Mixing
Numerical solution of Hamilton-Jacobi-Bellman equation with no boundary conditions
$begingroup$
I have a HJB equation that arises from a stochastic optimization problem.
$$u+apartial_{x}V+bpartial_{y}V+frac{sigma^{2}}{2}partial_{xx}V-rho V=0$$
Where $V(x,y)$ is the unkown function and $u$, $a$, $b$ are possibly functions of $x$, $y$ and $V$, and $sigma$ and $rho$ are constants. Moreover, $x$ is can be any real number, while $nin[0,1].$
I am trying to implement a upwind scheme numerically, approximating the derivatives $partial_{x}V$ with forward differences when $a$ is positive and backward differences otherwise. Similarly, I am using forward differences for $partial_{y}V$ iff $b>0$. For the second derivative, I am using central differences.
My question is: what should I do at the extreme of grid points? I tried forcing backward or forward differences, but the solution seems unstable and wrong, with some very small or very large derivatives near the boundaries of the grid.
I know that probably this equation has an unique viscosity solution (and that solution is the value function). But is there some finite difference scheme that is known to converge to the viscosity solution even when we do not have any boundary conditions?
pde numerical-methods finite-differences hamilton-jacobi-equation
edited Jan 4 at 7:09
Dylan
12.5k31026
asked Jan 4 at 0:29
Pcw.Pcw.
1008
$endgroup$
add a comment |
$begingroup$
I have a HJB equation that arises from a stochastic optimization problem.
$$u+apartial_{x}V+bpartial_{y}V+frac{sigma^{2}}{2}partial_{xx}V-rho V=0$$
Where $V(x,y)$ is the unkown function and $u$, $a$, $b$ are possibly functions of $x$, $y$ and $V$, and $sigma$ and $rho$ are constants. Moreover, $x$ is can be any real number, while $nin[0,1].$
I am trying to implement a upwind scheme numerically, approximating the derivatives $partial_{x}V$ with forward differences when $a$ is positive and backward differences otherwise. Similarly, I am using forward differences for $partial_{y}V$ iff $b>0$. For the second derivative, I am using central differences.
My question is: what should I do at the extreme of grid points? I tried forcing backward or forward differences, but the solution seems unstable and wrong, with some very small or very large derivatives near the boundaries of the grid.
I know that probably this equation has an unique viscosity solution (and that solution is the value function). But is there some finite difference scheme that is known to converge to the viscosity solution even when we do not have any boundary conditions?
pde numerical-methods finite-differences hamilton-jacobi-equation
edited Jan 4 at 7:09
Dylan
12.5k31026
asked Jan 4 at 0:29
Pcw.Pcw.
1008
$endgroup$
$begingroup$
In short, no. You need to deduce some boundary condition at $infty$ in order to solve numerically. Then impose this on the boundary of your computational domain. Sometimes some insight from the stochastic optimization problem can help.
$endgroup$
– Jeff
Jan 13 at 3:44
$begingroup$
But isn't there any result that guarantees the solution will converge to the value function far away from the boundaries of the grid even if I input the wrong boundaries?
$endgroup$
– Pcw.
Jan 13 at 12:31
$begingroup$
Unfortunately, no. You can get different solutions for each different boundary condition you place on your computational grid, no matter how far out. Just consider a simple ODE $u + u' + u'' = 0$ on the whole real line. The unique solution is $u=0$, but if you restrict to an interval $[-R,R]$ and set some arbitrary boundary condition at $x=pm R$ you can get almost anything nonzero you like. Of course, here, $u'(pm R)=0$ or $u(pm R) = 0$ give the right solution, but it's not so clear in general what to choose.
