2
$\begingroup$

I am having difficulty deriving the weak formulation of the Black-Scholes Equation.

I have multiplied it with a test function phi and integrated over Omega. But results on the internet suggest integration by parts are used on the second integral, and then some calculations are skipped, and I don't obtain the same result as rest of the world apparently.

Remark: I'm aware, it is a backward parabolic partial differential equation.

Edit: Here is what I got so far;

Black-Scholes Equation:

$\frac{\partial V}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS \frac{\partial V}{\partial S} - rV = 0$

Variational / weak form:

$\int_\Omega \left(\frac{\partial V}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS \frac{\partial V}{\partial S} - rV \right)\phi dx$

< = >

(**) $\int_\Omega \frac{\partial V}{\partial t} \phi dx + \int_\Omega \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} \phi dx - \int_\Omega rS \frac{\partial V}{\partial S} \phi dx- \int_\Omega rV \phi dx$

Then I found sources saying that by integration by parts on the second integral in the above equation, we will obtain the following:

(*) $\int_\Omega \frac{1}{2} \sigma^2 S^2 \frac{\partial^2}{\partial S^2} \phi dx = -\int_\Omega \frac{1}{2} \sigma^2 S^2 \frac{\partial V}{\partial S} \frac{\partial \phi}{\partial S} dx - \int_\Omega \sigma^2 S \frac{\partial V}{ \partial S} \phi dx$

That I simply don't understand. If I use partial integration on the right hand side of the equation above I get the following:

$\int_\Omega \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} \phi dx = \frac{1}{2} \sigma^2 S^2 \left(\frac{\partial^2 V}{\partial S^2} \frac{\partial \phi}{\partial S} - \int_\Omega \frac{\partial V}{\partial S} \frac{\partial \phi}{\partial S} dx\right)$

Question 1: What else is happening in (*) besides integrations by parts?

Considering (*) again, we can substitute the left hand side into (**):

$\int_\Omega \frac{\partial V}{\partial t} \phi dx -\int_\Omega \frac{1}{2} \sigma^2 S^2 \frac{\partial V}{\partial S} \frac{\partial \phi}{\partial S} dx - \int_\Omega \sigma^2 S \frac{\partial V}{\partial S} \phi dx + \int_\Omega rS \frac{\partial V}{\partial S} - rV \phi dx$

Reducing:

(***) $\int_\Omega \frac{\partial V}{\partial t} \phi dx -\int_\Omega \frac{1}{2} \sigma^2 S^2 \frac{\partial V}{\partial S}\frac{\partial \phi}{\partial S} dx - \int_\Omega (\sigma^2-r) S \frac{\partial V}{\partial S} \phi dx - \int_\Omega rS \frac{\partial V}{\partial S} - rV \phi dx$

question 2:

Is the equation (***) the correct variational form of the Black-Scholes equation? I ask since I found various sources giving different variational forms after applying integrations by parts (in neither of them I'm abble to dublicate since I end in a situation as above, where I'm missing a step in the integration by parts).

Afterwards I would switch the sign by replacing t by tau=T-t, and obtain the variational form of the forward in time parabolic equation.

Thank you in advance for your suggestions, and spelling correction.

$\endgroup$
  • 2
    $\begingroup$ could you edit your post showing your attempt then maybe someone can point out what might be wrong $\endgroup$ – develarist Nov 13 '19 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.