Example 3.1 Exponential Method for Linear Boundary Value Problem
Consider the conduciton of heat in a rectangular cooling fin. The governing differential equation[6] in dimensionless form is

This boundary value problem is solved below by following the procedure described earlier.
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with(linalg):with(plots): |
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eq:=diff(theta(x),x$2)-H^2*theta(x); |
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A:=matrix(N,N,[0,1,H^2,0]); |
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The exponentials are converted to trigonometric form for convenience.
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mat:=map(convert,mat,trig); |
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Y0:=matrix(N,1,[1,c[1]]); |
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The solution is obtained as a function of x, H and the unknown initial condition c1. The first row of sol corresponds to w0 and the second row corresponds to
![`+`(cosh(`*`(H, `*`(x))), `/`(`*`(sinh(`*`(H, `*`(x))), `*`(c[1])), `*`(H)))](images/Example3.1 Rev 1_13.gif) |
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The boundary condition at x = 1 is applied to solve for c
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bc2:=subs(x=1,dthetadx)=0; |
![`+`(`-`(`*`(`/`(1, 2), `*`(H, `*`(`+`(cosh(H), `-`(sinh(H))))))), `*`(`/`(1, 2), `*`(H, `*`(`+`(cosh(H), sinh(H))))), `*`(cosh(H), `*`(c[1]))) = 0](images/Example3.1 Rev 1_16.gif) |
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This value of the constant c
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The expression for θ can be simplified further by using Maple's 'combine' command:
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You can make plots for different values of the heat transfer coefficient, H, by using a 'pars' array and a' do' loop.
![[1, 2, 5, 10]](images/Example3.1 Rev 1_21.gif) |
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clr:=[red,green,gold,blue]; |
![[red, green, gold, blue]](images/Example3.1 Rev 1_22.gif) |
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for i from 1 to 4 do p[i]:=plot(subs(H=pars[i],theta),x=0..1,color=clr[i]):od: |
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display(seq(p[i],i=1..4),thickness=4,title="Figure 3.1.1",axes=boxed,labels=["X","theta"],labeldirections=[HORIZONTAL,VERTICAL]); |
A three dimensional plot for θ can be made for different values of H as x varies between zero and one.
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plot3d(theta,x=0..1,H=0..10,axes=boxed,title="Figure 3.1.2",orientation=[45,45]); |
The above picture shows that as H increases the temperature distribution becomes nonuniform.