I don't understand how to correctly calculate truncation error

Question

I am looking at the finite difference methods to solve simple $u_t=a(x,t)u_{xx}$.

There are explicit, implicit, Crank Nicolson.
The latter is said to be more accurate since the local truncation error is of second order provided all expansions are done around point $t^{n+0.5}$. However, local truncation error basically tells us how well the difference equation approximates the p.d.e. Thus, is I do expansion of CN scheme around any other point than I don't have second order anymore.
Question1 How I can trust the results of the scheme if there is only a single point where second order occurs?

On the other hand if I have explicit scheme, regardless around which point I do Taylor I keep having first order, so I get why it is first order but what I don't understand is:
Question2 Why is CN supposed to give global error at the point on the grid of second order when the local error of second order is estimated at the point which is not even on the mesh?

Thanks!
EDIT: Any scheme can be written as $\frac{u^{n+1}-u^n}{\tau}=L_hu^{\theta}$. So, regardless for which $\theta$ we have RHS=$u_{xx}(x_i,t^{\theta})+(h^2)...$ So it really boils down to approximation of the first derivative on the left hand side. And here we have choice. We can take $\theta=1$ and have implicit scheme with a derivative approximated at the point $t^{n+1}$, we can take the midpoint and have a derivative being approximated at that point with higher accuracy. However, the point is not on the grid! What I don't understand is that we expand around the point which is not on the grid but measure the error at the point which is on the grid and where the expansion of the difference equation gives only first order. I am confused about it.

Answer 1

I will simplify and take $a$ to be constant. We know that the scheme can be written \begin{eqnarray*} u_{i}^{n+1} & = & u_{i}^{n}+\frac{a\Delta t}{2}\left(F_{i}^{n+1}+F_{i}^{n}\right)\\ F_{i}^{k} & \approx & u_{xx}\left(x_{i},t_{k}\right) \end{eqnarray*} assuming that we can approximate $u_{xx}$ to high enough order (which is probably done using the usual three point discretization). We want to find $\left|u\left(x_{i},t_{n+1}\right)-u_{i}^{n+1}\right|$. We Taylor expand about a point on the grid \begin{eqnarray*} u\left(x_{i},t_{n+1}\right) & = & u_{i}^{n}+\Delta t\frac{\partial u}{\partial t}+\frac{\Delta t^{2}}{2}\frac{\partial^{2}u}{\partial t^{2}}+\frac{\Delta t^{2}}{6}\frac{\partial^{3}u}{\partial t^{3}}+\cdots\\ F\left(x_{i},t_{n+1}\right) & = & F_{i}^{n}+\Delta t\frac{\partial F}{\partial t}+\frac{\Delta t^{2}}{2}\frac{\partial^{2}F}{\partial t^{2}}+\frac{\Delta t^{2}}{6}\frac{\partial^{3}F}{\partial t^{3}}+\cdots \end{eqnarray*} then \begin{eqnarray*} u\left(x_{i},t_{n+1}\right)-u_{i}^{n+1} & = & u_{i}^{n}+\Delta t\frac{\partial u}{\partial t}+\frac{\Delta t^{2}}{2}\frac{\partial^{2}u}{\partial t^{2}}+\frac{\Delta t^{2}}{6}\frac{\partial^{3}u}{\partial t^{3}}+\cdots\\ & & -u_{i}^{n}+\frac{a\Delta t}{2}\left(2F_{i}^{n}+\Delta t\frac{\partial F}{\partial t}+\frac{\Delta t^{2}}{2}\frac{\partial^{2}F}{\partial t^{2}}+\frac{\Delta t^{2}}{6}\frac{\partial^{3}F}{\partial t^{3}}+\cdots\right)\\ & = & \Delta t\left[u_{t}+au_{xx}\right]_{i}^{n}+\frac{\Delta t^{2}}{2}\frac{\partial^{2}u}{\partial t^{2}}+\frac{\Delta t^{2}}{6}\frac{\partial^{3}u}{\partial t^{3}}+\cdots\\ & & +\frac{a\Delta t}{2}\left(\Delta t\frac{\partial F}{\partial t}+\frac{\Delta t^{2}}{2}\frac{\partial^{2}F}{\partial t^{2}}+\frac{\Delta t^{2}}{6}\frac{\partial^{3}F}{\partial t^{3}}+\cdots\right) \end{eqnarray*} So we have the $u_t + au_{xx}$ term which is of order $\Delta t^2$ since it is the PDE at the previous iteration and the other terms have $\Delta t^2$ at least.

Answer 2

The truncation error of a numerical scheme is a property of the method and (of course) is independent of how you do the analysis. If you believe you can show that the truncation error for Crank-Nicolson is first order at the grid points, then write your analysis here and we can explain where you're going wrong.