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单词 ExampleOfSolvingTheHeatEquation
释义

example of solving the heat equation


Let a .  Determine the temperature function  (x,y)u(x,y)  on the plate, when the faces of the plate are .

The equation of the heat flow (http://planetmath.org/HeatEquation) in this case is

2uuxx′′+uyy′′= 0(1)

under the boundary conditionsMathworldPlanetmath

u(0,y)=0,u(π,y)=0,u(x,π)=C,uy(x, 0)=0.

We first try to separate the variables, i.e. seek the solution of (1) of the form

u(x,y):=X(x)Y(y).

Then we get

ux=XY,uxx′′=X′′Y,uy=XY,uyy′′=XY′′,

and thus (1) gets the form

X′′Y+XY′′= 0(2)

and the boundary conditions

X(0)=X(π)= 0,X(x)=CY(π),Y(0)= 0.

We separate the variables in (2):

X′′X=-Y′′Y

This equation is not possible unless both sides are equal to a same negative -k2, which implies for  X′′=-k2X  the solution

X:=C1coskx+C2sinkx

and for  Y′′=k2Y  the solution

Y:=D1coshky+D2sinhky.

The two first boundary conditions give  0=X(0)=C1,  0=X(π)=0+C2sinkπ,  and since  C20,  we must have  sinkπ=0,  i.e.

0<k:=n= 1, 2, 3,

Therefore

X(x):=C2sinnx,Y(y)nD1sinhny+nD2coshny.

The fourth boundary condition now yields that  0=Y(0)=nD2; thus  D2=0  and  Y(y):=D1coshny.  So (1) has infinitely many solutions

un:=C2D1sinnxcoshny=Ansinnxcoshny(3)

with  n+  and they all satisfy the boundary conditions except the third.  Because of the linearity of (1), also the sum

u:=n=1Ansinnxcoshny

of the functions (3) satisfy (1) and those boundary conditions, provided that this series converges.  The third boundary condition requires that

C=u(x,π)=n=1Ansinnxcoshnπ=n=1(Ancoshnπ)sinhnx

on the interval  0xπ.  But this is the Fourier sine seriesMathworldPlanetmath of the constant function  xC  on the half-interval  [0,π],  whence

Ancoshnπ=2π0πCsinnxdx=2Cnπ(1-(-1)n)n+.

The even (http://planetmath.org/EvenNumber) n’s here give 0 and the odd (http://planetmath.org/EvenNumber) give

A2m+1:=4C(2m+1)πcosh(2m+1)π(m=0, 1, 2,)

Thus we obtain the solution

u(x,y):=4Cπm=0sin(2m+1)xcosh(2m+1)y(2m+1)cosh(2m+1)π.

It can be shown that this series converges in the whole of the plate.

Visualization of the solution

Figure 1: Surface plot of the solution u(x,y), for C=1
Figure 2: Color-coded plot of the temperature u(x,y)

Remark.  The function u has been approximated in the plot by computing a partial sum of the true infinite-series solution.  However,there is substantial numerical error in the approximate solution near y=π,  evident in the small oscillations observed in the surface plot, that should not be there in .  This phenomenon is actually inevitable given that the boundary conditions are actually discontinuousMathworldPlanetmath at the corners  (0,π)  and  (π,π).

More precisely, observe that when  y=π,  the for  u(x,y)  reduces to the Fourier series

4Cπ(sinx+sin3x3+sin5x5+)

for the discontinuous function on  [-π,π]:

x{C,0<x<π-C,-π<x<0

That means the Fourier will necessarily be subject to the Gibbs phenomenon.  Of course, the series also cannot converge absolutely; in other of the series decay too slowly in magnitude, adversely affecting the numerical solution.

  • http://gold-saucer.afraid.org/math/planetmath/ExampleOfSolvingTheHeatEquation/heat.pyPython program to compute  u(x,y)  and produce the two figures

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更新时间:2025/5/4 17:11:38