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?

(ps. there is no Maple 2018 product to select, so I selected Maple 2017 from the menu).

The new UNTIL keyword added to Maple 2018 seems to me to be confusing and a useless addition to the Maple language.

First of all, using it in a DO loop, one does not use an closing END DO as normal. This makes it harder to use, since one is used to seeing an END DO to close each DO. They align physically better, which makes the code easier to read.

In addition, I can't think of anything that can't be done using current language constructs that needs this new keyword. Can someone? 

Adding a whole new keyword to make one type maybe few less characters seems like  a bad language design. A language constructs should be orthogonal to each others for the language to remain simple and clean. Adding more keywords for the fun of it should be avoided if something can be written using exisiting language constructs.

n := 37:
do
  n := n + 1
until isprime(n);
n;

very confusing to read. It has no end do. At first, I did not see where the loop ends. And the above can be done in many other ways allready, one direct way could be

n := 37:
do
  n := n + 1;
  if isprime(n) then
     break;
  fi;
od;
n;

And it is more clear as well since the DO is aligned with the END DO and I find the logic clearer with no new keyword added. Another way could be

n:= 37:
found_prime := false:
while not found_prime do
   n:= n + 1;
   found_prime := isprime(n);
od:
n;

I am sure there are other ways to do this.

Could someone please show an example where UNTIL is really needed and will make the code much more clear that justifies adding it to the language?