probabilistic solutions to the dirichlet problem - theory ...stat650/oldprojects/stat 650 project...

OutlineHistory of the Dirichlet Problem

Analytic SolutionsProbabilistic Solutions

Simulated Solutions

Probabilistic Solutions to the Dirichlet ProblemTheory and Simulation

David KahleStat 650 : Stochastic Differential Equations

April 17, 2008

David Kahle Stat 650 : Stochastic Differential Equations Probabilistic Solutions to the Dirichlet Problem

Simulated Solutions

Outline

I History of the Dirichlet Problem

I Analytical Solutions

I Probabilistic Solutions

I Simulated Solutions

I Concluding Remarks

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Outline

Simulated Solutions

Outline

Simulated Solutions

Outline

Simulated Solutions

Outline

Simulated Solutions

Johann Peter Gustav Lejeune Dirichlet (1805-1859)

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I 1805 : Born in Liege, Belgium

I 1819 : Jesuit Gymnasium in Cologne (Ohm)

I 1821 : Paris and the Academie des Sciences (Fourier, Laplace,Legendre, Poisson)

I 1828 : University of Berlin (Advised Eisenstein, Kronecker,Lipschitz, worked with Riemann)

I 1855 : University of Gottingen (Replacing Gauss)

I 1859 : Death (Works collected by Dedekind)

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Contributions

I Number theory (analytic and algebraic)

I Analysis - first solutions to the Dirichlet problem

I Differential equationsI Fourier seriesI Laplaces’ problem on the stability of the solar system

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Contributions

I Number theory (analytic and algebraic)I Analysis - first solutions to the Dirichlet problem

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Contributions

I Number theory (analytic and algebraic)I Analysis - first solutions to the Dirichlet problem

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The Dirichlet Problem

I Given an open set G ⊂ Rk with sufficiently smooth boundary∂G , find the unique solution u : Rk → R such that

∆u = ∇ · ∇u = 0

on G and u = f on ∂G for some continuous function f .

I Note

I u is harmonic on G .I u is the equilibrium temperature distribution on G given the

boundary temperatures (which do not change with time).I The Dirichlet problem is a boundary value problem.

Simulated Solutions

∆u = ∇ · ∇u = 0

on G and u = f on ∂G for some continuous function f .I Note

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∆u = ∇ · ∇u = 0

I u is harmonic on G .

I u is the equilibrium temperature distribution on G given theboundary temperatures (which do not change with time).

I The Dirichlet problem is a boundary value problem.

Simulated Solutions

∆u = ∇ · ∇u = 0

boundary temperatures (which do not change with time).

I The Dirichlet problem is a boundary value problem.

Simulated Solutions

∆u = ∇ · ∇u = 0

Simulated Solutions

The Dirichlet Problem - Example

∆u(x , y) =∂2

∂x2u(x , y) +

∂y2u(x , y) = 0.

on G = (0, a)× (0, b) and f is defined and continuous on the boundary G , ∂G .

Figure: The General Dirichlet Problem on the Rectangle (0, a)× (0, b)

Simulated Solutions

The Dirichlet Problem - Example

∆u(x , y) =∂2

∂x2u(x , y) +

∂y2u(x , y) = 0.

on G = (0, a)× (0, b) and f is defined and continuous on the boundary G , ∂G .

Figure: The General Dirichlet Problem on the Rectangle (0, a)× (0, b)

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Analytic Solution on [0, a]× [0, b] - Slide 1

I Suppose f1(x) = g1(y) = g2(y) ≡ 0, and f2(x) is continuous with 0 endpoints.

Solve the boundary value problem.

Figure: A More Specific Dirichlet Problem on the Rectangle(0, a)× (0, b)

Simulated Solutions

I Suppose f1(x) = g1(y) = g2(y) ≡ 0, and f2(x) is continuous with 0 endpoints.

Solve the boundary value problem.

