Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to solve a ODE that arises from a Hamilton-Jacobi-Bellman (HJB) equation. The equation is $$\frac{1}{2}b^2(1-\rho_s^2)\psi''-\frac{1}{2}\left(\frac{\mu-r}{\sigma}\right)^2\frac{(\psi')^2}{\psi''}+[ru+\theta a+b\rho_s(\mu-r)(1-\frac{2}{\sigma})]\psi'=0,$$ where $\mu, r, \sigma, \theta, a, \rho_s, b$ are constant. I want to determine $\psi'(u)$ (so that I get an integral form for $\psi(u)$). I have tried guessing (trial and error method) forms of the solution but didn't get far. I also tried the Legendre transform, but could not get the linear form. These are the method that I have seen being used in with these problems.

share|improve this question
1  
Apart from $u$ and $\psi$, the other variables are constant? –  Aryabhata Sep 15 '10 at 21:02

1 Answer 1

up vote 2 down vote accepted

Assume $\psi' = f$, then the equation is of the form:

$$Af' + B\frac{f^2}{f'} + (cu+d)f = 0$$

By putting $$g = \frac{f'}{f} = (\log f)'$$ we see that

$$Ag + \frac{B}{g} + (cu+d) = 0$$

This is a quadratic in $g$ and can be easily solved.

We get $$ \psi' = f = e^{\int g}$$

Hope that helps.

share|improve this answer
    
thanx, you make it to be so simple. –  Vaolter Sep 16 '10 at 11:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.