$$ \newcommand{\uex}{{u_{\small\mbox{e}}}} \newcommand{\half}{\frac{1}{2}} \newcommand{\tp}{\thinspace .} \newcommand{\Oof}[1]{\mathcal{O}(#1)} \newcommand{\x}{\boldsymbol{x}} \newcommand{\dfc}{\alpha} % diffusion coefficient \newcommand{\Ix}{\mathcal{I}_x} \newcommand{\Iy}{\mathcal{I}_y} \newcommand{\If}{\mathcal{I}_s} % for FEM \newcommand{\Ifd}{{I_d}} % for FEM \newcommand{\basphi}{\varphi} \newcommand{\baspsi}{\psi} \newcommand{\refphi}{\tilde\basphi} \newcommand{\xno}[1]{x_{#1}} \newcommand{\dX}{\, \mathrm{d}X} \newcommand{\dx}{\, \mathrm{d}x} \newcommand{\ds}{\, \mathrm{d}s} $$

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How do we pick the right solution in this case?

Let's investigate the nature of the two roots: 

>>> import sympy as sp
>>> dt, u_1, u = sp.symbols('dt u_1 u')
>>> r1, r2 = sp.solve(dt*u**2 + (1-dt)*u - u_1, u)  # find roots
>>> r1
(dt - sqrt(dt**2 + 4*dt*u_1 - 2*dt + 1) - 1)/(2*dt)
>>> r2
(dt + sqrt(dt**2 + 4*dt*u_1 - 2*dt + 1) - 1)/(2*dt)
>>> print r1.series(dt, 0, 2)
-1/dt + 1 - u_1 + dt*(u_1**2 - u_1) + O(dt**2)
>>> print r2.series(dt, 0, 2)
u_1 + dt*(-u_1**2 + u_1) + O(dt**2)

The r1 root behaves as \( 1/\Delta t\rightarrow\infty \) as \( \Delta t\rightarrow 0 \)! Therefore, only the r2 root is of relevance.

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