Example 8.3.mw
Example 8.3 Heat Transfer in a Rectangle
Example (8.1.3) is solved in Maple below:
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restart:with(inttrans):with(plots): |
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eq:=diff(u(x,t),t)=diff(u(x,t),x$2); |
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(1) |
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(2) |
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(3) |
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(4) |
The governing equation and the boundary conditions are converted to the Laplace domain:
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eqs:=subs(laplace(u(x,t),t,s)=U(x),eqs); |
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(5) |
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bc1:=subs(laplace(u(x,t),t,s)=U(0),bc1); |
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(6) |
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bc2:=subs(laplace(u(x,t),t,s)=U(1),bc2); |
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(7) |
The governing equation in the Laplace domain is solved as:
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(8) |
The governing equation is solved with the boundary conditions as:
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U(x):=rhs(dsolve({eqs,bc1,bc2},U(x))); |
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(9) |
The solution obtained is converted to the time domain as:
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u:=invlaplace(U(x),s,t); |
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(10) |
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plot3d(u,x=1..0,t=0..0.2,axes=boxed,title="Figure Exp. 8.5.",labels=[x,t,"u"],orientation=[120,60]); |
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plot([subs(t=0,u),subs(t=0.05,u),subs(t=0.1,u),subs(t=0.2,u)],x=0..1,title="Figure 8.6.",axes=boxed,thickness=5,labels=[x,"u"]); |
In all the examples previously discussed, the boundary conditions did not involve derivatives until now. When there is a derivative in the boundary condition it has to be taken care of while applying the Laplace transform as shown in the next example.