Help understanding how to apply IMEX methods to one-dimensional PDEs

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I need to compute a solution of the following PDE:

$$\frac{\partial u}{\partial t} + v \frac{\partial u}{\partial x} = 0$$

For didactic purposes, I need to use an IMEX method. The point is no one ever illustrated such methods to me, so I did some research on the web and on chatgpt and I got to some conclusions. I am here to ask if my reasoning is correct and if not (as it probably is) what I'm doing wrong.

What I learned is

  1. Take your PDE and solve one of the two derivatives explicitly. For this I can use, for instance, Heun's method. What I obtain is

$$\frac{\partial u}{\partial t} + v \frac{u_{ex}^{n+1} - u^n}{\Delta x} = 0$$

where $u_{ex}^{n+1}$ is the solution of Heun's method, i.e. the explicit step.

  1. Use your new explicit solution to resolve implicitly the other derivative. In other words, use Backward Euler (for instance) to resolve

$$\frac{u^{n+1}-u_{ex}^{n+1}}{\Delta t} + v \frac{u^{n+1}-u_{ex}^{n+1}}{\Delta x} = 0$$

where $u^{n+1}$ is the solution of the implicit method and of the whole time step.

  1. Iterate.

Is the reasoning right? I already don't think so because if I develop the last equation I get

\begin{align} \Delta x \cdot (u^{n+1}-u_{ex}^{n+1}) &= -v \Delta t \cdot (u^{n+1}-u_{ex}^{n+1}) \\ u^{n+1}(v\Delta t + \Delta x) &= u_{ex}^{n+1}(v\Delta t + \Delta x) \\ u^{n+1} &= u_{ex}^{n+1} \end{align}

which makes the implicit step useless.

So please, what's that I couldn't get right? It's ok even if you only point a book or a text where to learn.

EDIT:

As Matthew Cassell replied, I used the same discretisation both for time and space. Dividing the two different discretisations on the index $n$ for time and $k$ for space, I get

$$\frac{u_{k+1}^{n+1}-u_{k+1}^n}{\Delta t}+v\frac{u_{k+1}^n-u_k^n}{\Delta x}=0$$

where I first derived on $k$ (space) with the Heun's method (explicit) and then on $n$ (time) using Backward Euler (implicit). But now two more doubts arise:

  1. In the last equation, everything but $u_{k+1}^{n+1}$ is known, so I can easily develop the equation and get to $$u_{k+1}^{n+1}=u_{k+1}^n-v\Delta t\frac{u_{k+1}^n-u_k^n}{\Delta x}$$ So where's the implicit step? I expected an equation less trivial, in which something of uncertain nonsingularity would end at denominator; but I guess that since my implicit term $\left(\frac{\delta u}{\delta t}\right)$ is linear i don't get that. Good!

  2. In this way, I get $u_{k+1}^{n+1}$, which is the value at the next time step, but also at the next space step. Isn't it a problem that my solution moves forward in space, if I have no periodic boundary condition? In other words, how do I get $u_k^{n+1}$? But do I really need $u_k^{n+1}$?