I have a simple recursive equation, and the idea is for a given starting value of x, trace the future value of x, to see if there any convergence. I wrote a MATLAB code below which is simple, and obviously, x converges to a steady-state value 0.0292 pretty fast.
x(1) = 0.2;
for t = 1:1:100
x(t+1) = 0.12*x(t).^0.40;
end
However, for more complex equations I do not think I can apply the above code. What if, for example, I have:
x(t+1) = 0.12*(x(t)./(x(t+1)+x(t))).^0.40
How do I “solve” it to see (without trying to do any algebraic manipulations to isolate x(t+1) on the right-hand-side) to see what the path of x is?
Thank you.
As mentioned in the comments a simple approach would be give an initial guess for x(t+1) and try to converge to a solution for x(t+1) and then continue iterating.
recEq = #(p,q) 0.12 * (p ./ (q+p))^0.4;
x = nan(1, 100);
x(1) = 0.2;
for t = 1:100
kold = x(t);
k = recEq(x(t), kold);
while abs(k-kold) > 1e-8
kold = k;
k = recEq(x(t), kold);
end
x(t+1) = k;
end
I would like to solve the following equation :
f''+tau(x)*f+f^3=0
f'(0)=0
f(inf)=1
Where tau(x) can be any function of x. The problem is to translate the boundary condition f(inf)=1 into matlab. My first solution, by reading different posts on the web was to approximate infinity by a large finite number but it doesn't give satisfying solution. Here is the code that I use :
function f = solve_GL(tau)
options = bvpset('RelTol', 1e-5);
Xstart = 0;
Xend = 1000;
solinit = bvpinit(linspace(Xstart, Xend, 50), [0, 1]);
sol = bvp4c(#twoode, #twobc, solinit, options);
x = linspace(Xstart,Xend);
y = deval(sol,x);
plot(x,y(1,:))
function dydx = twoode(x,y)
dydx = [y(2); (-heaviside(x-2)+1).*y(1)+y(1).^3];
function res = twobc(ya,yb)
res = [ya(2); yb(1)-1];
If somebody has a solution to this problem, I would be glad to hear it !
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.
Im trying to solve the transient heat equation in 1D and comparing the analytical and numerial solutions. The solutions have the same trend but are very different, also the relative error is coming out to be zero, even though its clearly not supposed to be. Im not sure if I have solved the PDE correctly. (the pde is du/dt = d^2u/dx^2) and bcs are u(0,t)=1, u(100,t) = 0, u(x,0)=0. Can someone please take a look at my code?
function he
m = 0;
x = linspace(0,100,500);
t = linspace(0,1000,500);
sol = pdepe(m,#hepde,#heic,#hebc,x,t);
u = sol(:,:,1);
y = erfc(x./(2*(t.^0.5)));
r=(y-u(70,:))/y;
figure;
plot(x,u(50,:),'.',x,u(150,:),'.',x,u(250,:),'.',x,u(end,:),'.',x,y,'.');
title('Numerical Solutions at different times.');
legend('t=100','t=300','t=500','t=700','y ana',0);
xlabel('Distance x');
ylabel('u(x,t)');
figure;
plot(x,r);
title('error in numerical and analytical solution');
legend('error',0);
xlabel('Distance x');
ylabel('error');
% --------------------------------------------------------------------------
function [c,f,s] = hepde(x,t,u,DuDx)
c = 1;
f = DuDx;
s = 0;
% --------------------------------------------------------------------------
function u0 = heic(x)
u0 = 0;
% --------------------------------------------------------------------------
function [pl,ql,pr,qr] = hebc(xl,ul,xr,ur,t)
pl = ul-1;
ql = 0;
pr = ur;
qr = 0;
You actually had the solution all set up correctly, you just used the results incorrectly. I'll go through the errors one by one.
sol = pdepe(m,#hepde,#heic,#hebc,x,t);
u = sol(:,:,1);
You found the right answer, but the line u=sol(:,:,1); is useless as size(sol)=[2 2], so you may as well just do u=pdepe(...);.
Now when you calculated your exact solution you did it very strangely. You want to find it at each x/t combination, but you did it only at some of them. You need to use meshgrid to get all the combinations of x and t then calculate your exact solution at each one.
[X,T]=meshgrid(x,t);
y = erfc(X./(2*(T.^0.5)));
Then you need to calculate the error differently.
r=(y-u)./y;
figure(3);
And plot differently.
plot(x,u(50,:),'b',x,u(150,:),'g',x,u(250,:),'r',x,u(end,:),'c',x,y(50,:),'b--',x,y(150,:),'g--',x,y(250,:),'r--',x,y(end,:),'c--');
Actually your exact solution doesn't satisfy the boundary condition at x=100......
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);