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Question: Maple gives solution to PDE which gives 1/0 for first term in sum

Question:Maple gives solution to PDE which gives 1/0 for first term in sum

nm 8552 Maple 2019
pdsolve differential-equations
July 11 2019
1

fyi, there seems to be a problem here. Maple 2019, Physics version 395 on windows 10.

The solution given to this wave PDE by Maple as sum that starts from zero, has "n" in the denominator. When n=0, this gives division by zero.  Is this a bug?

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restart;
L:=3: c:=4: h:=1/10: b:=Pi*c/L:
f:=piecewise(0<=x and x<=L/3,3*h/L*x,L/3<x and x<=L,h):
pde := diff(u(x,t),t$2) + b*diff(u(x,t),t) = c^2*diff(u(x,t),x$2):
bc  := u(0,t)=0,D[1](u)(L,t)=0:
ic  := D[2](u)(x,0)=0,u(x,0)=f:
sol:=pdsolve([pde,bc,ic],u(x,t));
subs(n=0,sol)

u(x, t) = Sum(-((3/10)*I)*sin((1/6)*(1+2*n)*Pi*x)*(3^(1/2)*sin((1/3)*Pi*n)+cos((1/3)*Pi*n))*((2*I)*exp((4/3)*t*(I*n^(1/2)*(n+1)^(1/2)-1/2)*Pi)*n^(1/2)*(n+1)^(1/2)+(2*I)*exp(-(4/3)*(I*n^(1/2)*(n+1)^(1/2)+1/2)*t*Pi)*n^(1/2)*(n+1)^(1/2)+exp((4/3)*t*(I*n^(1/2)*(n+1)^(1/2)-1/2)*Pi)-exp(-(4/3)*(I*n^(1/2)*(n+1)^(1/2)+1/2)*t*Pi))/(n^(1/2)*(n+1)^(1/2)*Pi^2*(1+2*n)^2), n = 0 .. infinity)

Error, numeric exception: division by zero

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