Solving $2\frac{\partial S}{\partial z}+\left(\frac{\partial S}{\partial r}\right)^2=0$

Question

I have been trying to solve this PDE $$2\frac{\partial S}{\partial z}+\left(\frac{\partial S}{\partial r}\right)^2=0$$ The solution of this equation corresponding to a spherical wave of radius of curvature $R$ at $z=0$ is $$S=\frac{r^2}{2(z+R)}$$ I am unable to get to the solution. I have tried Charpit's method but that doesn't lead me to the required solution. Cylindrical coordinates have been used here. I encountered this equation paraxial approximation while doing self focusing of a gaussian beam.

Answer 1

Use separation of variables: $s(r,z)=f(r)g(z)$. Then $$2f(r)g'(z)+(f'(r)g(z))^2=0 \tag{1}$$ that you can rearrange as $$\frac{f'^2(r)}{f}+2\frac{g'(z)}{g^2}=0 \tag{2}.$$

Now you have the sum of a function that depends on $r$ and another function that depends on $z$ equal zero, hence each term separately must be equal a constant, say $k$ and $-k$. resp.

$$\frac{f'^2(r)}{f} =k \\ 2\frac{g'(z)}{g^2}=-k \tag{3}.$$ Now, these can be integrated separately. Since $\frac{g'(z)}{g^2}=-\frac{d}{dz}\frac{1}{g(z)}=-k$ you get $g(z)=\frac{1}{\tfrac{k}{2}z+\ell}$ where $\ell$ is another integration constant.

The other can also be integrated, write $\frac{f'(r)}{\sqrt{f}} =\pm\sqrt{k}$ but $\frac{f'(r)}{\sqrt{f}} = 2\frac{d}{dr} {\sqrt{f}}$, therefore $$\frac{f'(r)}{\sqrt{f}} = 2\frac{d}{dr}\sqrt{f}\\ =\pm\sqrt{k} \tag{4}$$ which you can immediately integrate as $\sqrt{f}= r\sqrt{\tfrac{k}{4}}+m$ or $$ f(r)=(r\sqrt{\tfrac{k}{4}} + m)^2 \tag{5}.$$ Here $m$ can have either sign. The values of $k,\ell,m$ are to be determined from the boundary conditions.