I was just wondering if someone could verify whether these are the two boundary conditions for a Call Spread Black-Scholes PDE.

The first one I have is:

$max(S_{T} - K_{1}, 0) - max(S_{T}-K_{2},0)$

While the second boundary condition I have is:

$S_{t} - K_{1}e^{-r(T-t)} + K_{2}e^{-r(T-t)} - S_{t} = (K_{2}-K_{1})e^{-r(T-t)}$

Is this correct? Thanks in advance

EDIT: Rather than create a new question, I thought I should ask it here: Should a call spread at time $t_{0}$ always take on a symmetric shape? I have the graph of a call spread PDE at time $t_{0}$ for spot values between 0 to 20, with zero interest rate, and with strikes $K_{1} = 9$, $K_{2} = 11$Call Spread. Does the shape of this call spread look okay, or does it have to be perfectly symmetrical? Thanks!

  • $\begingroup$ Are you looking for a condition imposed on the time domain (absence or arbitrage means the option value at maturity = its payoff) or for a condition on the spatial domain (which is merely an artefact required by the numerical method and in which case zer0 convexity (ie gamma = 0) should work assuming your spot mesh is sufficiently wide. Dirichlet type conditions as in MJ73550'S answer work as well.)? $\endgroup$
    – Quantuple
    May 19, 2016 at 7:42

1 Answer 1


Time $T$ boundary condition is correct $u(T,x)=(x-K_1)^+-(x-K_2)^+$.

Time $x\to 0$ boundary condition is known and is equal to $0$.

Time $x\to\infty$ boundary condition is also known and is correct $\lim_{x\to\infty}u(t,x)=(K_2-K_1)e^{-r(T-t)}.$

You need to be precise if you want your boundary be "absorbing" or "reflecting".

  • $\begingroup$ Thanks for that, I just wanted to confirm I had the right idea. My call spread doesn't appear to be completely symmetrical when $t_{0}$ so I though one of the boundary conditions may have been off... $\endgroup$ May 20, 2016 at 12:24

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