Solving the Discounted Geometric Brownian Motion Martingale ODE

# Find the solutions of the ODE from SDE


# Find the solutions of the ODE from SDE












> Consider the SDE $$dS_t = rS_t dt + \sigma S_t dB_t \ \ \ \text{where} \ r \ \text{and} \ \sigma \ \text{are constants}$$ a.) Find the ODE for the function $V(x)$ such that $e^{-rt}V(S_t)$ is martingale. b.) Find all the solutions to the ODE in (a).

Attempted solution for a.) $V(t,S_t)$ is martingale if and only if $V(t,x)$ satisfies $$\partial_t V(t,x) + \partial_x V(t,x)\mu(t,x) + \frac{1}{2}\partial_{xx}V(t,x)\sigma^2(t,x) = 0$$ hence set $V(t,x) = V(x)e^{-rt}$ in the equation above, we then have $$V(x)\partial_t e^{-rt} + e^{-rt}\partial_x V(x)\mu + {\color{red}{\frac{1}{2}}} e^{-rt}\partial_{xx}V(x)\sigma^2 = 0$$ where $\mu = r S_t$ and $\sigma = \sigma S_t$. Therefore the ODE for the function $V(x)$ is $$-rV(x)e^{-rt} + re^{-rt}S_t\frac{d V}{dx} + {\color{red}{\frac{1}{2}}}\sigma^2e^{-rt}S_t^2\frac{d^2 V}{dx^2} = 0$$

Attempted solution for b.) We have a second order constant coefficent linear differential equation of the form $$aV^{''} + bV' + cV = 0$$ where $a = \sigma^2 S_t^{2}$, $b = rS_t$, and $c = -r$. Thus the solution to this ode is $$V = \begin{cases} Ae^{m_1x} + Be^{m_2x}, & \text{if} \ am^2 + bm + c = 0 \ \text{has distinct real roots}\\ e^{ax}( C cos(\beta x) + Dsin(\beta x), & \text{if} \ am^2 + bm + c = 0 \ \text{has roots equal to} \alpha\pm \beta i \\ (Ax + B)e^{mx}, & \text{if} \ am^2 + bm + c = 0 \ \text{has 1 repeated root} \end{cases}$$

Not sure is this is correct any suggestions is greatly appreciated.

## Answer by Gordon (score 0, accepted)

https://quant.stackexchange.com/a/25184

I will consider (b) only. From part (a), the ODE is of the form \begin{align*} -r V(x) +rx\frac{dV(x)}{dx} + \frac{1}{2}\sigma^2 x^2 \frac{d^2V(x)}{dx^2} = 0.\tag{1} \end{align*} Let $x=e^y$. Then, \begin{align*} \frac{dV(x)}{dx} &= \frac{dV(e^y)}{dy}\frac{dy}{dx}\\ &=\frac{1}{x}\frac{dV(e^y)}{dy},\\ \frac{d^2V(x)}{dx^2} &= -\frac{1}{x^2}\frac{dV(e^y)}{dy} + \frac{d^2V(e^y)}{dy^2}\frac{1}{x^2}. \end{align*} Moreover, Equation (1) is now of the form \begin{align*} -r V(e^y) +\Big(r-\frac{1}{2}\sigma^2\Big)\frac{dV(e^y)}{dy} + \frac{1}{2}\sigma^2 \frac{d^2V(e^y)}{dy^2} = 0.\tag{2} \end{align*} The technique you have shown in your question can now be used to solve this equation. Specifically, \begin{align*} V(e^y)= Ae^{y}+Be^{-\frac{2r}{\sigma^2} y}. \end{align*} That is, \begin{align*} V(x)= Ax+Bx^{-\frac{2r}{\sigma^2}}, \end{align*} where $A$ and $B$ are constants.