Basically I would like to use the fsolve command in order to find the roots of an equation.
I think I should create a function handle that evaluates this equation in the form "right hand side - left hand side =0", but I've been struggling to make this work. Does anyone know how to do this?
The equation itself is 1/sqrt(f) = -1.74log((1.254/((1.27310^8)sqrt(f)))+((110^-3)/3.708)). So I would like to find the point of intersection of the left and right side by solving for 1/sqrt(f)+(1.74log((1.254/((1.27310^8)sqrt(f)))+((110^-3)/3.708))) = 0 using fsolve.
Thanks a lot!
The code so far (not working at all)
f = #(x) friction(x,rho,mu,e,D,Q, tol, maxIter) ;
xguess = [0, 1];
sol = fsolve(x, xguess ) ;
function y = friction(x,rho,mu,e,D,Q, tol, maxIter)
D = 0.1;
L = 100
rho = 1000;
mu = 0.001;
e = 0.0001;
Q = 0.01;
U = (4*Q)/(pi*D^2);
Re = (rho*U*D)/mu ;
y = (1/sqrt(x))-(-1.74*log((1.254/(Re*sqrt(x)))+((e/D)/3.708)))
end
Error message:
Error using lsqfcnchk (line 80)
FUN must be a function, a valid character vector expression, or an inline function object.
Error in fsolve (line 238)
funfcn = lsqfcnchk(FUN,'fsolve',length(varargin),funValCheck,gradflag);
Error in Untitled (line 6)
sol = fsolve(x, xguess ) ;
opt = optimset('Display', 'Iter');
sol = fsolve(#(x) friction(x), 1, opt);
function y = friction(x)
D = 0.1;
L = 100; % note -- unused
rho = 1000;
mu = 0.001;
e = 0.0001;
Q = 0.01;
U = (4*Q)/(pi*D^2);
Re = (rho*U*D)/mu ;
y = (1/sqrt(x))-(-1.74*log((1.254/(Re*sqrt(x)))+((e/D)/3.708)));
end
sol = 0.0054
the first argument of fsolve should be the function not variable. Replace:
sol = fsolve(x, xguess );
with
sol = fsolve(f, xguess );
And define Rho, mu, e etc before you define f (not inside the friction function).
Related
h=0.005;
x = 0:h:40;
y = zeros(1,length(x));
y(1) = 0;
F_xy = ;
for i=1:(length(x)-1)
k_1 = F_xy(x(i),y(i));
k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1);
k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));
k_4 = F_xy((x(i)+h),(y(i)+k_3*h));
y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;
end
I have the following code, I think it's right. I know there's parts missing on the F_xy because this is my follow up question.
I have dx/dt = = −x(2 − y) with t_0 = 0, x(t_0) = 1
and dy/dt = y(1 − 2x) with t_0 = 0, y(t_0) = 2.
My question is that I don't know how to get these equations in to the code. All help appreciated
You are using both t and x as independent variable in an inconsistent manner. Going from the actual differential equations, the independent variable is t, while the dependent variables of the 2-dimensional system are x and y. They can be combined into a state vector u=[x,y] Then one way to encode the system close to what you wrote is
h=0.005;
t = 0:h:40;
u0 = [1, 2]
u = [ u0 ]
function udot = F(t,u)
x = u(1); y = u(2);
udot = [ -x*(2 - y), y*(1 - 2*x) ]
end
for i=1:(length(t)-1)
k_1 = F(t(i) , u(i,:) );
k_2 = F(t(i)+0.5*h, u(i,:)+0.5*h*k_1);
k_3 = F(t(i)+0.5*h, u(i,:)+0.5*h*k_2);
k_4 = F(t(i)+ h, u(i,:)+ h*k_3);
u(i+1,:) = u(i,:) + (h/6)*(k_1+2*k_2+2*k_3+k_4);
end
with a solution output
Is F_xy your derivative function?
If so, simply write it as a helper function or function handle. For example,
F_xy=#(x,y)[-x*(2-y);y*(1-2*x)];
Also note that your k_1, k_2, k_3, k_4, y(i) are all two-dimensional. You need to re-size your y and rewrite the indices in your iterating steps accordingly.
I'm trying to solve a system of ordinary differential equations from y_1 to y_2, say
G' = D
D' = f(y,G,D)
with the initial conditions
G(y_1) = 0
D(y_1) = 0
My problem is that I do not know y_1 and y_2, to counter for this I of course need two additional equations which is
F_1(y_1,y_2,G(y_2)) = 0
F_2(y_1,y_2,G(y_2)) = 0
So far I have tried to implement it using fsolve (i guess the new name is fzero) to find the zeros, then from the functions F_1 and F_2 i call the ode45 to solve from y_1 to y_2 in order to calculate the functions. However it does not work and I can not seem to find any mistakes. Therefore i looked for new methods/ideas and I found the method bvp4c, but I'm not sure whether i can use it in my case. Does anyone have experience with bvp4c and know whether and how I should use it, or do you have other ideas?
Any help is appreciated.
