Here is the Link of the paper for more informations.
There are two questions I really got stucked,
(i) The equation (20) says that $\frac{1}{K(1-K)}$ is aprroximated by Taylor's series around $0.5$
$\begin{equation} T^{\gamma}_{n}(K) = \sum_{i=0}^{n} \gamma^{(i)}(0.5) \frac{\left(K - 0.5 \right)^i}{i!} = 4 \left(1 + \sum_{i=2}^{\bar{n}} (2K-1)^{2(i-1)} \right), \end{equation}$
where $\bar{n}$ is the ceiling part of $n/2$.
And according to the proof in the paper,
$\begin{equation} \exp\left(-\int_{0}^{K} T_{n}^{\gamma}(\zeta) \, d\zeta\right) \end{equation}$ approximates $\begin{equation} \exp\left(-\int_{0}^{K} \gamma(\zeta) \, d\zeta\right) \end{equation}$ according to the Assumption 3.1
can we REPLACE $T_{n}^{\gamma}(K)$ by $\gamma(K)$? and use $\gamma(K)$ in the exponential form for the following proof instead? My professor said "It's not equal, you couldn't replace it when you are doing a proof."
(ii) In this paper, it states that Under Assumption 3.1 the solution of the following non-autonomous Cauchy problem
$\begin{equation} \left\{ \begin{array}{l} \frac{\partial \psi(K, S)}{\partial K} = \alpha(K)S^2 \frac{\partial^2 \psi(K, S)}{\partial S^2} + \beta(K)S \frac{\partial \psi(K, S)}{\partial S} - \gamma(K)\psi(K, S),\quad (K, S) \in (0, 1) \times (0, S^{\ast}) \quad (22)\\ \psi(0, S) = \psi_0(S),\quad S \in (0, S^{\ast}) \end{array} \right. \end{equation} $
and they also let the transformation
$\begin{align} Y &= \ln(S) \quad \text{and} \quad \phi = \exp\left(-\int_0^K T_{n}^{\gamma}(\zeta) \, d\zeta\right) \cdot \psi \\ \end{align}$
The result after letting the transformations is
$\begin{equation} \frac{\partial \phi(K, Y)}{\partial K} = \alpha(K)Y^2 \frac{\partial^2 \phi(K, Y)}{\partial Y^2} + (\beta(K) - \alpha(K))Y \frac{\partial \phi(K, Y)}{\partial Y}. \end{equation}$
Here is where I am stucking, I've tried finding the result of $\psi$ from the above transformation, and here are what I got,
$\frac{\partial \phi}{\partial K} = \exp\left(-\int_0^K T_{n}^{\gamma}(\zeta) \, d\zeta\right)\left[ -T_{n}^{\gamma}(K)\psi(K,S) + \frac{\partial \psi}{\partial K} \right] $
$\frac{\partial \phi}{\partial Y} = \exp\left(-\int_0^K T_{n}^{\gamma}(\zeta) \, d\zeta\right) S \frac{\partial \psi}{\partial S} \quad \text{(using chain's rule)}$
$\frac{\partial^2 \phi}{\partial Y^2} = 2\frac{\partial \phi}{\partial Y} + S^2 \exp\left(-\int_0^K T_{n}^{\gamma}(\zeta) \, d\zeta\right) \frac{\partial^2 \psi}{\partial S^2} \quad \text{(using chain's rule)}.$
After replacing $\psi$ term in the PDE equation, The result is not even close.
I need all your help to guide me what I should do to transform the PDE from $\psi$ to $\phi$ form, if anyone wants to check my differentiation work, please kindly tell me, I will edit the post. Thank you in advance