Here's the full question and my attempt at answering it by solving the differential equation.
Consider the following SDE
$$
d\sigma = a(\sigma,t)dt + b(\sigma,t)dW
$$
The Forward Equation (FKE) is given by
$$
\frac{\partial p}{\partial t} = \frac{1}{2}\frac{\partial^2}{\partial\sigma^2}(b^2p) – \frac{\partial}{\partial\sigma}(ap)
$$
The steady state solution is given by setting $\frac{\partial p}{\partial t}=0$. Considering the boundary conditions that as $\sigma\rightarrow\infty$, $p\rightarrow 0$ and $\frac{\partial p}{\partial \sigma}\rightarrow 0$, show that the steady state solution is given by
$$
p(\sigma) = \frac{A}{b^2}e^{\int^\sigma\frac{2a}{b^2}d\sigma '}
$$
Essentially, this is a differential equation problem. Thing is, I'm stuck on how to solve it.
This is what I've got so far:
So, as per the question, the steady state solution is:
$$
\frac{1}{2}\frac{d^2}{d\sigma^2}(b^2p) – \frac{d}{d\sigma}(ap) = 0
$$
Integrating to convert to a first order differential equation?:
$$
\frac{1}{2}\frac{d}{d\sigma}(b^2p) – ap = A
$$
Given the boundary conditions in the question, $A = 0$??
$$
\frac{1}{2}\frac{d}{d\sigma}(b^2p) = ap
$$
Re-arranging we have a variable separable D.E
$$
\frac{1}{p} dp = \frac{2a}{b^2} d\sigma
$$
Integrating:
$$
\log p = \int \frac{2a}{b^2} d\sigma + D
$$
$$
p = e^{\int \frac{2a}{b^2} d\sigma + D}
$$
….and this is where I'm stuck. It looks a little like the expected solution but I don't know where the extra $\frac{A}{b^2}$ coefficient comes from and why there is a $d\sigma '$ in the expected solution.
Where have I gone wrong?
Best Answer
From $$\frac{1}{2} \frac{d}{d\sigma} (b^2 \cdot p) = a \cdot p$$ you concluded $$\frac{1}{p} \, dp = \frac{2a}{b^2} \, d\sigma$$
But this is not correct: Note that $b$ does depend on $\sigma$. You have to apply the chain rule:
$$\begin{align} \frac{1}{2} \frac{d}{d\sigma} (b^2 \cdot p) &= a \cdot p \\ \Rightarrow b \cdot \frac{d b}{d \sigma} \cdot p + \frac{1}{2} b^2 \cdot \frac{dp}{d \sigma} &= a \cdot p \\ \Rightarrow \frac{1}{p} \, dp &= 2 \frac{ \left(a- b \cdot \frac{db}{d\sigma}\right) }{b^2} \, d\sigma = \frac{2a}{b^2} \, d\sigma - 2 \frac{1}{b} \cdot \frac{db}{d\sigma} \, d\sigma \end{align}$$
From the second term on the righthand-side you obtain the factor $\frac{1}{b^2}$ in the given solution.
Concerning $d\sigma'$: Since $p=p(\sigma)$ you shouldn't use $\sigma$ as variable of integration at the same time. That's why they called it $\sigma'$ instead (has nothing to do with the derivative!).