Creating and solving large systems of equations

Question

I am trying to follow a solution in a book so that I can build my own model. They produce the set of equations below. The book claims it to be a system of equations with 10 unknowns; however from my understanding, to solve a system of equations, each equation in the system must have all of the unknowns for the solver to work. So I am wondering whether I have to rearrange all the equations so that each unknown appears in every equation? This seems like a very long process and I will definitely make an error somewhere. Is there an alternative to doing this or is this the only way? Thanks for any help you can offer.

Answer 1

You definitely don't have to have each unknown in each equation (unless you consider it to be there with a coefficient of $0$). The system $x=1,y=2$ doesn't have each unknown in each equation and is easy to solve. If you are solving this system by hand they will ripple through. If you write it as a matrix equation, the nice thing is that it is tridiagonal. Each equation only links three neighboring unknowns. You can solve the system in two passes. Your last equation is $T_9=\frac 12(T_8+T_{10})$ and $T_{10}$ is a constant, not a variable. You can insert this into the next equation up, which is $T_8=\frac 12(T_7+T_9)=\frac 12(T_7+\frac 12(T_8+T_{10})),T_8=\frac 23T_7+\frac 13T_{10}$ and so on. When you get to the top you will get a value for $T_0$, which you can ripple back downward to get all the rest.

You can get a computer algebra system to do the work for you. The funny stuff that goes on at the ground surface will make it a bit messy.

Answer 2

All of the variables do appear in each linear equation, but some of them have coefficient zero. For example, we could write the first equation as $$ T_0 - 2T_1 + T_2 + 0T_3 + 0T_4 + 0T_5 + 0T_6 + 0T_7 + 0T_8 + 0T_9 + 0T_{10}. $$ When you write out the matrix that corresponds to this system of linear equations it will have many zeroes. I don't expect that the problem is easy to solve, but you should not have any trouble setting it up from a linear algebra perspective.

Answer 3

I finally used a Fortran program to solve the problem and if anyone is interested here it is:

PROGRAMME GAUSS
DATA T0,T1,T2,T3,T4,T5,T6,T7,T8/9*0./
DATA T9/50./

real,parameter:: b=0.1714E-8
e = 0.6

DO 20 K = 1,1000

    t9 = (T8+50)/2.
    t8 = (T7+T9)/2.
    t7 = (T6+T8)/2.
    t6 = (T5+T7)/2.
    t5 = ((86.4*t4)+(0.817*t6))/87.217
    t4 = (T3+T5)/2.
    t3 = (T2+T4)/2.
    t2 = (T1+T3)/2.
    t1 = (T0+T2)/2.
    t0 = (280 + (e*b*((510.**4)-(t0+460.)**4)) +(86.4*t1))/89.9

    WRITE (*,10)K,T0,T1,T2,T3,T4,T5,T6,T7,T8,T9
    10 FORMAT (' ',I3,10(F8.1))
    20 CONTINUE
END PROGRAM GAUSS