I am trying to implement a finite difference scheme for KdV equation in MATLAB, and I have most of the code ready, except for approximation at the first level using initial condition. It was suggested I use Euler's method to obtain 'u' at m=1, and then use the scheme for m>=2.
How does one apply Euler's method in this context? Even just a general answer for approximation at the first level would be appreciated.
I am including my code for reference
close all
clear
clc
% Generating grid with n points, with the space between two points being
%(x2-x1)/(n-1)
x = linspace(-5,5,1001);
N=1001;
h=x(2)-x(1); % grid size
dt=0.05;
%Soliton initial condition
Am=8; %Amplitude
mu=sqrt(Am/2);
x0=-15;
c=1;
syms H(x)
H(x)=piecewise(x < 0,0,x > 0,1);
u= Am*(sech(mu*(x'-x0))).^2+c^2*H(x);
% Creating a matrix A - First order
A = diag(ones(N-1,1),1)-diag(ones(N-1,1),-1);
Cvector = zeros(N, 1);
Cvector(end) = 1;
u_ic = Cvector;
% First order finite difference scheme
diff_first=A*u/(2*h)+1/(2*h)*u_ic;
% Weighted average matrix for the term 'u'
A_w = diag(ones(N-1,1),1)+diag(ones(N-1,1),-1)+diag(ones(N,1));
diff_w=2*A_w*u +2*u_ic;
% Matrix multiplication of first derivative and weighted average for 6uu_x
diff_middle=diff_first.*diff_w;
% Creating a Third Order Matrix
r = zeros(1,N);
r(2:3) = [-2,1];
c = -r;
A_third = toeplitz(c,r);
% Difference scheme for third order term
diff_third=A_third*u/(h*h*h)-1/(h*h*h)*u_ic;
%Computing finite difference method
u = u - 2*dt*diff_middle-dt*diff_third;
plot(u)
Related
Matlab has the function randn to draw from a normal distribution e.g.
x = 0.5 + 0.1*randn()
draws a pseudorandom number from a normal distribution of mean 0.5 and standard deviation 0.1.
Given this, is the following Matlab code equivalent to sampling from a normal distribution truncated at 0 at 1?
while x <=0 || x > 1
x = 0.5 + 0.1*randn();
end
Using MATLAB's Probability Distribution Objects makes sampling from truncated distributions very easy.
You can use the makedist() and truncate() functions to define the object and then modify (truncate it) to prepare the object for the random() function which allows generating random variates from it.
% MATLAB R2017a
pd = makedist('Normal',0.5,0.1) % Normal(mu,sigma)
pdt = truncate(pd,0,1) % truncated to interval (0,1)
sample = random(pdt,numRows,numCols) % Sample from distribution `pdt`
Once the object is created (here it is pdt, the truncated version of pd), you can use it in a variety of function calls.
To generate samples, random(pdt,m,n) produces a m x n array of samples from pdt.
Further, if you want to avoid use of toolboxes, this answer from #Luis Mendo is correct (proof below).
figure, hold on
h = histogram(cr,'Normalization','pdf','DisplayName','#Luis Mendo samples');
X = 0:.01:1;
p = plot(X,pdf(pdt,X),'b-','DisplayName','Theoretical (w/ truncation)');
You need the following steps
1. Draw a random value from uniform distribution, u.
2. Assuming the normal distribution is truncated at a and b. get
u_bar = F(a)*u +F(b) *(1-u)
3. Use the inverse of F
epsilon= F^{-1}(u_bar)
epsilon is a random value for the truncated normal distribution.
Why don't you vectorize? It will probably be faster:
N = 1e5; % desired number of samples
m = .5; % desired mean of underlying Gaussian
s = .1; % desired std of underlying Gaussian
lower = 0; % lower value for truncation
upper = 1; % upper value for truncation
remaining = 1:N;
while remaining
result(remaining) = m + s*randn(1,numel(remaining)); % (pre)allocates the first time
remaining = find(result<=lower | result>upper);
end
I was asked to do circular convolution between two functions by sampling them, using the functions cconv. A known result of this sort of convolution is: CCONV( sin(x), sin(x) ) == -pi*cos(x)
To test the above I did:
w = linspace(0,2*pi,1000);
l = linspace(0,2*pi,1999);
stem(l,cconv(sin(w),sin(w))
but the result I got was:
which is absolutely not -pi*cos(x).
Can anybody please explain what is wrong with my code and how to fix it?
