I want to find only real roots of the equation which is ;
4*sqrt((1-(z^2/f1^2))*(1-z^2))-(2-z^2)^2-(m*z^4*sqrt(1-z^2/f1^2)/ ...
sqrt(1-((z^2/f1^2)/y^2)))
I know that equation includes complex roots, but I do not want to see them. Moreover, my code fails and says that;
Error using fzero (line 242) Function values at interval endpoints
must be finite and real.
Error in scholte (line 21) x=fzero(fun,x0)
Here is my code;
rho2 = 1000; %kg/m3
rho1 = 2700; %kg/m3
cl2 = 1481; %m/s
cl1 = 5919; %m/s
m = rho2/rho1;
y = cl2/cl1;
poi = 0.25;
f1 = (sqrt((1-2*poi)/(2*(1-poi))))^-1;
fun = #(z) 4*sqrt((1-(z^2/f1^2))*(1-z^2))-(2-z^2)^2- ...
(m*z^4*sqrt(1-z^2/f1^2)/sqrt(1-((z^2/f1^2)/y^2)));
x0 = [1 10];
x = fzero(fun, x0)
I changed x0 interval many times, but it showed the same error. How can I fix my code?
Your problem, as Matlab tells you, is that Function values at interval endpoints must be finite and real, and in your case they are not real:
fun(x0(1))
ans =
-1.0000 + 0.1454i
Your function is probably just too complex for fzero to handle. However I am not an expert, lets see if someone with more knowledge than me can point you in the correct direction for solving that equation.
Related
I'm having some trouble with this code. My professor asked us to create a function "feuler.m" in MATLAB to solve the initial-value problem given by the differential equation u′(t) = (2+2t)e^t and the initial condition u(0) = 0 over the interval [0, 5] that uses (forward) Euler’s method to graph the exact solution along with the approximate solution.
The input should be: n, the number of subintervals into which the interval [0,5] should be divided.
The output should be a graph of the exact solution and the numerical solution and print the value of the maximum error between the true solution and the numerical solution.
Note that the exact solution is given by u(t) = 2tet.
So far I have written the code:
function myeuler(N)
t = linspace(0, 5, N+1)';
ua = zeros(N+1,1);
ue = 2*t.*exp(t);
h = 5/N;
A = zeros(N,N);
A(2:N,1:N-1) = -eye(N-1);
A = A + eye(N);
b = h*(2+2*t(1:N)).*exp(t(1:N));
b(1) = b(1) + ua(1);
ua(2:N+1) = A\b;
plot(t, ua, 'r', t, ue, 'g')
end
I'm unsure if this is right.
I'm trying to solve a system of two second order non linear Odes and since it is a boundary valued problem I suppose I need to use the bvp4c function.
The system I'm talking about is the following:
f''(x) = F(f,f',x);
s''(x) = G(f, f',s,s',x)
with the conditions f(0) = pi, f(inf = 35) = s(inf = 35) = 0. The F and G functions are known and I assumed that 35 would be a decent replacement for infinity.
It is separable and I have already solved for f but I don't know how to solve it for s either.
The code that allegedly solves for f is the following:
options = bvpset('RelTol', 1e-5);
Xstart = 0.01;
Xend = 35;
solinit = bvpinit(linspace(Xstart, Xend, 1000), [0, 1]);
sol = bvp4c(#twoode, #twobc, solinit, options);
x = linspace(Xstart,Xend);
y = deval(sol,x);
figure(1)
plot(x,y(1,:))
figure(2)
plot(x,y(2,:))
function dydx = twoode(x,y)
dydx = [y(2); ((-1/(x^2 + 2+sin(y(1))^2))*(2*x*y(2) + sin(2*y(1))*y(2)^2 -
2*sin(2*y(1)) - (sin(y(1)^2)*sin(2*y(1)))/x^2) )];
end
function res = twobc(ya,yb)
res = [ya(1) - pi
yb(2)];
end
So my question is how can I use the results I obtained for f in order to solve the equation for s? I have tried doing the same things I did for f but if I define a function for s that uses y(1,:) and y(2,:) it gives me an error message that says y is not defined.
