Maybe it's time to leave the realm of wordy speculations and do some math ;-)).
The most simple example is the 1D wave equation, and so let's do the example of the string of length ##L## with fixed ends. The wave equation reads
$$(\partial_0^2-\partial_1^2) u(x^0,x^1)=0.$$
For simplicity I use relativistic notation, i.e., ##x^0=c t##, and the string is along the ##x^1## axis; ##\partial_{\mu}=\partial/\partial x^{\mu}##.
We look for the general solution of this equation with an arbitrary initial condition
$$u(0,x^1)=f(x^1), \quad \partial_0 u(0,x^1)=g(x^1)$$
with the boundary conditions
$$u(x^0,0)=u(x^0,L)=0.$$
To solve this we look for a complete set of eigenmodes, which we can use to expand our solution in terms of these eigenmodes.
To find these, we make the "separation ansatz"
$$u=A(x^0) B(x^1).$$
This gives
$$A''(x^0) B(x^1)-A(x^0) B''(x^1)=0 \; \Rightarrow \; A''(x^0)/A(x_0) =B''(x^1)/B(x^1).$$
Since the left-hand side is independent of ##x^1## and the right-hand side is independent of ##x^0## it must be a common constant, which we'll call ##-k^2##:
$$A''(x^0)=-k^2 A(x^0), \quad B''(x^1)=-k^2 B(x^1). \qquad (*)$$
The general solution of the second equation is
$$B(x^1)=C_1 \cos(k x^1) + C_2 \sin(k x^1).$$
The boundary conditions now say
$$B(0)=0 \; \Rightarrow \; C_1=0$$
and
$$B(L)=C_2 \sin(k L)=0.$$
Since we cannot set ##C_2=0##, because then we only get the trivial solution ##u=0##, we must have
$$\sin(k L)=0 \; \Rightarrow \; k \in \frac{\pi}{L} \mathbb{N}.$$
The general solution for the 2nd equation then is most conveniently written as
$$A(x^0)= A_1 \cos(k_n x^0) + A_2 \sin(k_n x^0).$$
The general solution then is a superposition of these eigensolutions
$$u(x^0,x^1)=\sum_{n=1}^{\infty} \sin(k_n x^1) [A_{1n} \cos(k_n x^0) + A_{2n} \sin(k_n x^0)].$$
The ##A_{1n}## and ##A_{2n}## can be found from the initial conditions:
$$u(0,x^1)=f(x^1)=\sum_{n=1}^{\infty} A_{1n} \sin(k_n x^1).$$
Now the modes are a complete set of orthogonal functions in the interval ##x^1 \in [0,L]##:
$$\int_0^L \mathrm{d} x^1 \sin(k_n x^1) \sin(k_m x^1)=\frac{L}{2} \delta_{mn}, \quad m,n \in \mathbb{N}.$$
Thus you get
$$A_{1n} = \frac{2}{L} \int_0^{L} \mathrm{d} x^1 f(x^1) \sin(k_n x).$$
The 2nd initial condition reads
$$\partial_0 u(0,x^1)=g(x^1)=\sum_{n=1}^{\infty} A_{2n} k_n \sin(k_n x^1),$$
leading to
$$A_{2n}=\frac{2}{L k_n}\int_0^{L} g(x^1) \sin(k_n x^1).$$
The general solution thus can be uniquely expanded in terms of the standing-wave modes
$$u_n(x^0,x^1)=\cos(k_n x^0) \sin(k_n x), \quad v_n(x^0,x^1)=\sin(k_n x^0) \sin(k_n x).$$
