Question

(MATLAB 2010 A Student version)

Question:
I am trying to solve 9 symbolic equations with 9 symbolic variables using “solve” command:

Problem:

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.

1. What is ‘z’ in the solution? What’s the source of this?

2. How to circumvent this and get meaningful solution?

Thanks in advance

Answer 1

The z parameter means for whatever value of z you substitute in, your set of solutions is valid for your equations.

You have nine equations to find nine unknowns. However your sixth equation 2*V33=0 renders the following equation -2*D*V33=0 redundant…so you actually only have eight independent equations, hence the extra parameter z. I don’t know more about the details about your PDE to tell you why it isn’t well-posed, but basically you need another equation involving your nonzero variables if you expect a unique solution.