# NDSolve There are fewer dependent variables, {V[S,t]}, than equation

I am trying to solve the BlackScholes PDE for Barrier option. It works fine for european barrier, but errors out on american boundary condition

sol1 = NDSolve[{D[V[S, t], t] + r*S*D[V[S, t], S] +
1/2 sigma^2 S^2 D[V[S, t], {S, 2}] - r V[S, t] == 0,
V[S, t] ==    Piecewise[{{0, S >= 120}, {Max[S - K, 0], S < 120}}]},V, {S, 0.1,   1000}, {t, 0, T}]


If I change the boundary condition to capital "T", it works fine

V[S, T] ==    Piecewise[{{0, S >= 120}, {Max[S - K, 0], S < 120}}]}


If it's for any time t, then it complains about fewer dependent variables. Is there any other way to specify boundary condition for 2 variables? It seems a simple issue, but can't find anything in mathematica documentation!

• For those who are not familiar with the Black-Scholes equation with the barrier option, can you either edit your question to include the PDEs and the boundary conditions you're trying to implement, or link to a description of same? Jul 2 '15 at 14:45
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• The code as first written does not include a boundary condition but instead an equation defining V for all S and t, which eliminates it as a dependent variable; hence, the error. Changing t to T turns the second equation into a bouncary condition, and all is well. You need a boundary condition in t too, something like V[S,0] == 0. Also, note that sigma and K are undefined. Although it does not matter here, it is bad practice to begin variable names with capital letters, because Mathematica functions also begin with capital letters. Jul 2 '15 at 15:05
• Yes I realize the problem. I basically want to specify boundary condition as V[S,T] = Max(S-K,0) for S<120 and V[S,t] = 0 for S>=120. Not sure how to write this in pde format for dsolve Jul 2 '15 at 15:31

barriersolution =

It's not at all stable for V[100,0] ranging from 0.43 to 0.56 for slight change of S boundaries. I was trying to price binary digitals with a pde where payoff is 1 if Spot > 100 at expiry. Probably I should supply the mesh along with the pde problem to get a stable solution. Maybe a mesh which is finer near expiry time and very sparse before expiry time.