Re: Second Order ODE -Euler

Quote:

Originally Posted by

**Tribal5** I'm trying to work out the following Cauchy-Euler problem, was hoping someone could help

the equation is:

1/2*J*s^2 v'' + (r -D)S v' - rV =0, for 0<S<A

V(s) = S - E, for s>a

when V(0) = 0 , V(A) = A - E, dV/dS = 1

I've to show v(s) = (A - E)(S/A)^m

where m = 1/J * [-(r-D-J/2) + Sqrt( (r-D-J/2)^2 + 2rJ))

**note J = Sigma^2 **(Sorry don't know how to use LaTeX)

I can work out m using V = S^{m } and i guess the root is v = C_{1}S^{m} + C_{2}S^{m }

But i just can't tie it all together, any ideas?

Are you using S and s to represent the same thing? If so, please be consistent...

Re: Second Order ODE -Euler

Yes sorry s and S are the same I typed a bit quick.

Re: Second Order ODE -Euler

Quote:

Originally Posted by

**Tribal5** I'm trying to work out the following Cauchy-Euler problem, was hoping someone could help

the equation is:

1/2*J*s^2 v'' + (r -D)S v' - rV =0, for 0<S<A

V(s) = S - E, for s>a

when V(0) = 0 , V(A) = A - E, dV/dS = 1

I've to show v(s) = (A - E)(S/A)^m

where m = 1/J * [-(r-D-J/2) + Sqrt( (r-D-J/2)^2 + 2rJ))

**note J = Sigma^2 **(Sorry don't know how to use LaTeX)

I can work out m using V = S^{m } and i guess the root is v = C_{1}S^{m} + C_{2}S^{m }

But i just can't tie it all together, any ideas?

Why would you guess ? That simplifies to . So you only need to guess a single exponential function.

Let .

Substitute these into your DE and see if you can solve...

Re: Second Order ODE -Euler

yes solving gives the quadratic m^2 + (2(r-D)/J - 1) m -2r/J = 0

if i say b^2 - 4ac > 0

the roots will be

then m(+ve) = 1/J * [-(r-D-J/2) + Sqrt( (r-D-J/2)^2 + 2rJ))

solution would be A S^m(+ve) + B S^m(-ve)

however i've to move from this to V(S) = (A - E)(S/A)^m(+ve) - sorry should have put +ve and -ve in