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I'm working on the Black-Scholes equation, but I'm pretty new to financial modeling. Right now, I am trying to understand the Black-Scholes PDE. I understand that the Black-Scholes equation is given by \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} with initial condition \begin{equation*} C(S,T) = \max (S-K, 0) \end{equation*} and boundary conditions \begin{equation*} C(0,t) = 0 \hspace{35pt} C(S,t) \rightarrow S \text{ as } S \rightarrow \infty \end{equation*} and $C(S,t)$ is defined over $0 < S < \infty$, $0 \leq t \leq T$.

TheThis can be further transformed and simplified into a heat diffusion equation is as described here.

If we make the following change of variable \begin{equation*} u = e^{-r\tau}C \hspace{20pt} \text{ or } \hspace{20pt} C = ue^{r\tau} \end{equation*} and \begin{equation*} S = e^x \hspace{25pt} \text{ and } \hspace{25pt} t = T- \tau \hspace{20pt} \end{equation*} we get the transformed heat equation

\begin{equation*} \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2} + (k-1)\frac{\partial u}{\partial x} - ku \end{equation*}

Where $k = \frac{2r}{\sigma^2}$. The following matlab code implements this. My question is, what exactly is the form of the boundary conditions for the the transformed equation?My question is, what exactly is the form of the boundary conditions for the the transformed equation? I can't seem to understand the parameters (related to the boundary conditions) given in the Matlab code. Any related literature would be highly appreciated.

And as an additional question, for the following graph

plot,

you get the most payoff when you wait until t = 4 and S = $e^{0.5}$. Is this insight correct? Additionally, in the graph above, what is the implication? Since the payoff is greatest when time to go, $t$ is maximum, does this mean we should exercise the option early?

I'm working on the Black-Scholes equation, but I'm pretty new to financial modeling. Right now, I am trying to understand the Black-Scholes PDE. I understand that the Black-Scholes equation is given by \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} with initial condition \begin{equation*} C(S,T) = \max (S-K, 0) \end{equation*} and boundary conditions \begin{equation*} C(0,t) = 0 \hspace{35pt} C(S,t) \rightarrow S \text{ as } S \rightarrow \infty \end{equation*} and $C(S,t)$ is defined over $0 < S < \infty$, $0 \leq t \leq T$.

The transformed equation is \begin{equation*} \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2} + (k-1)\frac{\partial u}{\partial x} - ku \end{equation*}

The following matlab code implements this. My question is, what exactly is the form of the boundary conditions for the the transformed equation? I can't seem to understand the parameters (related to the boundary conditions) given in the Matlab code. Any related literature would be highly appreciated.

And as an additional question, for the following graph

plot,

you get the most payoff when you wait until t = 4 and S = $e^{0.5}$. Is this insight correct? Additionally, in the graph above, what is the implication? Since the payoff is greatest when time to go, $t$ is maximum, does this mean we should exercise the option early?

I'm working on the Black-Scholes equation, but I'm pretty new to financial modeling. Right now, I am trying to understand the Black-Scholes PDE. I understand that the Black-Scholes equation is given by \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} with initial condition \begin{equation*} C(S,T) = \max (S-K, 0) \end{equation*} and boundary conditions \begin{equation*} C(0,t) = 0 \hspace{35pt} C(S,t) \rightarrow S \text{ as } S \rightarrow \infty \end{equation*} and $C(S,t)$ is defined over $0 < S < \infty$, $0 \leq t \leq T$.

This can be further transformed and simplified into a heat diffusion equation as described here.

If we make the following change of variable \begin{equation*} u = e^{-r\tau}C \hspace{20pt} \text{ or } \hspace{20pt} C = ue^{r\tau} \end{equation*} and \begin{equation*} S = e^x \hspace{25pt} \text{ and } \hspace{25pt} t = T- \tau \hspace{20pt} \end{equation*} we get the transformed heat equation

\begin{equation*} \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2} + (k-1)\frac{\partial u}{\partial x} - ku \end{equation*}

Where $k = \frac{2r}{\sigma^2}$. The following matlab code implements this. My question is, what exactly is the form of the boundary conditions for the the transformed equation? I can't seem to understand the parameters (related to the boundary conditions) given in the Matlab code. Any related literature would be highly appreciated.

