We have to consider the two cases x1 = 0 and x1 = 0. If x1 = 0 we have
the trivial solution. Thus, in the following we choose xi = 0.
Assume that the coefficients of the PDE do not depend on xi. Then и can
be separated as follows: u(t,xι,x2) = ^(x1,x2)≠(t,x2). With the boundary
condition u(T, xi,x2) = xι it follows ^(xi,x2)≠(T,x2) = xι and because of
x1 = 0 : φ(T,x2) = 0 for all x2. Hence, the following relationship holds:
Hx, x2) = , <r.j∙., r
After some computation the partial differential equation can be written as
(without noting the variables of the functions a, β and 7)
∩ , /ɪ ʌ, ,/. ∖( /A2(T,x2) I 2≠≈a(T,x2) 72≠z2UT,x2Λ
0 = ≠, (i,x) + ≠(i,x) 7∙ - , + /■ - ɪ ≠(T,xa) )
, , /. √n 2≠UT,x)ʌ , У ∣ t. ʌ
+^x2(t,x2) (ʃ 7 ^(T x2) / + 2 ^x2æ2(.’x2)j
without loss of generality choose the boundary condition ≠(T, x2) = 1.
Now assume that all coefficients are only time-dependent.
In the case of constant volatility, that is uv ≡ 0 and b ≡ 0 and hence
β ≡ 0 and 7 ≡ 0, we have the following solution of the PDE
∙^(t,x2) = exp ^x2 ʃ a(s)ds^ .
With this knowledge, for the general PDE with time-dependent coefficients
we try the ansatz
∙^(t,x2) = exp —-X2 a a(s)ds^ A(t).
with A(T) = 1. This leads to an ordinary first-order differential equation for
A which has the solution
A(t) = exp ^2 ʃ 2β(r) ^ʃ α(s)ds^ + 72(r) ^ʃ a(s)ds^ dr^ .
Hence, the solution of the PDE for u is
u(t,xι,x2) = xι exp ^x2 a(s)ds^
exp ^2 2^(r) ^ʃ a(s)ds^ + 72(r) ^ʃ a(s)ds^ dr
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