After finding equations of motion using the Symbolic Toolbox (R2016b, Windows), I have the following form:
M(q)*qddot = b(q,qdot) + u
M and b were found using equationsToMatrix.
Now, I need to separate b into Coriolis and Potential terms such that
M(q)*qddot + C(q,qdot)*qdot + G(q) = u
It would be extremely convenient if I could apply
[C,G] = equationsToMatrix(b,qdot)
but unfortunately it will not factor out qdot when b is nonlinear. I do not care (and in fact it is necessary) that C is a function of q and qdot, even after factoring out the vector qdot. I have tried coeffs and factor without results.
Thanks.
Posting my own solution so there can be at least one answer...
This function works, but it is not heavily tested. It works exactly as I suggested in my original question. Feel free to rename it so it doesn't conflict with the MATLAB builtin.
function [A,b] = equationsToMatrix( eq, x )
%EQUATIONSTOMATRIX equationsToMatrix for nonlinear equations
% factors out the vector x from eq such that eq = Ax + b
% eq does not need to be linear in x
% eq must be a vector of equations, and x must be a vector of symbols
assert(isa(eq,'sym'), 'Equations must be symbolic')
assert(isa(x,'sym'), 'Vector x must be symbolic')
n = numel(eq);
m = numel(x);
A = repmat(sym(0),n,m);
for i = 1:n % loop through equations
[c,p] = coeffs(eq(i),x); % separate equation into coefficients and powers of x(1)...x(n)
for k = 1:numel(p) % loop through found powers/coefficients
for j = 1:m % loop through x(1)...x(n)
if has(p(k),x(j))
% transfer term c(k)*p(k) into A, factoring out x(j)
A(i,j) = A(i,j) + c(k)*p(k)/x(j);
break % move on to next term c(k+1), p(k+1)
end
end
end
end
b = simplify(eq - A*x,'ignoreanalyticconstraints',true); % makes sure to fully cancel terms
end
Related
The equations can be found here. As you can see it is set of 8 scalar equations closed to 3 matrix ones. In order to let Matlab know that equations are matrix - wise, I declare variable time dependent vector functions as:
syms t p1(t) p2(t) p3(t)
p(t) = symfun([p1(t);p2(t);p3(t)], t);
p = formula(p(t)); % allows indexing for vector p
% same goes for w(t) and m(t)...
Known matrices are declared as follows:
A = sym('A%d%d',[3 3]);
Fq = sym('Fq%d%d',[2 3]);
Im = diag(sym('Im%d%d',[1 3]));
The system is now ready to be modeled according to guide:
eqs = [diff(p) == A*w + Fq'*m,...
diff(w) == -Im*p,...
Fq*w == 0];
vars = [p; w; m];
At this point, when I try to reduce index (since it equals 2), I receive following error:
[DAEs,DAEvars] = reduceDAEIndex(eqs,vars);
Error using sym/reduceDAEIndex (line 95)
Expecting as many equations as variables.
The error would not arise if we had declared all variables as scalars:
syms A Im Fq real p(t) w(t) m(t)
Quoting symfun documentation (tips section):
Symbolic functions are always scalars, therefore, you cannot index into a function.
However it is hard for me to believe that it's not possible to solve these equations matrix - wise. Obviously, one can expand it to 8 scalar equations, but the multi body system concerned here is very simple and the aim is to be able to solve complex ones - hence the question: is it possible to solve matrix DAE in Matlab, and if so - what has to be fixed in order for this to work?
Ps. I have another issue with Matlab DAE solver: input variables (known coefficient functions) for my model are time variant. As far as example is concerned, they are constant in all domain, however for my problem they change in time. This problem has been brought out here. I would be grateful if you referred to it, should you have any solution.
Finally, I managed to find correct syntax for this problem. I made a mistake of treating matrix variables (such as A, Fq) as a single entity. Below I present code that utilizes matrix approach and solves this particular DAE:
% Define symbolic variables.
