= Solution
Put
$$
A=\frac1{D^2}+\frac{ik}{2F},\qquad
q(x)=1+\frac{2iAx}{k}=1-\frac xF+\frac{2ix}{kD^2}.
$$
In free space, the <parabolic wave equation> is $E_x=iE_{zz}/(2k)$. Under the <Fourier transform> convention $\widehat E(p)=\int E(z)e^{-ipz}\,dz$, it becomes
$$
\partial_x\widehat E=-\frac{ip^2}{2k}\widehat E,\qquad
\widehat E(0,p)=\sqrt{\frac\pi A}\exp\left(-\frac{p^2}{4A}\right).
$$
The <Gaussian integral> is legitimate because $\operatorname{Re}A=1/D^2>0$. Invert the transform after multiplying by $e^{-ip^2x/(2k)}$. A second <Gaussian integral>, or equivalently <one-dimensional transverse Fresnel propagation>, gives
$$
\boxed{E(x,z)=q(x)^{-1/2}\exp\left(-\frac{Az^2}{q(x)}\right).}
$$
Choose the square-root branch continuously from $q(0)^{-1/2}=1$; for real $x\geq0$ there is no zero of $q$. This gives the correct incident field at $x=0$, and direct differentiation verifies the free <parabolic wave equation>.
For clarity, the squared envelope magnitude is
$$
|E(x,z)|^2=\frac1{|q(x)|}\exp\left(-\frac{2z^2}{D^2|q(x)|^2}\right),\qquad
D(x)=D\sqrt{(1-x/F)^2+\left(\frac{2x}{kD^2}\right)^2}.
$$
The <Gaussian beam with one transverse coordinate> remains Gaussian, with one-transverse-coordinate amplitude factor $q^{-1/2}$, not the $q^{-1}$ of a beam with two transverse coordinates. The negative initial quadratic phase produces focusing for $F>0$; <diffraction> prevents a singularity at $x=F$. These expressions describe the <paraxial approximation> to free propagation, rather than an exact unrestricted <Helmholtz equation> beam.
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