Code for reference:
function [Fval,sol,t,G] = EntrepreneurialFinanceNondiversifiableRisk
global sigma rf tau b epsilon gamma theta1 theta2 alpha ya K I taug mu eta...
omega
sigma=0.2236; rf=0.03; tau=0.1129; b=0.85; epsilon=0.2; gamma=2;
theta1 = -0.704; alpha = 0.6; theta2=1.704; ya=0.1438; K=27; I = 10;
taug = 0.1; mu = 0.04; eta = 0.4; omega = 0.1;
tau = 0;
option = optimset('Display','iter');
sol = fsolve(#f,[0.1483,2.8],option);
Fval = f(sol);
[t,G] = solvediff(sol(1),sol(2));
function F = f(x)
yd = x(1);
yu = x(2);
global rf tau b theta1 alpha theta2 K I taug
Vstar =# (y) (1-tau+tau*(1-theta1-(1-alpha)*(1-tau)* theta1/tau)^...
(1/theta1))*y/rf;
VstarPrime = (1-tau+tau*(1-theta1-(1-alpha)*(1-tau)...
*theta1/tau)^(1/theta1))*1/rf;
qbar =#(yd,yu) (yd^theta1-yd^theta2)/(yu^theta2*yd^theta1-yu^theta1*...
yd^theta2);
qunderbar =# (yd,yu) (yu^theta2-yu^theta1)/(yu^theta2*yd^theta1-...
yu^theta1*yd^theta2);
A =# (y) (1-tau)*(y/rf);
V = #(y,yd) A(y) + (tau * b)/rf * (1 - (y/yd)^(theta2)) - (1-alpha)*A(yd)*...
(y/yd)^(theta2);
F0 =# (yd,yu) b/rf-(b/rf-alpha*A(yd))*qunderbar(yd,yu)/(1-qbar(yd,yu));
[t,G] = solvediff(yd,yu);
F = zeros(2,1);
F(1) = Vstar(yu)-F0(yd,yu)-K-taug*(Vstar(yu)-K-I)-G(end,2);
F(2) = (1-taug)*VstarPrime-G(end,2);
function [t,G] = solvediff(yd,yu)
[t,G] = ode45(#diff,[yd,yu],[0,0]);
Assuming the domain of definition is big enough and you have good initial values, a Newton implementation with divided differences for the construction of the Jacobian looks like this:
h=1e-5
for k=1:10 do
Fval = F(yd, yu)
dFdyd = (F(yd+h, yu)-F(yd-h, yu))/(2*h)
dFdyu = (F(yd, yu+h)-F(yd, yu-h))/(2*h)
dF = [ dFdyd dFdyu ]
ynext = [ yd; yu ] - dF^(-1)*Fval;
yd = ynext(1); yu = ynext(2);
end
Following code throws out an error.
syms z positive;
syms n;
syms m;
N = 10;
Ms = 10;
Es = 1;
pd = 0.9;
pd_dash = 1-pd;
pf = 0.1;
pf_dash = 1-pf;
pr = 0.1;
qr = 1-pr;
p = 0.005
pi = pf_dash*p;
pb = pd_dash*p;
qi = 1-pi;
qb = 1-pb;
sm = symsum( z^((n+1)*Es), n, 0, N-1 );
temp_sum = symsum(z^((n+m+1)*Es)*qr^(n+m)*pr, m, 0, N-1);
z=1; %assume a value of z
x = eval(sm); %works fine
y = eval(temp_sum);
% Error:The expression to the left of the equals sign is not a valid target for an assignment.
Please suggest a way to resolve this.
The problem that I suspect is: the temp_sum comes out to be in piecewise(...) which eval is not capable of evaluating.
What you actually did:
Create a symbolic expression
Create a variable z which is unused
Call a undocumented function sym/eval
I assume you wanted to:
Create a symbolic expression
Substitute z with 1: temp_sum=subs(temp_sum,z,1)
get the result. Here I don't know what you really have because I don't know which variables are symbolic unknowns and which constants. Try simplify(temp_sum). If you substituted all unknowns it should return a number.
I need help plotting a differential equation ... it keeps coming out all funky and the graph is not what it's supposed to look like.
function [dydt] = diff(y,t)
dydt = (-3*y)+(t*(exp(-3*t)));
end
tI = 0;
yI = -0.1;
tEnd = 5;
dt = 0.5;
t = tI:dt:tEnd;
y = zeros(size(t));
y(1) = yI;
for k = 2:numel(y)
yPrime = diff(t(k-1),y(k-1));
y(k) = y(k-1) + dt*yPrime;
end
plot(t,y)
grid on
title('Engr')
xlabel('Time')
ylabel('y(t)')
legend(['dt = ' num2str(dt)])
That's my code, but the graph is not anything like what it's supposed to look like. Am I missing something like an index for the for statement?
Edit
I am getting an error:
Error using diff
Difference order N must be a positive integer scalar.
