This turned my head around!
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Edited by stapel -- Reason for edit: Replacing graphic with text
link to screen-shotThe celebrated Black-Scholes model [20] for the pricing of stock options is central in mathematical finance; see, for example, [156]. The PDE is given by:
. . . . .\(\displaystyle \L u_t\, +\, \frac{1}{2}\sigma^2 x^2 u_{xx}\, +\, rxu_x\, -\, ru\, =\, 0,\, \mbox{ }\, 0\, <\, x\, <\, \infty,\, \mbox{ }\, t\, \leq\, T\, \, \mbox{ }(1.27)\)
For the sake of completeness let us add that u is the sought value of the option under consideration, t is time, x is the current value of the underlying asset, r is the interest rate, \(\displaystyle \sigma\) is the volatility of the underlying asset, T is the expiry date, and E is the exercise price. In general, r and \(\displaystyle \sigma\) may vary, but they are assumed to be known constants, as are E and T.
For the European call option, we have the terminal condition:
. . . . .\(\displaystyle \L u(T,\,x)\, =\, max(x\,-\,E,\,0)\,\,\mbox{ }\,(1.28a)\)
...and the boundary conditions:
. . . . .\(\displaystyle \L u(t,\,0)\, =\, 0,\,\,u(t,\,x)\,\sim \, x\, -\, Ee^{-r(T\,-\,t)}\,\,\mbox{as}\,x\, \rightarrow \,\infty\,\,\mbox{ }\,(1.28b)\)
a) Show that the transformation:
. . . . .\(\displaystyle \L x\,=\,Ee^y ,\,\,t\,=\, T\,-\, \frac{2s}{\sigma^2} ,\, \, u\,=\, Ev(s,\,y)\)
...results in the initial value PDE:
. . . . .\(\displaystyle \L v_s\, =\, v_{yy}\, +\, (\kappa \,-\,1)v_y \, -\, \kappa v,\,\, \mbox{ }\, -\infty\, <\, y\, < \infty\)
. . . . .\(\displaystyle \L v(0,\, y)\, =\, max(e^y \, -\, 1,\, 0)\)
...where:
. . . . .\(\displaystyle \L \kappa \, =\, \frac{2s}{\sigma^2}\)
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Edited by stapel -- Reason for edit: Replacing graphic with text