Why is assume(...) do_something();  gives an error when I run the code one more time, but do_something() assuming ...; do not give an error when everything else is the same?

Is there semantic difference between the two forms? I thought they should work the same way. Here is an example

#in separate cell
restart;

#in separate cell
interface(showassumed=0):
pde := diff(u(x, t), t)=k*diff(u(x, t), x$2):
ic:=u(x,0)=0: bc:=u(0,t)=t:
assume(x>0);assume(t>0);assume(k>0):
sol:= pdsolve({pde,ic,bc},u(x,t)):

#in separate cell. Now this gives error
interface(showassumed=0):
pde := diff(u(x, t), t)=k*diff(u(x, t), x$2):
ic:=u(x,0)=0: bc:=u(0,t)=t:
assume(x>0);assume(t>0);assume(k>0):
sol:= pdsolve({pde,ic,bc},u(x,t)):
#error message now

Here is screen shot

Now will do the same, but use assuming. Now there is no error

#in one cell
restart;

#in one cell
interface(showassumed=0):
pde := diff(u(x, t), t)=k*diff(u(x, t), x$2):
ic:=u(x,0)=0: bc:=u(0,t)=t:
sol:= pdsolve({pde,ic,bc},u(x,t)) assuming x>0,t>0,k>0:

#in one cell, no error
interface(showassumed=0):
pde := diff(u(x, t), t)=k*diff(u(x, t), x$2):
ic:=u(x,0)=0: bc:=u(0,t)=t:
sol:= pdsolve({pde,ic,bc},u(x,t)) assuming x>0,t>0,k>0:

Here is screen shot

Why does one give an error, but the second one does not?

I thought they work the same way. Which method is recommended to use?

assume(...); do_something(); 

or 

do_something() assuming ...;

Maple 2018 on windows.

Could some Maple expert help me understand why pdsolve gives me this error message from trying to solve this Schrödinger pde and if there is a work around?

restart;
pde:=I*diff(f(x,t),t)=-diff(f(x,t),x$2)+2*x^2*f(x,t);
bc:=f(-infinity,t)=0,f(infinity,t)=0;
sol:=pdsolve([pde,bc],f(x,t));

I must be doing something wrong, but do not see it.

Mathematica solves the above as follows

pde=I D[f[x,t],t]==-D[f[x,t],{x,2}] + 2 x^2 f[x,t];
bc={f[-Infinity,t]==0,f[Infinity,t]==0};
sol=DSolve[{pde,bc},f[x,t],{x,t}]

thank you

Does any one know of a trick to make Maple solve this PDE using pdsolve?

restart;
pde:=diff(u(x,t),t)+diff(u(x,t),x)=0;
bc:=u(0,t)=0;
ic:=u(x,0)=x;
sol:=pdsolve({pde,ic,bc},u(x,t));

I have tried all HINTS and assumptions and Maple just returns () as solution.

Mathematica solve this as follows

ClearAll[u,x,t];
pde=D[u[x,t],t]+D[u[x,t],x]==0;
bc=u[0,t]==0;
ic=u[x,0]==x;
sol=DSolve[{pde,ic,bc},u[x,t],{x,t}]

Heavidside theta is basically a unit step function.

Maple 2018.

I am surprised Maple pdsolve can't solve this basic heat PDE. it is heat PDE on bar, with left end boundary condition being time dependent is only difference from basic heat PDE's on a bar.

May be a Maple expert can find a work around? I tried all the HINTS I know about.

restart;
#infolevel[pdsolve] := 3:
pde:=diff(u(x,t),t)=diff(u(x,t),x$2);
bc:=u(0,t)=t,u(Pi,t)=0:
ic:=u(x,0)=0:
sol:=pdsolve([pde,bc,ic],u(x,t)) assuming t>0 and x>0;

I also hope this question of mine do not get deleted as well, like the question I posted last night asking why pdsolve ignores assumptions that showed number of examples, was deleted few hrs after I posted it. 

If this question gets deleted, I will get the message that posts showing any problem in Maple software are not welcome here by Maplesoft and I will stop coming here.

This is using Maple 2018.

I noticed when solving Laplace PDE on disk, that no condition is needed to tell Maple if one is asking for solution inside the disk or outside. So how does Maple know which one it is?

It turned out Maple gives the same solution for the PDE outside as inside, which is wrong.

The solution to Laplace PDE inside a disk of radius 1 is

Which Maple gives correctly. But it also give the same above solution for outside the disk. The solution outside the disk should have r^(-n) and not r(n).  Like this

Basically Maple ignores assumption on `r`. This is what I tried

restart;
pde := (diff(r*(diff(u(r, theta), r)), r))/r+(diff(u(r, theta), theta, theta))/r^2 = 0:
bc := u(1, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi):
sol_inside:=pdsolve([pde, bc], u(r, theta), HINT = boundedseries) assuming r<1;

restart;
pde := (diff(r*(diff(u(r, theta), r)), r))/r+(diff(u(r, theta), theta, theta))/r^2 = 0:
bc := u(1, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi):
sol_outside:=pdsolve([pde, bc], u(r, theta)) assuming r>1;

Both the above gives the same answer.

How then does one solve the Laplace PDE outside the disk using Maple? And why is assumptions on "r" seems to be ignored in the above?