Error in diff3 (line 12)
yPrime = diff(t(k-1),y(k-1));
After fixing the errors pointed out by Danil Asotsky and horchler in the comments:
avoiding name conflict with built-in function 'diff'
changing the order of arguments to t,y.
decreasing the time-step dt to 0.1
converting ODE right-hand side to an anonymous function
(and removing unnecessary parentheses in the function definition), your code could look like this:
F = #(t,y) -3*y+t*exp(-3*t);
tI = 0;
yI = -0.1;
tEnd = 5;
dt = 0.1;
t = tI:dt:tEnd;
y = zeros(size(t));
y(1) = yI;
for k = 2:numel(y)
yPrime = F(t(k-1),y(k-1));
y(k) = y(k-1) + dt*yPrime;
end
plot(t,y)
grid on
title('Engr')
xlabel('Time')
ylabel('y(t)')
legend(['dt = ' num2str(dt)])
which performs as expected:
I want to solve equations in matlab, eg.
100+a/2=173*cos(b)
sqrt(3)*a/2=173*sin(b）
and the code would be:
[a,b]=solve('100+a/2=173*cos(b)','sqrt(3)*a/2=173*sin(b)','a','b')
However, if I want to take 100 as a variable, like
for k=1:100
[a,b]=solve('k+a/2=173*cos(b)','sqrt(3)*a/2=173*sin(b)','a','b')
end
There would be an error, how to make it?
degree=140/1000000;
p=42164000;
a=6378136.5;
b=6356751.8;
x_1=0;
y_1=p;
z_1=0;
for i=451:550
for j=451:550
alpha=(1145-i)*degree;
beta=(1145-j)*degree;
x_2=p/cos(alpha)*tan(beta);
y_2=0;
z_2=p*tan(alpha);
syms x y z x_1 x_2 y_1 y_2 z_1 z_2 a b
eq = [(x-x_1)*(y2-y_1)-(x_2-x_1)*(y-y_1),(x-x_1)*(z_2-z_1)-(x_2-x_1)*(z-z_1), b^2*(x^2+y^2)+a^2*(y^2)-a^2*b^2 ];
sol = solve(eq(1),x,eq(2),y, eq(3),z);
sol.x
sol.y
sol.z
end
end
I got the expression value, how do I get the numeric value of x,y,z?
[['x(1)=';'x(2)='],num2str(double(sol.x))]
not work ,shows
??? Error using ==> mupadmex
Error in MuPAD command: DOUBLE cannot convert the input expression into a double array.
If the input expression contains a symbolic variable, use the VPA function instead.
Error in ==> sym.sym>sym.double at 927
Xstr = mupadmex('mllib::double', S.s, 0);
Error in ==> f2 at 38
[['x(1)=';'x(2)='],num2str(double(sol.x))]
If you have access to the Symbolic Toolkit then you do the following:
syms a b k
eq = [k+a/2-173*cos(b), sqrt(3)*a/2-173*sin(b)];
sol = solve(eq(1),a,eq(2),b);
sol.a = simplify(sol.a);
sol.b = simplify(sol.b);
% There are two solutions for 'a' and 'b'
% check residuals for example k=20
subs(subs(eq,{a,b},{sol.a(1),sol.b(1)}),k,20)
% ans = 0.2e-13
subs(subs(eq,{a,b},{sol.a(2),sol.b(2)}),k,20)
% ans = 0.2e-13
Edit 1
Based on new code by OP the matlab script to solve this is:
clear all
clc
syms alpha beta
degree=140/1000000;
p=42164000;
a=6378136.5;
b=6356751.8;
x_1=0;
y_1=p;
z_1=0;
x_2 = p/cos(alpha)*tan(beta);
y_2 = 0;
z_2 = p*tan(alpha);
syms x y z
eq = [(x-x_1)*(y_2-y_1)-(x_2-x_1)*(y-y_1);...
(x-x_1)*(z_2-z_1)-(x_2-x_1)*(z-z_1); ...
b^2*(x^2+y^2)+a^2*(y^2)-a^2*b^2 ];
sol = solve(eq(1),x,eq(2),y,eq(3),z);
sol.x = simplify(sol.x);
sol.y = simplify(sol.y);
sol.z = simplify(sol.z);
pt_1 = [sol.x(1);sol.y(1);sol.z(1)] % First Solution Point
pt_2 = [sol.x(2);sol.y(2);sol.z(2)] % Second Solution Point
x = zeros(100,100);
y = zeros(100,100);
z = zeros(100,100);
for i=451:550
disp(['i=',num2str(i)])
for j=451:550
res = double(subs(pt_1,{alpha,beta},{(1145-i)*degree,(1145-j)*degree}));
x(i-450, j-450) = res(1);
y(i-450, j-450) = res(2);
z(i-450, j-450) = res(3);
end
end
disp('x=');
disp(x);
disp('y=');
disp(x);
disp('z=');
disp(x);
I would try
for i=1:100
k=num2str(i)
[a,b]=solve('100+a/2=173*cos(b)','sqrt(3)*a/2=173*sin(b)','a','b')
end
and then solve the equation