evaluate E_{l}[e^{−ϵ^{4}λσ}f(ϵX_{σ})]:

E_{l}[e^{−ϵ^{4}λσ}f(ϵX_{σ})]

= f(−ϵ)H_{−1}(v, l) + f(−2ϵ)H_{−2}(v, l) + f(a + ϵ)H_{L+1}(v, l) + f(a + 2ϵ)H_{L+2}(v, l)
= f(−ϵ)\frac{x + y + u + w}{2} + f(−2ϵ)\frac{x − y + u − w}{2}

+f(a + ϵ)\frac{x + y − u − w}{2} + f(a + 2ϵ)\frac{x − y − u + w}{2}

= (f(−ϵ) + f(−2ϵ) + f(a + ϵ) + f(a + 2ϵ))\frac{x}{2} + (f(−ϵ) − f(−2ϵ) + f(a + ϵ) − f(a + 2ϵ))\frac{y}{2}

+(f(−ϵ) + f(−2ϵ) − f(a + ϵ) − f(a + 2ϵ))\frac{u}{2} + (f(−ϵ) − f(−2ϵ) − f(a + ϵ) + f(a + 2ϵ))\frac{w}{2}

≈ (f(0) + f(a))x + ϵ(f′(0) − f′(a))\frac{y}{2} + (f(0) − f(a))u + ϵ(f′(0) + f′(a))\frac{w}{2}

where we set

s = e^{−ϵ^{4}λ}.

18 Sadao Sato

We define two new functions

C(x) = e^{−νx} cos(νx), S(x) = e^{−νx} sin(νx).  (8.6)

It is immediate to check that those functions satisfy

g′′′′ = −4ν^{4}g = −8λg.

Since

α^{L+2} ≈ e^{−(1+i)νϵ(L+2)} ≈ e^{−(1+i)νa},
α^{l+1} ≈ e^{−(1+i)νx},
α^{L+1−l} ≈ e^{−(1+i)ν(a−x)},

we get

γ ≈ C(a), ρ ≈ −S(a),
Re((1 + i)(α^{l+1} + α^{L+1−l})) ≈ C(x) + S(x) + C(a − x) + S(a − x),
Im(α^{l+2} + α^{L+2−l}) ≈ −S(x) − S(a − x),
· · · · · · .

Finally, we obtain the following theorem:

Theorem 8.1. Let f be a bounded function in C^{1}, and set

w(λ, x) =^{def} Q_{x}[e^{−λσ}f(Z_{σ})]

= (f(0) + f(a))\frac{(C(x) + S(x) + C(a − x) + S(a − x))(1 − C(a) + S(a)) − 2(S(x) + S(a − x))S(a)}{2(1 + 2S(a) − C^{2}(a) − S^{2}(a))}

+(f(0) − f(a))\frac{(C(x) + S(x) − C(a − x) − S(a − x))(1 + C(a) − S(a)) + 2(S(x) − S(a − x))S(a)}{2(1 − 2S(a) − C^{2}(a) − S^{2}(a))}

+(f′(0) − f′(a))\frac{−(C(x) + S(x) + C(a − x) + S(a − x))S(a) + (S(x) + S(a − x))(1 + C(a) + S(a))}{2ν(1 + 2S(a) − C^{2}(a) − S^{2}(a))}

+(f′(0) + f′(a))\frac{(C(x) + S(x) − C(a − x) − S(a − x))S(a) + (S(x) − S(a − x))(1 − C(a) − S(a))}{2ν(1 − 2S(a) − C^{2}(a) − S^{2}(a))}.

Then w(λ, x) satisfies

−\frac{1}{8} \frac{∂^{4}w}{∂x^{4}} − λw = 0

and

w(λ, 0) = f(0), w(λ, a) = f(a)

\frac{∂w}{∂x} (λ, 0) = f′(0), \frac{∂w}{∂x} (λ, a) = f′(a).

Remark 4.

(i) As λ goes to zero, we obtain

u(x) =^{def} Q_{x}[f(Z_{σ})]
= \frac{f(0) + f(a)}{2} + \frac{f(0) − f(a)}{2}(1 + 4(\frac{x}{a})^{3} − 6(\frac{x}{a})^{2})

+\frac{a(f′(0) − f′(a))}{2}(−(\frac{x}{a})^{2} + \frac{x}{a})

+\frac{a(f′(0) + f′(a))}{2}(2(\frac{x}{a})^{3} − 3(\frac{x}{a})^{2} + \frac{x}{a})

which satisfies

u′′′′ = 0, u(0) = f(0), u(a) = f(a), u′(0) = f′(0), u′(a) = f′(a).

(ii) When a goes to infinity, for any fixed values f(a) and f′(a), we obtain

Q_{x}[e^{−λσ}f(Z_{σ})] = f(0)(C(x) + S(x)) + \frac{f′(0)}{ν}S(x)

which naturally coincides with Theorem 6.1.

Acknowledgement: The author would like to thank Professor K. Nishioka for his friendship and many valuable discussions.

Department of Natural Sciences

Tokyo Denki University

Hatoyama-machi, Saitama Hiki, 350-0394,

Japan

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