Linear algebraic equations can arise in the solution of differential equations.

Question

Linear algebraic equations can arise in the solution of differential equations. For example, the following differential equation results from a steady-state mass balance for a chemical in a one-dimensional canal:

0=D*(((d^2)c)/(d*x^2))-U((d*c)/(d*x))-k*c

where c=concentration, t=time, x=distance, D=diffusion coefficient, U=fluid velocity, and k=a first-order decay rate. Convert this differential equation to an equivalent system of simulations algebraic equations. Given D=2, U=1, k=0.2, c(0)=80 and c(10)=10, solve these equations from x=0 to 10 and develop a plot of concentration versus distance.

how do you transform the differential equation into a system into linear equations?

Explanation / Answer

%Problem 11.27 D = 2 U = 1 k = 0.2 c_0 = 80 c_10 = 20 A = [-2*D-k D-U/2 0 0 0 0 0 0 0; ... D+U/2 -2*D-k D-U/2 0 0 0 0 0 0; ... 0 D+U/2 -2*D-k D-U/2 0 0 0 0 0; ... 0 0 D+U/2 -2*D-k D-U/2 0 0 0 0; ... 0 0 0 D+U/2 -2*D-k D-U/2 0 0 0; ... 0 0 0 0 D+U/2 -2*D-k D-U/2 0 0; ... 0 0 0 0 0 D+U/2 -2*D-k D-U/2 0; ... 0 0 0 0 0 0 D+U/2 -2*D-k D-U/2; ... 0 0 0 0 0 0 0 D+U/2 -2*D-k] b = [(-D-U/2)*c_0; 0; 0; 0; 0; 0; 0; 0; (-D+U/2)*c_10] c = A b plot([c_0; c(:); c_10] )
%Problem 11.27 D = 2 U = 1 k = 0.2 c_0 = 80 c_10 = 20 A = [-2*D-k D-U/2 0 0 0 0 0 0 0; ... D+U/2 -2*D-k D-U/2 0 0 0 0 0 0; ... 0 D+U/2 -2*D-k D-U/2 0 0 0 0 0; ... 0 0 D+U/2 -2*D-k D-U/2 0 0 0 0; ... 0 0 0 D+U/2 -2*D-k D-U/2 0 0 0; ... 0 0 0 0 D+U/2 -2*D-k D-U/2 0 0; ... 0 0 0 0 0 D+U/2 -2*D-k D-U/2 0; ... 0 0 0 0 0 0 D+U/2 -2*D-k D-U/2; ... 0 0 0 0 0 0 0 D+U/2 -2*D-k] b = [(-D-U/2)*c_0; 0; 0; 0; 0; 0; 0; 0; (-D+U/2)*c_10] c = A b plot([c_0; c(:); c_10] )
%Problem 11.27 D = 2 U = 1 k = 0.2 c_0 = 80 c_10 = 20 A = [-2*D-k D-U/2 0 0 0 0 0 0 0; ... D+U/2 -2*D-k D-U/2 0 0 0 0 0 0; ... 0 D+U/2 -2*D-k D-U/2 0 0 0 0 0; ... 0 0 D+U/2 -2*D-k D-U/2 0 0 0 0; ... 0 0 0 D+U/2 -2*D-k D-U/2 0 0 0; ... 0 0 0 0 D+U/2 -2*D-k D-U/2 0 0; ... 0 0 0 0 0 D+U/2 -2*D-k D-U/2 0; ... 0 0 0 0 0 0 D+U/2 -2*D-k D-U/2; ... 0 0 0 0 0 0 0 D+U/2 -2*D-k] b = [(-D-U/2)*c_0; 0; 0; 0; 0; 0; 0; 0; (-D+U/2)*c_10] c = A b plot([c_0; c(:); c_10] )
%Problem 11.27 D = 2 U = 1 k = 0.2 c_0 = 80 c_10 = 20 A = [-2*D-k D-U/2 0 0 0 0 0 0 0; ... D+U/2 -2*D-k D-U/2 0 0 0 0 0 0; ... 0 D+U/2 -2*D-k D-U/2 0 0 0 0 0; ... 0 0 D+U/2 -2*D-k D-U/2 0 0 0 0; ... 0 0 0 D+U/2 -2*D-k D-U/2 0 0 0; ... 0 0 0 0 D+U/2 -2*D-k D-U/2 0 0; ... 0 0 0 0 0 D+U/2 -2*D-k D-U/2 0; ... 0 0 0 0 0 0 D+U/2 -2*D-k D-U/2; ... 0 0 0 0 0 0 0 D+U/2 -2*D-k] b = [(-D-U/2)*c_0; 0; 0; 0; 0; 0; 0; 0; (-D+U/2)*c_10] c = A b plot([c_0; c(:); c_10] )


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