I am faced with a problem that is more mathematical than electrodynamic. However, not having a clearer or shorter title available, I preferred to highlight where the problem came from. However, suppose we have the following function:
$$a(x,y,z)=\int _{z'=-l}^l\oint _{s'}J(s',z')\frac{e^{-jk\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}}{4\pi \sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}ds'dz'$$
where $k$ is a real constant and $J(s',z')$ is an unknown function to be determined. The integration is made with respect to the variables $x', y', z '$, on a cylindrical surface that develops in height with the variation of $ z' $ and with cross-section (the section involves $ x '$ and $ y '$) described by a particular path determined by the curvilinear coordinate $ s' $. Knowing that $ a (x, y, z) $, has the form written above, and knowing that it also satisfies the equation:
$$\frac{\partial^2 }{\partial z^2}a(x,y,z)+k^2 a(x,y,z)=f(z)$$
with $f(z)$ known function, can I argue in some way that $a(x,y,z) = a(z)$?
Bearing this in mind:
$\frac{\partial^2 a(t,z)}{\partial z^2}+ka(t,z)=g(z)\overbrace{\Rightarrow }^?a(t,z)=a(z)$
surely we must take advantage of the first equation, if it introduces some useful information, but I do not understand well how to do it.
The deduction I am trying to make is made in the text of Robert S. Elliott: Antenna Theory and Design (pag. 284 and 322). The author, using the fact that $a(x,y,z)=a(z)$, and supposing that $J(s',z')=f(s')I(z')$, writes the following equation:
$$a(z)=\left[\int_{z'=-l}^{z_0-b}+\int_{z'=z_0-b}^{z_0+b}+\int_{z'=z_0+b}^{l}\right]I(z')\oint_{s'}f(s')\frac{e^{-jkR}}{4\pi R}ds'dz'\approx$$ $$\approx \left[\int_{z'=-l}^{z_0-b}+\int_{z'=z_0+b}^{l}\right]I(z')\oint_{s'}f(s')\frac{e^{-jk|z-z'|}}{4\pi |z-z'|}ds'dz'+\int_{z'=z_0-b}^{z_0+b}I(z')\oint_{s'}\frac{f(s')}{4\pi R}ds'dz'$$
where $b$ is a small quantity and $R=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$. From this he deduces that:
$$\int_{z'=z_0-b}^{z_0+b}\oint_{s'}\frac{f(s')}{4\pi R}ds'dz' = \text{const}.$$
If one can show me how to arrive at this fundamental relation, even by other ways different than passing through the fact that $a(x,y,z) = a(z)$ I would be very grateful. This is a fundamental relation because from it one can prove the equivalence between strip and cylindrical dipole.
Take the equation $$ \frac{\partial^2 a(t,z)}{\partial z^2}+ka(t,z)=g(z) $$ We have $$\frac{\partial}{\partial t}\left(\frac{\partial^2 a(t,z)}{\partial z^2}+ka(t,z)\right)=\frac{\partial g(z)}{\partial t}=0$$ Thus if $\varphi(t,z)=\frac{\partial a(t,z)}{\partial t}$, we have using Schwarz's theorem (assuming the functions here are $\mathcal{C}^2$) $$ \frac{\partial^2 \varphi(t,z)}{\partial z^2}+k\varphi(t,z)=0 $$ We can easily solve it : $$ \varphi(t,z)=A(t)e^{\sqrt{k}z}+B(t)e^{-\sqrt{k}z} $$ assuming that $\sqrt{k}=i\sqrt{-k}$ where $k<0$ (we suppose that $k\neq 0$, otherwise it is easy to solve : $$a(t,z)=\left(\iint g\right)(z)+\alpha(t)z+\beta(t)$$ with any $\alpha(t),\beta(t)$). Since $A(t)$ and $B(t)$ are depending of $t$ we can't say anything, let us suppose $A(t)$ and $B(t)$ constant. Since $\varphi(t,z)=\frac{\partial a(t,z)}{\partial t}$, we can integrate this to get $$ a(t,z)=(Ae^{\sqrt{k}z}+Be^{-\sqrt{k}z})t+h(z) $$ Putting this in the inital equation gives $\frac{\partial^2 h}{\partial z^2}+kh(z)=0$ and thus $$ h(z)=Ce^{\sqrt{k}z}+De^{-\sqrt{k}z} $$ We can finally write $$a(t,z)=(At+C)e^{\sqrt{k}z}+(Bt+D)e^{-\sqrt{k}z}$$ The only way this doesn't depend of $t$ is to have $A=B=0$.
About your first equation, using what we said above, we get $$ \varphi(x,y,z)=A(x,y)e^{\sqrt{k}z}+B(x,y)e^{-\sqrt{k}z}. $$ Notice that since you integrate $z$ from $-\ell$ to $\ell$, you have $a(x,y,-z)=a(x,y,z)$ so that $A(x,y)=B(x,y)$ and $$\varphi(x,y,z)=A(x,y)(e^{\sqrt{k}z}+e^{-\sqrt{k}z})$$ Since you integrate $(x,y)$ on a disk $A(x,y)$, only depends of $r=\sqrt{x^2+y^2}$. I don't think that $A(r)$ doesn't depend of $r$ (or there is a good reason in physics but it would be restrictive), thus we can't say that $a(x,y,z)=a(z)$.