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I have been given a problem to code the heat equation which is transformed from B-S equation (European call option) .

Now the boundary conditions are for European call option: $$C(S,T)=\max(S-K,0)$$ $$C(0,t)=0$$ $$C(S,t) \sim S \space as \> S\to \infty$$ Transforming it to heat equation :$$\frac{\partial u}{\partial \tau}=\frac{\partial^2u}{\partial x^2} $$ with the initial condition: $$u(x,0)=max(e^{\frac{k+1}{2}x}-e^{\frac{k-1}{2}x},0)$$ what are the boundary conditions for this heat equation ?

JejeBelfort
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user1157
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    Apply the same transformation that you used to get the initial condition from the payoff function to get the transformed boundary conditions from the original ones. – LocalVolatility Mar 27 '19 at 18:02
  • The first boundary condition is $0$. I am confused about the second one. – user1157 Mar 28 '19 at 15:29
  • @LocalVolatility , do you have any link to any such paper related to my question? I am stuck. – user1157 Mar 29 '19 at 10:11
  • Please add the exact change of variables you used to get from $C(S, t)$ to $u(x, \tau)$ and what your desired boundary conditions on $C(S, t)$ are, then I'll answer the question. – LocalVolatility Mar 29 '19 at 11:45
  • https://quant.stackexchange.com/questions/84/transformation-from-the-black-scholes-differential-equation-to-the-diffusion-equ

    here the derivation was given. they used the same parameters but the boundary conditions of the heat equation is not given. I get weird boundary conditions.

     – user1157 Mar 29 '19 at 18:40
  • @LocalVolatility , I have gone through the book that was given in that thread and looked into quite a few solutions. I did not see anywhere the transformed boundary conditions – user1157 Mar 29 '19 at 19:02
  • @user1157 in link you gave for the derivation, it is described the function $u= e^{r \tau}C$ and decided on a change of variables $x=y+\tau(r−\frac{σ^2}{2}) $ – mirmo Mar 29 '23 at 22:56

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