The Black-Scholes operator is given by

$$L_{BS}u(x) = \frac{1}{2}\sigma^2x^2\frac{\partial^2}{\partial x^2}u(x) + rx\frac{\partial}{\partial x}u(x) - ru(x)$$

I want to prove that this operator has no eigenvalue with positive real part.

Unfortunately, I am not entirely sure what the domain of this operator is but I am inclined to believe that it is something like

$$\mathcal{D} = \{u \in C^2: u \geq 0, u(x) \leq Me^{\alpha x^2} \text{ for some }M, \alpha \}$$

Writing $L_{BS}u = \lambda u$, I do get eigenvalues with positive real part unless I am making an error somewhere. It is also possible that I am not in the right domain. Any help/suggestion appreciated.

  • $\begingroup$ You should list the domain you're interested in. Is it $\Bbb R$, $[0,1]$, etc? Boundary conditions are also important for eigenvalue purposes. $\endgroup$ – Cameron Williams May 29 '18 at 16:44
  • $\begingroup$ @CameronWilliams $u$ takes in two arguments actually, $t$ and $x$. On $t \in [0,T)$ $u$ is twice continuously differentiable in $x$ for $x \in [0,\infty)$. The boundary condition is $u(T,x) = f(x)$ for some function of $x$ (not necessarily continuous). I did not mention the $t$ argument because I am not sure if that plays a role here. The other boundary conditions $u(t,0)$ and $\lim_{x\to\infty}u(t,x)$ depend on $f$ and there is no general formula for them. $\endgroup$ – Calculon May 29 '18 at 16:51
  • $\begingroup$ Hmm I see. Well this as written is in a Cauchy Euler form so you can apply that typical trick to get eigenvalues. $\endgroup$ – Cameron Williams May 29 '18 at 17:34
  • $\begingroup$ @CameronWilliams I would really appreciate it if you could let me know the name of that typical trick or send me a link to it. $\endgroup$ – Calculon May 29 '18 at 17:41
  • $\begingroup$ en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation $\endgroup$ – Cameron Williams May 29 '18 at 18:11

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