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Finding relationship between solutions of two different first order linear PDEs (using Feynman-Kac)

Existence of first order PDEs?Relationship of SDE and Feynman-Kac PDECharacteristic Function of Geometric Brownian motion - PDE with terminal condition approachDifference Between Feynman-Kac PDE and Kolmogorov Backward EquationUse a stochastic representation result (feynman kac theorem) to solve boundary value problemStochastic Representation of Boundary Value ProblemEntire solutions of first order linear homogeneous evolution PDEsCritical points of solutions to linear PDEsStochastic differential equation with quadratic drift and volatilityCoupled first order PDEs using characteristics

0

1

$begingroup$

The first partial differential equation, using the shorthand $mathcalLu(t,x) = mu(x,t) fracpartial upartial x(x,t) + tfrac12 sigma^2(x,t) fracpartial^2 upartial x^2(x,t)$, is:
beginequation
fracpartial upartial t(x,t) + mathcalLu(x,t) -Vu(x,t) + f(x,t) = 0
tag1endequation

and the other is
beginequation
fracpartial vpartial t(x,t) + mathcalLv(x,t) -(V + W)* v(x,t) + f(x,t) = 0
tag2
endequation

both with terminal condition: $v(T,x) = u(T,x) = g(x)$ and with constants $V$ and $W$.

So the only difference is that the second equation has the term $-Wv(x,t)$ included. I want to find out how I can relate the two solutions, so that if I have a solution of $(1)$ I can multiply it by something and get a solution of $(2)$.

I am wondering if this is possible. So far, I have written, using Feynman-Kac, solutions to both equations:

beginequation
u(x,t) = E^Qleft[ int_t^T e^-V (r-t)f(X_r,r)dr + e^-V(T-t)g(X_T) Bigg| X_t=x right]
endequation

beginequation
v(x,t) = E^Qleft[ int_t^T e^-(V +G) (r-t)f(X_r,r)dr + e^-(V+G)(T-t)g(X_T) Bigg| X_t=x right]
endequation

I can't seem to think of a way to connect the two together, but according to a book I'm reading there should be a way to do this (but there's no other details). Any help would be greatly appreciated. Thanks!

pde stochastic-calculus stochastic-pde

share|cite|improve this question

asked Mar 30 at 2:52

SladeSlade

66111

$endgroup$

add a comment | 

0

1

$begingroup$

The first partial differential equation, using the shorthand $mathcalLu(t,x) = mu(x,t) fracpartial upartial x(x,t) + tfrac12 sigma^2(x,t) fracpartial^2 upartial x^2(x,t)$, is:
beginequation
fracpartial upartial t(x,t) + mathcalLu(x,t) -Vu(x,t) + f(x,t) = 0
tag1endequation

and the other is
beginequation
fracpartial vpartial t(x,t) + mathcalLv(x,t) -(V + W)* v(x,t) + f(x,t) = 0
tag2
endequation

both with terminal condition: $v(T,x) = u(T,x) = g(x)$ and with constants $V$ and $W$.

So the only difference is that the second equation has the term $-Wv(x,t)$ included. I want to find out how I can relate the two solutions, so that if I have a solution of $(1)$ I can multiply it by something and get a solution of $(2)$.

I am wondering if this is possible. So far, I have written, using Feynman-Kac, solutions to both equations:

beginequation
u(x,t) = E^Qleft[ int_t^T e^-V (r-t)f(X_r,r)dr + e^-V(T-t)g(X_T) Bigg| X_t=x right]
endequation

beginequation
v(x,t) = E^Qleft[ int_t^T e^-(V +G) (r-t)f(X_r,r)dr + e^-(V+G)(T-t)g(X_T) Bigg| X_t=x right]
endequation

I can't seem to think of a way to connect the two together, but according to a book I'm reading there should be a way to do this (but there's no other details). Any help would be greatly appreciated. Thanks!

pde stochastic-calculus stochastic-pde

share|cite|improve this question

asked Mar 30 at 2:52

SladeSlade

66111

$endgroup$

add a comment | 

$begingroup$

The first partial differential equation, using the shorthand $mathcalLu(t,x) = mu(x,t) fracpartial upartial x(x,t) + tfrac12 sigma^2(x,t) fracpartial^2 upartial x^2(x,t)$, is:
beginequation
fracpartial upartial t(x,t) + mathcalLu(x,t) -Vu(x,t) + f(x,t) = 0
tag1endequation

and the other is
beginequation
fracpartial vpartial t(x,t) + mathcalLv(x,t) -(V + W)* v(x,t) + f(x,t) = 0
tag2
endequation

both with terminal condition: $v(T,x) = u(T,x) = g(x)$ and with constants $V$ and $W$.

So the only difference is that the second equation has the term $-Wv(x,t)$ included. I want to find out how I can relate the two solutions, so that if I have a solution of $(1)$ I can multiply it by something and get a solution of $(2)$.

