Consider the Laplace transform of a vectorial ($\in\mathbb{C}^3$) function $\vec{v}(x)\exp(i k_y y + i k_z z)$ $$ \mathcal{L}(\vec{v}(x)\exp(i k_y y + i k_z z))(s) \equiv \exp(i k_y y + i k_z z) \int_0^{\infty} \vec{v}(x) \exp(- s x)dx $$ This has two convenient properties
- There exists a $3\times 3$ matrix $M(s,k_y,k_z)$ that lets us express the Laplace transform of the curl of $\vec{v}$ in terms of the Laplace transform of $\vec{v}$: $$\mathcal{L}(\nabla\times \vec{v}(x)\exp(i k_y y + i k_z z))(s) = M(s,k_y,k_z) \mathcal{L}(\vec{v}(x)\exp(i k_y y + i k_z z))(s)$$
- There exists a function $f(x,a)$ such that $$\mathcal{L}(f(x,a)\vec{v}(x)\exp(i k_y y + i k_z z))(s) = \mathcal{L}(\vec{v}(x)\exp(i k_y y + i k_z z))(s+a)$$ ($f(x,a)=\exp(-a x)$ for the standard Laplace transform)
Let us now consider a vectorial function $\vec{v}(r)\exp(i m \theta + i k_z z)$ in cylindrical coordinates. Is there a "Laplace-like" transform w.r.t. $r$, that retains the above two properties?
The presence of factors $1/r$ in the cylindrical curl causes the standard Laplace transform to lose the first property.