Warning: Matrix is singular to working precision Error

Hello! I'm having throuble solving this system of equations using the implicit method for this a PDE.

Could someone help me to figure out how to fix my code?

Thank you in advance!

%% 
clc;
clear;
close all;

%% Parameters

x0=0;
xn=1;
t=2;

tstep=0.04;
xstep=0.04;

nsteps_for_t=t/tstep;
nsteps_for_x=xn/xstep;

nx=nsteps_for_x+1;
nt=nsteps_for_t+1;

x = 0:xstep:1; % defining x-scale
t = 0:tstep:2; % defining t-scale

%% Defining the first three time-levels

%First time-level (Based on given initial Condition)
t_1= sin(pi*x)+0.5*sin(3*pi*x);
t_1(1,nsteps_for_x+1)=0; %from u(1,t)=0
%Second time-level
t_2= cos(pi*tstep)'*sin(pi*x)+0.5*cos(3*pi*tstep)'*sin(3*pi*x);
t_2(1,nsteps_for_x+1)=0; %from u(1,t)=0
%Third time-level
t_3= cos(pi*(2*tstep))'*sin(pi*x)+0.5*cos(3*pi*(2*tstep))'*sin(3*pi*x);
t_3(1,nsteps_for_x+1)=0; %from u(1,t)=0

%% Creating matrix A and B

%Inserting boundary conditions into matrix A
A=zeros(nx,nx);
A(1,:)=0;
A(nx,:)=0;

for i=2:nsteps_for_x
A(i,i-1)=(-tstep^2)/xstep^2;
A(i,i)=2+((2*tstep^2)/xstep^2);
A(i,i+1)=(-tstep^2)/xstep^2;
end

%Inserting boundary conditions into matrix B
B=zeros(nx,nx);
B(1,:)=0;
B(nx,:)=0;

for i=2:nsteps_for_x
B(i,i-1)=5;
B(i,i)=-4;
B(i,i+1)=1;
end
        
%% Finding the numerical solution (u_numer)
for i=1:nsteps_for_t
          
       b(:,1)=t_1;
       b(:,2)=t_2;
       b(:,3)=t_3;
           
       C=B*b;
       u_numer=A\C;
end 

%% Error
Difference_A_N = abs(u_numer-u_analit);
ERROR=(1/(n+1))*(sum(Error_A_N,'all'))

Answer 1

[Walter Roberson]

The first and the last rows of B are all zero, because

B=zeros(nx,nx);
B(1,:)=0;
B(nx,:)=0;

Nothing other than zeros in the first and last row. You do not even need those assignments of 0 because you initialized it all to 0.

WIth the first and last row being the same (no matter whether it was all 0 or all (say) 8.7234309, then you can be sure that the matrix will be singular. But even if one of the two rows was different (such as if you had stored something non-zero in the last row) then you know that a row or a column that is all 0 guarantees that the matrix will be singular.

The same logic holds for A as well: you can be sure that it is singular matrix.

With B being singular, C=B*b will be singular, and we know that A is singular. So whether do A\C or A/C we know that both sides are singular and so neither \ nor / will be able to give a meaningful result.

You could do

u_numer = pinv(A)*C;

but you have to ask yourself if that is meaningful.

You are doing this in for i loop that does not involve i in any way, so you are repeating exactly the same calculation each time, which is not useful.

[Tatiane Cristine Antonio]

Hi Walter,

Thanks for looking at my code.

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About:

B=zeros(nx,nx);

B(1,:)=0;

B(nx,:)=0;

Initially, I decided to write these lines because I intended to insert values different than zeros later on. Although, at first I didn't realize that was the reason for the error that was getting until you pointed that out. Thank you very much! I adjusted my matrix and know the results look great!

I appreciate your help!

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