In the documentation of cconv it says that:
c = cconv(a,b,n) circularly convolves vectors a and b. n is the length of the resulting vector. If you omit n, it defaults to length(a)+length(b)-1. When n = length(a)+length(b)-1, the circular convolution is equivalent to the linear convolution computed with conv.
I believe that the reason for your problem is that you do not specify the 3rd input to cconv, which then selects the default value, which is not the right one for you. I have made an animation showing what happens when different values of n are chosen.
If you compare my result for n=200 to your plot you will see that the amplitude of your data is 10 times larger whereas the length of your linspace is 10 times bigger. This means that some normalization is needed, likely a multiplication by the linspace step.
Indeed, after proper scaling and choice of n we get the right result:
res = 100; % resolution
w = linspace(0,2*pi,res);
dx = diff(w(1:2)); % grid step
stem( linspace(0,2*pi,res), dx * cconv(sin(w),sin(w),res) );
This is the code I used for the animation:
hF = figure();
subplot(1,2,1); hS(1) = stem(1,cconv(1,1,1)); title('Autoscaling');
subplot(1,2,2); hS(2) = stem(1,cconv(1,1,1)); xlim([0,7]); ylim(50*[-1,1]); title('Constant limits');
w = linspace(0,2*pi,100);
for ind1 = 1:200
set(hS,'XData',linspace(0,2*pi,ind1));
set(hS,'YData',cconv(sin(w),sin(w),ind1));
suptitle("n = " + ind1);
drawnow
% export_fig(char("D:\BLABLA\F" + ind1 + ".png"),'-nocrop');
end
What are the difference between the following two functions?
prepTransform.m
function [mu trmx] = prepTransform(tvec, comp_count)
% Computes transformation matrix to PCA space
% tvec - training set (one row represents one sample)
% comp_count - count of principal components in the final space
% mu - mean value of the training set
% trmx - transformation matrix to comp_count-dimensional PCA space
% this is memory-hungry version
% commented out is the version proper for Win32 environment
tic;
mu = mean(tvec);
cmx = cov(tvec);
%cmx = zeros(size(tvec,2));
%f1 = zeros(size(tvec,1), 1);
%f2 = zeros(size(tvec,1), 1);
%for i=1:size(tvec,2)
% f1(:,1) = tvec(:,i) - repmat(mu(i), size(tvec,1), 1);
% cmx(i, i) = f1' * f1;
% for j=i+1:size(tvec,2)
% f2(:,1) = tvec(:,j) - repmat(mu(j), size(tvec,1), 1);
% cmx(i, j) = f1' * f2;
% cmx(j, i) = cmx(i, j);
% end
%end
%cmx = cmx / (size(tvec,1)-1);
toc
[evec eval] = eig(cmx);
eval = sum(eval);
[eval evid] = sort(eval, 'descend');
evec = evec(:, evid(1:size(eval,2)));
% save 'nist_mu.mat' mu
% save 'nist_cov.mat' evec
trmx = evec(:, 1:comp_count);
pcaTransform.m
function [pcaSet] = pcaTransform(tvec, mu, trmx)
% tvec - matrix containing vectors to be transformed
% mu - mean value of the training set
% trmx - pca transformation matrix
% pcaSet - output set transforrmed to PCA space
pcaSet = tvec - repmat(mu, size(tvec,1), 1);
%pcaSet = zeros(size(tvec));
%for i=1:size(tvec,1)
% pcaSet(i,:) = tvec(i,:) - mu;
%end
pcaSet = pcaSet * trmx;
Which one is actually doing PCA?
If one is doing PCA, what is the other one doing?
The first function prepTransform is actually doing the PCA on your training data where you are determining the new axes to represent your data onto a lower dimensional space. What it does is that it finds the eigenvectors of the covariance matrix of your data and then orders the eigenvectors such that the eigenvector with the largest eigenvalue appears in the first column of the eigenvector matrix evec and the eigenvector with the smallest eigenvalue appears in the last column. What's important with this function is that you can define how many dimensions you want to reduce the data down to by keeping the first N columns of evec which will allow you to reduce your data down to N dimensions. The discarding of the other columns and keeping only the first N is what is set as trmx in the code. The variable N is defined by the prep_count variable in prepTransform function.
The second function pcaTransform finally transforms data that is defined within the same domain as your training data but not necessarily the training data itself (it could be if you wish) onto the lower dimensional space that is defined by the eigenvectors of the covariance matrix. To finally perform the reduction of dimensions, or dimensionality reduction as it is popularly known, you simply take your training data where each feature is subtracted from its mean and you multiply your training data by the matrix trmx. Note that prepTransform outputting the mean of each feature in the vector mu is important in order to mean subtract your data when you finally call pcaTransform.
