notes on schr¶dinger's equation, bessel bridges and first-passage time problems

OutlineOverview of the first passage time problem

Integral equationsPDE approach

Derivation of the density?

Notes on Schrodinger’s equation, Bessel bridgesand first-passage time problems

Gerardo Hernandez-del-Valle

March 3, 2010

Gerardo Hernandez-del-Valle Notes on Schrodinger’s equation, Bessel bridges

Overview of the first passage time problem

Integral equationsDiscrete caseContinuous case, Schrodinger (1915)Brownian motion case, Ricciardi, et. al. (1984)

PDE approachGirsanov’s Theorem (1960)Girsanov’s and boundary problems3-dimensional Bessel bridge, boundary problemFeynman-Kac, PDE’s vs. SDE’s

Derivation of the density?Guideline of the proof (backward equation)Sketch of the proof (backward equation)Solution of the forward equationGreen’s function

The problem

Given that X is a one-dimensional Markov process and f is a twicedifferentiable function. Let

T := inf {t ≥ 0|Xt = f (t)}

be the first time that X reaches the moving boundary f .

I What is the density ϕ of T ?

The problem

Given that X is a one-dimensional Markov process and f is a twicedifferentiable function. Let

T := inf {t ≥ 0|Xt = f (t)}

be the first time that X reaches the moving boundary f .

I What is the density ϕ of T ?

Discrete caseContinuous case, Schrodinger (1915)Brownian motion case, Ricciardi, et. al. (1984)

Let (Xn)n≥0 be discrete-time, homogeneous Markov process. Thenthe Chapman-Kolmogorov equation describes:

0 2 4 6 8 10

Px(Xn = z) =∑y

P(Xn−k = z)Px(Xk = y)

0 2 4 6 8 10

Px(Xn = z) =∑y

0 2 4 6 8 10

Px(Xn = z) =∑y

Chapman-Kolmogorov and first passage time τ

If (Xn)n≥0 is a discrete-time, homogeneous Markov process (takingvalues in a countable setS); x and z be given and fixed in S ,g : N→ S , and

τ := inf{k ≥ 1|Xk = g(k)}

be the first-passage time of X over g . Then by theChapman-Kolmogorov equation it follows that:

Chapman-Kolmogorov and first passage time τ

If (Xn)n≥0 is a discrete-time, homogeneous Markov process (takingvalues in a countable setS); x and z be given and fixed in S ,g : N→ S , and

τ := inf{k ≥ 1|Xk = g(k)}

be the first-passage time of X over g . Then by theChapman-Kolmogorov equation it follows that:

Chapman-Kolmogorov and first passage time τ (cont.)

0 2 4 6 8 10

Px(Xn = z) =n∑

Pg(k)(Xn−k = z)Px(τ = k).

0 2 4 6 8 10

Px(Xn = z) =n∑

0 2 4 6 8 10

Px(Xn = z) =n∑

Theorem. Let (Xt)t≥0 be a strong, time-homogenous Markovprocess with continuous sample paths started at x , letg : (0,∞)→ R be a continuous function satisfying g(0+) ≥ x ,and

τ := inf{t > 0|Xt ≥ g(t)}

be the first-passage time of X over g , and let F = Fx denote thedistribution of τ . Then:

Px(Xt ∈ G ) =

0Pg(s)(Xt−s ∈ G )F (ds)

for each measurable set G contained in [g(t),∞).

τ := inf{t > 0|Xt ≥ g(t)}

be the first-passage time of X over g , and let F = Fx denote thedistribution of τ .

Px(Xt ∈ G ) =

τ := inf{t > 0|Xt ≥ g(t)}

Px(Xt ∈ G ) =

τ := inf{t > 0|Xt ≥ g(t)}

Px(Xt ∈ G ) =

Observations

I The previous equation, which links the distribution of theprocess X with the distribution of the random time τ is aChapman-Kolmogorv equation of Volterra type.

I This equation may be related to a partial differential equationof the forward or backward type.

