Suppose $\tilde{u}(k,t)$ is the Fourier transform of a function $u(x,t)$.
Often, one makes the ansatz $$ \tilde{u}(k,t)=e^{-i\omega t}\tilde{u}(k),~~(*) $$ where $e^{-i\omega t}$ is an arbitrary expression for the time dependent part of $\tilde{u}(k,t)$.
I am trying to understand the idea behind this ansatz.
My conjecture is the following:
The Fourier operator $\mathcal{F}u\to\tilde{u}$ is often defined on $L^1(\mathbb{R}$), mapping into $L^1(\mathbb{R})$. So maybe the functions $e^{-i\omega t}$ are eigenfunctions of some operator $G$ on $L^1(\mathbb{R})$ and form an orthonormal basis of $L^1(\mathbb{R})$ with the consequence that the idea behind the ansatz is maybe that we can write each function in $L^1(\mathbb{R})$, e.g. the Fourier transform $\tilde{u}(k,t)=\mathcal{F}u(x,t)$ as some linear combination of the functions $e^{-i\omega t}$, in this case as exactly ONE $e^{-i\omega t}$?
Edit I should give more context!
Suppose we consider a function $u=u(x,t)$ defined on whole $\mathbb{R}$ and that, for each fixed $t$, can be expressed as Fourier integral: $$ u(x,t)=\int_{-\infty}^{+\infty}\, dk e^{ikx}\tilde{u}(k,x). $$
For the Fourier transform $\tilde{u}(k,x)$ we make the product-ansatz $$ \tilde{u}(k,t)=e^{-i\omega t}\tilde{u}(k) $$ as described above. That is, we try to write the solution as $$ u(x,t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\, dk e^{i(kx-\omega t)}\tilde{u}(k). $$ The functions $e^{i(kx-\omega t)}$ are called Fourier modes.
(By the way: I thought that functions $e^{i\omega t}$ are called Fourier modes?!)
So now again my question: Why is it "reasonable" to make the product ansatz for the Fourier transform $\tilde{u}(k,t)$?