= Solution
The coefficient changes sign where $1-\lambda^2/x^2=0$, so the single turning point is
$$
x_{\rm tp}=\lambda.
$$
Put $p(x)=(\lambda^2/x^2-1)^{1/2}$ and $q(x)=(1-\lambda^2/x^2)^{1/2}$. The <WKB approximation> satisfying the <boundary condition> at $x=1$ is
$$
y_1(x)\sim A p(x)^{-1/2}
\sin\left(\frac1\epsilon\int_1^x p(s)\,ds\right),
\qquad 1\leq x<\lambda,
$$
while decay at infinity selects
$$
y_2(x)\sim B q(x)^{-1/2}
\exp\left(-\frac1\epsilon\int_\lambda^xq(s)\,ds\right),
\qquad x>\lambda.
$$
Back to article page
