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Consider this initial-boundary-value problem: $$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2},\quad \Omega=(0,1)\times(0,1),\quad t>0.$$ $$u(x,0,t)=1,u(x,1,t)=0,u_x(0,y,t)=u_x(1,y,t)=0,\quad t>0$$ $$u(x,y,0)=0,\quad (x,y)\in\Omega.$$ For this code

f[t_] := Piecewise[{{0, t <= 0}, {1, t > 0}}]
NDSolveValue[{D[u[x, y, t], t] == 
   D[u[x, y, t], x, x] + D[u[x, y, t], y, y], u[x, 0, t] == f[t], 
  u[x, 1, t] == 0, Derivative[1, 0, 0][u][0, y, t] == 0, 
  Derivative[1, 0, 0][u][1, y, t] == 0, u[x, y, 0] == 0}, u, {x, 0, 
  1}, {y, 0, 1}, {t, 0, 10}]

I got error message

Encountered non-numerical value for a derivative at t == 0

Piecewise function works well on my other code. I don't know how to fix this one.

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  • $\begingroup$ Since you are only solving for t >= 0 anyways, why do you need the discontinuity in the time derivative? $\endgroup$
    – Marius Ladegård Meyer
    Mar 5, 2016 at 23:04
  • $\begingroup$ To avoid conflict with initial condition $u(x,y,0)=0$ $\endgroup$
    – Sukan
    Mar 5, 2016 at 23:12
  • $\begingroup$ I mean, does it make sense that $u$ jumps from 0 to 1 along $y=0$ as $t$ passes $t=0$...? Should it not be a derivative that makes this jump for instance? $\endgroup$
    – Marius Ladegård Meyer
    Mar 5, 2016 at 23:22
  • $\begingroup$ My best answer is: I cannot change the problem $\endgroup$
    – Sukan
    Mar 5, 2016 at 23:24
  • $\begingroup$ Ok. Is the second condition on $u$ supposed to be $u(x,1,t)=0$ instead of $u(0,1,t)$? $\endgroup$
    – Marius Ladegård Meyer
    Mar 5, 2016 at 23:28

1 Answer 1

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The problem is actually 1-dimensional. Hence let us drop the dependence of u on x.

Also we choose a slightly more smooth function instead of the jump

f[t_, a_] := (a t)/(a t + 1)

The parameter "a" gives the slope at t=0 which can be chosen arbitrarily steep.

The problem is then numerically solved by

With[{a = 10}, 
 soly = NDSolveValue[{Derivative[0, 1][u][y, t] == 
     Derivative[2, 0][u][y, t], u[0, t] == f[t, a], u[1, t] == 0, 
    u[y, 0] == 0}, u, 
       {y, 0, 1}, {t, 0, 10}]];

Plotting it

Plot3D[soly[y, t], {y, 0, 1}, {t, 0, 5}, AxesLabel -> {"y", "t"}]

enter image description here

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