Observations

Theorem Let B be one-dimensional standard Brownian motion, letf be a continuously differentiable function with f (0) > 0, and

T := inf {t ≥ |Bt = f (t)}

be the first passage time of B over the moving boundary f . Thenthe density ϕf of T satisfies the following Volterra integralequation of the second kind:

ϕf (t) =f (t)

(f (t)√

)−∫ t

f (t)− f (s)

(t − s)φ

(f (t)− f (s)√

t − s

)ϕf (s)ds

where φ is the density of the standard Normal r.v.

T := inf {t ≥ |Bt = f (t)}

be the first passage time of B over the moving boundary f .

Thenthe density ϕf of T satisfies the following Volterra integralequation of the second kind:

ϕf (t) =f (t)

(f (t)√

)−∫ t

f (t)− f (s)

(t − s)φ

(f (t)− f (s)√

t − s

)ϕf (s)ds

T := inf {t ≥ |Bt = f (t)}

ϕf (t) =f (t)

(f (t)√

)−∫ t

f (t)− f (s)

(t − s)φ

(f (t)− f (s)√

t − s

)ϕf (s)ds

T := inf {t ≥ |Bt = f (t)}

ϕf (t) =f (t)

(f (t)√

)−∫ t

f (t)− f (s)

(t − s)φ

(f (t)− f (s)√

t − s

)ϕf (s)ds

Comments

I Ricciardi et. al.’s equation is not the only integral equationwhich may be derived from Schrodinger’s generalrepresentation.

I Volterra type integral equations for the density of τ can beexplicitly solved only in the case in which the boundary is“linear”.

I We will show the density φ has a partial differentialrepresentation.

Comments

Girsanov’s Theorem (1960)Girsanov’s and boundary problems3-dimensional Bessel bridge, boundary problemFeynman-Kac, PDE’s vs. SDE’s

Girsanov’s theorem tells us how stochastic processes behave underchanges in measure.In particular, suppose that f is a twice differentiable function suchthat: f (0) > 0 and f ′′(t) ≥ 0, then it follows from Girsanov’stheorm that:

P PB· B.M. B.M. +

∫ ·0 f ′(u)du

B· B.M.−∫ ·0 f ′(u)du B.M.

Girsanov’s theorem tells us how stochastic processes behave underchanges in measure.

In particular, suppose that f is a twice differentiable function suchthat: f (0) > 0 and f ′′(t) ≥ 0, then it follows from Girsanov’stheorm that:

P PB· B.M. B.M. +

∫ ·0 f ′(u)du

B· B.M.−∫ ·0 f ′(u)du B.M.

P PB· B.M. B.M. +

∫ ·0 f ′(u)du

B· B.M.−∫ ·0 f ′(u)du B.M.

P PB· B.M. B.M. +

∫ ·0 f ′(u)du

B· B.M.−∫ ·0 f ′(u)du B.M.

Where the measure P and P are related through theRadon-Nikodym derivative:(

:= exp

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

which in fact is a ”martingale” and induces the followingrelationship:

P(Bt ∈ A) = E[(

I(Bt∈A)

:= exp

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

which in fact is a ”martingale”

and induces the followingrelationship:

P(Bt ∈ A) = E[(

I(Bt∈A)

:= exp

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

P(Bt ∈ A) = E[(

I(Bt∈A)

:= exp

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

P(Bt ∈ A) = E[(

I(Bt∈A)

In particular: (back to our problem)

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

]where

P PT inf {t ≥ 0|Bt = f (t)} inf

{t ≥ 0|Bt = f (0)

Note that under P the density of T is known, since the boundary isconstant!

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

]where

{t ≥ 0|Bt = f (0)

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

]where

{t ≥ 0|Bt = f (0)

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

]where

{t ≥ 0|Bt = f (0)

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

]where

{t ≥ 0|Bt = f (0)

Example: Linear boundary

Suppose the boundary is linear, i.e. f (t) = a + bt, for a, b > 0 andt ≥ 0 (f (0) = a). Then:

P(T < t) = E[

{−bBt −

}I(T<t)

{−bBT −

}I(T<t)

{−ba− 1

}I(T<t)

{−ba− 1

}ϕa(s)ds

where ϕa is the density of the first passage time of a BM over thefixed boundary a = f (0).

