I have a doubt regarding the signal of the dissipative term in a finite difference solution for an equation of the form
$$ \frac{\partial u}{\partial x}+f(x)=0, u(0)=0 $$
In which $f$ is an odd function and the solution (judging by the modelled phoenomena) should be an even function in a symmetric domain (say, $D=[-1; 1]$).
If I'm using the finite difference approximation
$$ \frac{\partial u}{\partial x}=\Delta_h u_i=\frac{u_i-u_{i-1}}{x_i-x_{i-1}} $$
The second order Taylor series decomposition of $u_i$ and $u_{i-1}$ results in a dissipative term proportional to $\Delta x \frac{\partial^2 u}{\partial x^2}$. Thus I'm trying to solve the equation by defining the residual as
$$ r_i=\Delta_hu_i+f(x_i)+\alpha \Delta^2_hu_i \times \Delta x $$
With $\alpha$ being a hyperparameter by which I can regulate how much artificial dissipation I want to apply.
I'm confused about the parity of the artificial dissipation term, however. Without it, the residual is an odd function in $D$; however, if $u$ is an even function as I said above, $\Delta x \frac{\partial^2 u}{\partial x^2}$ should also be one: therefore after it is added to the residual function to account for dissipation, the residual will not be odd when provided with an even $u$. In my specific case, this leads to $u$ converging to a physically impossible, uneven funcion after some Newton iterations when $\alpha\neq 0$.
Am I getting the idea of the dissipation term wrong?