I am wondering if this is possible. So far, I have written, using Feynman-Kac, solutions to both equations:

beginequation
u(x,t) = E^Qleft[ int_t^T e^-V (r-t)f(X_r,r)dr + e^-V(T-t)g(X_T) Bigg| X_t=x right]
endequation

beginequation
v(x,t) = E^Qleft[ int_t^T e^-(V +G) (r-t)f(X_r,r)dr + e^-(V+G)(T-t)g(X_T) Bigg| X_t=x right]
endequation

I can't seem to think of a way to connect the two together, but according to a book I'm reading there should be a way to do this (but there's no other details). Any help would be greatly appreciated. Thanks!

pde stochastic-calculus stochastic-pde

share|cite|improve this question

asked Mar 30 at 2:52

SladeSlade

66111

$endgroup$

share|cite|improve this question

asked Mar 30 at 2:52

SladeSlade

66111

share|cite|improve this question

asked Mar 30 at 2:52

SladeSlade

66111

asked Mar 30 at 2:52

SladeSlade

66111

asked Mar 30 at 2:52

SladeSlade

66111

1 Answer
1

active

oldest

votes

1

$begingroup$

Suppose $u$ and $v$ solve
$$
left[partial_t+mathcalL-cright]u+f=0
qquad textand qquad
left[partial_t+mathcalL-left(c+dright)right]v+e^dtf=0.
$$
Moreover, suppose $u(T,cdot)=g$ and $v(T,cdot)=e^dTg$. If we
assume enough regularity on $u$ and $v$ along with $mu$, $sigma$,
and $g$, the Feynman-Kac formula tells us that
$$
u(t,x)=mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-cleft(T-tright)g(X_T)mid X_t=xright]
$$
and
beginalign*
v(t,x) & =mathbbEleft[int_t^Te^-(c+d)(r-t)e^drf(r,X_r)dr+e^-(c+d)(T-t)e^dTg(X_T)right]\
& =mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-c(T-t)g(X_T)right]e^dt\
& =e^dtu(t,x)
endalign*
where $X$ satisfies an Ito SDE with infinitesimal generator $mathcalL$

Are you sure this is not what you/the text meant?

Note: a simpler argument to go from one PDE to the other is by a change of variables.

share|cite|improve this answer

answered Mar 30 at 3:50

parsiadparsiad

18.7k32453

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was it! Thanks a lot! The text was unclear so I didn't realize the $f$ term could be adjusted as well. Can you elaborate on the change of variables method? Is it just using $left[partial_t+mathcalL-left(c+dright)right]v+e^dtf=0$ with $v = e^dtu(t,x)$ and then expanding out $partial_tv = partial_te^dtu = partial_tu * e^dt + d*e^dtu$, and so this results in $left[partial_t+mathcalL-cright]u+f=0$ when dividing out $e^dt$. (Or doing the reverse by multiplying the $u$ equation by $e^dt$.)
  $endgroup$
  – Slade
  Mar 30 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's right (+1). If $u$ is a $C^1,2$ solution of the first PDE, then you can define $v equiv e^dt u$ to be a solution of the second. Vice versa, if $v$ is a $C^1,2$ solution of the second PDE, you can define $u equiv e^-dt v$ as a solution of the first.
  $endgroup$
  – parsiad
  Mar 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nit: Try not to use $*$ to mean multiplication. This operator is commonly used for convolutions.
  $endgroup$
  – parsiad
  Mar 30 at 19:34
  

  
  
  
  

add a comment | 

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1

$begingroup$

Suppose $u$ and $v$ solve
$$
left[partial_t+mathcalL-cright]u+f=0
qquad textand qquad
left[partial_t+mathcalL-left(c+dright)right]v+e^dtf=0.
$$
Moreover, suppose $u(T,cdot)=g$ and $v(T,cdot)=e^dTg$. If we
assume enough regularity on $u$ and $v$ along with $mu$, $sigma$,
and $g$, the Feynman-Kac formula tells us that
$$
u(t,x)=mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-cleft(T-tright)g(X_T)mid X_t=xright]
$$
and
beginalign*
v(t,x) & =mathbbEleft[int_t^Te^-(c+d)(r-t)e^drf(r,X_r)dr+e^-(c+d)(T-t)e^dTg(X_T)right]\
& =mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-c(T-t)g(X_T)right]e^dt\
& =e^dtu(t,x)
endalign*
where $X$ satisfies an Ito SDE with infinitesimal generator $mathcalL$

Are you sure this is not what you/the text meant?