Figure: A More Specific Dirichlet Problem on the Rectangle(0, a)× (0, b)

Simulated Solutions

I Separation of variables : Suppose u(x , y) = X (x)Y (y), then

∆u(x , y) = 0 ⇐⇒ ∂2

∂x2u(x , y) +

∂y2u(x , y) = 0

⇐⇒ Y (y)X ′′(x) + X (x)Y ′′(y) = 0

⇐⇒ −Y (y)X ′′(x) = X (x)Y ′′(y)

⇐⇒ −X ′′(x)

X (x)=

Y ′′(y)

Y (y)= k.

X ′′(x) + kX (x) = 0 and Y ′′(y)− kY (y) = 0.

Simulated Solutions

I Solve the ODE’s subject to boundary conditions :(Remember u(x , y) = X (x)Y (y)!)

I Note the boundary conditions

g1(y) = 0 =⇒ X (0) = 0

f1(x) = 0 =⇒ Y (0) = 0

g2(y) = 0 =⇒ X (a) = 0

f2(x) = f2(x) =⇒ Y (b) = f2(x)

Simulated Solutions

I Solve the ODE’s subject to boundary conditions :(Remember u(x , y) = X (x)Y (y)!)

I Note the boundary conditions

g1(y) = 0 =⇒ X (0) = 0

f1(x) = 0 =⇒ Y (0) = 0

g2(y) = 0 =⇒ X (a) = 0

f2(x) = f2(x) =⇒ Y (b) = f2(x)

Simulated Solutions

I Solve for X (x) - imposing boundary conditions :

I Begin with X ′′(x) + kX (x) = 0.

I For nontrivial solutions, k = µ2 > 0.I General solution : X (x) = c1 cosµx + c2 sinµx .I X (0) = 0 =⇒ c1 = 0.I X (a) = 0 =⇒ c2 sinµx = 0 =⇒ µ = µn = nπ

a , n ∈ N.

Conclusion : Xn(x) = sin(nπ

ax), n ∈ N.

Simulated Solutions

a , n ∈ N.

ax), n ∈ N.

Simulated Solutions

I For nontrivial solutions, k = µ2 > 0.

I General solution : X (x) = c1 cosµx + c2 sinµx .I X (0) = 0 =⇒ c1 = 0.I X (a) = 0 =⇒ c2 sinµx = 0 =⇒ µ = µn = nπ

a , n ∈ N.

ax), n ∈ N.

Simulated Solutions

I For nontrivial solutions, k = µ2 > 0.I General solution : X (x) = c1 cosµx + c2 sinµx .

I X (0) = 0 =⇒ c1 = 0.I X (a) = 0 =⇒ c2 sinµx = 0 =⇒ µ = µn = nπ

a , n ∈ N.

ax), n ∈ N.

Simulated Solutions

I For nontrivial solutions, k = µ2 > 0.I General solution : X (x) = c1 cosµx + c2 sinµx .I X (0) = 0 =⇒ c1 = 0.

I X (a) = 0 =⇒ c2 sinµx = 0 =⇒ µ = µn = nπa , n ∈ N.

ax), n ∈ N.

Simulated Solutions

a , n ∈ N.

ax), n ∈ N.

Simulated Solutions

a , n ∈ N.

ax), n ∈ N.

Simulated Solutions

I Solve for Y (y) - imposing boundary conditions :

I Begin with Y ′′(x)− kY (x) = 0.

I For nontrivial solutions, k = µ2n > 0.

I General solution : Y (y) = An coshµny + Bn sinhµny .I Y (0) = 0 =⇒ An = 0.

Conclusion : Yn(y) = Bn sinh(nπ

ay), n ∈ N.

I Still have one boundary condition!

Simulated Solutions

ay), n ∈ N.

Simulated Solutions

ay), n ∈ N.

Simulated Solutions

I General solution : Y (y) = An coshµny + Bn sinhµny .

I Y (0) = 0 =⇒ An = 0.

ay), n ∈ N.

Simulated Solutions

ay), n ∈ N.

Simulated Solutions

ay), n ∈ N.

Simulated Solutions

ay), n ∈ N.

