I would like to solve the PDE$$P_{t}\left(t, r{(t)}\right)+ \Lambda r{(t)} P_{r}\left(t,r{(t)}\right)+\frac{1}{2}r{(t)}^2{\sigma}^2P_{rr}\left(t, r{(t)}\right)=r(t)P\left(t, r{(t)}\right), $$ where $$\Lambda :=\alpha_{1}+\alpha_{2}\ln r{(t)},$$ and $r(t)$ evolves according to the black-karasinki model. I want to solve this using the finite difference method. However the problem is I only know one boundary condition: $$P\left(T ,r; T\right)=1.$$ I would like to know the other possible boundary condition in order for the equation to be solved numerically.
Boundary conditions for zero coupond bond PDE
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0Can I ask which other boundary condition do you need? You should solve the equation in the domain $]-\infty, \infty[\times [0, T]$, so you need only one BC – 2017-01-28
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0@7raiden7 I was thinking I need two BCs because the PDE is second order equation. – 2017-01-29
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0You mean that you need an initial *and* a set of boundary conditions (it's an IVBP)? In your case the space domain should be $\mathbbm{R}_+$, you just need to impose the condition $P(\cdot, 0, T) = 1$. Since you would anyway use a computational domain such as $[r_m, r_M] $, and assuming a continuous compounding, I would impose $P(t, r_m, T) = e^{-r_m\cdot t} $ and $P(t, r_M, T) = e^{-r_M \cdot t} $. Otherwise, you can always impose null Neumann conditions and taking your space domain very large, assuring that the effects of the BC will not alter the solution significantly – 2017-01-29