q = sym('q%d',[3 1]); % state variables
a = sym('a'); k = sym('k'); % constant parameters
t = sym('t','real'); % independent variable
% Define system variables and group them in vectors:
p1(t) = sym('p1(t)'); p2(t) = sym('p2(t)'); p3(t) = sym('p3(t)');
w1(t) = sym('w1(t)'); w2(t) = sym('w2(t)'); w3(t) = sym('w3(t)');
m1(t) = sym('m1(t)'); m2(t) = sym('m2(t)');
pvect = [p1(t); p2(t); p3(t)];
wvect = [w1(t); w2(t); w3(t)];
mvect = [m1(t); m2(t)];
% Define matrices:
mass = diag(sym('ms%d',[1 3]));
Fq = [0 -1 a;
0 0 1];
A = [1 0 0;
0 1 a;
0 a -q(1)*a] * k;
% Define sets of equations and aggregate them into one set:
set1 = diff(pvect,t) == A*wvect + Fq'*mvect;
set2 = mass*diff(wvect,t) == -pvect;
set3 = Fq*wvect == 0;
eqs = [set1; set2; set3];
% Close all system variables in one vector:
vars = [pvect; wvect; mvect];
% Reduce index of the system and remove redundnat equations:
[DAEs,DAEvars] = reduceDAEIndex(eqs,vars);
[DAEs,DAEvars] = reduceRedundancies(DAEs,DAEvars);
[M,F] = massMatrixForm(DAEs,DAEvars);
We receive very simple 2x2 ODE for two variables p1(t) and w1(t). Keep in mind that after reducing redundancies we got rid of all elements from state vector q. This means that all left variables (k and mass(1,1)) are not time dependent. If there had been time dependency of some variables within the system, the case would have been much harder to solve.
% Replace symbolic variables with numeric ones:
M = odeFunction(M, DAEvars,mass(1,1));
F = odeFunction(F, DAEvars, k);
k = 2000; numericMass = 4;
F = #(t, Y) F(t, Y, k);
M = #(t, Y) M(t, Y, numericMass);
% set the solver:
opt = odeset('Mass', M); % Mass matrix of the system
TIME = [1; 0]; % Time boundaries of the simulation (backwards in time)
y0 = [1 0]'; % Initial conditions for left variables p1(t) and w1(t)
% Call the solver
[T, solution] = ode15s(F, TIME, y0, opt);
% Plot results
plot(T,solution(:,1),T,solution(:,2))
The GMRES algorithm and its matlab implementation are supposed to solve linear equations system, such as
%Ax = b
A = rand(4);
b = rand(4,1);
x = gmres(A,b);
One can also use a function handle
foo = #(x) A*x + conj(A)*5*x;
y = gmres(foo,b);
What I want is to solve the following
B = rand(4);
H = rand(4);
foo2 = H*B + B*H;
X = gmres(foo2, B) %Will not run!
--Error using gmres (line 94)
--Right hand side must be a column vector of length 30 to match the coefficient matrix.
Mathematically speaking I don't see why gmres couldn't apply to this problem as well.
Note: What I'm really trying to solve is an implicit euler method for a PDE dB/dt = B_xx + B_yy, so H is in fact a second derivative matrix using finite difference.
Thank you
Amir
If I've understood right you want to use GMRES to solve an a sylvester equation
A*X + X*A = C
for n-by-n matrices A, X and C.
(I asked a related question yesterday over at SciComp and got this great answer.)
To use GMRES you can express this matrix-matrix equation as a size n^2 matrix-vector equation. For convenience we can use the Kronecker product, implemented in MATLAB with kron:
A = randn(5);
X = randi(3,[5 5]);
C = A*X + X*A;
% Use the Kronecker product to form an n^2-by-n^2 matrix
% A*X + X*A
bigA = (kron(eye(5),A) + kron(A.',eye(5)));
% Quick check that we're getting the same answer
norm(bigA*X(:) - C(:))
% Use GMRES to calculate X from A and C.
vec_X_gmres = gmres(bigA,C(:));
X_gmres = reshape(vec_X_gmres,5,5);
I'm writing a function of the Gauss Seidel method of solving a linear system of equations of the form Ax=b, x being the unknown we are looking for.
I am having a problem with the while loop in my function, it seems that it runs infinitely. I can't seem to figure out why.
This is my function for creating the coefficient matrix A and the column vectors x and b, all with the same number of rows of course. No problem with this one.
function [A, b, x0] = test_system(n)
u = ones(n, 1);
A = spdiags([u 4*u u], [-1 0 1], n, n);
b = zeros(n, 1);
b(1) = 3;
b(2 : 2 : end-2) = -2;
b(3 : 2 : end-1) = 2;
b(end) = -3;
x0 = ones(n, 1);
This is my function for solving the system. I have included all of it just in case, but I believe the real problem is within the while loop at the very end which runs infinitely when I execute the function. The counter doesn't break away from it either. I can't really see what its problem is. Any clues?
