about the boundary condition of B-S PDE, for a Euopean call option V(S,t), they are, 1) V(0, t) =0, for all t.2) V(S,t)-->S, for S-->\infty3) V(S,T)=max(S-K,0).Question: is 2) a general condition? For example, if the payoff function is V(S,T)=max(S^2-K,0),shouldn't 2) change to V(S,t)--->S^2, as S--->\infty?thanks.

The condition 2) is based on the fact that as the stock price approaches infinity, it is ever more likely that the option will be exercised. So, it's based on financial reasoning, not mathematical.Stale

No, 2 is not a general condition. In your example, V=S^2 is not a solution of the BSE so won't be a suitable boundary condition at infinity. In this case look for a separable solution of the form V=f(t)S^2.P

QuoteOriginally posted by: StaleThe condition 2) is based on the fact that as the stock price approaches infinity, it is ever more likely that the option will be exercised. So, it's based on financial reasoning, not mathematical.StaleWhy do people look at C, not P? Just curious.

QuoteOriginally posted by: bquantabout the boundary condition of B-S PDE, for a Euopean call option V(S,t), they are, 1) V(0, t) =0, for all t.2) V(S,t)-->S, for S-->\infty3) V(S,T)=max(S-K,0).Question: is 2) a general condition? For example, if the payoff function is V(S,T)=max(S^2-K,0),shouldn't 2) change to V(S,t)--->S^2, as S--->\infty?thanks.Note: S^2-K isn't OK because of the units of measure.$^2 minus $ equals ????

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: StaleThe condition 2) is based on the fact that as the stock price approaches infinity, it is ever more likely that the option will be exercised. So, it's based on financial reasoning, not mathematical.StaleWhy do people look at C, not P? Just curious.Hard to say.Back in the olde days, the CBOE listed only Call contracts. Short selling, or buying put options was considered unpatriotic. Ha!

QuoteOriginally posted by: PaperCutNote: S^2-K isn't OK because of the units of measure. $^2 minus $ equals ????Go easy on him! K has dimensions of $^2, and multiply the whole thing by a factor of 1/$. This is a "power option."P

thanks for pointing out this, for example take the payoff as (S^2-K)/S_0, it will be fine in units.I am still confused, Stale says the boundary condition is decided based on financial point of view,then if we excercise it as S-->\infty, does not the payoff approach S^2?I can understand S^2 will not be a suitable bondary condtion if I look at the analytical solution,but generally how to decide the bondary condition? Is it derived purely from financial orlook at the analytical solution first? I doubt the latter.QuoteOriginally posted by: PaulQuoteOriginally posted by: PaperCutNote: S^2-K isn't OK because of the units of measure. $^2 minus $ equals ????Go easy on him! K has dimensions of $^2, and multiply the whole thing by a factor of 1/$. This is a "power option."P

Technically, S=infty is a natural boundary of the underlying diffusion, so no boundary conditions are necessary or allowed.However, for numerical PDE truncations, it is nice to put in something reasonable.If you put in the wrong behavior V(t,Smax) = Smax^2, you will still get the correct answer with a PDE solverfor any finite S as Smax ->infty, IMO.The right behavior is really (*) V(t,S) ~ S^2 exp{(r + sig^2) t)}, as S >> 1To derive the right behavior for any payoff, use Paul's idea, or simply write the payoff as phi(y), using y = Log S(T). Then, the exact answer isV(t,S) = v(t,x), wherev(t,x) = exp(-r t) int exp{-(y - x - mu t)^2/(2 sig^2 t)} phi(y) dy/Sqrt[2 Pi sig^2 t] . Now x = log S(0) = log S. Let x >> 1, and the integral will reduce to something easy: you can simply take phi(y) ~ exp(2 y),which is how I got (*). Same thing works for any (integrable) payoff. You couldanalyze payoff's that grow as S^p for some power once and for all by this method and be done with it.regards,

QuoteTechnically, S=infty is a natural boundary of the underlying diffusion, so no boundary conditions are necessary or allowed.Indeed. It's a PDE in a 1/4 plane. The BC are pure numerical. Strict continuous BC are not correct (or prove that the truncated PDE is correct!). I have never seen a proof of this.PCThanks for the background.My question on C versus P was numerical. At Smax, P = 0 and is much easier than an exploding S at Smax?

QuoteOriginally posted by: bquantabout the boundary condition of B-S PDE, for a Euopean call option V(S,t), they are, 1) V(0, t) =0, for all t.2) V(S,t)-->S, for S-->\infty3) V(S,T)=max(S-K,0).Question: is 2) a general condition? For example, if the payoff function is V(S,T)=max(S^2-K,0),shouldn't 2) change to V(S,t)--->S^2, as S--->\infty?thanks.I would claim that this problem is not correctly posed (point 2) should be absent, as Alan mentions).

You can get non uniqueness on infinite domains if you allow 'exponential' growth at infinity. (I can't believe I've just mentioned uniqueness!)P

QuoteOriginally posted by: PaulYou can get non uniqueness on infinite domains if you allow 'exponential' growth at infinity. (I can't believe I've just mentioned uniqueness!)PYou must prove existence as well (in B space, it's getting awful pure maths now )My Laplace and Fourier transform manipulations are a bit rusty but what about doing LT in t (possible because we are on (0, infinity)) to get an ODE, solve the ODE and do an inverse LT? It should be possible because of x = log (S) and in your book you show how to transform to heat equationU_t = U_xx Now take LT with respect to t and then get a 2nd order ODE with 1 BC at s = 0 and a constraint U < M / s for s > 0, t > 0. The solution in Schaum PDE book.Maybe.

IMO, if a non-negative payoff function is not integrable w.r.t. the Gaussian I wrote, then the PDE problem has no solution for t>0.Alternatively, you can say the PDE solution instantaneously explodes. If there is some other interpretation of this case, I would love to hear it!

If you just want to solve for the value of the power option put x=S^2, you get BSE with different coeffs from usual, etc.P