= Solution
Since $x=r\cos\theta$, the function on the left of the defining expansion is $e^{ix}$. Therefore
$$
(\nabla^2+1)e^{ir\cos\theta}
=(\partial_x^2+\partial_y^2+1)e^{ix}=0,
$$
which is the two-dimensional <Helmholtz equation>.
In polar coordinates,
$$
\nabla^2=\frac{\partial^2}{\partial r^2}
+\frac1r\frac\partial{\partial r}
+\frac1{r^2}\frac{\partial^2}{\partial\theta^2}.
$$
Substitute the defining <Fourier cosine series> and equate the coefficient of each $\cos(n\theta)$. Its angular second derivative contributes $-n^2$, so every <Bessel function> $J_n$ satisfies
$$
J_n''+\frac1rJ_n'+\left(1-\frac{n^2}{r^2}\right)J_n=0,
$$
or equivalently the <Bessel differential equation>
$$
\boxed{r^2J_n''+rJ_n'+(r^2-n^2)J_n=0}.
$$
Expand the exponential through cubic order and use
$$
\cos^2\theta=\frac{1+\cos2\theta}{2},
\qquad
\cos^3\theta=\frac{3\cos\theta+\cos3\theta}{4}.
$$
This gives
$$
e^{ir\cos\theta}
=1-\frac{r^2}{4}
+i\left(r-\frac{r^3}{8}\right)\cos\theta
-\frac{r^2}{4}\cos2\theta
-\frac{ir^3}{24}\cos3\theta+O(r^4).
$$
Comparing with $J_0+2\sum_{n\geq1}i^nJ_n\cos(n\theta)$ yields
$$
\boxed{J_0(r)=1-\frac{r^2}{4}+O(r^4)},
$$
$$
\boxed{J_1(r)=\frac r2-\frac{r^3}{16}+O(r^5),
\quad J_2(r)=\frac{r^2}{8}+O(r^4),
\quad J_3(r)=\frac{r^3}{48}+O(r^5)}.
$$
Back to article page
