Governing Equation

$ \def \half {\frac{1}{2}} \def \kwart {\frac{1}{4}} \def \OF {\quad \mbox{or} \quad} \def \hieruit {\quad \Longrightarrow \quad} $ Imagine a grid which is extremely fine. Actually, the grid is so fine that it can by no physical means be distinguished from a "true" continuum. Then it may be assumed that also the numerical solution at such a fine grid can by no means be distinguished from the analytical one. Imagine now that we use our Newton-Raphson MultiGrid Solver for coarsening, step by step, this immensely fine grid. Also suppose that, at some step, the matrix coefficients correspond with some persistent scheme. If this is the case, only once, then the matrix-coefficients will always be changed in such a way that the scheme persists through all of the coarser and finer grids of the solver. However, an "exact" solution will be obtained by using a persistent scheme at a immensely fine grid. Coarsening does not change the persistent scheme, nor does it change the "exact" solution, as it is specified at the points of the successive grids. These considerations lead to the following Corollary:
Even if we use a persistent scheme at a very coarse grid, the solution will always be nothing else but just a specification of the exact solution at the grid points. To put it in a simple and straightforward way: any one-dimensional persistent scheme will give rise to numerical solutions which are {\bf exact} at the grid points. Hence such a scheme in fact will be identical to an Exact or Analytical Scheme.
Consider again the finite difference equation which is associated with a tri-diagonal system of equations: $$ - b . T_{i-1} + T_i - a . T_{i+1} = 0 $$ Meanwhile, we have found explicit formulas for the coefficients $a$ and $b$: $$ b = \frac{e^{+P/2.dx}}{e^{+Q/2.dx} + e^{-Q/2.dx}} \qquad a = \frac{e^{-P/2.dx}}{e^{+Q/2.dx} + e^{-Q/2.dx}} $$ Let's proceed now with the F.D. equation, working towards a continuous version: $$ - e^{+P/2.dx} . T(x-dx) + \left( e^{+Q/2.dx} + e^{-Q/2.dx} \right) T(x) - e^{-P/2.dx} . T(x+dx) = 0 $$ The exponential functions are expanded into a Taylor series, while carefully preserving all terms of second order: $$ - \left[ 1 + P/2.dx + (P/2.dx)^2/2 \right] T(x-dx) $$ $$ + \left[ 2 + (Q/2.dx)^2 \right] T(x) $$ $$ - \left[ 1 - P/2.dx + (P/2.dx)^2/2 \right] T(x+dx) = 0 $$ Collect terms with respect to different powers of $(dx)$: $$ - \left[ T(x-dx) - 2.T(x) + T(x+dx) \right] \; + \: dx . P \left[ T(x+dx) - T(x-dx) \right] / 2 $$ $$ - \: dx^2 \left[(P/2)^2/2.T(x-dx)-(Q/2)^2.T(x)+(P/2)^2/2.T(x+dx) \right] = 0 $$ The next step is to divide everything by $(-dx^2)$ and to work out every term.
First (order) derivative: $$ \frac{T(x+dx) - T(x-dx)}{2 . dx} = \frac{dT}{dx} \quad \mbox{for} \quad dx \rightarrow 0 $$ Second (order) derivative: $$ \frac{ \left[T(x+dx) - T(x) \right] / dx - \left[T(x) - T(x-dx) \right] / dx }{dx} = $$ $$ \frac{ dT/dx |_{x + \half dx} - dT/dx |_{x - \half dx} }{dx} = \frac{d^2 T}{dx^2} \quad \mbox{for} \quad dx \rightarrow 0 $$ The second order derivative is also recognized in the term with $dx^2$ which, after division by $(-dx^2)$, has become: $$ \left[(P/2)^2/2.T(x-dx)-(Q/2)^2.T(x)+(P/2)^2/2.T(x+dx) \right] $$ $$ = (P/2)^2 \half \left[ T(x-dx) - 2.T(x) + T(x+dx) \right] + \left[ (P/2)^2 - (Q/2)^2 \right] T(x) = $$ $$ = (P/2)^2 \frac{dx^2}{2} \left[ \frac{T(x-dx)-2.T(x)+T(x+dx)}{dx^2} \right] + \left[ (P/2)^2 - (Q/2)^2 \right] T(x) $$ Yielding a more or less continuous representation: $$ \left[ 1 + \half (P/2.dx)^2 \right] \frac{d^2T}{dx^2} - P \frac{dT}{dx} + \left[ (P/2)^2 - (Q/2)^2 \right] T(x) = 0 $$ We have deliberately left in the term $(P/2.dx)^2$, though it will go to zero with $dx \rightarrow 0$. This term is commonly known as "false diffusion" and it should be emphasized here that such a false diffusion term is can not be neglected if one wants to preserve all terms of a second order approximation for the governing differential equation. (The term especially becomes important for large P\'echlet numbers $|P|$.)
It is noted that "false diffusion", as mentioned here, should be well distinguished from the false diffusion in the book by Patankar (1980). The term is used there in connection with directional dependence of convection in a 2-D grid, which is quite a different matter.
If we finally do $dx\rightarrow 0$, then we find for the governing differential equation: $$ \frac{d^2T}{dx^2} - P \frac{dT}{dx} + \kwart (P^2 - Q^2) T(x) = 0 $$ This linear ODE (Ordinary Differential Equation) can be solved by conventional or by less conventional mathematical means. Finding the characteristic equation belongs to the former category: $$ \lambda^2 - P.\lambda + \kwart (P^2 - Q^2) = 0 $$ Giving: $$ \left[ \lambda - \half (P + Q) \right] \left[ \lambda - \half (P - Q) \right] = 0 \quad \hieruit $$ $$ \lambda = \half (P + Q) \OF \lambda = \half (P - Q) $$ And the accompanying solution: $$ T(x) = \lambda . e^{(P+Q)/2.x} + \mu . e^{(P-Q)/2.x} $$ Which has been found at an earlier stage.
The fact that oscillating (complex) solutions cannot be found in this manner is quite remarkable. This is in agreement with the finding that the discriminant of the characteristic equation is always positive. Write the differential equation in the form: $$ A \frac{d^2T}{dx^2} + B \frac{dT}{dx} + C.T = 0 $$ Then we find for the discriminant of the characteristic equation: $$ (B/2A)^2 - (C/A) = (P/2)^2 - \kwart (P^2 - Q^2) = \kwart Q^2 $$ $$ \hieruit Q/2 = \sqrt{(B/2A)^2-(C/A)} $$ The dependence on $Q$ is also apparent in an expression that we have found for the discriminant of the accompanying tri-diagonal matrix: $$ 1 - 4.a.b = tanh^2(Q/2.dx) \hieruit \sqrt{1 - 4.a.b} = tanh\left[ \sqrt{(B/2A)^2-(C/A)}.dx \right] $$ Last but not least, the factor $P$ is found to be equal to: $$ P/2 = - B / 2.A $$