I have to solve this equation
(tan(qh))/(tan(ph))=-(4k^2 pq)/(q^2-k^2)^2
here
d = 6.35;
h = d/2;
ct = 3076.4;
cl = 6207.7;
I want to give k values and each k value I get many omega (w) values. Just like sin(x)=0 means x values are nπ
I tried in this way.
syms w
d = 0.0065;
h = d/2;
ct = 3.0764;
cl = 6.2077;
k = 1:50:1000;
for i=1:length(k)
p = sqrt((w/cl)^2-k(i)^2); % p
q = sqrt((w/ct)^2-k(i)^2); % q
f = (tan(q*h)*(q^2-k(i)^2)^2)==-(4*k(i)^2*p*q*tan(p*h));
z = zeros(1,30);
for j=1:30
z(j) = vpasolve(f,w,[0 1000],'Random',true);
end
z=z(z>=0);
b = double(z);
c = unique(b);
cp = c/k(i); %phase velocity
f = c/(2*pi); %frequency
plot(f,cp,'bo')
hold on
end
In my case, answers are not matching with this image. My answers are in 10^-18 that is too small.
citation for this image. pg#7-9
https://smartech.gatech.edu/handle/1853/37158
Related
I am currently working on solving the problem $-\alpha u'' + \beta u = f$ with Neumann conditions on the edge, with the finite element method in MATLAB.
I managed to set up a code that works for P1 and P2 Lagragne finite elements (i.e: linear and quadratic) and the results are good!
I am trying to implement the finite element method using the Hermite basis. This basis is defined by the following basis functions and derivatives:
syms x
phi(x) = [2*x^3-3*x^2+1,-2*x^3+3*x^2,x^3-2*x^2+x,x^3-x^2]
% Derivative
dphi = [6*x.^2-6*x,-6*x.^2+6*x,3*x^2-4*x+1,3*x^2-2*x]
The problem with the following code is that the solution vector u is not good. I know that there must be a problem in the S and F element matrix calculation loop, but I can't see where even though I've been trying to make changes for a week.
Can you give me your opinion? Hopefully someone can see my error.
Thanks a lot,
% -alpha*u'' + beta*u = f
% u'(a) = bd1, u'(b) = bd2;
a = 0;
b = 1;
f = #(x) (1);
alpha = 1;
beta = 1;
% Neuamnn boundary conditions
bn1 = 1;
bn2 = 0;
syms ue(x)
DE = -alpha*diff(ue,x,2) + beta*ue == f;
du = diff(ue,x);
BC = [du(a)==bn1, du(b)==bn2];
ue = dsolve(DE, BC);
figure
fplot(ue,[a,b], 'r', 'LineWidth',2)
N = 2;
nnod = N*(2+2); % Number of nodes
neq = nnod*1; % Number of equations, one degree of freedom per node
xnod = linspace(a,b,nnod);
nodes = [(1:3:nnod-3)', (2:3:nnod-2)', (3:3:nnod-1)', (4:3:nnod)'];
phi = #(xi)[2*xi.^3-3*xi.^2+1,2*xi.^3+3*xi.^2,xi.^3-2*xi.^2+xi,xi.^3-xi.^2];
dphi = #(xi)[6*xi.^2-6*xi,-6*xi.^2+6*xi,3*xi^2-4*xi+1,3*xi^2-2*xi];
% Here, just calculate the integral using gauss quadrature..
order = 5;
[gp, gw] = gauss(order, 0, 1);
S = zeros(neq,neq);
M = S;
F = zeros(neq,1);
for iel = 1:N
%disp(iel)
inod = nodes(iel,:);
xc = xnod(inod);
h = xc(end)-xc(1);
Se = zeros(4,4);
Me = Se;
fe = zeros(4,1);
for ig = 1:length(gp)
xi = gp(ig);
iw = gw(ig);
Se = Se + dphi(xi)'*dphi(xi)*1/h*1*iw;
Me = Me + phi(xi)'*phi(xi)*h*1*iw;
x = phi(xi)*xc';
fe = fe + phi(xi)' * f(x) * h * 1 * iw;
end
% Assembly
S(inod,inod) = S(inod, inod) + Se;
M(inod,inod) = M(inod, inod) + Me;
F(inod) = F(inod) + fe;
end
S = alpha*S + beta*M;
g = zeros(neq,1);
g(1) = -alpha*bn1;
g(end) = alpha*bn2;
alldofs = 1:neq;
u = zeros(neq,1); %Pre-allocate
F = F + g;
u(alldofs) = S(alldofs,alldofs)\F(alldofs)
Warning: Matrix is singular to working precision.