Since I am quite new to solving dfferential equations with Matlab and to using Matlab in general I am probably making some trivial mistake but I have been looking for answers and couldn't find any. I hope someone with enough patience can help me.
Thanks in advance for any useful advice.
I'm having difficulty using the fsolve function to solve a set of 5 equations in Matlab.
Here are the 5 equations:
y = a + d + e
y + x = c + d + 2e
2x = 4a + 2b + 2c
k1 = (d * b^3 / (a * c) ) * ((P/Pref)/(a+b+c+d+e))^2
k2 = be/(dc)
y,x,k1,k2,P,Pref are all parameters that I set, but would like to leave them in the function so that I can change them quickly in my code to find new answers. a,b,c,d,e are the variables that I'd like to solve for (they are compositions of a reaction equilibrium equation)
I tried to hard code the parameters in the function, but that didn't work. I'm just not sure what to do. Every thing I change creates a new error. The most common is that the data type has to be "double".
Edit: adding code
first the function:
function F = myfun(Q,I)
a = Q(1);
b = Q(2);
c = Q(3);
d = Q(4);
e = Q(5);
x = I(1);
y = I(2);
k1 = I(3);
k2 = I(4);
P = I(5);
Pref = I(6);
F(1) = a + d + e - y;
F(2) = c + d + 2*e - y - x;
F(3) = 4*a + 2*b + 2*c - 2*x;
F(4) = ((d * b^3)/(a*c))*((P/Pref)/(a+b+c+d+e))^2 - k1;
F(5) = (b*e)/(c*d);
next is the program:
%Q = [a,b,c,d,e]
%I = [x,y,k1,k2,P,Pref]
%The values for the inputs will be changed to vary the output
%Inputs:
x=5;
y=1;
k1=5;
Pref=1;
P=1;
k2=-0.01;
syms K
k1 = solve(log10(k1) - k1);
syms L
k2 = solve(log10(k2) - k2);
x = double(x);
y = double(y);
Pref = double(Pref);
P = double(P);
k1 = double(k1);
k2 = double(k2);
%Solving:
I = [x,y,k1,k2,P,Pref];
q = [0,0,0,0,0]; %initial guess
Q = fsolve(#myfun,[q,I])
when I run this, these errors comes up:
Error using myfun (line 7)
Not enough input arguments.
Error in fsolve (line 218)
fuser = feval(funfcn{3},x,varargin{:});
Error in Coal (line 27)
Q = fsolve(#myfun,[q,I])
Caused by:
Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue.
Edit 2: changed the fsolve line, but still got errors:
Error using trustnleqn (line 28)
Objective function is returning undefined values at initial point. FSOLVE cannot continue.
Error in fsolve (line 376)
[x,FVAL,JACOB,EXITFLAG,OUTPUT,msgData]=...
Error in Coal (line 27)
fsolve(#(q) myfun(q,I),q)
Edit3: changed a couple parameters and the initial guess, I am now getting an answer, but it also comes up with this:
Solver stopped prematurely.
fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunEvals = 500 (the default value).
ans =
0.0000 2.2174 3.7473 1.4401 3.8845
how can I get it to not stop prematurely?
Re: how can I get it to not stop prematurely?
Pass non-default options in your call to fsolve.
Refer to the signature, fsolve(myfun,q,options) here,
http://www.mathworks.com/help/optim/ug/fsolve.html
And read about creating the options using optimoptions here,
http://www.mathworks.com/help/optim/ug/optimoptions.html
You should be able to get it to "not stop prematurely" by increasing the values of convergence criteria such as TolFun and TolX.
