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Given constants $r,\sigma, K>0$, considering the Black-Scholes PDE of a Europoean call: \begin{cases} \frac{\partial c}{\partial t}+rs\frac{\partial c}{\partial s}+\frac{1}{2}\sigma^2s^2 \frac{\partial^2 c}{\partial s^2}-rc = 0\\ c(T,s) =(s-K)^+\\ c(t,0) = 0 \end{cases} I learned that we can do the substitution like this: \begin{cases} \tau=T-t\\ u=ce^{r\tau}\\ x=\ln\frac{s}{K}+(r-\frac{1}{2}\sigma^2)(T-t) \end{cases} Set $u=u(\tau,x)$, then $c(t,s)=e^{-r\tau}u(\tau,x)$. Then we have: \begin{gather*} \frac{\partial c}{\partial t} = re^{-r\tau}u+e^{-r\tau}(\frac{\partial u}{\partial \tau} \frac{d \tau}{dt} + \frac{\partial u}{\partial x} \frac{\partial x}{\partial t}) = e^{-r\tau}[ru-\frac{\partial u}{\partial \tau}- (r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x}]\\ \frac{\partial c}{\partial s} = e^{-r\tau}\frac{\partial u}{\partial x} \frac{\partial x}{\partial s} = e^{-r\tau}\frac{1}{s}\frac{\partial u}{\partial x}\\ \frac{\partial^2 c}{\partial s^2} = e^{-r\tau}[-\frac{1}{s^2}\frac{\partial u}{\partial x}+\frac{1}{s^2}\frac{\partial^2 u}{\partial x^2}]=e^{-r\tau}\frac{1}{s^2}(\frac{\partial^2 u}{\partial x^2}-\frac{\partial u}{\partial x}) \end{gather*} Thus: \begin{align*} \frac{\partial c}{\partial t}+rs\frac{\partial c}{\partial s}+\frac{1}{2}\sigma^2s^2 \frac{\partial^2 c}{\partial s^2}-rc &= e^{-r\tau}[-\frac{\partial u}{\partial \tau}- (r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x} + r\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2(\frac{\partial^2 u}{\partial x^2}-\frac{\partial u}{\partial x})]\\ &= e^{-r\tau}[-\frac{\partial u}{\partial \tau} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2}]\\ &=0 \end{align*} Namely we get the heat equation with boundary condition: \begin{cases} \frac{\partial u}{\partial \tau} = \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}\\ u(0,x)=K(e^x - 1)^+\\ u(\tau, -\infty) = 0 \end{cases} I am stuck here. Since the boundary condition is not integrable, I cannot use Fourier transformation with regard to $x$ here. I know there is a probabilistic way of doing it, by computing the conditional expectation. I want to solve the PDE directly. After getting the heat equation here, what should I do next to solve it?

Besides, is there any way to start directly from the PDE of $c$, without change of variable? And, how do we know that change of variable work here? It seems to me very coincidentally the PDE becomes a heat equation in the end.

Thank you so much!

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    $\begingroup$ It is well known that the Black-Scholes PDE can be transformed into the heat equation. Your change-of-variables looks OK, but to check here is something similar. The problem with the integrability of $u(0,x)$ is manageable as shown in the answer. $\endgroup$ – RRL Jul 24 '18 at 4:05
  • $\begingroup$ @RRL Thank you! But I have another question. How do we know that way of change of variable? $\endgroup$ – Edward Wang Jul 25 '18 at 2:19
  • $\begingroup$ Is the question how would you know to use these transformations? These are just standard tricks, like solving an ODE with an integrating factor: $x' +x = f \implies (e^t x)' = e^tf$ $\endgroup$ – RRL Jul 25 '18 at 4:02
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You can apply the Fourier transform with complex wavenumber, $k = k_r + ik_i$.

Since $u(0,x) = \mathcal{O}(e^x)$ as $x \to +\infty$ and $u(0,x) = 0$ for $x \leqslant 0$ the Fourier integral (applied to the initial condition) converges when $\Im{(k)} > 1$,

$$\hat{u}(0,k) = \int_{-\infty}^\infty u(0,x) e ^{ikx} \, dx = \int_{-\infty}^\infty \underbrace{u(0,x) e^{-k_i x}}_{\in L^1(\mathbb{R})}e ^{ik_rx} \, dx \\ = \int_{0}^\infty K(e^x-1)e ^{ikx}\,dx = -\frac{K}{1 + ik} + \frac{K}{ik} \\ = \frac{K}{ik-k^2}$$

Apply the transform to the PDE in the usual way and obtain an ODE for the transform $\hat{u}(\tau,k)$ of the form

$$\frac{\partial \hat{u}}{\partial \tau} = - \frac{\sigma^2k^2}{2} \hat{u},$$

with the solution

$$\hat{u}(\tau,k) = \hat{u}(0,k) e^{-\sigma^2k^2\tau/2} = \frac{Ke^{-\sigma^2k^2\tau/2}}{ik - k^2}$$

The inverse transform takes the form of a contour integral in the complex plane

$$u(\tau,x) = \frac{1}{2\pi}\int_{i\beta-\infty}^{i\beta+\infty} \hat{u}(\tau,x) e ^{-ikx} \, dk, $$

for any $\beta > 1$, leading ultimately to the Black-Scholes solution.

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