General solutions for HJB equations in a special case.

2011-07-14T21:59:38Z

I am reading the book of Wendell Flemming in control theorem to learn the HJB equation

Here is the setting that interests me: Let $g_i: i=1,2$ be $C^2 =C^{2}(-\infty,\infty)$ functions such that $0\leq g_1(x)\leq C_1$ and $0\leq g_2(x)\leq e^{x}C_2$ for some FIXED constants $C_1,C_2$. If $f_i : i=1,2$ are $C^2$ functions then A FIXED $\lambda$, define

$Af_i(x)=(\mu -\frac{\sigma^2}{2})\frac{\partial f_i(x)}{\partial x}+\frac{\sigma^2}{2}\frac{\partial^2 f_i(x)}{\partial x^2} +\lambda(g_i(x)-f_i(x))$.

I am interested in finding two functions $f_1,f_2\in C^2$ such that for each $x\in\mathbb{R}$ :

$\min \{ \rho f_1(x) - Af_1(x), f_1(x)- f_2(x) +e^x D_1\} =0$

$\min \{ \rho f_2(x) - Af_2(x), f_2(x)- f_1(x) - e^x D_2\} =0$

where $\mu,\sigma,\rho, D_1>D_2$ are fixed positive constants satisfying: $\rho<\mu<\rho+\lambda$

If consider the equation $\rho f_i(x) - Af_i(x)=0$ ALONE, using Laplace transform,I came

up with the following solutions for $\rho f_i(x) - Af_i(x)=0$ where $i=1,2$

$f_i(x)=B_1^i e^{\nu_1x}+B_2^ie^{\nu_2x} - \frac{2\lambda}{(\nu_2 - \nu_1)\sigma^2}\int_0^x g_i(x-t)(e^{\nu_2t} - e^{\nu_1 t})dt$

with $\nu_{1,2} =(\mu/\sigma^2 -1/2)\pm\sqrt{(\mu/\sigma^2 -1/2)^2+ 2(\rho+\lambda)/\sigma^2}$ and $B^i_{1,2}$ are some real numbers

My question( hope that it does not violate the forum's rules) is that: how can I use these solutions to solve the above HJB equations? Thanks so much for your interests.

General solutions for HJB equations in a special case. - MathOverflow

General solutions for HJB equations in a special case.