We have the following PDE
$$\frac{\partial U}{\partial \tau}=0.5\sigma^2\frac{\partial^2 U}{\partial \epsilon^2}+(r-0.5\sigma^2)\frac{\partial U}{\partial \epsilon}$$
where U
is a function of $\tau$ and $\epsilon$.
Now if we write $x=\epsilon+(r-0.5\sigma^2)\tau$ and $U=W(x,\tau)$ the book says that we get to
$\frac{\partial W}{\partial \tau}=0.5\sigma^2\frac{\partial^2 W}{\partial x^2}$
I couldn't figure out how to get to the transformed PDE. Tried the chain rule for partial derivatives but I didn't get to the correct answer.
Any input will be much appreciated.
Assuming that $r $ and $\sigma$ are constants, we can use the suggested $$x=\epsilon+(r-0.5\sigma^2)\tau \iff \epsilon = x - (r-0.5\sigma^2)\tau$$ to compute the partial derivatives of $W(x,\tau) := U(\epsilon(x,\tau),\tau)$: \begin{align} \frac{\partial}{\partial x} W(x,\tau) &= \frac{\partial U}{\partial \epsilon} \cdot \frac{\partial \epsilon}{\partial x}\\ &= \frac{\partial U}{\partial \epsilon}\\ \Rightarrow \frac{\partial^2}{\partial x^2} W(x,\tau) &= \frac{\partial U^2}{\partial \epsilon^2} \end{align} and \begin{align} \frac{\partial}{\partial \tau} W(x,\tau) &= \frac{\partial U}{\partial \epsilon} \cdot \frac{\partial \epsilon}{\partial \tau} + \frac{\partial U}{\partial \tau}\\ &= -(r-0.5\sigma^2)\frac{\partial U}{\partial \epsilon} + \frac{\partial U}{\partial \tau} \end{align}
Now, multiplying the first equation by $0.5\sigma^2$ and subtracting the second equation from it: \begin{align} \Rightarrow 0.5\sigma^2\frac{\partial^2 W}{\partial x^2} - \frac{\partial W}{\partial \tau} &= 0.5\sigma^2\frac{\partial U^2}{\partial \epsilon^2} + (r-0.5\sigma^2)\frac{\partial U}{\partial \epsilon} - \frac{\partial U}{\partial \tau}\\ &= 0 \end{align} The key step here is that the RHS above is equal to zero because of the original PDE. The desired result now follows.