How do I expand absolute values?

Question

If we have this expression:

$$f = uu-\left( u + \frac{\partial u}{\partial x} \delta x \right) \left( u + \frac{\partial u}{\partial x} \delta x \right)$$

we can expand it to this:

$$f = u^2-\left( u^2 + 2u\frac{\partial u}{\partial x} \delta x + \left( \frac{\partial u}{\partial x} \delta x \right)^2 \right) $$

if we assume that $\delta x$ is very small, then the last term is negligible and the expression becomes simply:

$$f = - 2u\frac{\partial u}{\partial x} \delta x$$

Now assume that part of the expression is really an absolute value. If we return to the original expression, it becomes:

$$f = u|u|-\left( u + \frac{\partial u}{\partial x} \delta x \right) \left(\bigg| u + \frac{\partial u}{\partial x} \delta x \bigg| \right)$$

Can this be simplified any further?

Answer 1

After consideration I think I understand how to proceed. We have four different cases:

Case 1: $~~u < 0~~$ and $~~u + \frac{\partial u}{\partial x} \,\delta x < 0$

Case 2: $~~u > 0~~$ and $~~u + \frac{\partial u}{\partial x} \,\delta x > 0$

Case 3: $~~u < 0~~$ and $~~u + \frac{\partial u}{\partial x} \,\delta x > 0$

Case 4: $~~u > 0~~$ and $~~u + \frac{\partial u}{\partial x} \,\delta x < 0$

If we assume that $u \gg \frac{\partial u}{\partial x} \, \delta x$, then we can ignore cases (3) and (4), so:

Case 1:

\begin{align} f_{u<0} & = u (-u) - \left( u + \frac{\partial u}{\partial x} \, \delta x \right) \left( -u - \frac{\partial u}{\partial x} \, \delta x \right) \nonumber \\[1ex] & = -u^2 - \left( -u^2 - 2\frac{\partial u}{\partial x} \, \delta x - \left( \frac{\partial u}{\partial x} \, \delta x \right)^2 \right) \nonumber \\[1ex] & = 2\frac{\partial u}{\partial x} \, \delta x \nonumber \end{align}

Case 2:

\begin{align} f_{u>0} & = u u - \left( u + \frac{\partial u}{\partial x} \delta x \right) \left( u + \frac{\partial u}{\partial x} \, \delta x \right) \nonumber \\[1ex] & = u^2 - \left( u^2 + 2\frac{\partial u}{\partial x} \, \delta x + \left( \frac{\partial u}{\partial x} \, \delta x \right)^2 \right) \nonumber \\[1ex] & = -2\frac{\partial u}{\partial x} \, \delta x \nonumber \end{align}

These cases can now be combined together:

$$f = \text{sgn}(-u)~2\frac{\partial u}{\partial x} \, \delta x$$