MATLAB: “Solve” command in MATLAB returns solution in terms of ‘z’? What is ‘z’ here

solve

(MATLAB 2010 A Student version)
Question: I am trying to solve 9 symbolic equations with 9 symbolic variables using "solve" command:
Problem Code:
    N=2
    V = sym(zeros(N+2+1)); % 5*5 symbolic matrix
    syms D x; %D is diffusion constant
    %Creating 3*3 non-zero symbolic entries
    for row = 0:N
        for col = 0:N
            V(row+1, col+1) = sym(sprintf('V%d%d', row+1, col+1));
        end
    end
    count=1;
    for m=0:N
        for n=0:N
            % Diffusion PDE written in Differential Transform domain
            eqn(count) = (m+1)*V(m+1+1,n+1)-D*(n+1)*(n+2)*V(m+1,n+2+1);  
            count=count+1;
        end
    end
    %length(eqn) is 9
    eqn=eqn(1:7);      %as last 2 terms are [0 0] in eqn
Initial condition for PDE are given by Dirac Delta func ("delta(x)"). Thus, it is composed of 2 conditions (one for x=0 and one for x not equal to 0)
Thus, at x=0, it is "1" giving:
    eqn(length(eqn)+1)=V(1,1)-V(2,1)+V(3,1)-1; % ini.condn
And at x NOT equal to 0, it is "0" giving: (does it make sense?)
    eqn(length(eqn)+1)=(V(1,1)-V(2,1)+V(3,1))+(V(1,2)-V(2,2)+V(3,2))*x + 
    (V(1,3) - V(2,3)+ V(3,3))*x^2; %ini. condn
    S=solve(eqn,'V11' ,'V12' ,'V13' ,'V21' ,'V22' ,'V23' ,'V31' ,'V32' ,'V33');
Answer: S.V11, S.V12, S.V13, S.V21 are functions of 'z'. Rest are 0
I am getting 'z' in my solutions.
What is 'z' in the solution? What's the source of this?
How to circumvent this and get meaningful solution?
Thanks in advance.
PS: I have posted this question on "Stackoverflow" too.

Best Answer

Are the "z" inside RootOf() ? If so you should read the documentation for that.
Briefly,
RootOf(expression, z)
means that the value of the RootOf() is the set of values for z such that the expression becomes 0. For example, since there is no analytical form for general polynomials of degree 5 or higher, MuPAD would insert a RootOf() showing that the values at that point are the values that solve that polynomial.
If it is RootOf() a quartic (order 4) polynomial in z, then you can use the degree option of solve() to get the explicit algebraic solution, but that solution will be very long and you are unlikely to be able to make much intuitive sense out of it.
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