Consider the Black-Scholes problem $$\frac{\partial A}{\partial t}+\frac{\sigma^2B^2}{2}\frac{\partial^2A}{\partial B^2}+rB\frac{\partial A}{\partial B}-rA=0 \hspace{3mm}\textrm{and}\hspace{3mm} A(B,T)=f(b),$$ with $B>0$ and $t<T$, and where $\sigma>0$, $T>0$, $r\in \mathbb{R}$ are constants.

Now, let $A(B,t)$ be a solution of the above problem and let it be infinitely differentiable in $B>0$ and $t<T$.

Show that $$A_1(B,t)=B\frac{\partial A}{\partial B}(B,t)$$ satisfies the partial differential equation in the above problem.

So I thought the best thing to do would be to look at the two identities:

$$\frac{\partial}{\partial B}B\left(\frac{\partial A}{\partial B}\right)=\frac{\partial A}{\partial B}+S\frac{\partial^2 A}{\partial B^2}$$ and $$\frac{\partial^2}{\partial B^2}B\left(\frac{\partial A}{\partial B}\right)=2\frac{\partial^2 A}{\partial B^2}+S\frac{\partial^3 A}{\partial B^3}.$$

Then taking the derivative of the Black-Scholes problem (with respect to $B$), I obtain: $$\frac{\partial}{\partial t}\frac{\partial A}{\partial B}+\frac{\sigma^2}{2}\left( 2B\frac{\partial^2A}{\partial B^2}+B^2\frac{\partial^3A}{\partial B^3}\right)+r\left(\frac{\partial A}{\partial B}+B\frac{\partial^2 A}{\partial B^2}\right)-r\frac{\partial A}{\partial B}=0.$$

Then I use the two aforementioned identities and multiply by $B$ to get $$\frac{\partial}{\partial t}\left(S\frac{\partial A}{\partial B}\right)+\frac{\sigma^2}{2}S^2\frac{\partial^2}{\partial B^2}\left(S\frac{\partial A}{\partial B}\right)+rB\frac{\partial}{\partial B}\left(S\frac{\partial A}{\partial B}\right)-rB\frac{\partial}{\partial B}=0.$$ After this point I get rather confused as I can't seem to cancel out the terms? Have I made any mistakes with my method?


One possible solution is as follows:

Replacing $A_{1}(B,t)$ in the left hand side of the black-schole equation we obtain

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Expanding the last equation we have that

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Now from the black-schole equation we derive that enter image description here

Taking the derivative respect to $B$ for the both hands of the last equation we obtain

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Replacing this last equation in the second equation we obtain


  • $\begingroup$ I understand, thank you! $\endgroup$ – Levi Apr 16 '15 at 21:46

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