I'm talking about Volterra integral equations of second order:
In my case g is an ugly integral also between a and x, also a=0 (for both g and the integral above). K is equal to 1.
I found some information about Fredholm equations, but they are not exactly the same (fixed intervals, they don't have x on the integral sign), I wonder if maybe I can reconduct my analysis to a Fredholm equation? And if so, how can I solve it in Matlab?
You have a Volterra equation of the second kind, and in the case when the kernel takes the form k(x-t) it can be solved an operational way(method).
Since your task is not unique, will be convenient to use a ready solution:
http://www.mathworks.com/matlabcentral/fileexchange/49721-volterra-integral-equations-solver
Solve your equation (if I understood correctly problem statement):
isolve(1,10*x,1)
ans =
10*exp(x) - 10
Will check the correctness of the solution by substituting in the source:
10*x + int(y,0,x)
ans =
10*exp(x) - 10
Solved correctly.
P.S. Sorry for my English.
Related
I'm trying to solve this numerical differential equations, can someone help?
clc;
clear;
syms A1(z) A2(z)
lamda1 = 1560*(10^-9);
c=3*(10^8);
d_eff=27*(10^-12);
omga1=(2*pi*c)/(lamda1);
omga2=omga1*2;
n=2.2;
k1=(n*omga1)/c;
k2=(n*omga2)/c;
ode1 = diff(A1) == (2*i*(omga1^2)*d_eff*A2*conj(A1)*exp(-i*(2*k1-k2)*z))/(k1*(c^2));
ode2 = diff(A2) == (i*(omga2^2)*d_eff.*(A1.^2).*exp(i*(2*k1-k2)*z))/(k2*(c^2));
odes = [ode1; ode2];
cond1 = A1(0) == 1;
cond2 = A2(0) == 0;
conds = [cond1; cond2];
M = matlabFunction(odes)
sol = ode45(M(1),[0 20],[2 0]);
in this question both ODE are coupled, hence there's only 1 ODE to solve:
1.- use 1st equation to write
A1=f(A2,dA2/dz)
and feed this expression into 2nd equation.
2.- regroup
n1=1j*k1^4/k2^2*1/deff*.25
n2=1j*3*(k1-k2)
now the ODE to solve is
y'=n1/y^2*exp(n2*z)
3.- It can obviously be done in MATLAB, but for this particular ODE in my opinion the Wolfram online ODE Symbolic Solver does a better job.
Input the obtained ODE of previous point into the ODE solver available in this link
https://www.wolframalpha.com/input?i=y%27%27+%2B+y+%3D+0
and solve
4.- The general (symbolic) solutions for A2 are
Note that I used k1 instead of n1 and k2 instead of n2 just in the Wolfram ODE Solver.
Rewording ; the k1 k2 expressions of the general solutions are not the wave numbers k1 k2 of the equations in the question. Just replace accordingly.
5.- Now get A1 using expression in point 2.
6.- I have spotted 2 possible errors in the MATLAB code posted in the question that shouldn't be ignored by the question originator:
In the far right side of the posted MATLAB code, the exponential expressions
6.1.- both show
exp(-i(2*k1-k2)z))/(k1(c^2))
there's this (k1*(c^2)) dividing the exponent wheras in the question none of the exponentials show such denominator their respective exponents.
6.2.- the dk or delta k expression in the exponentials of the question are obviously k2-k1 or k1-k2 , here there may be room for a sign ambiguity, that may shoot a wave solution onto the opposite direction, yet the point here is where
*exp(-1i*(2*k1-k2)*z)
should probably be
*exp(-1i*2*(k1-k2)*z)
or just
exp(-1i*(k1-k2)*z)
6.3.- and yes, in MATLAB (-1)^.5 can be either expressed with 1j or 1i but as written in the MATLAB code made available in the question, since only a chunk of code has been made available, it's fair to assume that no such i=1j has been done.
I got this equation after I solved that problem here
I would like to solve it using Matlab.
The function roots returns polynomial roots. Simply pass the coefficients:
roots([1,-6,-36,216,-324])
Use solve from MATLAB's symbolic mathematics toolbox:
>> syms x;
>> y = solve('x^4-6*x^3-36*x^2+216*x-324==0')
y =
7.7446378738164683022795580182987
-6.3360292312480789716536487435108
2.2956956787158053346870453626061 - 1.1543655214730370697054534567177*i
2.2956956787158053346870453626061 + 1.1543655214730370697054534567177*i
The first line of code declares that x is a symbolic mathematics variable that we can use with the toolbox. Next, we use solve and we put in a string that describes the equation that is seen in your post. Bear in mind that when we multiply coefficients, you need to use the * operator, and for equality, we need to use double equals, or ==. The output should give you four roots, as dictated by the fundamental theorem of algebra. You'll see that you have two real roots, as well as two imaginary roots.
