Example 4.4
In the previous examples, the initial condition was a constant and independent of x. The same technique can be applied for the case where there is an initial temperature profile. Consider the transient heat conduction problem in a slab
=α
u(x,0) =sin(πx)
(0,t) = 0 and u(∞,t) is defined (4.1.4)
Equation (4.1.4) is solved in Maple below. The programs given for the previous examples have to be modified to solve equation (4.1.4) by only changing the initial condition:
| > | restart:with(linalg):with(inttrans):with(plots): |
| > | eq:=diff(u(x,t),t)=alpha*diff(u(x,t),x$2); |
| (1) |
| > | u(x,0):=sin(Pi*x); |
| (2) |
| > | bc1:=u(0,t)=0; |
| (3) |
| > | bc2:=u(infinity,t)=defined; |
| (4) |
The following solution and plots are obtained:
| > | eqs:=laplace(eq,t,s); |
| (5) |
| > | eqs:=subs(laplace(u(x,t),t,s)=U(x),eqs); |
| (6) |
| > | bc1:=laplace(bc1,t,s); |
| (7) |
| > | bc1:=subs(laplace(u(0,t),t,s)=U(0),bc1); |
| (8) |
| > | U(x):=rhs(dsolve({eqs,bc1},U(x))); |
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(9) |
| > | eqc:=coeff(U(x),exp(1/alpha^(1/2)*s^(1/2)*x)); |
| (10) |
| > | _C2:=solve(eqc,_C2); |
| (11) |
| > | U(x):=eval(U(x)); |
| (12) |
| > | u:=invlaplace(U(x),s,t); |
| (13) |
| > | pars:={alpha=0.001}; |
| (14) |
| > | plot3d(subs(pars,u),x=10..0,t=100..0,axes=boxed,title="Figure Exp. 4.1.9.",labels=[x,t,"u"],orientation=[-60,60]); |
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| > | plot([subs(pars,t=1,u),subs(pars,t=10,u),subs(pars,t=100,u),subs(pars,t=200,u)],x=0..1,axes=boxed,title="Figure Exp. 4.1.10.",thickness=5,labels=[x,"u"]); |
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| > |