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2DHeat.mws

Section 11.5. The 2D-Heat Equation

by Alain Goriely, goriely@math.arizona.edu,
(http://www.math.arizona.edu/~goriely)

Abstract: This section illustrates Section 11.5 in Kreyszig 's book (8th ed.)

Application Areas/Subjects: Engineering, Applied Mathematics

Keywords: Heat equation, 2D, steady flows, Fourier series
Other Worksheets in the same package.

Prerequisites:  plots

Note:  Send me an e-mail (comments-criticisms) if you use this worksheet.

 > restart;assume(n,integer):with(plots): setoptions(thickness=2): #set the tickness of the lines in the plots

Warning, the name changecoords has been redefined

Introduction

In this worksheet, I use Maple to illustrate Section 11.5 of Kreyszig 's book: Advanced Engineering Mathematics.

Let be the temperature in a two dimensional media. The heat profile obeys the following PDEs
(t
he so-called 2D heat equation):

where  is the diffusion constant  ( : themal conductivity/ (specific heat *density) )

We consider stationary profiles, that is time-independent solutions of the heat equations. These represent steady heat flows in 2D. Hence, we have, the LAPLACE EQUATION:

We are looking for a steady flow in a rectangle in the ( ) plane with the following boundary conditions

and

and

Here we explore different steady solutions of the heat equation in 2D, starting with initial heat profile on one side.

Look at the 3D Plots!

Section 1: The eigenfunctions

 > Su:=u=(A[n]*sin(n*Pi*x/a)*sinh(n*Pi*y/a));

Let us verify that this is indeed a solution of the equation:

 > Diff(u,x\$2) + Diff(u,`\$`(y,2))=eval(subs(Su,diff(u,`\$`(x,2))+diff(u,`\$`(y,2))));

We want to see what these modes look like: Start with n=1, the FUNDAMENTAL solution:

The fundamental eigenfunction

 > M1:=subs(n=1,a=2*Pi,b=Pi,c=2,subs(Su,A[n]=1,u));

The other eigenfunctions

 > M1:=subs(n=2,a=2*Pi,b=Pi,c=2,subs(Su,A[n]=1,u));

 > M1:=subs(n=3,a=2*Pi,b=Pi,c=2,subs(Su,A[n]=1,u));

The modes > all have a negative part. Therefore, as such they are not physical solutions of the heat equations. However, the superposition of these eigenfunctions are solutions (as long as ).

Section 2: An example of a symmetric profile

We now take a simple profile for and look at the steady flow"

 > f(x)=(1-cos(x));plot((1-cos(x)),x=0..2*Pi,thickness=3);

 > A:=2/(2*Pi*sinh(m*Pi*Pi/2/Pi))*int((1-cos(x))*sin(m*x/2),x=0..2*Pi);

 > S:=(N,x)->subs(m=1,A)*sin(x/2)*sinh(y/2)+          sum(subs(m=k,A)*sin(k*x/2)*sinh(k*y/2),k=3..N): St:=S(10,x):

Section 3: An asymmetric profile

 > f:=(sin(x/2)-1/2*sin(x));plot(f,x=0..2*Pi,thickness=3,color=blue);

 > A1:=1/sinh(Pi/2):A2:=-1/2/sinh(Pi):S:=A1*sin(x/2)*sinh(y/2)+A2*sin(x)*sinh(y);

Section 4: Another asymmetric profile

 > f:=(sin(x/2)-1/4*sin(4*x/2));plot(f,x=0..2*Pi,thickness=3,color=blue);

 > A1:=1/sinh(Pi/2):A4:=-1/4/sinh(4*Pi/2):A3:=-1/5/sinh(3*Pi/2): A5:=1/5/sinh(5*Pi/2): S:=A1*sin(x/2)*sinh(y/2)+A4*sin(4*x/2)*sinh(4*y/2);