I thought in Maple the standard was to use _C1, and _C2, etc... for constants in the solutions returned.

Sometimes Maple mixes _C1 and c[2] in the same result. Is this common, to be expected sometimes and is OK? I noticed this only recently. 

I was thinking may be some part of Maple code still was not updated to use _C1 notation? Here is an example

restart;
pde:=diff(u(x,t),t)+ diff( u(x,t),x )^3 + 6 * u(x,t)* diff(u(x,t),x) = 0;
sol:=pdsolve(pde,u(x,t));

which gives

sol := u(x, t) = -(3/2)*_C1^2+3*(t*_c[2]+x)*_C1-(3/2)*(t*_c[2]+x)^2-(1/6)*_c[2]

With latest Physics updates  268

Is there an option, like AllSolutions used with solve, so that pdsolve would return all solutions to a PDE when it is nonlinear?

I looked at pdsolve help and do not see a HINT that looks like might do this.

For example, this PDE, Maple returns one solution. But Mathematica returns 2 solutions

restart;
pde:= diff(u(x,t),t) = diff(u(x,t),x$5)+10*diff(u(x,t),x$3)*u(x,t)+25*diff(u(x,t),x$2)*diff(u(x,t),x)+
             20*u(x,t)^2*diff(u(x,t),x);
sol:=pdsolve(pde,u(x,t));

#sol := u(x, t) = -12*tanh(176*_C2^5*t+_C2*x+_C1)^2*_C2^2+8*_C2^2

But there is another solution

sol1:=u(x,t)=-(1/2)* _C1^2*(-2 + 3*tanh(x*_C1+ t*_C1^5 + _C2)^2)
pdetest(sol1,pde)
#0

Here is another example. Maple returns one solution and Mathematica 7 solutions

restart;
pde:= diff(u(x,t),t)= u(x,t)*(1-u(x,t))+ diff(u(x,t),x$2);
sol:=pdsolve(pde,u(x,t));

#sol := u(x, t) = (1/4)*tanh(-5*t*(1/12)+(1/12)*sqrt(6)*x+_C1)^2-
              (1/2)*tanh(-5*t*(1/12)+(1/12)*sqrt(6)*x+_C1)+1/4

But there are other solutions

pde = D[u[x, t], t] == u[x, t] (1 - u[x, t]) + D[u[x, t], {x, 2}];
DSolve[pde, u[x, t], {x, t}]

I've tested some (not all) of these 7 solutions in Maple using pdetest and Maple agrees they are solutions:

restart;
pde:= diff(u(x,t),t)= u(x,t)*(1-u(x,t))+ diff(u(x,t),x$2);
sol:=pdsolve(pde,u(x,t));
with(MmaTranslator);
sol2:=FromMma(`-(1/4) (-3 + Tanh[(5 t)/12 - (I x)/(2 Sqrt[6]) - C[3]]) (1 + 
   Tanh[(5 t)/12 - (I x)/(2 Sqrt[6]) - C[3]])`);
pdetest(u(x,t)=sol2,pde);
#0

I tried setting 

       _AllSolutions:=true

But it had no effect. Is there other options?

Maple 2018.2.1, using Physicsupdates 266.

I undertsand method=Fourier needs boundary conditions to work, but I do not think this error message is right. Compare

restart;
pde := diff(u(r, theta), r, r)+diff(u(r, theta), theta, theta) = 0;
iv := u(2, theta) = 3*sin(2*theta)+1;
pdsolve([pde,iv], u(r,theta), method = Fourier)

With

restart;
pde := diff(u(r, theta), r, r)+diff(u(r, theta), theta, theta) = 0;
pdsolve(pde, u(r,theta), method = Fourier)

Error, (in pdsolve/info) wrong extra arguments: {method = Fourier}
 

pdsolve should return no solution instead. The way it is above, I thought at first I had wrong syntax with the "method = Fourier" settings and I think this error message can be misleading to a user.
 

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.