Simulated Solutions

I Find u(x , y), impose last boundary condition :

I Superposition principle :

u(x , y) =∞∑

Xn(x)Yn(y) =∞∑

Bn sin(nπ

sinh(nπ

I Last boundary condition

u(x , b) = f2(x) =∞∑

(Bn sinh

sin(nπ

(Bn sinh

(nπa b))

= Fourier sine coefs of f2(x) on (0, a).

Bn sinh(nπ

f2(x) sin(nπ

Simulated Solutions

I Find u(x , y), impose last boundary condition :I Superposition principle :

u(x , y) =∞∑

Xn(x)Yn(y) =∞∑

Bn sin(nπ

sinh(nπ

u(x , b) = f2(x) =∞∑

(Bn sinh

sin(nπ

(Bn sinh

(nπa b))

Bn sinh(nπ

f2(x) sin(nπ

Simulated Solutions

u(x , y) =∞∑

Xn(x)Yn(y) =∞∑

Bn sin(nπ

sinh(nπ

u(x , b) = f2(x) =∞∑

(Bn sinh

sin(nπ

(Bn sinh

(nπa b))

Bn sinh(nπ

f2(x) sin(nπ

Simulated Solutions

u(x , y) =∞∑

Xn(x)Yn(y) =∞∑

Bn sin(nπ

sinh(nπ

u(x , b) = f2(x) =∞∑

(Bn sinh

sin(nπ

(Bn sinh

(nπa b))

Bn sinh(nπ

f2(x) sin(nπ

Simulated Solutions

I Final solution! (Finally...)

Solution : u(x , y) =∞∑

Bn sin(nπ

sinh(nπ

a sinh(

nπa b) ∫ a

0f2(x) sin

Simulated Solutions

I Final solution! (Finally...)

Solution : u(x , y) =∞∑

Bn sin(nπ

sinh(nπ

a sinh(

nπa b) ∫ a

0f2(x) sin

Simulated Solutions

Problems with Analytic Solutions to the Dirichlet Problem

I This is HARD!!

I What do you do if...

I ...you can’t compute the integral (hard boundary functions)?I ...you have a strangely shaped G (complicated boundary)?

Figure: The Dirichlet Problem on the 2-d Bat Symbolwith f (x , y) = exp {Γ (?(x + y))} for (x , y) ∈ ∂G

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I This is HARD!!I What do you do if...

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I ...you can’t compute the integral (hard boundary functions)?

I ...you have a strangely shaped G (complicated boundary)?

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A Necessary Stopping Time

I Begin with a stopping time.

τ = inf{

t > 0 : ~Bt /∈ G}.

This is the time at which the (k-dimensional) Brownianmotion reaches the boundary ∂G .

Simulated Solutions

A Necessary Stopping Time

I Begin with a stopping time.

τ = inf{

t > 0 : ~Bt /∈ G}.

This is the time at which the (k-dimensional) Brownianmotion reaches the boundary ∂G .

Simulated Solutions

Theorem 1 : Solution Under 4 Conditions

I Theorem 1 : Assume

1. G is an open set,2. G is bounded,3. ∆u exists and is 0 in G ,4. u = f , a continuous function, on ∂G .

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

where the notation is interpreted as a Brownian motion whichoriginates from ~x .

Simulated Solutions

1. G is an open set,

2. G is bounded,3. ∆u exists and is 0 in G ,4. u = f , a continuous function, on ∂G .

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

1. G is an open set,2. G is bounded,

3. ∆u exists and is 0 in G ,4. u = f , a continuous function, on ∂G .

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

1. G is an open set,2. G is bounded,3. ∆u exists and is 0 in G ,

4. u = f , a continuous function, on ∂G .

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Theorem 1 : Proof - Overview

I Step 1 : P~x [τ <∞] = 1.

I Step 2 : The main result,

u(~x) = E~x [f (Bτ )] .

Simulated Solutions

Theorem 1 : Proof - Overview

I Step 1 : P~x [τ <∞] = 1.

I Step 2 : The main result,

u(~x) = E~x [f (Bτ )] .

