I'm trying to derive the following boundary conditions for heston's stochastic volatility model.
This is p. 289 of Shreve's Stochastic calculus for finance
\begin{align} c(T, s, v) &=(s-K)^{+} \text {for all } s \geq 0, v \geq 0 \\ c(t, 0, v) &=0 \text { for all } 0 \leq t \leq T, v \geq 0 \\ c(t, s, 0) &=\left(s-e^{-r(T-t)} K\right)^{+} \text {for all } 0 \leq t \leq T, s \geq 0 \\ \lim _{s \rightarrow \infty} \frac{c(t, s, v)}{s-K} &=1 \text { for all } 0 \leq t \leq T, v \geq 0 \\ \lim _{v \rightarrow \infty} c(t, s, v) &=s \text { for all } 0 \leq t \leq T, s \geq 0 \end{align}
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A bit of background for the heston's stochastic volatility model
$$ d S(t)=r S(t) d t+\sqrt{V(t)} S(t) d \widetilde{W}_{1}(t) $$
$$ d V(t)=(a-b V(t)) d t+\sigma \sqrt{V(t)} d \widetilde{W}_{2}(t) $$
There is a function c that satisfies
$$ c(t, S(t), V(t))=\widetilde{\mathbb{E}}\left[e^{-r(T-t)}(S(T)-K)^{+} \mid \mathcal{F}(t)\right], \quad 0 \leq t \leq T $$
subject to
$$ c_{t}+r s c_{s}+(a-b v) c_{v}+\frac{1}{2} s^{2} v c_{s s}+\rho \sigma s v c_{s v}+\frac{1}{2} \sigma^{2} v c_{v v}=r c $$
I eventually show that
$c(t, s, v)=s \mathbb{E}^{t, x, v} \mathbb{I}_{\{X(T) \geq \log K\}}-e^{-r(T-t)} K \mathbb{E}^{t, x, v} \mathbb{I}_{\{X(T) \geq \log K\}}$
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I'm trying to derive the boundary conditions
$c(T, s, v) =(s-K)^{+} \text {for all } s \geq 0, v \geq 0$ comes from evaluating at $t = T$
$c(t, 0, v) =0 \text { for all } 0 \leq t \leq T, v \geq 0$ Heuristically it's 0 as the strike price will always be greater than the stock price. If I evaluate the PDE at $s= 0$, I get $c_t + (a-bv)c_v + \frac{1}{2}\sigma^2vc_{vv} = rc$, which looks like the black-scholes pde, not sure where I can go after this though.
$\lim _{s \rightarrow \infty} \frac{c(t, s, v)}{s-K} =1 \text { for all } 0 \leq t \leq T, v \geq 0 $, I understand that as the stock price is approaches infinity, it will almost surely finish above the strike price. Taking the limit solves this.
$\lim _{v \rightarrow \infty} c(t, s, v) =s \text { for all } 0 \leq t \leq T, s \geq 0$ confuses me completely.
Thank you.