Solution from NDSolveValue contradicts the boundary condition

Question

I have a PDE problem, and here is my code:

solN = Module[{k = 1, A = 1, u0 = 1, l = 1, \[Beta] = 1, T = 10}, 
  NDSolveValue[{D[u[x, t], t] - k D[u[x, t], x, x] == 
     A Exp[-\[Beta] t] + NeumannValue[0, x == 0], u[l, t] == 0, 
    u[x, 0] == u0}, u, {x, 0, l}, {t, 0, T}, 
   Method -> {"PDEDiscretization" -> {"MethodOfLines", 
       "SpatialDiscretization" -> {"FiniteElement"}}}]]

Manipulate[Plot[{solN[x, t]}, {x, 0, 1}, PlotRange -> {0, 1.3}], {t, 0, 1}]

From the manipulation we can see that the solution contradicts the boundary condition at x==1, so how to fix that?

Edit: I know the initial conditions contradict the boundary conditions at point x==1,t==0, however this PDE has an analytical solution. The following codes show the analytical solution and its plot.

nsol = Compile[{x, t, n}, Module[{k = 1, A = 1, u0 = 1, l = 1, \[Beta] = 1}, 
   Sum[(Sqrt[2]*Cos[(Pi*x*(-1 + 2*i))/(2*l)]*((2*(-1)^i*Sqrt[2]*Sqrt[l]*u0)/(E^((k*Pi^2*t*(-1 + 2*i)^2)/(4*l^2))*(Pi - 2*Pi*i)) + 
       (8*(-1)^i*Sqrt[2]*A*(E^((-t)*\[Beta]) - E^(-((k*Pi^2*t*(1 - 2*i)^2)/(4*l^2))))*l^(5/2))/((-4*l^2*\[Beta] + k*Pi^2*(1 - 2*i)^2)*(Pi - 2*Pi*i))))/Sqrt[l], {i, 1, n}]]]
Manipulate[
 Plot[nsol[x, t, n], {x, 0, 1}, 
  PlotRange -> {0, 1.3}], {t, 0, 1}, {n, Range[100]}]

Answer 1

btw, you have this

        Warning: boundary and initial conditions are inconsistent.

You can change IC to use Piecewise to make BC and IC agree.

k = 1; A = 1; u0 = 1; L = 1; β = 1; T = 10;
solNFiniteElements = 
 NDSolveValue[{D[u[x, t], t] - k D[u[x, t], x, x] == 
    A Exp[-β t] + NeumannValue[0, x == 0],
   u[L, t] == 0,
   u[x, 0] == Piecewise[{{u0, 0 <= x < L}, {0, x == L}}]
   },
  u, {x, 0, L}, {t, 0, T},
  Method -> {"PDEDiscretization" -> {"MethodOfLines", 
      "SpatialDiscretization" -> {"FiniteElement"}}}
  ]

And now

Manipulate[
 Plot[solNFiniteElements[x, t0], {x, 0, 1}, PlotRange -> {0, 1.3}],
 {{t0, 0, "time"}, 0, 0.5, .01, Appearance -> "Labeled"},
 TrackedSymbols :> {t0}
 ]

Due to abrupt change in solution at t=0 from 1 to zero at x=L, solution at t=0 will not be smooth. But will be at any time after t=0