We have an ODE on [a,b] as follows. \begin{equation}\left\{\begin{array}{l} -u^{\prime \prime}(x)+p(x) u^{\prime}(x)+q(x) u(x)=f(x), \quad x \in[a, b] \\ u(a)=u(b)=0 \end{array}\right.\end{equation}
I want to extend to [a,b] to the real line by a conform mapping $\psi(t)$.
For $l$ which is a non-negative real number, define the transformation of the variable $$ v(t)=\left(\left(\phi^{\prime}\right)^{l} u\right) \circ \psi(t)=\left(\phi^{\prime}(\psi(t))\right)^{l} u(\psi(t)) $$ A calculation by using the chain rule and $x=\psi(t)$ leads to the equalities $$ \begin{array}{l} u(x)=\frac{1}{\left(\phi^{\prime}(x)\right)^{l}} v(t) \\ \frac{\mathrm{d} u(x)}{\mathrm{d} x}=\left(\phi^{\prime}(x)\right)^{1-l} \frac{\mathrm{d} v}{\mathrm{d} t}+\left(\left(\phi^{\prime}(x)\right)^{-l}\right)^{\prime} v(t) \\ \frac{\mathrm{d}^{2} u(x)}{\mathrm{d} x^{2}}=\left(\phi^{\prime}(x)\right)^{2-l} \frac{\mathrm{d}^{2} v}{\mathrm{d} t^{2}}+(1-2 l)\left(\phi^{\prime}(x)\right)^{-l} \phi^{\prime \prime}(x) \frac{\mathrm{d} v}{\mathrm{d} t}+\left(\phi^{\prime}(x)^{-l}\right)^{\prime \prime} v(t) \end{array} $$
- Substituting the equations to an original ODE, we have a new ODE on $(-\infty,\infty)$ $$ \left(\phi^{\prime}(x)\right)^{2-l} \frac{\mathrm{d}^{2} v}{\mathrm{d} t^{2}}+(1-2 l)\left(\phi^{\prime}(x)\right)^{-l} \phi^{\prime \prime}(x) \frac{\mathrm{d} v}{\mathrm{d} t}+\left(\phi^{\prime}(x)^{-l}\right)^{\prime \prime} v(t)=f_{2}(x) $$ In addition, $v(t)$ satisfies the following conditions: $$ \lim _{t \rightarrow \pm \infty} v(t)=0 $$ which follow from $\psi(-\infty)=0, \psi(\infty)=1$
My questions:
In fact, I have a solution above, just I need some explanation for the following questions.
1- What is the logic behind selecting the $v(t)$ as follows $$ v(t)=\left(\left(\phi^{\prime}\right)^{l} u\right) \circ \psi(t) $$ and how to get the following $\left(\left(\phi^{\prime}\right)^{l} u\right) \circ \psi(t)=\left(\phi^{\prime}(\psi(t))\right)^{l} u(\psi(t))$
2- Could you explain more explicitly how to get $\frac{du(x)}{dx}$ and $\frac{d^2u(x)}{dx^2}$ by using chain rule?
Best regards,
from a Ph.D. student at mathematics.