My guess is that this works as long as the coeff b(x) is benign. 

Alan,
NY = 500 = NT using Roberts Weiss fdm (max err= 3,52e-5)
NY = NT = 1000, err = 1.1e-5
NY = NT = 2000, err = 3.21e-6

// copy from Excel not always works

X MOC exact solution X FDM solution
0 0.385821 0 0.385824
0.002 0.38582 0.002 0.385822
0.004 0.385818 0.004 0.385821
0.006 0.385816 0.006 0.385819
0.008 0.385814 0.008 0.385817
0.01 0.385812 0.01 0.385815
0.012 0.385811 0.012 0.385813
0.014 0.385809 0.014 0.385811
0.016 0.385807 0.016 0.385808
0.018 0.385805 0.018 0.385806
0.02 0.385803 0.02 0.385804
0.022 0.385802 0.022 0.385802
0.024 0.3858 0.024 0.3858
0.026 0.385798 0.026 0.385797
0.028 0.385796 0.028 0.385795
0.03 0.385794 0.03 0.385792
0.032 0.385792 0.032 0.38579
0.034 0.385791 0.034 0.385788
0.036 0.385789 0.036 0.385786
0.038 0.385787 0.038 0.385784
0.04 0.385785 0.04 0.385782
0.042 0.385783 0.042 0.385781
0.044 0.385781 0.044 0.385779

OK. Well at T=20 you are only examining the initial data from y=20/21 to y=1. So, next put a discontinuity halfway in-between at y=41/42. For example:

[$]f(y) = e^{-y} \times 1 \left(y > \frac{41}{42}\right) + 3 \times 1 \left(y \le \frac{41}{42}\right)[$], 
where [$]1(condition)=1[$] if the condition is true and 0 otherwise. 

Now your output (for your X=0 to 1) should start at 3 for about the first half and then jump to your previous results for the remainder of it. The exact [$]y[$]-jump point in the solution is where, at your T, [$]\xi = \frac{41}{42}[$] -- something like the solution to

(*) [$]\frac{20 - 19 y}{21 - 20 y} = \frac{41}{42}[$].

I really wish you hadn't switched from y to X, as it just makes it that much harder to discuss! Anyway, solve (*) and see if your solution jumps at that value of your X.

What's the pde you re looking at with these conditions?

So

[$]u(y,t)= 1 [$] when [$]|y-t| \le 1/3[$]

[$]u(y,t)= 0 [$] when [$]|y-t| \gt 1/3[$].

When you solve numerically does the discontinuity disappear, e.g. spurious diffusion?