As a simpler example, consider the following
>> c=3;
>> kfunc=@(a,b) a*b-c;
The @(a,b) says that the expression to the right will act as a function with input arguments a and b while everything else in the expression (in particular, c) will be viewed as a constant. The values of the constant variables will be taken from their values in the workspace at the moment kfunc is defined. Thus, for example,
>> kfunc(2,4) %2*4-3=5
ans =
5
It is important to understand, though, that kfunc is not aware of any changes you later make to the constant stuff. Thus, for example, this doesn't result in different output:
>> c=5; kfunc(2,4)
ans =
5
If I want kfunc to be aware of the new value of c, I have to rebuild it:
>> c=5; kfunc=@(a,b) a*b-c;
>> kfunc(2,4) %2*4-5=3
ans =
3
With that in mind, it should be clear that the output of your function,
kFunc = @(region,state) k+(k*(Nu-1)*(state.u<crackT));
can only vary with the "state" input argument. The "region" input variable does not appear in the defining expresssion k+(k*(Nu-1)*(state.u<crackT)) and everything else in the expression is a constant from the point of view of kFunc.