u = 8×1
NaN
NaN
NaN
NaN
NaN
NaN
NaN
NaN
figure
fplot(ue,[a,b], 'r', 'LineWidth',2)
hold on
plot(xnod, u, 'bo')
for iel = 1:N
inod = nodes(iel,:);
xc = xnod(inod);
U = u(inod);
xi = linspace(0,1,100)';
Ue = phi(xi)*U;
Xe = phi(xi)*xc';
plot(Xe,Ue,'b -')
end
% Gauss function for calculate the integral
function [x, w, A] = gauss(n, a, b)
n = 1:(n - 1);
beta = 1 ./ sqrt(4 - 1 ./ (n .* n));
J = diag(beta, 1) + diag(beta, -1);
[V, D] = eig(J);
x = diag(D);
A = b - a;
w = V(1, :) .* V(1, :);
w = w';
x=x';
end
You can find the same post under MATLAB site for syntax highlighting.
Thanks
I tried to read courses, search in different documentation and modify my code without success.
I'm trying to solve a system of ode's using Runge-kutta, i made a function for RK2(f,h,x0,y0,xfinal) and tried to solve the system shown below with specified IC's. Could someone help fix the code as I get errors and code doesn't work.
ode set
beta = 1/3;
gamma = 1/7;
syms R S I % Symbolic Math Toolbox
N = S+I+R;
ode1 = -(beta*I*S)/N;
ode2 = -(beta*I*S)/N-gamma*I;
ode3 = gamma*I;
odes = [ode1,ode2,ode3];
for j = odes
RK2(j,0.2,0,8e6,7);
end
function [xs,ys] = RK2(f,h,x0,y0,xfinal)
ffnc = matlabFunction(f);
fprintf('\n x y ');
o = 1;
while x0 <= xfinal
fprintf('\n%4.3f %4.3f ',x0,y0); %values of x and y
xs(o) = x0;
ys(o) = y0;
k1 = h*ffnc (x0,y0);
x1 = x0+h;
k2 = h*ffnc (x1,y0+k1);
y1 = y0+(k1+k2)/2;
x0 = x1;
y0 = y1;
o = o+1;
end
end
I'm trying to implement Divide and Conquer SVD of an upper bidiagonal matrix B, but my code is not working. The error is:
"Unable to perform assignment because the size of the left side is
3-by-3 and the size of the right side is 2-by-2.
V_bar(1:k,1:k) = V1;"
Can somebody help me fix it? Thanks.
function [U,S,V] = DivideConquer_SVD(B)
[m,n] = size(B);
k = floor(m/2);
if k == 0
U = 1;
V = 1;
S = B;
return;
else
% Divide the input matrix
alpha = B(k,k);
beta = B(k,k+1);
e1 = zeros(m,1);
e2 = zeros(m,1);
e1(k) = 1;
e2(k+1) = 1;
B1 = B(1:k-1,1:k);
B2 = B(k+1:m,k+1:m);
%recursive computations
[U1,S1,V1] = DivideConquer_SVD(B1);
[U2,S2,V2] = DivideConquer_SVD(B2);
U_bar = zeros(m);
U_bar(1:k-1,1:k-1) = U1;
U_bar(k,k) = 1;
U_bar((k+1):m,(k+1):m) = U2;
D = zeros(m);
D(1:k-1,1:k) = S1;
D((k+1):m,(k+1):m) = S2;
V_bar = zeros(m);
V_bar(1:k,1:k) = V1;
V_bar((k+1):m,(k+1):m) = V2;
u = alpha*e1'*V_bar + beta*e2'*V_bar;
u = u';
D_tilde = D*D + u*u';
% compute eigenvalues and eigenvectors of D^2+uu'
[L1,Q1] = eig(D_tilde);
eigs = diag(L1);
S = zeros(m,n)
S(1:(m+1):end) = eigs
U_tilde = Q1;
V_tilde = Q1;
%Compute eigenvectors of the original input matrix T
U = U_bar*U_tilde;
V = V_bar*V_tilde;
return;
end
With limited mathematical knowledge, you need to help me a bit more -- as I cannot judge if the approach is correct in a mathematical way (with no theory given;) ). Anyway, I couldn't even reproduce the error e.g with this matrix, which The MathWorks use to illustrate their LU matrix factorization
A = [10 -7 0
-3 2 6
5 -1 5];
So I tried to structure your code a bit and gave some hints. Extend this to make your code clearer for those people (like me) who are not too familiar with matrix decomposition.