Simulated Solutions

Theorem 1 : Proof - Step 1

I Step 1 : P~x [τ <∞] = 1.

I Define K = diameter of G

K := sup {||~x − ~y ||k : ~x , ~y ∈ G} .

I Note for any point ~x in G∣∣∣∣∣∣~B1 − ~x∣∣∣∣∣∣

k> K =⇒ ~B1 /∈ G =⇒ τ < 1,

P~x [τ < 1]≥P~x

[∣∣∣∣∣∣~B1 − ~x∣∣∣∣∣∣

[∣∣∣∣∣∣~B1

∣∣∣∣∣∣k> K

]=:εK > 0,

sosup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k , if k = 1.

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1.

K := sup {||~x − ~y ||k : ~x , ~y ∈ G} .

k> K =⇒ ~B1 /∈ G =⇒ τ < 1,

P~x [τ < 1]≥P~x

[∣∣∣∣∣∣~B1 − ~x∣∣∣∣∣∣

[∣∣∣∣∣∣~B1

∣∣∣∣∣∣k> K

]=:εK > 0,

sosup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k , if k = 1.

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1.

K := sup {||~x − ~y ||k : ~x , ~y ∈ G} .

k> K =⇒ ~B1 /∈ G =⇒ τ < 1,

P~x [τ < 1]≥P~x

[∣∣∣∣∣∣~B1 − ~x∣∣∣∣∣∣

[∣∣∣∣∣∣~B1

∣∣∣∣∣∣k> K

]=:εK > 0,

sosup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k , if k = 1.

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1.

K := sup {||~x − ~y ||k : ~x , ~y ∈ G} .

k> K =⇒ ~B1 /∈ G =⇒ τ < 1,

P~x [τ < 1]≥P~x

[∣∣∣∣∣∣~B1 − ~x∣∣∣∣∣∣

[∣∣∣∣∣∣~B1

∣∣∣∣∣∣k> K

]=:εK > 0,

sosup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k , if k = 1.

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1.

K := sup {||~x − ~y ||k : ~x , ~y ∈ G} .

k> K =⇒ ~B1 /∈ G =⇒ τ < 1,

P~x [τ < 1]≥P~x

[∣∣∣∣∣∣~B1 − ~x∣∣∣∣∣∣

[∣∣∣∣∣∣~B1

∣∣∣∣∣∣k> K

]=:εK > 0,

sosup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k , if k = 1.

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1 (cont).

I Generalize to all k ∈ N

sup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k

to all k ∈ N. Suppose pt(~x , ~y) is the transition probabilitydensity of the Brownian motion.Then the Markov property implies

P~x [τ ≥ k] ≤∫

p1(~x , ~y)P~y [τ ≥ k − 1] d~y

≤ (1− εK )k−1P~x [B1 ∈ G ] (Inductive hypothesis)

≤ (1− εK )k .

I Conclusion : P~x [τ <∞] = 1.

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1 (cont).

sup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k

to all k ∈ N. Suppose pt(~x , ~y) is the transition probabilitydensity of the Brownian motion.

Then the Markov property implies

P~x [τ ≥ k] ≤∫

p1(~x , ~y)P~y [τ ≥ k − 1] d~y

≤ (1− εK )k .

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1 (cont).

sup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k

P~x [τ ≥ k] ≤∫

p1(~x , ~y)P~y [τ ≥ k − 1] d~y

≤ (1− εK )k .

Simulated Solutions

I Step 1 : P~x [τ <∞] = 1 (cont).

sup~x∈G

P~x [τ ≥ k] ≤ (1− εK )k

P~x [τ ≥ k] ≤∫

p1(~x , ~y)P~y [τ ≥ k − 1] d~y

≤ (1− εK )k .

Simulated Solutions

I Step 2 : u(~x) = E~x [f (Bτ )].

I By Ito’s formula,

u(~Bt)− u(~B0) =

∇u(~Bs) · d~Bs +1

∆u(~Bs) ds,

but 12

0∆u(~Bs) ds = 0 since ∆u = 0 on G , and, since the

first term is a local martingale,

u(~Bt) is a local martingale on [0, τ).

