I've been using the following syntax to set boundary condition which is a derivative, when passing it to pdsolve. Say we want to set u(r,theta,t) to have insulated boundary conditions at r=1. So the BC will be

For example, to set derivative of u w.r.t. "r" to zero when r=1

       eval(  diff(u(r,theta,t),r), r=1) = 0;  #(1)

or using this syntax

       D[1](u)(1, theta, t) = 0;  #(2)

But now I find, on one example below, that the above no longer works.  I have to use this syntax (which I did not know about) for it to work

        D[1]*u(1, theta, t) = 0;  #(3)

Has something changed? why when using (3) pdsolve now gives result, but when using (2) or (1) it returns unevaluated? are the three semantically equivalent? when to use which syntax?

I am using Physics updates 265, Latest Maple 2018.2 

Here is an example showing the (1,2)  syntax no longer works, but the (3) syntax works

#articolo example 6.9.2
restart;

#using (1) syntax
pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r);
bc_on_r := eval(diff(u(r,theta,t),r), r=1) = 0;
bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0;
ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta);
pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))

does not solve it.

restart;

#using (2) syntax
pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r);
bc_on_r := D[1](u)(1, theta, t) = 0; 
bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0;
ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta);
pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))

does not solve it.

restart;

#using(3) syntax
pde := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r);
bc_on_r := D[1]*u(1,theta,t)=0;
bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0;
ic := u(r,theta,0)=(r-1/3*r^3)*sin(theta);
pdsolve([pde, bc_on_r,bc_on_theta,ic], u(r, theta, t), HINT = boundedseries(r = [0]))

I've used syntax (1) before many times and it works. Here is an example where all three syntax work

pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=eval(diff(u(x,t),x),x=0)=0,u(L,t)=0;
ic:=u(x,0)=f(x);
sol:=pdsolve([pde,bc,ic],u(x,t));

pdsolve gives

 Using syntax (2)

pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=D[1](u)(0,t)=0,u(L,t)=0;
ic:=u(x,0)=f(x);
sol:=pdsolve([pde,bc,ic],u(x,t));

gives same answer as (1) and using syntax (3)

pde:=diff(u(x,t),t)=k*diff(u(x,t),x$2);
bc:=D[1]*u(0,t)=0,u(L,t);
ic:=u(x,0)=f(x);
sol:=pdsolve([pde,bc,ic],u(x,t));

The answer also looks like different and simpler, but I assume they are equivalent for now without looking too much into it.

Which syntax should one use as now I am really confused.

It looks like (3) is the one that should be used? Why the others did not work on first example? i.e. pdsolve did not give an answer at all?  And if (1,2,3) syntax are supposed to be equivalent, whysecond example gives slightly different looking answer when using one syntax vs. the other?

And finally to make things more confusing, here is an example where syntax (3) does not work, but syntax 1 and 2 work:

#articolo example 8.4.3
restart;
pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t;
bc := u(0, t) = 5, (u(1, t)+ D[1](u)(1, t)) = 10;
ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5;
pdsolve([pde, bc,ic], u(x, t))

gives answer.

restart;
pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t;
bc := u(0, t) = 5, (u(1, t)+eval(diff(u(x,t),x),x=1))  = 10;
ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5;
pdsolve([pde, bc,ic], u(x, t))

gives same answer. But

restart;
pde := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t;
bc := u(0, t) = 5, (u(1, t)+ D[1]*u(1, t)) = 10;
ic:= u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5;
pdsolve([pde, bc,ic], u(x, t))

does not work.

Clearly there is something I do not understand between these 3 syntaxes and when to use which.

Using 265 version

Physics:-Version()
 "C:\Maple_updates\Physics+Updates.maple", 2018, December 22, 

    10:41 hours, version in the MapleCloud: 265, version 

    installed in this computer: not installed

downloaded today.

When I try to install (as an example) physics update using Maple 2018.2.1 on windows 10, it keeps hanging in the middle as shown above.

I closed Maple, starting again and tried again, same thing happens. I waited for more than 10 minutes and nothing happens.  I click on the install button in the clouds windows from Maple itself to install it as I always did.

Do others have problem installing this?  I'll try again in few hrs, it might be the Maple server is having some issues but thought to ask.

This is a Wave PDE inside disk. with fixed edge of disk,  and no theta dependency. initial position and velocity  given.

When using "c^2" for the  wave propagation speed, Maple only gives solution when also using assumptions to tell that c is positive.

restart; 
Physics:-Version()[2]; 
`2018, November 28, 1:35 hours, version in the MapleCloud: 224, version installed in this computer: 224`

And the PDE is 

pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2)  + (1/r)* diff(u(r,t),r)  ) ; 
ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; 
bc:=u(1,t)=0; 
sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0;

sol:=()

Now adding assuming c>0, or "c::positive", or "c>=0", it solves it 

restart;
pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2)  + (1/r)* diff(u(r,t),r)  ) ; 
ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; 
bc:=u(1,t)=0; 
sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0,c>=0;

Any idea why the assumption "c>0"  is needed when the speed is given as "c^2" ? I also tried assumption "c::real", but it still did not solve it.   It seems related to Maple using Laplace transform to solve it.

It seems to me the assumption c>0 should not needed here. But if it is, I'd like to learn why.

Maple 2018.2, windows 10

Update

Thanks to comment below

   " so u(r, 0)=1 (for all r), and u(1, t)=0 (for all t). So what is u(1,0)??? "

I should have added "r>0,r<1".  The reason I did not, is that I copied the code to solve this from Mathematica. In Mathematica, it did not need this assumption to solve it. It seems to depend on the method of solution used by Mathematica vs. Maple to cause this difference.

But now I see it helps to have it there. With the above assumption, now, there is no need to do any assumption on "c" at all and it gives solution. adding c>0 or t>0 or c>=0, makes no change now.

restart;
pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2)  + (1/r)* diff(u(r,t),r)  ) ; 
ic:=u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; 
bc:=u(1,t)=0; 
sol:=pdsolve([pde, ic,bc], u(r, t)) assuming t>0,r>0,r<1;

convert(...,Int) in Maple 2018.2 works for fourier, invfourier, laplace, but does not work for invlaplace.  

Why is that? Is there a workaround?

expr:=fourier(f(x), x, w):
convert(expr,Int);

expr:=invfourier(f(w), w, x):
convert(expr,Int);

expr:=laplace(f(s),s,t):
convert(expr,Int)

expr:=invlaplace(f(s),s,t):
convert(expr,Int)

Was expecting to see the Mellin's inverse formula.

Maple 2018.2 on windows 10

Using Latest Physics updates (I am not sure when this started), pdsolve gives Error, (in PDEtools/eval/2) numeric exception: division by zero on the following problem from a HW from text book.

restart;
PackageTools:-IsPackageInstalled("Physics Updates");
                             "220"


pde:=diff(w(x,t),t)+3*t*diff(w(x,t),x)=w(x,t);
ic:=w(x,0)=f(x);
sol:=pdsolve([pde,ic],w(x,t));

Error, (in PDEtools/eval/2) numeric exception: division by zero
 

Mathematica answer btw is 

pde = D[w[x, t], t] + 3 t D[w[x, t], x] == w[x, t];
ic = w[x, 0] == f[x];
sol = Simplify[DSolve[{pde, ic}, w[x, t], {x, t}]]

This is on Maple 2018.2 on windows 10 64 bit.

Any idea what is causing this and any workaround? Do others get the same exception?