Different answer using Dsolve or NDSolve to solve a PDE

Question

I'm working on getting a numerical solution for the following PDE: $$ u_t-u_{xx}=0,\ x\in[0,1], t\in[0,1],\\ u(x,0)=\sin(2\pi x),\\ u(0,t)=0,\\ u_x(1,t)=2\pi e^{-t}. $$ The hand solution for this PDE is $u(x,t)=e^{-t}\sin(2\pi x)$.

I tried DSolve to get the analytical solution.

DSolveValue[{D[u[x, t], {t, 1}] - D[u[x, t], x, x] == 0, 
  u[x, 0] == Sin[2 \[Pi] x], u[0, t] == 0, 
  Derivative[1, 0][u][1, t] == 2 \[Pi] E^-t}, u[x, t], {x, t}]

$$\underset{K[1]=1}{\overset{\infty }{\sum }}\sqrt{2} \sin \left(\frac{1}{2} \pi x (2 K[1]-1)\right) \left(-\frac{32 (-1)^{K[1]} \left(-\cosh (t)+e^{-\frac{1}{4} \pi ^2 t (1-2 K[1])^2}+\sinh (t)\right) \sqrt{2}}{\pi \left(4-\pi ^2 (1-2 K[1])^2\right) (1-2 K[1])^2}-\frac{128 (-1)^{K[1]} e^{-\frac{1}{4} \pi ^2 t (2 K[1]-1)^2} \sqrt{2}}{\pi (1-2 K[1])^2 \left(4 K[1]^2-4 K[1]-15\right)}\right)+2 \pi e^{-t} x$$

However, DSolve gives an answer containing Inactive[Sum].

I also tried NDSolve:

sol = NDSolveValue[{D[u[x, t], {t, 1}] - D[u[x, t], x, x] == 0, 
    u[x, 0] == Sin[2 \[Pi] x], 
    u[0, t] == 0, (D[u[x, t], x] /. x -> 1) == 2 \[Pi] E^-t}, 
   u[x, t], {x, 0, 1}, {t, 0, 1}, 
   Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"TensorProductGrid", 
       "MinPoints" -> 50}}];
Plot3D[{sol, E^-t Sin[2 \[Pi] x]}, {x, 0, 1}, {t, 0, 1}]

It turns out that the numerical solution differs from the analytical one.

I'm confused about this, does my code have any problem or am I missing something here?

Answer 1

I verified it. Your hand solution exp(-t)*sin(2*Pi*x) is wrong. It does not verify the pde at all. It only verifies the BC/IC

pde = D[u[x, t], t] - D[u[x, t], {x, 2}] == 0
ic = u[x, 0] == Sin[2 Pi x]
bc = {u[0, t] == 0, (D[u[x, t], x] /. x -> 1) == 2 Pi  Exp[-t] }
DSolve[{pde, ic, bc}, u[x, t], {x, t}]

Lets check your solution

solNeeded = u -> Function[{x, t}, Exp[-t]*Sin[2*Pi*x]]
pde /. solNeeded

You see, it is not an identity.

most solutions for pde's come out as Fourier series and so the sum there is what one would expect. (from applying separation of variables, that is what should come out).

Mathematica's solution also agrees with Maple's

pde:=diff(u(x,t),t)-diff(u(x,t),x$2)=0;
ic:=u(x,0)=sin(2*Pi*x);
bc:=u(0,t)=0,D[1](u)(1,t)=2*Pi*exp(-t);
pdsolve([pde,ic,bc])

does my code have any problem or am I missing something here?

Unless both Mathematica and Maple (the two most advanced CAS systems in the world) are both wrong and you are right, I would say the error is in your hand solution :)