Simulated Solutions

I Step 2 : u(~x) = E~x [f (Bτ )].

I By Ito’s formula,

u(~Bt)− u(~B0) =

∇u(~Bs) · d~Bs +1

∆u(~Bs) ds,

but 12

0∆u(~Bs) ds = 0 since ∆u = 0 on G , and, since the

first term is a local martingale,

u(~Bt) is a local martingale on [0, τ).

Simulated Solutions

I Step 2 : u(~x) = E~x [f (Bτ )].

I There exists a time-change γ : R+ → R+ such that

Xt = u(Bγ(t)) is a bounded martingale.

(We already knew it was a continuous bounded local martingale on [0, τ).)

I =⇒ Xta.s.−→ X∞, Xt

Lp−→ X∞ (MGCT)with Xt = E~x [X∞|Gt ], Gt = Fγ(t)

=⇒ E~x [Xt ] = E~x [X∞].I τ <∞ =⇒ X∞ = f (Bτ ).I t = 0 =⇒

u(~x) = E~x [X0] = E~x [X∞] = E~x [f (Bτ )].

Simulated Solutions

I Step 2 : u(~x) = E~x [f (Bτ )].

u(~x) = E~x [X0] = E~x [X∞] = E~x [f (Bτ )].

Simulated Solutions

I Step 2 : u(~x) = E~x [f (Bτ )].

=⇒ E~x [Xt ] = E~x [X∞].

I τ <∞ =⇒ X∞ = f (Bτ ).I t = 0 =⇒

u(~x) = E~x [X0] = E~x [X∞] = E~x [f (Bτ )].

Simulated Solutions

I Step 2 : u(~x) = E~x [f (Bτ )].

=⇒ E~x [Xt ] = E~x [X∞].I τ <∞ =⇒ X∞ = f (Bτ ).

I t = 0 =⇒

u(~x) = E~x [X0] = E~x [X∞] = E~x [f (Bτ )].

Simulated Solutions

I Step 2 : u(~x) = E~x [f (Bτ )].

u(~x) = E~x [X0] = E~x [X∞] = E~x [f (Bτ )].

Simulated Solutions

1. G is an open set,2. ∆u exists and is 0 in G ,3. u = f , a continuous function, on ∂G .

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

1. G is an open set,

2. ∆u exists and is 0 in G ,3. u = f , a continuous function, on ∂G .

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

1. G is an open set,2. ∆u exists and is 0 in G ,

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Then for ~x ∈ G ,

u(~x) = E~x [f (Bτ )],

Simulated Solutions

Theorem 3 : Vast Generalization

I Theorem 3 : Suppose

bi (x)∂

∂xi+

nXi,j=1

aij (x)∂2

∂xi∂xj,

where aij (x) = aji (x) and bi (x) are continuous functions. L is said to be a

semi-elliptic if a(x) = [aij (x)] is positive semi-definite for all x . Then if

1. G is an open set,2. Lu exists and is 0 in G ,

3. u = f , a continuous function, on ∂G .Then for ~x ∈ G ,

u(~x) = E~x[f (Xτ )1[τ<∞]

where Xt is a diffusion satisfying

dXt = b(Xt) dt + σ(Xt) dBt

with b(x) = [bi (x)] and σ(x) such that 12σ(x)σ∗(x) = [aij (x)].

Simulated Solutions

bi (x)∂

∂xi+

nXi,j=1

aij (x)∂2

∂xi∂xj,

semi-elliptic if a(x) = [aij (x)] is positive semi-definite for all x . Then if1. G is an open set,

2. Lu exists and is 0 in G ,

u(~x) = E~x[f (Xτ )1[τ<∞]

Simulated Solutions

bi (x)∂

∂xi+

nXi,j=1

aij (x)∂2

∂xi∂xj,

semi-elliptic if a(x) = [aij (x)] is positive semi-definite for all x . Then if1. G is an open set,2. Lu exists and is 0 in G ,

u(~x) = E~x[f (Xτ )1[τ<∞]