How to use these functions
To use these functions effectively, first determine the trmx matrix, which contain the principal components of your data by first defining how many dimensions you want to reduce your data down to as well as the mean of each feature stored in mu:
N = 2; % Reduce down to two dimensions for example
[mu, trmx] = prepTransform(tvec, N);
Next you can finally perform dimensionality reduction on your data that is defined within the same domain as tvec (or even tvec if you wish, but it doesn't have to be) by:
pcaSet = pcaTransform(tvec, mu, trmx);
In terms of vocabulary, pcaSet contain what are known as the principal scores of your data, which is the term used for the transformation of your data to the lower dimensional space.
If I can recommend something...
Finding PCA through the eigenvector approach is known to be unstable. I highly recommend you use the Singular Value Decomposition via svd on the covariance matrix where the V matrix of the result already gives you the eigenvectors sorted which correspond to your principal components:
mu = mean(tvec, 1);
[~,~,V] = svd(cov(tvec));
Then perform the transformation by taking the mean subtracted data per feature and multiplying by the V matrix, once you subset and grab the first N columns of V:
N = 2;
X = bsxfun(#minus, tvec, mu);
pcaSet = X*V(:, 1:N);
X is the mean subtracted data which performs the same thing as doing pcaSet = tvec - repmat(mu, size(tvec,1), 1);, but you are not explicitly replicating the mean vector over each training example but letting bsxfun do that for you internally. However, taking advantage of MATLAB R2016b, this repeating can be done without the explicit call to bsxfun:
X = tvec - mu;
Further Reading
If you fully want to understand the code that was written and the theory behind what it's doing, I recommend the following two Stack Overflow posts that I have written that talk about the topic:
What does selecting the largest eigenvalues and eigenvectors in the covariance matrix mean in data analysis?
How to use eigenvectors obtained through PCA to reproject my data?
The first post brings the code you presented into light which performs PCA using the eigenvector approach. The second post touches base on how you'd do it using the SVD towards the end of the answer. This answer I've written here is a mix between the two posts above.
My approach
fun = #(y) (1/sqrt(pi))*exp(-(y-1).^2).*log(1 + exp(-4*y))
integral(fun,-Inf,Inf)
This gives NaN.
So I tried plotting it.
y= -10:0.1:10;
plot(y,exp(-(y-1).^2).*log(1 + exp(-4*y)))
Then understood that domain (siginificant part) is from -4 to +4.
So changed the limits to
integral(fun,-10,10)
However I do not want to always plot the graph and then know its limits. So is there any way to know the integral directly from -Inf to Inf.
Discussion
If your integrals are always of the form
I would use a high-order Gauss–Hermite quadrature rule.
It's similar to the Gauss-Legendre-Kronrod rule that forms the basis for quadgk but is specifically tailored for integrals over the real line with a standard Gaussian multiplier.
Rewriting your equation with the substitution x = y-1, we get
.
The integral can then be computed using the Gauss-Hermite rule of arbitrary order (within reason):
>> order = 10;
>> [nodes,weights] = GaussHermiteRule(order);
>> f = #(x) log(1 + exp(-4*(x+1)))/sqrt(pi);
>> sum(f(nodes).*weights)
ans =
0.1933
I'd note that the function below builds a full order x order matrix to compute nodes, so it shouldn't be made too large.
There is a way to avoid this by explicitly computing the weights, but I decided to be lazy.
Besides, event at order 100, the Gaussian multiplier is about 2E-98, so the integrand's contribution is extremely minimal.
And while this isn't inherently adaptive, a high-order rule should be sufficient in most cases ... I hope.
Code
function [nodes,weights] = GaussHermiteRule(n)
% ------------------------------------------------------------------------------
% Find the nodes and weights for a Gauss-Hermite Quadrature integration.
%
if (n < 1)
error('There is no Gauss-Hermite rule of order 0.');
elseif (n < 0) || (abs(n - round(n)) > eps())
error('Given order ''n'' must be a strictly positive integer.');
else
n = round(n);
end
% Get the nodes and weights from the Golub-Welsch function
n = (0:n)' ;
b = n*0 ;
a = b + 0.5 ;
c = n ;
[nodes,weights] = GolubWelsch(a,b,c,sqrt(pi));
end
function [xk,wk] = GolubWelsch(ak,bk,ck,mu0)
%GolubWelsch
% Calculate the approximate* nodes and weights (normalized to 1) of an orthogonal
% polynomial family defined by a three-term reccurence relation of the form
% x pk(x) = ak pkp1(x) + bk pk(x) + ck pkm1(x)
%
% The weight scale factor mu0 is the integral of the weight function over the
% orthogonal domain.
