Ax = 0 with 2 values of x known

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kimfet
kimfet 2022년 3월 31일
답변: Fabio Freschi 2022년 5월 16일
I have a matrix in the form Ax = 0, and for x I know the first and last position. How can I solve this system?

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Riccardo Scorretti
Riccardo Scorretti 2022년 3월 31일
편집: Riccardo Scorretti 2022년 4월 1일
Basically, you are asking how to impose Dirichlet boundary conditions, right? If so, there are several methods:
  • Substitution method: replace the first and last equation by: x = known_value. For instance
A(1,:) = 0 ; A(1,1) = 1; b(1) = known_value_1;
A(n,:) = 0 ; A(n,n) = 1; b(n) = known_value_2;
  • Penalty method: "overwrite" the existing equation with the boundary condition, weighted with a huge value, for instance:
TGV = 1E30; % = Terribly Giant Value
A(1,1) = TGV ; b(1) = TGV*known_value_1;
A(n,n) = TGV ; b(n) = TGV*known_value_2;
Assuming that b is the known vector, then you can solve the linear system with an appropriate direct or iterative linear solver.
The penalty method is approximate, but it has the advantage that it is quite simple and (most importantly), it preserves eventual symmetry properties of the matrix. This is the way (I don't know if it is the only way) Dirichlet boundary conditions are implemented in FreeFem++ (see https://doc.freefem.org/references/types.html):
See also this discussion:
By the way, I recently posted a complete program where Dirichlet boundary conditions are imposed by using the first method. The program solves a PDE by using the Finite Difference method, but the way of imposing Dirichlet boundary conditions is exactly the same:
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kimfet
kimfet 2022년 4월 2일
Hello,
This worked for me! It was Dirichlett BC for a FEM problem with a Galerkin approach, to solve the Steady-State Convection-Diffusion equation in 1D. A pretty easy CFD problem, but working on the solution of the system was a bit tricky. It worked for me (I used the substitution method because it was easier for me to implement).
Thank's for your time!

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추가 답변(3개)

Torsten
Torsten 2022년 3월 31일
Use lsqlin for the problem
min : ||A*x||_2
s.c.
x(1) = known_value_1
x(n) = known_value_2
if A has n columns.

Matt J
Matt J 2022년 4월 1일
편집: Matt J 2022년 4월 2일
b=-A(:,[1,end])*[xInitial;xFinal];
x=A(:,2:end-1)\b;
x=[xInitial, x', xFinal]';
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Matt J
Matt J 2022년 4월 2일
Right.

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Fabio Freschi
Fabio Freschi 2022년 5월 16일
The post is old but the question arises many times, so I post here my solution. It is a generalization of @Matt J solution.
Suppose you have a matrix with unknowns partitioned in free x and dirichlet . And by chance the partition is such that the free variables come first
The free unknowns can be obtained by solving the first block row:
The second block row is of no interest, since is known.
In the general case when are not at the end of the unknown vector, but they are identified by the index vector idFix and the corresponding Dirichlet values xFix, you can write the function
function x = solveproblem(A,b,idFix,xFix)
% get number of unknowns
n = length(b);
% move to rhs the Dirichlet values
b = b-A(:,idfix)*xfix;
% get indices of free variables
iFree = setdiff(1:n,idFix);
% reduce stiffness and rhs
A = A(:,iFree);
A = A(iFree,:);
b = b(iFree,:);
% solution
xRed = A\b;
% re-assemble the unknown vector
x(idFree) = xRed;
x(idFix) = xFix;
end
This method preserves the original properties of the original matrix (in particular if it's SPD). There is also an interesting way to include in this approach that a selection of the vector of unknowns has the same value, for example a floating potential of an equipotential object, but it goes beyond the original question

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