P(T < t) = E[

{−bBt −

}I(T<t)

{−bBT −

}I(T<t)

{−ba− 1

}I(T<t)

{−ba− 1

}ϕa(s)ds

P(T < t) = E[

{−bBt −

}I(T<t)

{−bBT −

}I(T<t)

{−ba− 1

}I(T<t)

{−ba− 1

}ϕa(s)ds

P(T < t) = E[

{−bBt −

}I(T<t)

{−bBT −

}I(T<t)

{−ba− 1

}I(T<t)

{−ba− 1

}ϕa(s)ds

P(T < t) = E[

{−bBt −

}I(T<t)

{−bBT −

}I(T<t)

{−ba− 1

}I(T<t)

{−ba− 1

}ϕa(s)ds

P(T < t) = E[

{−bBt −

}I(T<t)

{−bBT −

}I(T<t)

{−ba− 1

}I(T<t)

{−ba− 1

}ϕa(s)ds

P(T < t) = E[

{−bBt −

}I(T<t)

{−bBT −

}I(T<t)

{−ba− 1

}I(T<t)

{−ba− 1

}ϕa(s)ds

In general, if a := f (0), and f ′′ ≥ 0. Then:

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

[e−f ′(t)Bt+

R t0 f ′′(u)Budu− 1

R t0 (f ′(u))2duI(T<t)

[e−f ′(T )BT +

R T0 f ′′(u)Budu− 1

R T0 (f ′(u))2duI(T<t)

0E[e−f ′(s)a+

R s0 f ′′(u)Budu− 1

R s0 (f ′(u))2du

∣∣∣T = s]ϕa(s)ds

0e−f ′(s)a− 1

R s0 (f ′(u))2du E

R s0 f ′′(u)Budu

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

= E[e−f ′(t)Bt+

[e−f ′(T )BT +

0E[e−f ′(s)a+

R s0 (f ′(u))2du

0e−f ′(s)a− 1

R s0 (f ′(u))2du E

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

[e−f ′(t)Bt+

= E[e−f ′(T )BT +

0E[e−f ′(s)a+

R s0 (f ′(u))2du

0e−f ′(s)a− 1

R s0 (f ′(u))2du E

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

[e−f ′(t)Bt+

[e−f ′(T )BT +

0E[e−f ′(s)a+

R s0 (f ′(u))2du

0e−f ′(s)a− 1

R s0 (f ′(u))2du E

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

[e−f ′(t)Bt+

[e−f ′(T )BT +

0E[e−f ′(s)a+

R s0 (f ′(u))2du

0e−f ′(s)a− 1

R s0 (f ′(u))2du E

P(T < t) = E[

{−∫ t

0f ′(u)dBu −

0(f ′(u))2du

}I(T<t)

[e−f ′(t)Bt+

[e−f ′(T )BT +

0E[e−f ′(s)a+

R s0 (f ′(u))2du

0e−f ′(s)a− 1

R s0 (f ′(u))2du E

What is P(Bu ∈ A|T = s))?

0.0 0.2 0.4 0.6 0.8

Such a process X = a− BsI{T=s} has the following dynamics:

dXt = dWt

dt − Xt

s − tdt, 0 ≤ t ≤ s, X0 = a

∣∣∣T = s]

{∫ s

0f ′′(u)(a− Xu)du

}]= ef ′(s)a−f ′(0)a E

{−∫ t

0f ′′(u)Xudu

dXt = dWt +1

− Xt

s − tdt, 0 ≤ t ≤ s, X0 = a

∣∣∣T = s]

{∫ s

}]= ef ′(s)a−f ′(0)a E

{−∫ t

0f ′′(u)Xudu

dXt = dWt +1

dt − Xt

s − tdt,

0 ≤ t ≤ s, X0 = a

∣∣∣T = s]

{∫ s

}]= ef ′(s)a−f ′(0)a E

{−∫ t

0f ′′(u)Xudu

dXt = dWt +1

dt − Xt

s − tdt, 0 ≤ t ≤ s, X0 = a

∣∣∣T = s]