Alternatively, you can use roots in MATLAB by specifying a vector of coefficients starting from the highest order down to the lowest as per Daniel's answer.
What is the least computational time consuming way to solve in Matlab the equation:
exp(ax)-ax+c=0
where a and c are constants and x is the value I'm trying to find?
Currently I am using the in built solver function, and I know the solution is single valued, but it is just taking longer than I would like.
Just wanting something to run more quickly is insufficient for that to happen.
And, sorry, but if fzero is not fast enough then you won't do much better for a general root finding tool.
If you aren't using fzero, then why not? After all, that IS the built-in solver you did not name. (BE EXPLICIT! Otherwise we must guess.) Perhaps you are using solve, from the symbolic toolbox. It will be more slow, since it is a symbolic tool.
Having said the above, I might point out that you might be able to improve by recognizing that this is really a problem with a single parameter, c. That is, transform the problem to solving
exp(y) - y + c = 0
where
y = ax
Once you know the value of y, divide by a to get x.
Of course, this way of looking at the problem makes it obvious that you have made an incorrect statement, that the solution is single valued. There are TWO solutions for any negative value of c less than -1. When c = -1, the solution is unique, and for c greater than -1, no solutions exist in real numbers. (If you allow complex results, then there will be solutions there too.)
So if you MUST solve the above problem frequently and fzero was inadequate, then I would consider a spline model, where I had precomputed solutions to the problem for a sufficient number of distinct values of c. Interpolate that spline model to get a predicted value of y for any c.
If I needed more accuracy, I might take a single Newton step from that point.
In the event that you can use the Lambert W function, then solve actually does give us a solution for the general problem. (As you see, I am just guessing what you are trying to solve this with, and what are your goals. Explicit questions help the person trying to help you.)
solve('exp(y) - y + c')
ans =
c - lambertw(0, -exp(c))
The zero first argument to lambertw yields the negative solution. In fact, we can use lambertw to give us both the positive and negative real solutions for any c no larger than -1.
X = #(c) c - lambertw([0 -1],-exp(c));
X(-1.1)
ans =
-0.48318 0.41622
X(-2)
ans =
-1.8414 1.1462
Solving your system symbolically
syms a c x;
fx0 = solve(exp(a*x)-a*x+c==0,x)
which results in
fx0 =
(c - lambertw(0, -exp(c)))/a
As #woodchips pointed out, the Lambert W function has two primary branches, W0 and W−1. The solution given is with respect to the upper (or principal) branch, denoted W0, your equation actually has an infinite number of complex solutions for Wk (the W0 and W−1 solutions are real if c is in [−∞, 0]). In Matlab, lambertw is only implemented for symbolic inputs and thus is very slow method of solving your equation if you're interested in numerical (double precision) solutions.
If you wish to solve such equations numerically in an efficient manner, you might look at Corless, et al. 1996. But, as long as your parameter c is in [−∞, 0], i.e., -exp(c) in [−1/e, 0] and you're interested in the W0 branch, you can use the Matlab code that I wrote to answer a similar question at Math.StackExchange. This code should be much much more efficient that using a naïve approach with fzero.
If your values of c are not in [−∞, 0] or you want the solution corresponding to a different branch, then your solution may be complex-valued and you won't be able to use the simple code I linked to above. In that case, you can more fully implement the function by reading the Corless, et al. 1996 paper or you can try converting the Lambert W to a Wright ω function: W0(z) = ω(log(z)), W−1(z) = ω(log(z)−2πi). In your case, using Matlab's wrightOmega, the W0 branch corresponds to:
fx0 =
(c - wrightOmega(log(-exp(c))))/a
and the W−1 branch to:
fxm1 =
(c - wrightOmega(log(-exp(c))-2*sym(pi)*1i))/a
If c is real, then the above reduces to
fx0 =
(c - wrightOmega(c+sym(pi)*1i))/a
and
fxm1 =
(c - wrightOmega(c-sym(pi)*1i))/a
Matlab's wrightOmega function is also symbolic only, but I have written a double precision implementation (based on Lawrence, et al. 2012) that you can find on my GitHub here and that is 3+ orders of magnitude faster than evaluating the function symbolically. As your problem is technically in terms of a Lambert W, it may be more efficient, and possibly more numerically accurate, to implement that more complicated function for the regime of interest (this is due to the log transformation and the extra evaluation of a complex log). But feel free to test.