Simulated Solutions

bi (x)∂

∂xi+

nXi,j=1

aij (x)∂2

∂xi∂xj,

Then for ~x ∈ G ,

u(~x) = E~x[f (Xτ )1[τ<∞]

Simulated Solutions

bi (x)∂

∂xi+

nXi,j=1

aij (x)∂2

∂xi∂xj,

u(~x) = E~x[f (Xτ )1[τ<∞]

Simulated Solutions

bi (x)∂

∂xi+

nXi,j=1

aij (x)∂2

∂xi∂xj,

u(~x) = E~x[f (Xτ )1[τ<∞]

Simulated Solutions

2d Dirichlet [0, 1]× [0, 1] Example

I a = b = 1,

g1(y) = 0 g2(y) = 0

f1(x) = 0 f2(x) = 100 sinπx

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Figure: Simulated Solution to the Specified Dirichlet Problem, N =100, ∆t = 1/202, h = .0025

Simulated Solutions

I a = b = 1,

g1(y) = 0 g2(y) = 0

f1(x) = 0 f2(x) = 100 sinπx

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Figure: Simulated Solution to the Specified Dirichlet Problem, N =100, ∆t = 1/202, h = .0025

Simulated Solutions

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Figure: Simulated Solution to the Specified Dirichlet Problem with True ContoursDavid Kahle Stat 650 : Stochastic Differential Equations Probabilistic Solutions to the Dirichlet Problem

Simulated Solutions

Bibliography

I Asmar, Nakhle H.Partial Differential Equations with Fourier Series and Boundary Value Problems.2nd. Upper Saddle River: Pearson Prentice Hall, 2005.

I Durrett, Richard. Stochastic Calculus : A Practical Introduction. Boca Raton:CRC Press, 1996.

I Evans, Lawrence C. Partial Differential Equations. Providence: AmericanMathematical Society, 1998.

I Feynman, Richard P., Leighton, Robert, and Matthew Sands.The Feynman Lectures on Physics, Vol 2. Reading: Addison Wesley PublishingCompany, Inc., 1964.

I O’Connor, J. J. and Robertson, E. F. “Johann Peter Gustav Lejeune Dirichlet.”Dirichlet Biography. May 2007. University of St. Andrews. 12 Apr 2008〈http://www-groups.dcs.st-and.ac.uk/ history/Biographies/Dirichlet.html〉.

I Oskendal, Bernt.Stochastic Differential Equations : An Introduction with Applications. 5th.New York: Springer, 1998.

I Steele, J. Michael. Stochastic Calculus and Financial Applications. New York:Springer, 2001.

I Thompson, James R.Simulation : A Modeler’s Approach. New York: John Wiley & Sons, Inc., 2000.

Simulated Solutions

Analytic Solution on [0, a]× [0, b] - Slide 4 (Appendix)

I Why is k = µ2 > 0 in the equation X ′′ + kX = 0?

I Suppose k = −µ2 < 0. Then

X ′′+kX = X ′′−µ2X = 0 =⇒ X (x) = c1 coshµx+c2 sinhµx .

I Impose X (0) = 0 : X (0) = 0 =⇒ c1 = 0

I Impose X (a) = 0 : X (a) = 0 =⇒ c2 sinhµa = 0 =⇒ c2 = 0(sinh(x) = 0 ⇐⇒ x = 0)

k = −µ2 < 0 =⇒ The solution is trivial.

I k = 0 leads to an obviously trivial solution.

Simulated Solutions

I Impose X (0) = 0 : X (0) = 0 =⇒ c1 = 0

Simulated Solutions

I Impose X (0) = 0 : X (0) = 0 =⇒ c1 = 0

Simulated Solutions

I Impose X (0) = 0 : X (0) = 0 =⇒ c1 = 0

Simulated Solutions

I Impose X (0) = 0 : X (0) = 0 =⇒ c1 = 0

probabilistic solutions to the dirichlet problem - theory ...stat650/oldprojects/stat 650 project...

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