{∫ s

}]= ef ′(s)a−f ′(0)a E

{−∫ t

0f ′′(u)Xudu

dXt = dWt +1

dt − Xt

s − tdt, 0 ≤ t ≤ s, X0 = a

∣∣∣T = s]

{∫ s

}]= ef ′(s)a−f ′(0)a E

{−∫ t

0f ′′(u)Xudu

dXt = dWt +1

dt − Xt

s − tdt, 0 ≤ t ≤ s, X0 = a

∣∣∣T = s]

{∫ s

= ef ′(s)a−f ′(0)a E[

{−∫ t

0f ′′(u)Xudu

dXt = dWt +1

dt − Xt

s − tdt, 0 ≤ t ≤ s, X0 = a

∣∣∣T = s]

{∫ s

}]= ef ′(s)a−f ′(0)a E

{−∫ t

0f ′′(u)Xudu

Let g(a) = 1 for all a ∈ R and k(u, a) := f ′′(u)a,k(t, a) : [0, s]× R+ → [0,∞).

{−∫ t

0f ′′(u)Xudu

}]= E0,a

[g(Xs) exp

{−∫ s

0k(u, Xu)du

}]Theorem. (Feynman-Kac) Under some “conditions” supposethat we have the following stochastic representation:

v(t, a) = Et,a[g(XT ) exp

{−∫ s

tk(u,Xu)du

th(u,Xu) exp

{−∫ u

tk(θ,Xθ)dθ

Let g(a) = 1 for all a ∈ R and k(u, a) := f ′′(u)a,k(t, a) : [0, s]× R+ → [0,∞). Then:

{−∫ t

0f ′′(u)Xudu

[g(Xs) exp

{−∫ s

0k(u, Xu)du

{−∫ s

tk(u,Xu)du

th(u,Xu) exp

{−∫ u

tk(θ,Xθ)dθ

{−∫ t

0f ′′(u)Xudu

}]= E0,a

[g(Xs) exp

{−∫ s

0k(u, Xu)du

Theorem. (Feynman-Kac) Under some “conditions” supposethat we have the following stochastic representation:

{−∫ s

tk(u,Xu)du

th(u,Xu) exp

{−∫ u

tk(θ,Xθ)dθ

{−∫ t

0f ′′(u)Xudu

}]= E0,a

[g(Xs) exp

{−∫ s

0k(u, Xu)du

{−∫ s

tk(u,Xu)du

th(u,Xu) exp

{−∫ u

tk(θ,Xθ)dθ

Then v(t, a) satisfies the following Cauchy problem

−∂v

∂t+ kv = At + h; in [0, s)× R+,

v(s, a) = g(a); a ∈ R+,

Definition. A fundamental solution of the second-order partialdifferential equation

− ∂u

∂t+ ku = Atu (1)

is a nonnegative function G (t, a; τ, b) defined on 0 ≤ t < τ < s,a, b ∈ R+, with the property that for every g ∈ C0(R), τ ∈ (0, s],the function

u(t, a) :=

∫ ∞0

G (t, a; τ, b)g(b)db; 0 ≤ t < τ, a ∈ R+

Then v(t, a) satisfies the following Cauchy problem

−∂v

∂t+ kv = At + h; in [0, s)× R+,

v(s, a) = g(a); a ∈ R+,

Definition. A fundamental solution of the second-order partialdifferential equation

− ∂u

∂t+ ku = Atu (1)

is a nonnegative function G (t, a; τ, b) defined on 0 ≤ t < τ < s,a, b ∈ R+, with the property that for every g ∈ C0(R), τ ∈ (0, s],the function

u(t, a) :=

∫ ∞0

G (t, a; τ, b)g(b)db; 0 ≤ t < τ, a ∈ R+

is bounded, satisfies (1) and

limt↑τ

u(t, a) = g(a); a ∈ R+,

for fixed (τ, b) ∈ (0, s]× R+, the function

ϕ(t, a) := G (t, a; τ, b)

satisfies the backward Kolmogorov equation in the backwardvariables (t, a). And, for fixed (t, a) ∈ [0, s)× R+ the function

ψ(τ, b) := G (t, a; τ, b)

satisfies the forward Kolmogorov equation in the forward variables(τ, b).