%
% Calculate the terms for the orthonormal version of the polynomials
alpha = sqrt(ak(1:end-1) .* ck(2:end));
% Build the symmetric tridiagonal matrix
T = full(spdiags([[alpha;0],bk,[0;alpha]],[-1,0,+1],length(alpha),length(alpha)));
% Calculate the eigenvectors and values of the matrix
[V,xk] = eig(T,'vector');
% Calculate the weights from the eigenvectors - technically, Golub-Welsch requires
% a normalization, but since MATLAB returns unit eigenvectors, it is omitted.
wk = mu0*(V(1,:).^2)';
end
I've had success with transforming such infinite-bounded integrals using a numerical variable transformation, as explained in Numerical Recipes 3e, section 4.5.3. Basically, you substitute in y=c*tan(t)+b and then numerically integrate over t in (-pi/2,pi/2), which sweeps y from -infinity to infinity. You can tune the values of c and b to optimize the process. This approach largely dodges the question of trying to determine cutoffs in the domain, but for this to work reliably using quadrature you have to know that the integrand does not have features far from y=b.
A quick and dirty solution would be to look for a position, where your function is sufficiently small enough and then taking it as limits. This assumes that for x>0 the function fun decreases montonically and fun(x) is roughly the same size as fun(-x) for all x.
%// A small number
epsilon = eps;
%// Stepsize for searching bound
stepTest = 1;
%// Starting position for searching bound
position = 0;
%// Not yet small enough
smallEnough = false;
%// Search bound
while ~smallEnough
smallEnough = (fun(position) < eps);
position = position + stepTest;
end
%// Calculate integral
integral(fun, -position, position)
If your were happy with plotting the function, deciding by eye where you can cut, then this code will suffice, I guess.
I have a problem with numerical derivative of a vector that is x: Nx1 with respect to another vector t (time) that is the same size of x.
I do the following (x is chosen to be sine function as an example):
t=t0:ts:tf;
x=sin(t);
xd=diff(x)/ts;
but the answer xd is (N-1)x1 and I figured out that it does not compute derivative corresponding to the first element of x.
is there any other way to compute this derivative?
You are looking for the numerical gradient I assume.
t0 = 0;
ts = pi/10;
tf = 2*pi;
t = t0:ts:tf;
x = sin(t);
dx = gradient(x)/ts
The purpose of this function is a different one (vector fields), but it offers what diff doesn't: input and output vector of equal length.
gradient calculates the central difference between data points. For an
array, matrix, or vector with N values in each row, the ith value is
defined by
The gradient at the end points, where i=1 and i=N, is calculated with
a single-sided difference between the endpoint value and the next
adjacent value within the row. If two or more outputs are specified,
gradient also calculates central differences along other dimensions.
Unlike the diff function, gradient returns an array with the same
number of elements as the input.
I know I'm a little late to the game here, but you can also get an approximation of the numerical derivative by taking the derivatives of the polynomial (cubic) splines that runs through your data:
function dy = splineDerivative(x,y)
% the spline has continuous first and second derivatives
pp = spline(x,y); % could also use pp = pchip(x,y);
[breaks,coefs,K,r,d] = unmkpp(pp);
% pre-allocate the coefficient vector
dCoeff = zeroes(K,r-1);
% Columns are ordered from highest to lowest power. Both spline and pchip
% return 4xn matrices, ordered from 3rd to zeroth power. (Thanks to the
% anonymous person who suggested this edit).
dCoeff(:, 1) = 3 * coefs(:, 1); % d(ax^3)/dx = 3ax^2;
dCoeff(:, 2) = 2 * coefs(:, 2); % d(ax^2)/dx = 2ax;
dCoeff(:, 3) = 1 * coefs(:, 3); % d(ax^1)/dx = a;
dpp = mkpp(breaks,dCoeff,d);
dy = ppval(dpp,x);
The spline polynomial is always guaranteed to have continuous first and second derivatives at each point. I haven not tested and compared this against using pchip instead of spline, but that might be another option as it too has continuous first derivatives (but not second derivatives) at every point.
The advantage of this is that there is no requirement that the step size be even.
There are some options to work-around your issue.
First: you can make your domain larger. Instead of N, use N+1 gridpoints.
Second: depending on the end-point of interest, you can use
Forward difference: F(x + dx) - F(x)
Backward difference: F(x) - F(x - dx)