- #1
jend23
- 12
- 0
Essentially, this looks like a differential equation problem but being rusty on differential equations I am a little stuck.
Consider the following SDE
[tex]
d\sigma = a(\sigma,t)dt + b(\sigma,t)dW
[/tex]
The Forward Equation (FKE) is given by
[tex]
\frac{\partial p}{\partial t} = \frac{1}{2}\frac{\partial^2}{\partial\sigma^2}(b^2p) - \frac{\partial}{\partial\sigma}(ap)
[/tex]
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
[tex]
p(\sigma) = \frac{A}{b^2}e^{\int^\sigma\frac{2a}{b^2}d\sigma '}
[/tex]
So, as per the question, the steady state solution is:
[tex]
\frac{1}{2}\frac{d^2}{d\sigma^2}(b^2p) - \frac{d}{d\sigma}(ap) = 0
[/tex]
Integrating to convert to a first order differential equation?:
[tex]
\frac{1}{2}\frac{d}{d\sigma}(b^2p) - ap = A
[/tex]
Given the boundary conditions in the question, ##A = 0##??
[tex]
\frac{1}{2}\frac{d}{d\sigma}(b^2p) = ap
[/tex]
Re-arranging we have a variable separable D.E
[tex]
\frac{1}{p} dp = \frac{2a}{b^2} d\sigma
[/tex]
Integrating:
[tex]
\log p = \int \frac{2a}{b^2} d\sigma + D
[/tex]
[tex]
p = e^{\int \frac{2a}{b^2} d\sigma + D}
[/tex]
...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.
Any help appreciated. Please bear in mind that my differential equation and integration skills are very poor so any hint as to where I've gone wrong would be great, thanks!
Homework Statement
Consider the following SDE
[tex]
d\sigma = a(\sigma,t)dt + b(\sigma,t)dW
[/tex]
The Forward Equation (FKE) is given by
[tex]
\frac{\partial p}{\partial t} = \frac{1}{2}\frac{\partial^2}{\partial\sigma^2}(b^2p) - \frac{\partial}{\partial\sigma}(ap)
[/tex]
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
[tex]
p(\sigma) = \frac{A}{b^2}e^{\int^\sigma\frac{2a}{b^2}d\sigma '}
[/tex]
Homework Equations
The Attempt at a Solution
So, as per the question, the steady state solution is:
[tex]
\frac{1}{2}\frac{d^2}{d\sigma^2}(b^2p) - \frac{d}{d\sigma}(ap) = 0
[/tex]
Integrating to convert to a first order differential equation?:
[tex]
\frac{1}{2}\frac{d}{d\sigma}(b^2p) - ap = A
[/tex]
Given the boundary conditions in the question, ##A = 0##??
[tex]
\frac{1}{2}\frac{d}{d\sigma}(b^2p) = ap
[/tex]
Re-arranging we have a variable separable D.E
[tex]
\frac{1}{p} dp = \frac{2a}{b^2} d\sigma
[/tex]
Integrating:
[tex]
\log p = \int \frac{2a}{b^2} d\sigma + D
[/tex]
[tex]
p = e^{\int \frac{2a}{b^2} d\sigma + D}
[/tex]
...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.
Any help appreciated. Please bear in mind that my differential equation and integration skills are very poor so any hint as to where I've gone wrong would be great, thanks!