function [U,S,V] = DivideConquer_SVD(B)
% m x n matrix
[m,n] = size(B);
k = floor(m/2);
if k == 0
disp('if') % for debugging
U = 1;
V = 1;
S = B;
% return; % net necessary as you don't do anything afterwards anyway
else
disp('else') % for debugging
% Divide the input matrix
alpha = B(k,k); % element on diagonal
beta = B(k,k+1); % element on off-diagonal
e1 = zeros(m,1);
e2 = zeros(m,1);
e1(k) = 1;
e2(k+1) = 1;
% divide matrix
B1 = B(1:k-1,1:k); % upper left quadrant
B2 = B(k+1:m,k+1:m); % lower right quadrant
% recusrsive function call
[U1,S1,V1] = DivideConquer_SVD(B1);
[U2,S2,V2] = DivideConquer_SVD(B2);
U_bar = zeros(m);
U_bar(1:k-1,1:k-1) = U1;
U_bar(k,k) = 1;
U_bar((k+1):m,(k+1):m) = U2;
D = zeros(m);
D(1:k-1,1:k) = S1;
D((k+1):m,(k+1):m) = S2;
V_bar = zeros(m);
V_bar(1:k,1:k) = V1;
V_bar((k+1):m,(k+1):m) = V2;
u = (alpha*e1.'*V_bar + beta*e2.'*V_bar).'; % (little show-off tip: '
% is the complex transpose operator; .' is the "normal" transpose
% operator. It's good practice to distinguish between them but there
% is no difference for real matrices anyway)
D_tilde = D*D + u*u.';
% compute eigenvalues and eigenvectors of D^2+uu'
[L1,Q1] = eig(D_tilde);
eigs = diag(L1);
S = zeros(m,n);
S(1:(m+1):end) = eigs;
U_tilde = Q1;
V_tilde = Q1;
% Compute eigenvectors of the original input matrix T
U = U_bar*U_tilde;
V = V_bar*V_tilde;
% return; % net necessary as you don't do anything afterwards anyway
end % for
end % function
Consider the following calculation of the tangent tangent correlation which is performed in a for loop
v1=rand(25,1);
v2=rand(25,1);
n=25;
nSteps=10;
mean_theta = zeros(nSteps,1);
for j=1:nSteps
theta=[];
for i=1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta = [theta acosd(d/n1/n2)];
end
mean_theta(j)=mean(theta);
end
plot(mean_theta)
How can matlab matrix calculations be utilized to make this performance better?
There are several things you can do to speed up your code. First, always preallocate. This converts:
theta = [];
for i = 1:(n-j)
%...
theta = [theta acosd(d/n1/n2)];
end
into:
theta = zeros(1,n-j);
for i = 1:(n-j)
%...
theta(i) = acosd(d/n1/n2);
end
Next, move the normalization out of the loops. There is no need to normalize over and over again, just normalize the input:
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
%...
theta(i) = acosd(dot(v(i,:),v(i+j,:)));
This does change the output very slightly, within numerical precision, because the different order of operations leads to different floating-point rounding error.
Finally, you can remove the inner loop by vectorizing that computation:
i = 1:(n-j);
theta = acosd(dot(v(i,:),v(i+j,:),2));
Timings (n=25):
Original: 0.0019 s
Preallocate: 0.0013 s
Normalize once: 0.0011 s
Vectorize: 1.4176e-04 s
Timings (n=250):
Original: 0.0185 s
Preallocate: 0.0146 s
Normalize once: 0.0118 s
Vectorize: 2.5694e-04 s
Note how the vectorized code is the only one whose timing doesn't grow linearly with n.