However, it's advisable to read up on the underlying algorithms you're relying upon to perform this numerical solution here,
(EDIT: I tried to fix non-linky links, but I'm not allowed to provide more than two ... Boo) www.mathworks.com/help/optim/ug/fsolve.html#moreabout
The error you received simply indicates that after 500 function evaluations the algorithm has not yet converged on an acceptable solution conforming to the default solver options. Just increasing MaxFunEvals may allow the algorithm extra iterations needed to converge within default tolerances. For example,
options = optimoptions('MaxFunEvals',1000); % try something bigger than 500
fsolve(#(q) myfun(q,I),q,options);
I am having difficulty in finding roots of a nonlinear equation. I have tried Matlab and Maple both, and both give me the same error which is
Error, (in RootFinding:-NextZero) can only handle isolated zeros
The equation goes like
-100 + 0.1335600000e-5*H + (1/20)*H*arcsinh(2003.40/H)
The variable is H in the equation.
How do I find the roots (or the approximate roots) of this equation?
Matlab Code:
The function file:
function hor_force = horizontal(XY, XZ, Lo, EAo, qc, VA)
syms H
equation = (-1*ZZ) + (H/qc)*(cosh((qc/H)*(XZ- XB))) - H/qc + ZB;
hor_force = `solve(equation);`
The main file:
EAo = 7.5*10^7;
Lo = 100.17;
VA = 2002;
XY = 0;
ZY = 0;
XB = 50;
ZB = -2;
XZ = 100;
ZZ = 0;
ql = 40;
Error which Matlab shows:
Error using sym/solve (line 22)
Error using maplemex
Error, (in RootFinding:-NextZero) can only handle isolated zeros
Error in horizontal (line 8)
hor_force = solve(equation);
Error in main (line 34)
h = horizontal(XY, XZ, Lo, EAo, ql, VA)
http://postimg.org/image/gm93z3b7z/
You don't need the symbolic toolbox for this:
First, create an anonymous function that can take vectors at input (use .* and ./:
equation = #(H) ((-1*ZZ) + (H./qc).*(cosh((qc./H).*(XZ- XB))) - H./qc + ZB);
Second, create a vector that you afterwards insert into the equation to find approximately when the sign of the function changes. In the end, use fzero with x0 as the second input parameter.
H = linspace(1,1e6,1e4);
x0 = H(find(diff(sign(equation(H))))); %// Approximation of when the line crosses zero
x = fzero(equation, x0) %// Use fzero to find the crossing point, using the initial guess x0
x =
2.5013e+04
equation(x)
ans =
0
To verify:
You might want to check out this question for more information about how to find roots of non-polynomials.
In Maple, using the expression from your question,
restart:
ee := -100 + 0.1335600000e-5*H + (1/20)*H*arcsinh(2003.40/H):
Student:-Calculus1:-Roots(ee, -1e6..1e6);
[ 5 ]
[-1.240222868 10 , -21763.54830, 18502.23816]
#plot(ee, H=-1e6..1e6, view=-1..1);
I am using the following code to produce a numerical solution to a system of ODE's with 6 boundary conditions.
I am using the initial conditions to obtain a solution but I must vary three other conditions in order to find the true solution. The function I have is as follows:
function diff = prob5diff(M,Fx,Fy)
u0 = [pi/2 0 0]';
sSpan = [0 13];
p = #(t,u) prob5(t,u,M,Fx,Fy);
options = odeset('reltol',1e-6,'abstol',1e-6);
[s,u] = ode45(p,sSpan,u0,options);
L = length(s);
x = u(:,2); y = u(:,3); theta = u(:,1);
diff(1) = x(L) - 5;
diff(2) = y(L);
diff(3) = theta(L) + pi/2;
end
Ultimately, different values of M,Fx, and Fy will produce different solutions and I would like a solution such that the values in diff are as close to zero as possible so I want fsolve to iterate through different values of M,Fx, and Fy
I am receiving the following error: when I call it in this way:
opt = optimset('Display','iter','TolFun',1e-6);
guess = [1;1;1];
soln = fsolve(#prob5diff,guess,opt);
Error in line:
soln = fsolve(#prob5diff,guess,opt);
Caused by:
Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue.
Thanks!
One problem is that you have to call fsolve on prob5diff which takes a single vector input since your guess is a single vector:
prob5diff(x)
M = x(1);
Fx = x(2);
Fy = x(3);