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Finite difference method for 1D heat equation?

Okkes Dulgerci

Posted 9 years ago

Hi All, I am learning Finite Difference Method and try to solve following equation. for f(x)=sin(pi x), ub=0=uf, L=1, alpha=1/20 we know exact solution is u[x_, t_] := Sin[(\[Pi] x)/L] E^(-((\[Alpha] \[Pi]^2 t)/L^2)) Here what I did so far L = 1; n = 50;(num interior points) \[CapitalDelta]x = L/n; \[CapitalDelta]t = 0.002;(cfl:dt/dx^2\[LessEqual].5) \[Alpha] = 1./20; \[Alpha] \[CapitalDelta]t/\[CapitalDelta]x^2 0.25 T[i_, 0] := Sin[\[Pi] i \[CapitalDelta]x](initial value) T[0, j_] := 0;(boundaries) T[1, j_] := 0.;(boundaries) T[i_, j_] :=T[i, j] = T[i, j - 1] + \[Alpha] \[CapitalDelta]t/\[CapitalDelta]x^2 (T[i + 1, j - 1] - 2 T[i, j - 1] + T[i - 1, j - 1]); s = Table[T[i, j], {j, 1, n}, {i, 1, n}]; ListPointPlot3D[s] My second approach is n = 6; \[Alpha] = 1/20; L = 1; BC1 = 0; BC2 = 0; \[CapitalDelta]x = L/(n - 1.) \[CapitalDelta]t = 4/(n - 1.) r = \[Alpha] \[CapitalDelta]t/\[CapitalDelta]x ^2 U[t_] = Table[Subscript[U, i], {i, 0, n}] IC = Thread[U[0] == (Sin[\[Pi] #] & /@ Range[0., 1, (1 - 0)/n])] A = SparseArray[{{n , n } -> 1., {i_, i_} -> (1 - 2. r), {i_, j_} /; Abs[i - j] == 1 -> r}, {n - 1, n - 1}, 0.]; But I am lost, don't know how to apply IC and BC for the matrix set up to solve the system.. Any suggestion? Thanks in advance..

Hi All, I am learning Finite Difference Method and try to solve following equation.

enter image description here

for f(x)=sin(pi x), ub=0=uf, L=1, alpha=1/20

we know exact solution is

u[x_, t_] := Sin[(\[Pi] x)/L] E^(-((\[Alpha] \[Pi]^2 t)/L^2))

Here what I did so far

L = 1;

n = 50;(*num interior points*)
\[CapitalDelta]x = L/n;
\[CapitalDelta]t = 0.002;(*cfl:dt/dx^2\[LessEqual].5*)
\[Alpha] = 1./20;

\[Alpha] \[CapitalDelta]t/\[CapitalDelta]x^2

0.25

T[i_, 0] := Sin[\[Pi] i \[CapitalDelta]x](*initial value*)

T[0, j_] := 0;(*boundaries*)
T[1, j_] := 0.;(*boundaries*)

T[i_, j_] :=T[i, j] = T[i, j - 1] + \[Alpha] \[CapitalDelta]t/\[CapitalDelta]x^2 (T[i + 1, j - 1] - 2 T[i, j - 1] + T[i - 1, j - 1]);

s = Table[T[i, j], {j, 1, n}, {i, 1, n}];

ListPointPlot3D[s]

My second approach is

n = 6; \[Alpha] = 1/20; L = 1; BC1 = 0; BC2 = 0;

\[CapitalDelta]x = L/(n - 1.)

 \[CapitalDelta]t = 4/(n - 1.)


r = \[Alpha]   \[CapitalDelta]t/\[CapitalDelta]x ^2


U[t_] = Table[Subscript[U, i], {i, 0, n}]
IC = Thread[U[0] == (Sin[\[Pi] #] & /@ Range[0., 1, (1 - 0)/n])]



A = SparseArray[{{n , n } -> 1., {i_, i_} -> (1 - 2. r), {i_, j_} /; Abs[i - j] == 1 ->    r}, {n - 1, n - 1}, 0.];

But I am lost, don't know how to apply IC and BC for the matrix set up to solve the system..

Any suggestion? Thanks in advance..

POSTED BY: Okkes Dulgerci

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Okkes Dulgerci

Posted 9 years ago

Hi All. I finally figure it out. Here is the code.. Note that I am taking alpha=c=1 a = 1.; b = 0.2; c = 1.; n = 6; m = 12; h = a/(n - 1); k = b/(m - 1); r = (c^2 k)/h^2; T[n - 1, j_] := 0.; T[0, j_] := 0.; T[i_, 0] := Sin[h i \[Pi]] T[i_, j_] := T[i, j] = r T[i + 1, j - 1] + (1 - 2 r) T[i, j - 1] + r T[i - 1, j - 1]; (s = Table[T[i, j], {i, 0, n - 1}, {j, 0, m - 1}]) // Transpose // MatrixForm ListPointPlot3D[s] ListPlot3D[s, Mesh -> None, InterpolationOrder -> 3, ColorFunction -> "SouthwestColors"] exact[x_, t_] := Sin[\[Pi] x] E^(-\[Pi]^2 t) Table[exact[x h, t k], {x, 0, n-1}, {t, 0, m-1}] // Chop // Transpose // MatrixForm

Hi All. I finally figure it out. Here is the code.. Note that I am taking alpha=c=1

a = 1.;
b = 0.2;
c = 1.;
n = 6;
m = 12;
h = a/(n - 1);
k = b/(m - 1);
r = (c^2 k)/h^2;
T[n - 1, j_] := 0.;
T[0, j_] := 0.;
T[i_, 0] := Sin[h i \[Pi]]
T[i_, j_] :=  T[i, j] = r T[i + 1, j - 1] + (1 - 2 r) T[i, j - 1] + r T[i - 1, j - 1];

(s = Table[T[i, j], {i, 0, n - 1}, {j, 0, m - 1}]) //  Transpose // MatrixForm
ListPointPlot3D[s]
ListPlot3D[s, Mesh -> None, InterpolationOrder -> 3, ColorFunction -> "SouthwestColors"]

exact[x_, t_] := Sin[\[Pi] x] E^(-\[Pi]^2 t)
Table[exact[x h, t k], {x, 0, n-1}, {t, 0, m-1}] // Chop // Transpose // MatrixForm

POSTED BY: Okkes Dulgerci

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