By Mark S. Gockenbach (siam, 2010)
Chapter 6: Heat flow and diffusion
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Chapter 6: Heat flow and diffusionSection 6.1: Fourier series methods for the heat equationExample 6.2: An inhomogeneous exampleConsider the IBVP 
The constant A has value 0.208 cm2/s. clear A=0.208 A = 0.2080
where the coefficient an(t) satisfies the IVP 
The values c1, c2, c3, ... are the Fourier sine coefficients of the constant function 10-7: syms n x pi 2/100*int(10^(-7)*sin(n*pi*x/100),x,0,100) ans = -(cos(pi*n) - 1)/(5000000*pi*n) mysubs(ans,n) ans =
-((-1)^n - 1)/(5000000*pi*n) c=ans c = -((-1)^n - 1)/(5000000*pi*n) The coefficient an(t) is given by the following formula: syms t s int(exp(-A*n^2*pi^2*(t-s)/100^2)*c,s,0,t) ans = ((-1)^n - 1)/(104*pi^3*n^3*exp((13*pi^2*n^2*t)/625000)) - ((-1)^n - 1)/(104*pi^3*n^3) a=simplify(ans) a =
((1/exp((13*pi^2*n^2*t)/625000) - 1)*((-1)^n - 1))/(104*pi^3*n^3) Next I define the (partial) Fourier series of the solution: S=symsum(a*sin(n*pi*x/100),n,1,10); I can now look at some "snapshots" of the solution. For example, I will show the concentration distribution after 10 minutes (600 seconds). Some trial and error may be necessary to determine how many terms in the Fourier series are required for a qualitatively correct solution. (As discussed in the text, this number decreases as t increases.) S600=subs(S,t,600); xx=linspace(0,100,201); plot(xx,subs(S600,x,xx)) 
The wiggles in the graph suggest that 10 terms is not enough for a correct graph (after 10 minutes, the concentration distribution ought to be quite smooth). Therefore, I will try again with 20 terms: S=symsum(a*sin(n*pi*x/100),n,1,20); S600=subs(S,t,600); plot(xx,subs(S600,x,xx)) 
Now with 40 terms: S=symsum(a*sin(n*pi*x/100),n,1,40); S600=subs(S,t,600); plot(xx,subs(S600,x,xx)) 
It appears that 20 terms is enough for a qualitatively correct graph at t=600. Download 2.45 Mb. Share with your friends:
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