@Joe Riel So I'm not sure how to set up the equtions and ibc's.  Say after 300s how would we do it?

Sorry to hear you won't be getting your down thumb badge, perhaps the administrators could honor it to you anyway.  I, for one, am glad to see the down thumb votes finally gone.  Ironically does the down thumb vote actually add a thumbs down to your favorites?  That would be funny.

It also appears the spam has stopped (yay .. a brief celebration ensues).  So there are two apparent silent fixes, NICE!  And the new mapleprimes news brief section should tell us what's going on ... when it's added.  

@Joe Riel That's what I thought I might have to do - model from a new point boundary. 

As a side question would this be the correct ibc's to model the bar at an initial temp all cooling down in air?
ibc := T(x, 0) = 200+273.15, D[1](T)(0, t) = 0,D[1](T)(2, t) = 0

@Joe Riel That's what I thought I might have to do - model from a new point boundary. 

As a side question would this be the correct ibc's to model the bar at an initial temp all cooling down in air?
ibc := T(x, 0) = 200+273.15, D[1](T)(0, t) = 0,D[1](T)(2, t) = 0

@JoeRiel @Robert Thank you both!

Suppose at some later time I take away the input heat at x=0? 

@JoeRiel @Robert Thank you both!

Suppose at some later time I take away the input heat at x=0? 

@JoeRiel @Robert Thank you both!

Suppose at some later time I take away the input heat at x=0. 

@JoeRiel @Robert Thank you both!

Suppose at some later time I take away the input heat at x=0. 

@Robert Israel thanks.  So does D[1](T)(2,t)=0 mean there that the rate of heat change there is zero?

One thing I find is that if I use a 2d plot at a specific time say t=100000, Maple plods through calculations up to that point before it presents me with the graph.  Shouldn't that be faster, I mean nearly instantaneous? 

And how can I get the value at a specific time and point on the rod, say at x=2 and t=10800 (3 hours)?  Skin burns at around 130 F or about 50 C (323 K) and after 3 hours, from the graph, it is roughly just below 330K probably still too hot to touch the end of the rod.  I couldn't make sense of the help pages to determine a value at a specific point in time. 

@Robert Israel thanks.  So does D[1](T)(2,t)=0 mean there that the rate of heat change there is zero?

One thing I find is that if I use a 2d plot at a specific time say t=100000, Maple plods through calculations up to that point before it presents me with the graph.  Shouldn't that be faster, I mean nearly instantaneous? 

And how can I get the value at a specific time and point on the rod, say at x=2 and t=10800 (3 hours)?  Skin burns at around 130 F or about 50 C (323 K) and after 3 hours, from the graph, it is roughly just below 330K probably still too hot to touch the end of the rod.  I couldn't make sense of the help pages to determine a value at a specific point in time. 

@Joe Riel Okay, then I should leave my pde as I had it.  And replace my D[1](T)(2,t)=T2 boundary condition with just T(2,t)=T2. ..... no I guess that doesn't work either as that condition holds that end at T2 throughout all of t.

hmmm ....

@Joe Riel Okay, then I should leave my pde as I had it.  And replace my D[1](T)(2,t)=T2 boundary condition with just T(2,t)=T2. ..... no I guess that doesn't work either as that condition holds that end at T2 throughout all of t.

hmmm ....

So does the cooling in air across the whole length of the rod still occur with that as a boundary condition at the end of the rod?  That was the reason I left the cooling term it in the main pde equation.  I just want to make sure it would still apply across the whole length of the bar as well (I think it does, but I'm not too sure).

So does the cooling in air across the whole length of the rod still occur with that as a boundary condition at the end of the rod?  That was the reason I left the cooling term it in the main pde equation.  I just want to make sure it would still apply across the whole length of the bar as well (I think it does, but I'm not too sure).

Wow!  Okay sorry for the errors, thanks for clearing that up. 

I understand why Maple picks the value at (1,0) to be 1.  Okay I see now, it just doesn't work for the temperature distribution function x that I chose.  A contradiction arises as I approach x=1.  I should have chose something like u(x,0)=x*(1-x) for the temperature distribution instead.