$endgroup$
– Jeff
Jan 14 at 1:17
add a comment |
$begingroup$
I have a HJB equation that arises from a stochastic optimization problem.
$$u+apartial_{x}V+bpartial_{y}V+frac{sigma^{2}}{2}partial_{xx}V-rho V=0$$
Where $V(x,y)$ is the unkown function and $u$, $a$, $b$ are possibly functions of $x$, $y$ and $V$, and $sigma$ and $rho$ are constants. Moreover, $x$ is can be any real number, while $nin[0,1].$
I am trying to implement a upwind scheme numerically, approximating the derivatives $partial_{x}V$ with forward differences when $a$ is positive and backward differences otherwise. Similarly, I am using forward differences for $partial_{y}V$ iff $b>0$. For the second derivative, I am using central differences.
My question is: what should I do at the extreme of grid points? I tried forcing backward or forward differences, but the solution seems unstable and wrong, with some very small or very large derivatives near the boundaries of the grid.
I know that probably this equation has an unique viscosity solution (and that solution is the value function). But is there some finite difference scheme that is known to converge to the viscosity solution even when we do not have any boundary conditions?
pde numerical-methods finite-differences hamilton-jacobi-equation
edited Jan 4 at 7:09
Dylan
12.5k31026
asked Jan 4 at 0:29
Pcw.Pcw.
1008
$endgroup$
I have a HJB equation that arises from a stochastic optimization problem.
$$u+apartial_{x}V+bpartial_{y}V+frac{sigma^{2}}{2}partial_{xx}V-rho V=0$$
Where $V(x,y)$ is the unkown function and $u$, $a$, $b$ are possibly functions of $x$, $y$ and $V$, and $sigma$ and $rho$ are constants. Moreover, $x$ is can be any real number, while $nin[0,1].$
I am trying to implement a upwind scheme numerically, approximating the derivatives $partial_{x}V$ with forward differences when $a$ is positive and backward differences otherwise. Similarly, I am using forward differences for $partial_{y}V$ iff $b>0$. For the second derivative, I am using central differences.
My question is: what should I do at the extreme of grid points? I tried forcing backward or forward differences, but the solution seems unstable and wrong, with some very small or very large derivatives near the boundaries of the grid.
I know that probably this equation has an unique viscosity solution (and that solution is the value function). But is there some finite difference scheme that is known to converge to the viscosity solution even when we do not have any boundary conditions?
pde numerical-methods finite-differences hamilton-jacobi-equation
pde numerical-methods finite-differences hamilton-jacobi-equation
edited Jan 4 at 7:09
Dylan
12.5k31026
asked Jan 4 at 0:29
Pcw.Pcw.
1008
edited Jan 4 at 7:09
Dylan
12.5k31026
asked Jan 4 at 0:29
Pcw.Pcw.
1008
edited Jan 4 at 7:09
Dylan
12.5k31026
edited Jan 4 at 7:09
Dylan
12.5k31026
edited Jan 4 at 7:09
Dylan
12.5k31026
12.5k31026
asked Jan 4 at 0:29
Pcw.Pcw.
1008
asked Jan 4 at 0:29
Pcw.Pcw.
1008
asked Jan 4 at 0:29
Pcw.Pcw.
1008
1008
$begingroup$
In short, no. You need to deduce some boundary condition at $infty$ in order to solve numerically. Then impose this on the boundary of your computational domain. Sometimes some insight from the stochastic optimization problem can help.
$endgroup$
– Jeff
Jan 13 at 3:44
$begingroup$
But isn't there any result that guarantees the solution will converge to the value function far away from the boundaries of the grid even if I input the wrong boundaries?
$endgroup$
– Pcw.
Jan 13 at 12:31
$begingroup$
Unfortunately, no. You can get different solutions for each different boundary condition you place on your computational grid, no matter how far out. Just consider a simple ODE $u + u' + u'' = 0$ on the whole real line. The unique solution is $u=0$, but if you restrict to an interval $[-R,R]$ and set some arbitrary boundary condition at $x=pm R$ you can get almost anything nonzero you like. Of course, here, $u'(pm R)=0$ or $u(pm R) = 0$ give the right solution, but it's not so clear in general what to choose.
$endgroup$
– Jeff
Jan 14 at 1:17
add a comment |
$begingroup$
In short, no. You need to deduce some boundary condition at $infty$ in order to solve numerically. Then impose this on the boundary of your computational domain. Sometimes some insight from the stochastic optimization problem can help.
