Plotting a non-linear graph in Matlab

Hi guys,
I have this equation f(g)=1+cos(g)*cosh(g)-a*g*(cos(g)*sinh(g)-sin(g)*cosh(g))=0. I am trying to plot it for (a,g).
I have the initial condition that when a=0 then g=1.8751.
I am not sure how to code it as an equation with two unknowns and two initial values ??
My suggestion would be to plot the graph by using the solve function by using a=a+da (where da is small increment of a) and using the previous value of g as an initial value to find g for the new value of a.
Regards, Rihab

3 Commenti

f(g)=1+cos(g)*cosh(g)-a*g*(cos(g)*sinh(g)-sin(g)*cosh(g))=0.
there are two equality sign in this equation write correct one plz
@Muhammad Usman Saleem I did mean to put two equal signs because the function is equal to zero. I want to plot (a,g) when f(g)=0.
good night, tomorrow will guide u in this regards

Accedi per commentare.

 Risposta accettata

a = linspace(-4, 2); %the interesting part of the graph
g = zeros(size(a));
oldguess = 1.8751;
for K = 1 : length(a)
fa = @(g) 1+cos(g)*cosh(g)-a(K)*g*(cos(g)*sinh(g)-sin(g)*cosh(g));
this_g = fzero(fa, oldguess);
g(K) = this_g;
oldguess = this_g;
end
plot(a, g)
However, there are many solutions. You will, for example, get a different plot if you start with oldguess = 50

1 Commento

That works! However, what if I want to plot for a larger range of a? When I change the vector size from [-4,2] to [0,100] the plot seems to look different.

Accedi per commentare.

Più risposte (1)

Star Strider
Star Strider il 15 Ott 2016
Modificato: Star Strider il 15 Ott 2016
When in doubt, plot first:
[A,G] = meshgrid(-0.5:0.01:0.5, -20:0.1:20);
fga = @(a,g) 1+cos(g).*cosh(g)-a.*g.*(cos(g).*sinh(g)-sin(g).*cosh(g))
F = fga(A,G);
figure(1)
meshc(A, G, F)
grid on
There appear to be a possibly infinite number of solutions. Which one do you want?

Categorie

Scopri di più su 2-D and 3-D Plots in Centro assistenza e File Exchange

Tag

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by