changing the scale on the Y axis

Good evening sir

Sir I need -1 to 1 on Y-axis but i got this on X-axis. Now I request you please help me how to getting -1 to 1 on Y-axis. I attached my code and graph relating to my code.

Code:

xmesh1 = linspace(-1,0,10);
xmesh2 = linspace(0,1,10);
xmesh = [xmesh1,xmesh2];
yinit = [0 1 0 1 0 1 0 1];
init = bvpinit(xmesh,yinit);
sol = bvp5c(@f, @bc, init);
plot(sol.x,sol.y(2,:),'b-o',sol.x,sol.y(3 ,:),'r-o')

Warning: Imaginary parts of complex X and/or Y arguments ignored.

%line([1 1], [0 1], 'Color', 'k')
% legend('y1','y3')
% % title('A Three-Point BVP Solved with bvp5c')
% xlabel({'x', '\lambda = 2, \kappa = 5'})
% ylabel('v(x) and C(x)')
hold on

function dydx = f(x,y,region) % equations being solved

W=1;alpha=1;M=2;k1=1;k2=1;k3=1;k4=1;k=1; Ps=1;Po=1;R1=0;R2=2;
omega=1; 
us = y(1);
usy = y(2);
uo = y(3);
u0y = y(4);
thetas = y(5);
thetasy = y(6);
thetao = y(7);
thetaoy = y(8);
dydx = zeros(8,1);
switch region
    case 1    % x in [-1 0]
        dydx(1) = usy;
        dydx(2) = (M^2 + 1/k2)*us - alpha*Ps*R2;  
        dydx(3) = u0y;
        dydx(4) = (M^2 + 1/k1 + 1i*omega*R2)*uo - alpha*Po*R2;  % 1/K2 ?
        dydx(5) = thetasy+thetas;
        dydx(6) = k1*usy^2;
        dydx(7) = thetaoy;
        dydx(8) = k*thetao - k3*thetasy*thetaoy;
    case 2    % x in [0 1]
        dydx(1) = usy;
        dydx(2) = (M^2 + 1/k1)*us - Ps*R1;  
        dydx(3) = u0y;
        dydx(4) = (M^2 + 1/k1 + 1i*omega*R1)*uo - Po*R1;
        dydx(5) = thetasy;
        dydx(6) = k2*usy^2;
        dydx(7) = thetaoy;
        dydx(8) = k*thetao - k4*thetasy*thetaoy;
end
end
function res = bc(YL,YR)

mu1=1; mu2=1; k1=1; k2=1;
res = [YL(1,1)
    YL(3,1)
    YL(5,1)
    YL(7,1)
    YR(1,2)-1
    YR(3,2)-1
    YR(5,2)-1
    YR(7,2)
    YR(1,1)-YL(1,2)
    mu1*YR(2,1)-mu2*YL(2,2)
    YR(3,1)-YL(3,2)
    mu1*YR(4,1)-mu2*YL(4,2)
    YR(5,1)-YL(5,2)
    k1*YR(6,1)-k2*YL(6,2)
    YR(7,1)-YL(7,2)
    k1*YR(8,1)-k2*YL(8,2)];     
end

The graph relating to code is

But I need the -1 to 1 range on Y-axis like the following graph:

[Torsten]

What is u, what is y and what is R_I in your code ?

[mallela ankamma rao]

These are the equations relating to code sir

[Torsten]

So y is the independent variable x in your code. But what function is plotted and which parameter is varied ?

Once you have found this, you can simply plot independent and dependent variables reversed:

plot(sol.y(2,:),sol.x,'b-o',sol.y(3,:),sol.x,'r-o')

instead of

plot(sol.x,sol.y(2,:),'b-o',sol.x,sol.y(3 ,:),'r-o')

But note that the solutions are complex-valued; thus it is only possible to plot abs(sol.y), real(sol.y) or imag(sol.y).

[mallela ankamma rao]

Thank you Mr Torsten sir.I got the below graph sir

But the graphs are reversed. I need graphs like the following graph sir and X range is 0 to 1 sir. Please Give me some suggestions sir.

[Torsten]

But the graphs are reversed. I need graphs like the following graph sir and X range is 0 to 1 sir. Please Give me some suggestions sir.

If you know that your sol.y(2,:) and sol.y(3,:) are plotted in the article (real or imaginary part or absolute value) and your plot is different, then there must be something different in your code compared to the code that the authors used. So the only advice I can give is to compare your code to the mathematical equations.

[mallela ankamma rao]

Ok sir thank you

[mallela ankamma rao]

Thank you very much Torsten sir

Answer 1

[ScottB]

ylim([-1 1]);
y_values = [-1:0.2:1];
ha = gca;
set(ha, 'ytick', y_values);
yticklabels('manual');
yticklabels(y_values);

[mallela ankamma rao]

I tried but its not working sir. Please tell me where should I put this code in my code sir