Guideline of the proof (backward equation)Sketch of the proof (backward equation)Solution of the forward equationGreen’s function

As has been pointed out, to derive the density of T we must firstcompute the expected value of the 3-dimensional Bessel bridgefunctional. We will proceed by first studying solutions of thefollowing PDE’s:

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

a− a

s − t

)∂ϕ

∂a(t, a);

(f )∂ψ

∂τ(τ, b) + f ′′k (τ)bψ(τ, b) =

∂2ψ

∂b2(τ, b)

− ∂

b− b

s − τ

)ψ(τ, b)

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

a− a

s − t

)∂ϕ

∂a(t, a);

(f )∂ψ

∂τ(τ, b) + f ′′k (τ)bψ(τ, b) =

∂2ψ

∂b2(τ, b)

− ∂

b− b

s − τ

)ψ(τ, b)

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

a− a

s − t

)∂ϕ

∂a(t, a);

(f )∂ψ

∂τ(τ, b) + f ′′k (τ)bψ(τ, b) =

∂2ψ

∂b2(τ, b)

− ∂

b− b

s − τ

)ψ(τ, b)

(Deriving a solution) Step 1):

(b) − ∂ϕ1

∂t(t, a) + f ′′(t)aϕ1(t, a) =

∂2ϕ1

∂a2(t, a)

)∂ϕ1

∂a(t, a);

(f )∂ψ1

∂τ(τ, b) + f ′′k (τ)bψ1(τ, b) =

∂2ψ1

∂b2(τ, b)

− ∂

)ψ1(τ, b)

(Deriving a solution) Step 1):

(b) − ∂ϕ1

∂t(t, a) + f ′′(t)aϕ1(t, a) =

∂2ϕ1

∂a2(t, a)

)∂ϕ1

∂a(t, a);

(f )∂ψ1

∂τ(τ, b) + f ′′k (τ)bψ1(τ, b) =

∂2ψ1

∂b2(τ, b)

− ∂

)ψ1(τ, b)

Step 2) (Related to Schrodinger’s eq. for time dependent “linear”potential):

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

(f )∂ψ

∂τ(τ, b) + f ′′(τ)bψ(τ, b) =

∂2ψ

∂b2(τ, b)

Step 2) (Related to Schrodinger’s eq. for time dependent “linear”potential):

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

(f )∂ψ

∂τ(τ, b) + f ′′(τ)bψ(τ, b) =

∂2ψ

∂b2(τ, b)

Example. Quadratic boundary If f (t) = a + bt2, fora, b > 0,

then f ′′(t) = 2b. Hence

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

−∂ϕ∂t

(t, a) + 2baϕ(t, a) =1

∂2ϕ

∂a2(t, a)

Applying the Laplace transform L with respect to t we have

−λL[ϕ(·, a)] + 2baL[ϕ(·, a)]− 1

∂a2L[ϕ(·, a)] = ϕ(0, a).

The independent solutions to this second order non-homogeneousO.D.E. are Airy’s functions [Martin-Lof (1998)].

Example. Quadratic boundary If f (t) = a + bt2, fora, b > 0,then f ′′(t) = 2b. Hence

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

−∂ϕ∂t

(t, a) + 2baϕ(t, a) =1

∂2ϕ

∂a2(t, a)

−λL[ϕ(·, a)] + 2baL[ϕ(·, a)]− 1

∂a2L[ϕ(·, a)] = ϕ(0, a).

Example. Quadratic boundary If f (t) = a + bt2, fora, b > 0,then f ′′(t) = 2b. Hence

(b) − ∂ϕ

∂t(t, a) + f ′′(t)aϕ(t, a) =

∂2ϕ

∂a2(t, a)

−∂ϕ∂t

(t, a) + 2baϕ(t, a) =1

∂2ϕ

∂a2(t, a)

−λL[ϕ(·, a)] + 2baL[ϕ(·, a)]− 1

∂a2L[ϕ(·, a)] = ϕ(0, a).

