In his TASI notes Oliver DeWolfe starts with the KG equation on the Poincaré patch metric $$ds^2=\frac{r^2}{L^2}(-dt^2+dx^2)+\frac{L^2}{r^2}dr^2.$$ When we use the ansatz$$\phi(r\rightarrow\infty,x,t)=\frac{\alpha(x,t)L^{2\Delta_-}}{r^{\Delta_-}}+\dots+\frac{\beta(x,t)L^{2\Delta_+}}{r^{\Delta_+}}+\dots$$ on the KG equation $$\left(-\frac{1}{\sqrt{-g}}\partial_\mu\sqrt{-g}g^{\mu\nu}\partial_\nu+m^2\right)\phi=0$$ we end up with an equation which is an ODE in r since the $$\partial_t^2$$,$$\partial_x^2$$ terms have a factor $${1}/{r^2}$$ which we can ignore.
The equation I end up with is $$-\frac{3r}{L^2}\partial_r\phi-\frac{r^2}{L^2}\partial_r^2\phi+m^2\phi=0.$$
When I substitute the ansatz I don't confirm that the equation is satisfied since I have a lot of terms with different powers of $$r$$. How can I show that this is indeed satisfied?
To solve equations of the form $$ar^2 \partial^2_{r}\phi + b r \partial_r\phi +c \phi=0$$ with constants $$a$$, $$b$$, $$c$$ (which is what you have) one sets $$\phi(r) =r^\lambda$$ so the equation becomes $$(a\lambda(\lambda-1) +b\lambda +c)r^\lambda=0.$$ Thus one solves the quadradic equation $$a\lambda(\lambda-1) +b\lambda +c=0$$ to get the two values of $$\lambda$$.
• The ansatz given is identical to mine! The $\Delta_{\pm}$ in the ansatz are just my $-\lambda$.I don't see what your problem is. When one includes the $1/r^2$ terms it is more complicated, but it's still the usual Frobenius indicial equation at a regular singular point. May 9 '20 at 12:56