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I'm working on the Black-Scholes equation, but I'm pretty new to financial modeling. Right now, I am trying to understand the Black-Scholes PDE. I understand that the Black-Scholes equation is given by \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} with initial condition \begin{equation*} C(S,T) = \max (S-K, 0) \end{equation*} and boundary conditions \begin{equation*} C(0,t) = 0 \hspace{35pt} C(S,t) \rightarrow S \text{ as } S \rightarrow \infty \end{equation*} and $C(S,t)$ is defined over $0 < S < \infty$, $0 \leq t \leq T$.

This can be further transformed and simplified into a heat diffusion equation as described here.

If we make the following change of variable \begin{equation*} u = e^{-r\tau}C \hspace{20pt} \text{ or } \hspace{20pt} C = ue^{r\tau} \end{equation*} and \begin{equation*} S = e^x \hspace{25pt} \text{ and } \hspace{25pt} t = T- \tau \hspace{20pt} \end{equation*} we get theThe transformed heat equation

  is \begin{equation*} \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2} + (k-1)\frac{\partial u}{\partial x} - ku \end{equation*}

The following matlab code implements this. My question is, what exactly is the form of the boundary conditions for the the transformed equation?My question is, what exactly is the form of the boundary conditions for the the transformed equation? I can't seem to understand the parameters (related to the boundary conditions) given in the Matlab code. Any related literature would be highly appreciated.

And as an additional question, for the following graph

plot,

you get the most payoff when you wait until t = 4 and S = $e^{0.5}$. Is this insight correct? Additionally, in the graph above, what is the implication? Since the payoff is greatest when time to go, $t$ is maximum, does this mean we should exercise the option early?

I'm working on the Black-Scholes equation, but I'm pretty new to financial modeling. Right now, I am trying to understand the Black-Scholes PDE. I understand that the Black-Scholes equation is given by \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} with initial condition \begin{equation*} C(S,T) = \max (S-K, 0) \end{equation*} and boundary conditions \begin{equation*} C(0,t) = 0 \hspace{35pt} C(S,t) \rightarrow S \text{ as } S \rightarrow \infty \end{equation*} and $C(S,t)$ is defined over $0 < S < \infty$, $0 \leq t \leq T$.

This can be further transformed and simplified into a heat diffusion equation as described here.

If we make the following change of variable \begin{equation*} u = e^{-r\tau}C \hspace{20pt} \text{ or } \hspace{20pt} C = ue^{r\tau} \end{equation*} and \begin{equation*} S = e^x \hspace{25pt} \text{ and } \hspace{25pt} t = T- \tau \hspace{20pt} \end{equation*} we get the transformed heat equation

 \begin{equation*} \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2} + (k-1)\frac{\partial u}{\partial x} - ku \end{equation*}

The following matlab code implements this. My question is, what exactly is the form of the boundary conditions for the the transformed equation? I can't seem to understand the parameters (related to the boundary conditions) given in the Matlab code. Any related literature would be highly appreciated.

I'm working on the Black-Scholes equation, but I'm pretty new to financial modeling. Right now, I am trying to understand the Black-Scholes PDE. I understand that the Black-Scholes equation is given by \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} with initial condition \begin{equation*} C(S,T) = \max (S-K, 0) \end{equation*} and boundary conditions \begin{equation*} C(0,t) = 0 \hspace{35pt} C(S,t) \rightarrow S \text{ as } S \rightarrow \infty \end{equation*} and $C(S,t)$ is defined over $0 < S < \infty$, $0 \leq t \leq T$.

The transformed equation is \begin{equation*} \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2} + (k-1)\frac{\partial u}{\partial x} - ku \end{equation*}

The following matlab code implements this. My question is, what exactly is the form of the boundary conditions for the the transformed equation? I can't seem to understand the parameters (related to the boundary conditions) given in the Matlab code. Any related literature would be highly appreciated.

And as an additional question, for the following graph

plot,

you get the most payoff when you wait until t = 4 and S = $e^{0.5}$. Is this insight correct? Additionally, in the graph above, what is the implication? Since the payoff is greatest when time to go, $t$ is maximum, does this mean we should exercise the option early?

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Black-Scholes Boundary Conditions for the given PDE

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