# Black Scholes Separable Solutions

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!

• 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. Commented Feb 12, 2020 at 19:51
• 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. Commented Feb 12, 2020 at 19:57
• 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? Commented Feb 12, 2020 at 20:11
• 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. Commented Feb 12, 2020 at 20:17
• $f_{t}(t, x)=-\frac{1}{2} \sigma^{2} x^{2} f_{x x}(t, x)-r x f_{x}(t, x)+r f(t, x)$ I think this is actually the formula i want to work with Commented Feb 12, 2020 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 $$\phi=h$$