How to solving and plot only one non-linear equation with two or one variables without using ezplot

I want to solve and plot one non linear equation with two variables k and f, the equation is shown here, where k varies from 0 to 1. The other algebras are given.

(((2*pi*f).^2)/((2*pi*fv).^2)-1)*((2-(((2*pi*f).^2)/((k.^2)*(cs.^2))).^2)-4*((1-(((2*pi*f).^2)/((k.^2)*(cs.^2)))).^0.5)*((1-(((2*pi*f).^2)/((k.^2)*(cl.^2)))).^0.5)-m*(((2*pi*f).^4)/(A*(cs.^4)*den*(k.^3)))*((1-(((2*pi*f).^2)/((k.^2)*(cl.^2)))).^0.5))=0

For every k value, there are more than one possible solutions for f value, where the f value can be in form complex number and is limited from 0 to 20.

However, it gives 'The input to FSOLVE should be either a structure with valid fields or consist of at least two arguments.' after running the command.

Apart from that, i actually want to plot a continous function rather than scattering the point if it is possible. I don't want to use ezplot because the plot is not correct. Thankyou

clc

clear all

syms k f

cs = 121;

cl = 227;

A = (1.3^2)*(3^0.5)/2;

den = 1300;

m = 6700;

fv = 4.9;

k=0:0.01:1;

fun = @(f) ((2*pi*f).^2)/((2*pi*fv).^2)-1)*((2-(((2*pi*f).^2)/((k.^2)*(cs.^2))).^2)-4*((1-(((2*pi*f).^2)/((k.^2)*(cs.^2)))).^0.5)*((1-(((2*pi*f).^2)/((k.^2)*(cl.^2)))).^0.5)-m*(((2*pi*f).^4)/(A*(cs.^4)*den*(k.^3)))*((1-(((2*pi*f).^2)/((k.^2)*(cl.^2)))).^0.5)));

fsolve(fun, [0, 20])

scatter(k,f)

ylim([0 20]);

xlim([0 1])

Answer 1

[darova]

Numerical way

[K,F] = mehsgrid(-10:10);
Z = your_function
contour(K,F,Z,[0 0])            % contour at 0 level

[Tak Yeung POON]

It gets some error when running. 

Error using contour (line 55)
DOUBLE cannot convert the input expression into a double array.
If the input expression contains a symbolic variable, use VPA.

Error in analytical (line 20)
contour(K,F,Z,[0 0])


Is it a 1D graph? Because i want it to be 1D rather than 2D or 3D. Thankyou

[Tak Yeung POON]

my full code is shown here

clc
clear all
syms f k
cs = 121 %ok
cl = 227 %ok 
A = (1.3^2)*(3^0.5)/2 %ok
den = 1300 %ok
m = 6700 %ok
fv = 4.9

[K,F] = meshgrid(-10:10);
Z = ((((2*pi*f).^2)/((2*pi*fv).^2)-1)*((2-(((2*pi*f).^2)/((k.^2)*(cs.^2))).^2)-4*((1-(((2*pi*f).^2)/((k.^2)*(cs.^2)))).^0.5)*((1-(((2*pi*f).^2)/((k.^2)*(cl.^2)))).^0.5)-m*(((2*pi*f).^4)/(A*(cs.^4)*den*(k.^3)))*((1-(((2*pi*f).^2)/((k.^2)*(cl.^2)))).^0.5)))
contour(K,F,Z,[0 0]) 


ylim([0 20])
xlim([0 1])

[darova]

What do you see?

[Tak Yeung POON]

Although there is no error, it doesnt give a correct plot. For every K, there are more than one F values, is there other approach to get the plot for the nonlinear equation?

clc
clear all
syms F K

cs = 250 
cl = 468 
A = 1.3
den = 1600 
m = 10500
fv = 2.43
sz = 10


[K,F] = meshgrid(-10:10);

Z =((((2.*pi.*F).^2)./((2.*pi.*fv).^2)-1).*((2-(((2.*pi.*F).^2)./((K.^2).*(cs.^2))).^2)-4.*((1-(((2.*pi.*F).^2)./((K.^2).*(cs.^2)))).^0.5).*((1-(((2.*pi.*F).^2)./((K.^2).*(cl.^2)))).^0.5)-m.*(((2.*pi.*F).^4)./(A.*(cs.^4).*den.*(K.^3))).*((1-(((2.*pi.*F).^2)./((K.^2).*(cl.^2)))).^0.5)))

contour(K,F,Z,[0,0])

ylim([0 20])
xlim([0 1])

[darova]

This is how your plot looks like if

surf(K,F,Z)

So contour at level 0 is correct

[Tak Yeung POON]

how about if i want to obtain the data the f corresponding to k for the equation

i.e. k f

0.1 abc

0.2 def

0.3 ghi

0.4 jkl

[darova]

Do you have polyxpoly?

[Tak Yeung POON]

ys

[darova]

try this script

[Tak Yeung POON]

but can you use fsolve to find the solution of f value where k=0:0.01:1

[darova]

What about interp1? fsolve is used for equations