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I am trying to solve the following one-dimensional problem:

(to better understand and extend the FEM for a more complex problem),

Needs["NDSolve`FEM`"]

f = 1000;
x0 = 1;
e = 2;
l = 100;
d = 10;

u[x_] = x0 Sin[(2*\[Pi]*f) * x];

ufun = NDSolveValue[{Laplacian[u[x], {x}] == (e/l) d u'[x],
u[0] == u[x], u[1] == 0}, u, {x, 0, 1}];


Plot[ufun[x], {x, 0, 1}]

The output is supposed to be an exponential function. @x=0 I am applying a sine wave and @x=1, the value is zero, (which I can't figure out how to do it). Any suggestions or recommendations?

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  • $\begingroup$ What do you mean by "@x=0 I am applying a sine wave"? Have you mixed up t and x? Can you show us the equation you're trying to solve with traditional math notation? $\endgroup$
    – xzczd ♦
    Commented Jan 12, 2023 at 13:27
  • $\begingroup$ @xzczd ∇^2 u=ε/(l d) (du/dt) , ε=constant value, l=100, d=1. so I am trying to get distribution of u(x) across the distance x. $\endgroup$
    – a019
    Commented Jan 12, 2023 at 13:41
  • $\begingroup$ …So you've mixed up t and x. The equation should be Laplacian[u[t, x], {x}] == e/(l d) D[u[t, x],t]. And this is a spatially 1D problem i.e. 1+1D problem. There're many related examples in document and this site, please read them carefully. $\endgroup$
    – xzczd ♦
    Commented Jan 12, 2023 at 13:47
  • $\begingroup$ @xzczd oh I see, if possible can you share any example or documents, so, I know that I am looking at a right example. $\endgroup$
    – a019
    Commented Jan 12, 2023 at 14:01
  • $\begingroup$ Reading examples in document of NDSolve and DSolve by pressing F1 should be quite enough for your problem. For more advanced introduction see e.g. PDEModels/tutorial/HeatTransfer/HeatTransfer. $\endgroup$
    – xzczd ♦
    Commented Jan 12, 2023 at 14:19

1 Answer 1

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Almost:

f = 1000;
x0 = 1;
e = 2;
l = 100;
d = 10;
uBC[x_] = x0 Sin[(2*\[Pi]*f)*x];
ufun = NDSolveValue[{Laplacian[u[x], {x}] == (e/l) d u'[x], 
    u[0] == uBC[0], u[1] == 1}, u, {x, 0, 1}];
Plot[ufun[x], {x, 0, 1}]

enter image description here

You'd need to rename your boundary condition function to uBC and evaluate that at 0. I also changed the second BC, as the first is also 0 at x=0 and that makes for an uninteresting solution.

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