Timing code:
function so
n = 25;
v1 = rand(n,1);
v2 = rand(n,1);
nSteps = 10;
mean_theta1 = method1(v1,v2,nSteps);
mean_theta2 = method2(v1,v2,nSteps);
fprintf('diff method1 vs method2: %g\n',max(abs(mean_theta1(:)-mean_theta2(:))));
mean_theta3 = method3(v1,v2,nSteps);
fprintf('diff method1 vs method3: %g\n',max(abs(mean_theta1(:)-mean_theta3(:))));
mean_theta4 = method4(v1,v2,nSteps);
fprintf('diff method1 vs method4: %g\n',max(abs(mean_theta1(:)-mean_theta4(:))));
timeit(#()method1(v1,v2,nSteps))
timeit(#()method2(v1,v2,nSteps))
timeit(#()method3(v1,v2,nSteps))
timeit(#()method4(v1,v2,nSteps))
function mean_theta = method1(v1,v2,nSteps)
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta=[];
for i=1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta = [theta acosd(d/n1/n2)];
end
mean_theta(j) = mean(theta);
end
function mean_theta = method2(v1,v2,nSteps)
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta = zeros(1,n-j);
for i = 1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta(i) = acosd(d/n1/n2);
end
mean_theta(j) = mean(theta);
end
function mean_theta = method3(v1,v2,nSteps)
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta = zeros(1,n-j);
for i = 1:(n-j)
theta(i) = acosd(dot(v(i,:),v(i+j,:)));
end
mean_theta(j) = mean(theta);
end
function mean_theta = method4(v1,v2,nSteps)
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
i = 1:(n-j);
theta = acosd(dot(v(i,:),v(i+j,:),2));
mean_theta(j) = mean(theta);
end
Here is a full vectorized solution:
i = 1:n-1;
j = (1:nSteps).';
ij= min(i+j,n);
a = cat(3, v1(i).', v2(i).');
b = cat(3, v1(ij), v2(ij));
d = sum(a .* b, 3);
n1 = sum(a .^ 2, 3);
n2 = sum(b .^ 2, 3);
theta = acosd(d./sqrt(n1.*n2));
idx = (1:nSteps).' <= (n-1:-1:1);
mean_theta = sum(theta .* idx ,2) ./ sum(idx,2);
Result of Octave timings for my method,method4 from the answer provided by #CrisLuengo and the original method (n=250):
Full vectorized : 0.000864983 seconds
Method4(Vectorize) : 0.002774 seconds
Original(loop) : 0.340693 seconds
Is there an algorithm that computes the Drazin inverse of a singular matrix? I would like to apply it either in MATLAB or Mathematica.
Read this article:
Fanbin Bu and Yimin Wei, The algorithm for computing the Drazin inverses of two-variable polynomial matrices, Applied mathematics and computation 147.3 (2004): 805-836.
in appendix there are several MATLAB code. The first one is this:
function DrazinInverse1a = DrazinInverse1(a)
%-----------------------------------------
%Compute the Drazin Inverse of a matrix 'a' using the limited algorithm.
%Need computing the index of 'a'.
global q1 q2 s1 s2
[m,n] = size(a);
if m~= n
disp('Matrix is must be square!')
end
%-----------------------------------------
% Computer the index of A and note r = rank(A^k).
[k,r,a1,a] = index(a);
F = eye(n);
g = -trace(a);
g = collect(g);
for i = 1:r-1
G = g*eye(n);
F = a*F+G;
g = -trace(a*F)/(i+1);
g = collect(g);
end
DrazinInverse1a = a1*F;
DrazinInverse1a = -1/g*DrazinInverse1a;
DrazinInverse1a = simplify(DrazinInverse1a);
end
function [k,r,a1,a] = index(a)
%To compute the index of 'a'.
k = 0;
n = length(a);
r = n;
a0 = a;
r1 = rank(a);
a1 = eye(n);
while r ~= r1
r = r1;
a1 = a;
a = a*a0;
r1 = rank(a);
k = k+1;
end
r = sym2poly(r);
end