I am trying to solve the following differential equations on matlab. (They are the equations obtained from the yang-mills-higgs lagrangian for the hoofy polyakov monopole ansatz). This is my function file. I have two variables h and k and their derivatives w.r.t to a variable t. My x(1)=h, x(2)=k, x(3)=dh\dt, x(4)=dk\dt. All the functions have initial value 0.
function xprime = monopole( t,x )
%UNTITLED Summary of this function goes here
% Detailed explanation goes here
xprime(1)=x(3);
xprime(2)=x(4);
xprime(4)=(1/(t.^2)).*((x(2).^2)-1).*x(2) + 4.*(x(1).^2).*x(2);
xprime(3)=(2/(t.^2)).*(x(2).^2).*x(1)-(1-(x(1)).^2).*x(1)-(2/t).*x(3);
xprime=xprime(:);
end
Now when I run the following code
>
> t0=0;
>> tf=10;
>> x0=[0 0 0 0];
>> [t,s]=ode45(#monopole,[t0,tf],x0);
>> plot(t,s(:,1));
I am not getting anything. The graph window appears but it doesnt contain anything. This equations are supposed to have solutions. The dotted curves is what one should get with the curve starting from 1 is k, and from 0 is h.
What is my mistake?
When this happens the first thing you should do is look at the values in the t and s vectors. In this case s(1,1) contains 0, and s(:,2:end) are all NaN. Hence nothing on the plot.
As to why this is happening, a few thoughts
Are you sure that your definition of monopole is correct?
Why are you showing a plot where k(0) = 1, but passing it an initial condition of k=0?
Why are you using h^prime(0) = 0 initial conditions, but in the plot it looks like h^prime(0) has a non-zero slope?
The term with the 1./t^2 sure looks suspicious; just think about it, at the very first step you are going to divide 0 by 0, hence NaN. Perhaps the ode solver is having a hard time with this, and another solver would work better (note: I have very little experience with ODE solvers, so take this with a big grain of salt).
Finally, to make sure you really understand how to use the ODE solver, why not start with a very simple ODE where you know the exact answer (i.e. harmonic oscillator).
I have a system of 2 equations in 2 unknowns that I want to solve using MATLAB but don't know exactly how to program. I've been given some information about a gamma distribution (mean of 1.86, 90% interval between 1.61 and 2.11) and ultimately want to get the mean and variance. I know that I could use the normal approximation but I'd rather solve for A and B, the shape and scale parameters of the gamma distribution, and find the mean and variance that way. In pseudo-MATLAB code I would want to solve this:
gamcdf(2.11, A, B) - gamcdf(1.61, A, B) = 0.90;
A*B = 1.86;
How would you go about solving this? I have the symbolic math toolbox if that helps.
The mean is A*B. So can you solve for perhaps A in terms of the mean(mu) and B?
A = mu/B
Of course, this does no good unless you knew B. Or does it?
Look at your first expression. Can you substitute?
gamcdf(2.11, mu/B, B) - gamcdf(1.61, mu/B, B) = 0.90
Does this get you any closer? Perhaps. There will be no useful symbolic solution available, except in terms of the incomplete gamma function itself. How do you solve a single equation numerically in one unknown in matlab? Use fzero.
Of course, fzero looks for a zero value. But by subtracting 0.90, that is resolved.
Can we define a function that fzero can use? Use a function handle.
>> mu = 1.86;
>> gamfun = #(B) gamcdf(2.11, mu/B, B) - gamcdf(1.61, mu/B, B) - 0.90;
So try it. Before we do that, I always recommend plotting things.
>> ezplot(gamfun)
Hmm. That plot suggests that it might be difficult to find a zero of your function. If you do try it, you will find that good starting values for fzero are necessary here.
Sorry about my first try. Better starting values for fzero, plus some more plotting does give a gamma distribution that yields the desired shape.
>> B = fzero(gamfun,[.0000001,.1])
B =
0.0124760672290871
>> A = mu/B
A =
149.085442218805
>> ezplot(#(x) gampdf(x,A,B))
In fact this is a very "normal", i.e, Gaussian, looking curve.