@ChrisK You're welcome & thanks for the bounty. @C.E. Thanks for the support. There are five upvotes now, FWIW. Partly it competes with mathematica.stackexchange.com/q/28337, which was posted much earlier.

Hi everyone! I'm not able to color the stuffa properly. My idea was that there two eigenvalues function of(x,y) but may look four but they are just two due to periodicty. I just need two colors to color them. I did that but they kind of mixing

(to make it more clear, you can think the box to be periodic, means all opposite faces are the same)

Just a small question from a newbie: why does ContourPlot[y == (k x /. k -> 1), {x, -1, 1}, {y, -1, 1}] work only with parens around the substitution? Why does it matter to the engine what to apply the substitution to?

@monk-time, probably something to do with evaluation order, as Plot and related functions have HoldAll, but seems surprising to me.

@Shamina Might be best to open a regular question.

@monk-time For what it's worth, you can use With to inject values into held expressions: With[{k = 1}, ContourPlot[y == k x, {x, -1, 1}, {y, -1, 1}]]. This works reliably, and it is apparent that it will work. It is not clear from your code what's going to happen because y == (k x /. k -> 1) is passed to the function unevaluated, and we don't know what the code on the inside looks like.

@monk-time The Head of y == (k x /. k -> 1) is Equal, which ContourPlot interprets as an equation to be plotted. The Head of y == k x /. k -> 1 is ReplaceAll, which ContourPlot interprets as a numeric function whose contour lines are to be plotted; of course, it's a boolean function that returns False almost all the time, so ContourPlot finds no contour lines to be plotted. It's designed to ignore parts of a domain in which the expression is not numeric.

You could use ContourPlot[Evaluate[y == k x /. k -> 1], {x, -1, 1}, {y, -1, 1}] to force the first argument to be evaluated before being passed to ContourPlot; then the expression passed has the head Equal.

good points

@C.E., @MichaelE2: Thank you both, that was very helpful! I really need to look into internals deeper to get a better understanding of what gets evaluated and when.

P.S. I'm not exactly young, but I still get blown away by the fact that one can just hop on the web and get an expert reply almost instantly (and for free!). It's just so cool.

@Nasser If I remember correctly you have an experience with older versions, right? mathematica.stackexchange.com/q/190205/5478