Note: a simpler argument to go from one PDE to the other is by a change of variables.

share|cite|improve this answer

answered Mar 30 at 3:50

parsiadparsiad

18.7k32453

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was it! Thanks a lot! The text was unclear so I didn't realize the $f$ term could be adjusted as well. Can you elaborate on the change of variables method? Is it just using $left[partial_t+mathcalL-left(c+dright)right]v+e^dtf=0$ with $v = e^dtu(t,x)$ and then expanding out $partial_tv = partial_te^dtu = partial_tu * e^dt + d*e^dtu$, and so this results in $left[partial_t+mathcalL-cright]u+f=0$ when dividing out $e^dt$. (Or doing the reverse by multiplying the $u$ equation by $e^dt$.)
  $endgroup$
  – Slade
  Mar 30 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's right (+1). If $u$ is a $C^1,2$ solution of the first PDE, then you can define $v equiv e^dt u$ to be a solution of the second. Vice versa, if $v$ is a $C^1,2$ solution of the second PDE, you can define $u equiv e^-dt v$ as a solution of the first.
  $endgroup$
  – parsiad
  Mar 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nit: Try not to use $*$ to mean multiplication. This operator is commonly used for convolutions.
  $endgroup$
  – parsiad
  Mar 30 at 19:34
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose $u$ and $v$ solve
$$
left[partial_t+mathcalL-cright]u+f=0
qquad textand qquad
left[partial_t+mathcalL-left(c+dright)right]v+e^dtf=0.
$$
Moreover, suppose $u(T,cdot)=g$ and $v(T,cdot)=e^dTg$. If we
assume enough regularity on $u$ and $v$ along with $mu$, $sigma$,
and $g$, the Feynman-Kac formula tells us that
$$
u(t,x)=mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-cleft(T-tright)g(X_T)mid X_t=xright]
$$
and
beginalign*
v(t,x) & =mathbbEleft[int_t^Te^-(c+d)(r-t)e^drf(r,X_r)dr+e^-(c+d)(T-t)e^dTg(X_T)right]\
& =mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-c(T-t)g(X_T)right]e^dt\
& =e^dtu(t,x)
endalign*
where $X$ satisfies an Ito SDE with infinitesimal generator $mathcalL$

Are you sure this is not what you/the text meant?

Note: a simpler argument to go from one PDE to the other is by a change of variables.

share|cite|improve this answer

answered Mar 30 at 3:50

parsiadparsiad

18.7k32453

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This was it! Thanks a lot! The text was unclear so I didn't realize the $f$ term could be adjusted as well. Can you elaborate on the change of variables method? Is it just using $left[partial_t+mathcalL-left(c+dright)right]v+e^dtf=0$ with $v = e^dtu(t,x)$ and then expanding out $partial_tv = partial_te^dtu = partial_tu * e^dt + d*e^dtu$, and so this results in $left[partial_t+mathcalL-cright]u+f=0$ when dividing out $e^dt$. (Or doing the reverse by multiplying the $u$ equation by $e^dt$.)
  $endgroup$
  – Slade
  Mar 30 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's right (+1). If $u$ is a $C^1,2$ solution of the first PDE, then you can define $v equiv e^dt u$ to be a solution of the second. Vice versa, if $v$ is a $C^1,2$ solution of the second PDE, you can define $u equiv e^-dt v$ as a solution of the first.
  $endgroup$
  – parsiad
  Mar 30 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nit: Try not to use $*$ to mean multiplication. This operator is commonly used for convolutions.
  $endgroup$
  – parsiad
  Mar 30 at 19:34
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $u$ and $v$ solve
$$
left[partial_t+mathcalL-cright]u+f=0
qquad textand qquad
left[partial_t+mathcalL-left(c+dright)right]v+e^dtf=0.
$$
Moreover, suppose $u(T,cdot)=g$ and $v(T,cdot)=e^dTg$. If we
assume enough regularity on $u$ and $v$ along with $mu$, $sigma$,
and $g$, the Feynman-Kac formula tells us that
$$
u(t,x)=mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-cleft(T-tright)g(X_T)mid X_t=xright]
$$
and
beginalign*
v(t,x) & =mathbbEleft[int_t^Te^-(c+d)(r-t)e^drf(r,X_r)dr+e^-(c+d)(T-t)e^dTg(X_T)right]\
& =mathbbEleft[int_t^Te^-c(r-t)f(r,X_r)dr+e^-c(T-t)g(X_T)right]e^dt\
& =e^dtu(t,x)
endalign*
where $X$ satisfies an Ito SDE with infinitesimal generator $mathcalL$

Are you sure this is not what you/the text meant?

Note: a simpler argument to go from one PDE to the other is by a change of variables.

share|cite|improve this answer

answered Mar 30 at 3:50

parsiadparsiad

18.7k32453

$endgroup$

share|cite|improve this answer

answered Mar 30 at 3:50

parsiadparsiad

18.7k32453

answered Mar 30 at 3:50

parsiadparsiad

18.7k32453

answered Mar 30 at 3:50

parsiadparsiad

18.7k32453