Be gentle, I'm new at Matlab :)
function [x] = GaussSeidel(A,b,x0,tol)
% implementation of the GaussSeidel iterative method
% for solving a linear system of equations Ax = b
%INPUTS:
% A: coefficient matrix
% b: column vector of constants
% x0: setup for the unknown vector (using vector of ones)
% tol: result must be within 'tol' of correct answer.
%OUTPUTS:
% x: unknown
%check that A is a matrix
if ~(ismatrix(A))
error('A is not a matrix');
end
%check that A is square
[m,n] = size(A);
if m ~= n
error('Matrix A is not square');
end
%check that b is a column vector
if ~(iscolumn(b))
error('b is not a column vector');
end
%check that x0 is a column vector
if ~(iscolumn(x0))
error('x0 is not a column vector');
end
%check that A, b and x0 agree in size
[rowA,colA] = size(A);
[rowb,colb] = size(b);
[rowx0,colx0] = size(x0);
if ~isequal(colA,rowb)||~isequal(rowb,rowx0)
error('matrix dimensions of A, b and xo do not agree');
end
%check that A and b have real entries
if ~isreal(A) || ~isreal(b)
error('matrix A or vector b do not have real entries');
end
%check that the provided tolerance is positive
if tol <= 0
error('tolerance must be positive');
end
%check that A is strictly diagonally dominant
absoluteA = abs(A);
row_sum=sum(absoluteA,2);
diagonal=diag(absoluteA);
if ~all(2*diagonal > row_sum)
warning('matrix A is not strictly diagonally dominant');
end
L = tril(A,-1);
U = triu(A,+1);
D = diag(diag(A));
x = x0;
M1 = inv(D).*L;
M2 = inv(D).*U;
M3 = D\b;
k = 0; %iterations counter
disp(size(M1));
disp(size(M2));
disp(size(M3));
disp(size(x));
while (norm(A*x - b) > tol)
for i=1:n
x(i) = - M1(i,:).*x - M2(i,:).*x + M3(i,:);
end
k=k+1;
if(k >= 10e4)
error('too many iterations carried out');
end
end
end %end function
Coding Discussion
The source of the coding error is the use of element-wise operations instead of matrix-matrix and matrix-vector operations.
These operations produce zero-matrices, so no progress is made.
The M matrices should be defined by
M1 = inv(D)*L; % Note that D\L is more efficient
M2 = inv(D)*U; % Note that D\U is more efficient
M3 = D\b;
and the iterator performing the update should be
x(i) = - M1(i,:)*x - M2(i,:)*x + M3(i,:);
Method Discussion
I also think it is worth mentioning that the code is currently implementing the Jacobi Method since the updater has the form
while a Gauss-Seidel update has the form
.
I don’t have 50 reputation so I can not comment this.
The line if(k >= 10e4), I think this does not make what you think it would. 10e4 is 100,000 and 1e4 is 10,000. This will be the reason why you think your counter does not work. Matlab ist still running because it’s running further than you think it will. Also I run in the same Problem knedlsepp already pointed out.
I am trying to solve equation systems, which contain algebraic as well as differential equations. To do this symbolically I need to combine dsolve and solve (do I?).
Consider the following example:
We have three base equations
a == b + c; % algebraic equation
diff(b,1) == 1/C1*y(t); % differential equation 1
diff(c,1) == 1/C2*y(t); % differential equation 2
Solving both differential equations, eliminating int(y,0..t) and then solving for c=f(C1,C2,a) yields
C1*b == C2*c or C1*(a-c) == C2*c
c = C1/(C1+C2) * a
How can I convince Matlab to give me that result? Here is what I tried:
syms a b c y C1 C2;
Eq1 = a == b + c; % algebraic equation
dEq1 = 'Db == 1/C1*y(t)'; % differential equation 1
dEq2 = 'Dc == 1/C2*y(t)'; % differential equation 2
[sol_dEq1, sol_dEq2]=dsolve(dEq1,dEq2,'b(0)==0','c(0)==0'); % this works, but no inclusion of algebraic equation
%[sol_dEq1, sol_dEq2]=dsolve(dEq1,dEq2,Eq1,'c'); % does not work
%solve(Eq1,dEq1,dEq2,'c') % does not work
%solve(Eq1,sol_dEq_C1,sol_dEq_C2,'c') % does not work
No combination of solve and/or dsolve with the equations or their solutions I tried gives me a useful result. Any ideas?
Now I assumed that you wanted the code to be rather general, so I made it to be able to work with any given number of equations and any given number of variables, and I did no calculation by hand.