Remark. Let us introduce the Airy solution to the heat equation

u(t, x) := exp

3− xt

}Ai(t2 − x)

v(t, x) := exp

3− (x − zj)t

}Ai(t2 − (x − zj))

[You may consult for instance: Airy functions and applications tophysics by Olivier Vallee & Manuel Soares] where zj is a zero of Ai .Then:

v(t, t2) = exp

3− (t2 − zj)t

}Ai(zj)

The usefulness of this Remark with become evident after a fewslides.

Step 4) (Heat equation):

(b) − ∂ϕ

∂t(t, a) =

∂2ϕ

∂a2(t, a)

(f )∂ψ

∂τ(τ, b) =

∂2ψ

∂b2(τ, b)

Step 4) (Heat equation):

(b) − ∂ϕ

∂t(t, a) =

∂2ϕ

∂a2(t, a)

(f )∂ψ

∂τ(τ, b) =

∂2ψ

∂b2(τ, b)

Step 1.

Proposition. Equation (b) satisfies the following relationship

ϕ(t, a; s) = ϕ1(t, a)[A(t; s) exp

(B(t; s)a2

B(t; s) =1

2(s − t)

A(t; s) = c · (s − t)3/2

af ′′(t)ϕ1(t, a)− 1

aa(t, a)− 1

a(t, a)− ϕ1t (t, a) = 0

Step 2.

Letting ϕ1 = 1/aϕ2, we have

ϕ1t =

ϕ1a =

a −1

ϕ1aa =

aa −2

a3ϕ2.

af ′′(t)ϕ2(t, a)− 1

aa(t, a)− ϕ2t (t, a) = 0.

Step 2.

ϕ1t =

ϕ1a =

a −1

ϕ1aa =

aa −2

a3ϕ2.

af ′′(t)ϕ2(t, a)− 1

aa(t, a)− ϕ2t (t, a) = 0.

Step 2.

ϕ1t =

ϕ1a =

a −1

ϕ1aa =

aa −2

a3ϕ2.

af ′′(t)ϕ2(t, a)− 1

aa(t, a)− ϕ2t (t, a) = 0.

Step 3.

Theorem. The backward Kolmogorov equation

af ′′(t)ϕ2(t, a)− 1

aa(t, a)− ϕ2t (t, a) = 0.

has a solution:

ϕ2(t, a) = exp

0(f ′(u))2du − f ′(t)a

}ω(τ − t, a +

∫f ′(t)dt)

where ω is any solution of the heat equation.

Step 3.

Theorem. The backward Kolmogorov equation

af ′′(t)ϕ2(t, a)− 1

aa(t, a)− ϕ2t (t, a) = 0.

has a solution:

ϕ2(t, a) = exp

0(f ′(u))2du − f ′(t)a

}ω(τ − t, a +

∫f ′(t)dt)

where ω is any solution of the heat equation.

Sketch of Proof (Theorem)

I First set ϕ2(t, a) = λ(t, a)eβ(t)a, (where β is determinedlater) and substitute into equation.

We get

−λt(t, a) =1

2λaa(t, a) +

2β2(t)λ(t, a) + β(t)λa(t, a)

+a(βt(t)− f ′′(t))λ(t, a)

I (Change of variable.) Let y = a− v(t) and letλ(t, a) = u(t, y)

−∂u

∂t(t, y) + vt(t)

∂y(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+β(t)∂u

∂y(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

I First set ϕ2(t, a) = λ(t, a)eβ(t)a, (where β is determinedlater) and substitute into equation.