$endgroup$
– Jeff
Jan 13 at 3:44
$begingroup$
But isn't there any result that guarantees the solution will converge to the value function far away from the boundaries of the grid even if I input the wrong boundaries?
$endgroup$
– Pcw.
Jan 13 at 12:31
$begingroup$
Unfortunately, no. You can get different solutions for each different boundary condition you place on your computational grid, no matter how far out. Just consider a simple ODE $u + u' + u'' = 0$ on the whole real line. The unique solution is $u=0$, but if you restrict to an interval $[-R,R]$ and set some arbitrary boundary condition at $x=pm R$ you can get almost anything nonzero you like. Of course, here, $u'(pm R)=0$ or $u(pm R) = 0$ give the right solution, but it's not so clear in general what to choose.
$endgroup$
– Jeff
Jan 14 at 1:17
$begingroup$
In short, no. You need to deduce some boundary condition at $infty$ in order to solve numerically. Then impose this on the boundary of your computational domain. Sometimes some insight from the stochastic optimization problem can help.
$endgroup$
– Jeff
Jan 13 at 3:44
$begingroup$
In short, no. You need to deduce some boundary condition at $infty$ in order to solve numerically. Then impose this on the boundary of your computational domain. Sometimes some insight from the stochastic optimization problem can help.
$endgroup$
– Jeff
Jan 13 at 3:44
$begingroup$
But isn't there any result that guarantees the solution will converge to the value function far away from the boundaries of the grid even if I input the wrong boundaries?
$endgroup$
– Pcw.
Jan 13 at 12:31
$begingroup$
But isn't there any result that guarantees the solution will converge to the value function far away from the boundaries of the grid even if I input the wrong boundaries?
$endgroup$
– Pcw.
Jan 13 at 12:31
$begingroup$
Unfortunately, no. You can get different solutions for each different boundary condition you place on your computational grid, no matter how far out. Just consider a simple ODE $u + u' + u'' = 0$ on the whole real line. The unique solution is $u=0$, but if you restrict to an interval $[-R,R]$ and set some arbitrary boundary condition at $x=pm R$ you can get almost anything nonzero you like. Of course, here, $u'(pm R)=0$ or $u(pm R) = 0$ give the right solution, but it's not so clear in general what to choose.
$endgroup$
– Jeff
Jan 14 at 1:17
$begingroup$
Unfortunately, no. You can get different solutions for each different boundary condition you place on your computational grid, no matter how far out. Just consider a simple ODE $u + u' + u'' = 0$ on the whole real line. The unique solution is $u=0$, but if you restrict to an interval $[-R,R]$ and set some arbitrary boundary condition at $x=pm R$ you can get almost anything nonzero you like. Of course, here, $u'(pm R)=0$ or $u(pm R) = 0$ give the right solution, but it's not so clear in general what to choose.
$endgroup$
– Jeff
Jan 14 at 1:17
add a comment |
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$begingroup$
In short, no. You need to deduce some boundary condition at $infty$ in order to solve numerically. Then impose this on the boundary of your computational domain. Sometimes some insight from the stochastic optimization problem can help.
$endgroup$
– Jeff
Jan 13 at 3:44
$begingroup$
But isn't there any result that guarantees the solution will converge to the value function far away from the boundaries of the grid even if I input the wrong boundaries?
$endgroup$
– Pcw.
Jan 13 at 12:31
$begingroup$
Unfortunately, no. You can get different solutions for each different boundary condition you place on your computational grid, no matter how far out. Just consider a simple ODE $u + u' + u'' = 0$ on the whole real line. The unique solution is $u=0$, but if you restrict to an interval $[-R,R]$ and set some arbitrary boundary condition at $x=pm R$ you can get almost anything nonzero you like. Of course, here, $u'(pm R)=0$ or $u(pm R) = 0$ give the right solution, but it's not so clear in general what to choose.
$endgroup$
– Jeff
Jan 14 at 1:17