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Question: possible wrong solution by pdsolve

I solved this PDE by hand to verify Maple's solution. I think Maple solution is wrong. This PDE is the heat PDE on a bar (1D) with boundary coditions on both ends are function of time and zero initial conditions.

unassign('A,B,x,t,L,k,f');
pde := diff(u(x,t),t)= diff(u(x,t),x$2):
bc := u(0, t) = A(t), u(1, t) = B(t):
ic := u(x, 0) = 0:
sol1:=pdsolve([pde, ic, bc], u(x, t));

#now try when A(t)=sin(t),B(t)=t, use 20 terms for the sum
sol2:=simplify(subs([infinity=20,B(tau)=tau,A(tau)=sin(tau),A(0)=0,B(0)=0,A(t)=sin(t),B(t)=t],sol1)):
sol3:=simplify(value(subs(t=1,sol2))):
evalf(subs(x=0.5,sol3))

 

Also doing pdetest(sol1,pde); on the above solution does ot return zero as expected.

To verify more, I solved the same PDE again, but now using an explicit values for the boundary conditions A(t), B(t). Using A(t)=sin(t), B(t)=t. Then found the value again of the solution u at x=0.5 and t=1 like in the above, and it gives different value:

unassign('A,B,x,t,L,k,f');
pde := diff(u(x,t),t)= diff(u(x,t),x$2):
bc := u(0, t) = sin(t), u(1, t) = t:
ic := u(x, 0) = 0:
sol4:=pdsolve([pde, ic, bc], u(x, t));
sol5:=simplify(subs(infinity=20,sol4)):
sol6:=simplify(value(subs(t=1,sol5))):
evalf(subs(x=0.5,sol6))

Then I typed my hand solution into Maple and for the same values x=0.5, t=1 and same number of terms, I also get the same value 0.819. 

I do not see at all where the function sin integral should come into play in this solution. 

Could some Maple expert please check to see what is going on with this solution to Maple? 

Using Maple 2019.1 and Physics version 370

 

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