I want to find all the solutions of the Black Scholes PDE that are of the form f(x,t)=theta(x) or f(x,t)=phi(t).

Can someone explain and help with this? I know the PDE formula is

$f_{t}(t, x)=-\frac{1}{2} \sigma^{2}x^2 f_{x x}(t, x)-r x f_{x}(t, x)+rf(t, x)$

$f(T, x)=h(x)$

Thank you!

  • $\begingroup$ I also need to find solutions that are of the form of phi(t)*theta(x) and figure out what terminal value problems I can solve with these solutions? Can you assist? I am so lost. $\endgroup$ – galeshapley Feb 12 at 19:51
  • 1
    $\begingroup$ If the solution is of the form theta(x) that means it is not a function of t. So the derivative with respect to t namely $f_t(t,x)$ can be set to zero; the equation becomes a ODE. Similarly if the solution is of the form phi(t), you can set all the derivatives with respect to x to zero, resulting in an extremely simple equation, easy to solve. $\endgroup$ – noob2 Feb 12 at 19:57
  • $\begingroup$ So, I will end up with $\theta(x)=-\frac{1}{2} \sigma^{2}(x) f_{x x}(x)-x$ and with r(t)f(t)=phi(t). Is this right? $\endgroup$ – galeshapley Feb 12 at 20:11
  • $\begingroup$ And for the part that I asked in the comments, should I just be multiplying these solutions together? I am not sure what terminal vlaue problem I can solve with these. $\endgroup$ – galeshapley Feb 12 at 20:17
  • 1
    $\begingroup$ $f_{t}(t, x)=-\frac{1}{2} \sigma^{2} x^{2} f_{x x}(t, x)-r x f_{x}(t, x)+r f(t, x)$ I think this is actually the formula i want to work with $\endgroup$ – galeshapley Feb 12 at 20:28

$f(t,x)=\theta(x)$ means that the actual price deos not depend on time so: $$f(t,x)=f(T,x)=h(x)$$ so the only solution is $h(x)$.

$f(t,x)=\phi(t)$ means that the price deos not depend on the price.it means that at time T: $$f(T,x)=\phi(t)=h(x)$$ so if h depend on x, there is no solution and if h is constant, there is one solution $\phi=h$

  • $\begingroup$ Thank you! this is helpful. So when it asks what can be solved if f(T,x)=theta(x)*phi(t) what does this mean? Does this mean we can solve stuff with any of the solutions above? $\endgroup$ – galeshapley Feb 15 at 23:56

Your Answer

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

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