A question on WKB method.

Question

If the differential equation is given by: $\epsilon^2y''+\epsilon xy'-y=-1$ where $y(0)=0$ and $y(1)=3$. I know we have to take $z=y-1$ in order to make the differential equation homogeneous but I' m struggling to solve it after. It would be great if someone could help me with this WKB approximation. Thanks

Answer 1

With the WKB method you approximate basis solutions of the homogeneous equation $ϵ^2y''+ϵxy'-y=0$. Using $y=\exp(S/\delta)$ you get the equation $$ ϵ^2(S''/δ+S'^2/δ^2)+ϵxS'/δ-1=0 $$ Using $δ=ϵ$ gives $$ ϵS''+S'^2+xS'-1=0 $$ Setting $s=S'$ and expanding $s=s_0+ϵs_1+...$ gives the first equation for $s_0$ as $$ (2s_0+x)^2=4+x^2\implies s_0=\frac12(-x\pm\sqrt{4+x^2}) $$ then in the next scale $$ (2s_0+x)s_1=-s_0'\implies s_1=\frac12\frac{\pm\sqrt{4+x^2}-x}{4+x^2} $$ and so on.

Next integrate to get the components of the exponents, combine to the solution and solve for the boundary conditions. With the anti-derivatives \begin{align} \int\sqrt{4+x^2}dx&=\frac x2\sqrt{4+x^2}+2\sinh^{-1}(\frac x2)\\ \int\frac{dx}{\sqrt{4+x^2}}&=2\sinh^{-1}(\frac x2)\\ \int \frac{x\,dx}{4+x^2}&=\frac12\ln(4+x^2) \end{align} we get \begin{align} y_1(x)&=\frac1{\sqrt[4]{4+x^2}}\exp\left(\frac{x(\sqrt{4+x^2}-x)+4\sinh^{-1}(\frac x2)}{4ϵ}+\sinh^{-1}(\frac x2)\right) \\[1em] y_2(x)&=\frac1{\sqrt[4]{4+x^2}}\exp\left(-\frac{x(\sqrt{4+x^2}+x)+4\sinh^{-1}(\frac x2)}{4ϵ}-\sinh^{-1}(\frac x2)\right) \\[1em] y(x)&\approx 1+Ay_1(x)+By_2(x) \end{align} The first constant $A$ has to be a multiple of $1/y_1(1)$ to get finite values for $ϵ\to 0$. Then the basis functions generate the boundary layers, $y_1$ at $x=1$ and $y_2$ at $x=0$, taking extremely small values outside their boundary layers. Thus $A=2/y_1(1)$ and $B=-1/y_2(0)$.

Implementing this with a boundary value solver shows good coincidence with this WKB approximation for sufficiently small $ϵ$.