We get

−λt(t, a) =1

2λaa(t, a) +

2β2(t)λ(t, a) + β(t)λa(t, a)

+a(βt(t)− f ′′(t))λ(t, a)

−∂u

∂t(t, y) + vt(t)

∂y(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+β(t)∂u

∂y(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

I First set ϕ2(t, a) = λ(t, a)eβ(t)a, (where β is determinedlater) and substitute into equation. We get

−λt(t, a) =1

2λaa(t, a) +

2β2(t)λ(t, a) + β(t)λa(t, a)

+a(βt(t)− f ′′(t))λ(t, a)

−∂u

∂t(t, y) + vt(t)

∂y(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+β(t)∂u

∂y(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

I First set ϕ2(t, a) = λ(t, a)eβ(t)a, (where β is determinedlater) and substitute into equation. We get

−λt(t, a) =1

2λaa(t, a) +

2β2(t)λ(t, a) + β(t)λa(t, a)

+a(βt(t)− f ′′(t))λ(t, a)

−∂u

∂t(t, y) + vt(t)

∂y(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+β(t)∂u

∂y(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

Set vt(t) = β(t). Then,

−∂u

∂t(t, y) + vt(t)

∂y(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+β(t)∂u

∂y(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

becomes

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

−∂u

∂t(t, y) + vt(t)

∂y(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+β(t)∂u

∂y(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

becomes

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

−∂u

∂t(t, y) + vt(t)

∂y(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+β(t)∂u

∂y(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

becomes

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

Set βt(t) = f ′′(t). Then,

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

becomes

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

becomes

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

+(y + v(t))(βt(t)− f ′′(t))u(t, y)

becomes

−∂u

∂t(t, y) =

∂y2(t, y) +

2β2(t)u(t, y)

Finally

I Let G (u) = 1/2β2(u) and

ϕ3(t, y) = u(t, y)eR t0 G(u)du

−∂ϕ3

∂t(t, y) =

∂2ϕ3

∂y2(t, y) �

Finally

−∂ϕ3

∂t(t, y) =

∂2ϕ3

∂y2(t, y) �

Finally

−∂ϕ3

∂t(t, y) =

∂2ϕ3

∂y2(t, y) �

Proposition. Equation (f) satisfies the following relationship

ψ(τ, b, s) = ψ1(τ, b; s)[A(τ ; s) exp

(B(τ ; s)b2

B(τ ; s) = − 1

2(s − τ)

A(t; s) = c · (s − τ)−1/2(bf ′′(τ)− 1

b2− 1

s − τ

)ψ1(τ, b)− 1

bb(τ, b) +1

b(τ, b, s)

+ψ1τ (τ, a, s) = 0.

We proceed by first setting ψ1 = b · ψ2: ⇒

(bf ′′(τ)− 1

s − τ

)ψ2(τ, b)− 1

bb(τ, b) + ψ2τ (τ, b) = 0.

Next ψ2 = 1/(s − τ) · ψ3: ⇒

bf ′′(τ)ψ3(τ, b)− 1

bb(τ, b) + ψ3τ (τ, b) = 0.

We proceed by first setting ψ1 = b · ψ2: ⇒(bf ′′(τ)− 1

s − τ

)ψ2(τ, b)− 1

bb(τ, b) + ψ2τ (τ, b) = 0.

Next ψ2 = 1/(s − τ) · ψ3: ⇒

bf ′′(τ)ψ3(τ, b)− 1

bb(τ, b) + ψ3τ (τ, b) = 0.

s − τ

)ψ2(τ, b)− 1

bb(τ, b) + ψ2τ (τ, b) = 0.

Next ψ2 = 1/(s − τ) · ψ3: ⇒

bf ′′(τ)ψ3(τ, b)− 1

bb(τ, b) + ψ3τ (τ, b) = 0.

s − τ

)ψ2(τ, b)− 1

bb(τ, b) + ψ2τ (τ, b) = 0.

Next ψ2 = 1/(s − τ) · ψ3: ⇒

bf ′′(τ)ψ3(τ, b)− 1

bb(τ, b) + ψ3τ (τ, b) = 0.

Theorem. Given the forward equation as in previous slide:

∂ψ3

∂τ(τ, b) =

∂2ψ3

∂b2(τ, b)− bf ′′(τ)ψ3(τ, b)

we have the following solution

ψ3(τ, b) = exp

∫ τ

0(f ′(u))2du + f ′(τ)b

}ω(τ, b +

∫f ′(τ)dτ)

(recall that ω is any solution of the heat equation).

The derivation and verification of a Green function will bepostponed to upcoming notes

notes on schr¶dinger's equation, bessel bridges and first-passage time problems

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