@tomleslie 

Thank you again. I appreciate your help.

@tomleslie 

Thank you for your considerations.
It seems that I was confused last night.
Probably your first response was what I asked for.
I am currently without Maple, and will get back to you later.

@Rouben Rostamian  

Thank you.  The solution also makes sense from the physical point of view.  As you noticed, this is a system of mass transfer.  The emitted mass at boundary at x = L during time 0 and T equals to what is lost from the inside (the right-hand side of the equation).

@Takeo2801 

One thing I am wondering now is that from BC2,

-d*diff(C(L, t), x) = hm * C(L, t) / K.

But the left-side and right-hand sides do not seem to agree (plots p6 and p7).

This may not be a Maple question but I would also appreciate a quick help on this (maybe some formulation mistakes?).

MaplePrime20171006.txt

@tomleslie 

Thank you for the advices and suggestions.  I also tried including the target differential in the system, but it did not work.  Using fdiff in the first place also did not work. 

I never thought of using a fitted function for this purpose.  I followed your code and obtained results on my side.  It really helped, although I think it will take some time to digest what I learned.  My main interest was on diff(C(x, t)) at x = L, so I made a changes on the variable from your solution (e.g., v1:=pds:-value(x=L, t=0..T,output=listprocedure);.