Note that the way that the symbolic Toolbox works changes drastically from year to year, but hopefully this will work for you. Now one can add the equation Eq1 to the list of inputs of dSolve but there are two problems with that: One is that dSolve seems to prefer character inputs and the second is that dSolve doesn't seem to realize that there are 3 independent variables a, b, and c (it only sees 2 variables, b and c).
To solve the second problem, I differentiated the original equation to get a new differential equation, there were three problems with that: the first is Matlab evaluated the derivative of a with respect to t as 0, so I had to replace a with a(t) and such like for b and c (I called a(t) the long version of a). The second problem was Matlab used inconsistent notation, instead of representing the derivative of a as Da, it represented it as diff(a(t), t) thus I had to replace the latter with the former and such like for b and c; this gave me Da = Db + Dc. The final problem with that is that the system is now under determined, so I had to get the initial values, here I could have solved for a(0) but Matlab seemed happy with using a(0) = b(0) + c(0).
Now back to the original first problem, to solve that I had to convert every sym back into a char.
Here is the code
function SolveExample
syms a b c y C1 C2 t;
Eq1 = sym('a = b + c');
dEq1 = 'Db = 1/C1*y(t)';
dEq2 = 'Dc = 1/C2*y(t)';
[dEq3, initEq3] = ...
TurnEqIntoDEq(Eq1, [a b c], t, 0);
% In the most general case Eq1 will be an array
% and thus DEq3 will be one too
dEq3_char = SymArray2CharCell(dEq3);
initEq3_char = SymArray2CharCell(initEq3);
% Below is the same as
% dsolve(dEq1, dEq2, 'Da = Db + Dc', ...
% 'b(0)=0','c(0)=0', 'a(0) = b(0) + c(0)', 't');
[sol_dEq1, sol_dEq2, sol_dEq3] = dsolve(...
dEq1, dEq2, dEq3_char{:}, ...
'b(0)=0','c(0)=0', initEq3_char{:}, 't')
end
function [D_Eq, initEq] = ...
TurnEqIntoDEq(eq, depVars, indepVar, initialVal)
% Note that eq and depVars
% may all be vectors or scalars
% and they need not be the same size.
% eq = equations
% depVars = dependent variables
% indepVar = independent variable
% initialVal = initial value of indepVar
depVarsLong = sym(zeros(size(depVars)));
for k = 1:numel(depVars)
% Make the variables functions
% eg. a becomes a(t)
% This is so that diff(a, t) does not become 0
depVarsLong(k) = sym([char(depVars(k)) '(' ...
char(indepVar) ')']);
end
% Next make the equation in terms of these functions
eqLong = subs(eq, depVars, depVarsLong);
% Now find the ODE corresponding to the equation
D_EqLong = diff(eqLong, indepVar);
% Now replace all the long terms like 'diff(a(t), t)'
% with short terms like 'Da'
% otherwise dSolve will not work.
% First make the short variables 'Da'
D_depVarsShort = sym(zeros(size(depVars)));
for k = 1:numel(depVars)
D_depVarsShort(k) = sym(['D' char(depVars(k))]);
end
% Next make the long names like 'diff(a(t), t)'
D_depVarsLong = diff(depVarsLong, indepVar);
% Finally replace
D_Eq = subs(D_EqLong, D_depVarsLong, D_depVarsShort);
% Finally determine the equation
% governing the initial values
initEq = subs(eqLong, indepVar, initialVal);
end
function cc = SymArray2CharCell(sa)
cc = cell(size(sa));
for k = 1:numel(sa)
cc{k} = char(sa(k));
end
end
Some minor notes, I changed the == to = as that seems to be a difference between our versions of Matlab. Also I added t as the independent variable in dsolve. I also assumed that you know about cells, numel, linear indexes, ect.
I am using the matlab code from this book: http://books.google.com/books/about/Probability_Markov_chains_queues_and_sim.html?id=HdAQdzAjl60C
Here is the Code:
function [pi] = GE(Q)
A = Q';
n = size(A);
for i=1:n-1
for j=i+1:n
A(j,i) = -A(j,i)/A(i,i);
end
for j =i+1:n
for k=i+1:n
A(j,k) = A(j,k)+ A(j,i) * A(i,k);
end
end
end
x(n) = 1;
for i = n-1:-1:1
for j= i+1:n
x(i) = x(i) + A(i,j)*x(j);
end
x(i) = -x(i)/A(i,i);
end
pi = x/norm(x,1);
Is there a faster code that I am not aware of? I am calling this functions millions of times, and it takes too much time.
MATLAB has a whole set of built-in linear algebra routines - type help slash, help lu or help chol to get started with a few of the common ways to efficiently solve linear equations in MATLAB.
