Fundamental system of solutions and Green`s function

Question

Suppose we have the fundamental system of solutions of equation $$y''(x)+(u_0x+u_1)y(x)=f(x)$$ with initial conditions $$y(0)=y_0,~ y'(0)=y_1.$$ Its fundamental system of solutions is ${\rm Ai}\left(-\frac{u_0x+u_1}{(-u_0)^{2/3}}\right)$ and ${\rm Bi}\left(-\frac{u_0x+u_1}{(-u_0)^{2/3}}\right)$ (Airy functions). How to obtain Green`s function of the problem?

Answer 1

The Green's function $g(x,\xi)$ should satisfy the ODE, $$ g(x,\xi)''+(u_{0}x+u_{1})g(x, \xi)=\delta(x-\xi) $$ with boundary conditions $g(0, \xi)=0$ and $g'(0, \xi)=0$. Then the Green's function $g(x, \xi)$ is given by $$g(x, \xi)=H(x-\xi)u_{\xi}(x)$$ where $u_{\xi}(x)$ satisfies $$u_{\xi}''(x)+(u_{0}x+u_{1})u_{\xi}(x)=0$$ with boundary conditions $u_{\xi}(\xi)=0$ and $u_{\xi}'(\xi)=1$. Suppose that $$u_{\xi}(x)=\alpha\operatorname*{Ai}\left(-\frac{u_{0}x+u_{1}}{(-u_{0})^{2/3}}\right)+\beta\operatorname*{Bi}\left(-\frac{u_{0}x+u_{1}}{(-u_{0})^{2/3}}\right)$$ then we have $$u_{\xi}(\xi)=\alpha\operatorname*{Ai}\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right)+\beta\operatorname*{Bi}\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right)=0$$and $$u_{\xi}'(\xi)=-\alpha u_{0}{\operatorname*{Ai}}'\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right)-\beta u_{0}{\operatorname*{Bi}}'\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right)=1$$ then we can solve the above two equations and get $$ \alpha=\frac{\pi}{u_{0}}\operatorname*{Bi}\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right), \qquad \beta=-\frac{\pi}{u_{0}}\operatorname*{Ai}\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right) $$ note that the Wronkian for Airy functions are $$W(\operatorname*{Ai}, \operatorname*{Bi};x)=\operatorname*{Ai}(x){\operatorname*{Bi}}'(x)-{\operatorname*{Ai}}'(x)\operatorname*{Bi}(x)=\frac{1}{\pi}$$ Thus the solution $u_{\xi}(x)$ can be represented using Airy functions as $$u_{\xi}(x)=\frac{\pi}{u_{0}}\operatorname*{Bi}\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right)\operatorname*{Ai}\left(-\frac{u_{0}x+u_{1}}{(-u_{0})^{2/3}}\right)-\frac{\pi}{u_{0}}\operatorname*{Ai}\left(-\frac{u_{0}\xi+u_{1}}{(-u_{0})^{2/3}}\right)\operatorname*{Bi}\left(-\frac{u_{0}x+u_{1}}{(-u_{0})^{2/3}}\right)$$ and finaly $$g(x, \xi)=H(x-\xi)u_{\xi}(x)$$ with $u_{\xi}(x)$ given as above.

the algorithm is from the book Green's Functions and Boundary Value Problems, by Ivar Stakgold and Michael J. Holst