Under the hood these functions are generally calling optimised LAPACK/BLAS library routines, which are generally the fastest way to do linear algebra in any programming language. Compared with a "slow" language like MATLAB it would not be unexpected if they were orders of magnitude faster than an m-file implementation.
Hope this helps.
Unless you are specifically looking to implement your own, you should use Matlab's backslash operator (mldivide) or, if you want the factors, lu. Note that mldivide can do more than Gaussian elimination (e.g., it does linear least squares, when appropriate).
The algorithms used by mldivide and lu are from C and Fortran libraries, and your own implementation in Matlab will never be as fast. If, however, you are determined to use your own implementation and want it to be faster, one option is to look for ways to vectorize your implementation (maybe start here).
One other thing to note: the implementation from the question does not do any pivoting, so its numerical stability will generally be worse than an implementation that does pivoting, and it will even fail for some nonsingular matrices.
Different variants of Gaussian elimination exist, but they are all O(n3) algorithms. If any one approach is better than another depends on your particular situation and is something you would need to investigate more.
function x = naiv_gauss(A,b);
n = length(b); x = zeros(n,1);
for k=1:n-1 % forward elimination
for i=k+1:n
xmult = A(i,k)/A(k,k);
for j=k+1:n
A(i,j) = A(i,j)-xmult*A(k,j);
end
b(i) = b(i)-xmult*b(k);
end
end
% back substitution
x(n) = b(n)/A(n,n);
for i=n-1:-1:1
sum = b(i);
for j=i+1:n
sum = sum-A(i,j)*x(j);
end
x(i) = sum/A(i,i);
end
end
Let's assume Ax=d
Where A and d are known matrices.
We want to represent "A" as "LU" using "LU decomposition" function embedded in matlab thus:
LUx = d
This can be done in matlab following:
[L,U] = lu(A)
which in terms returns an upper triangular matrix in U and a permuted lower triangular matrix in L such that A = LU. Return value L is a product of lower triangular and permutation matrices. (https://www.mathworks.com/help/matlab/ref/lu.html)
Then if we assume Ly = d where y=Ux.
Since x is Unknown, thus y is unknown too, by knowing y we find x as follows:
y=L\d;
x=U\y
and the solution is stored in x.
This is the simplest way to solve system of linear equations providing that the matrices are not singular (i.e. the determinant of matrix A and d is not zero), otherwise, the quality of the solution would not be as good as expected and might yield wrong results.
if the matrices are singular thus cannot be inversed, another method should be used to solve the system of the linear equations.
For the naive approach (aka without row swapping) for an n by n matrix:
function A = naiveGauss(A)
% find's the size
n = size(A);
n = n(1);
B = zeros(n,1);
% We have 3 steps for a 4x4 matrix so we have
% n-1 steps for an nxn matrix
for k = 1 : n-1
for i = k+1 : n
% step 1: Create multiples that would make the top left 1
% printf("multi = %d / %d\n", A(i,k), A(k,k), A(i,k)/A(k,k) )
for j = k : n
A(i,j) = A(i,j) - (A(i,k)/A(k,k)) * A(k,j);
end
B(i) = B(i) - (A(i,k)/A(k,k)) * B(k);
end
end
function Sol = GaussianElimination(A,b)
[i,j] = size(A);
for j = 1:i-1
for i = j+1:i
Sol(i,j) = Sol(i,:) -( Sol(i,j)/(Sol(j,j)*Sol(j,:)));
end
end
disp(Sol);
end
I think you can use the matlab function rref:
[R,jb] = rref(A,tol)
It produces a matrix in reduced row echelon form.
In my case it wasn't the fastest solution.
The solution below was faster in my case by about 30 percent.
function C = gauss_elimination(A,B)
i = 1; % loop variable
X = [ A B ];
[ nX mX ] = size( X); % determining the size of matrix
while i <= nX % start of loop
if X(i,i) == 0 % checking if the diagonal elements are zero or not
disp('Diagonal element zero') % displaying the result if there exists zero
return
end
X = elimination(X,i,i); % proceeding forward if diagonal elements are non-zero
i = i +1;
end
C = X(:,mX);
function X = elimination(X,i,j)
% Pivoting (i,j) element of matrix X and eliminating other column
% elements to zero
[ nX mX ] = size( X);
a = X(i,j);
X(i,:) = X(i,:)/a;
for k = 1:nX % loop to find triangular form
if k == i
continue
end
X(k,:) = X(k,:) - X(i,:)*X(k,j);
end