how to get data from an interval

Hi, I need to use the y values between the x=50 and x=180 values to do another plot. How can I get these vaules from the current plot? I've readed some answers using the 'findobj' function but I don't know if it's the correct function for my task and I don't understand how to use it in my specific situation.

Could someone help me please?

here's my code:

clc;
clear;
close all;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


P=0.5*800000;%kN                                -carico agente
R=20;%mm                                        -raggio medio
L=3*75;%mm                                      -lunghezza tubo
s=1;%mm                                         -spessore di parete
ni=0.3;%                                        -coeff poisson
E=210000;%MPa                                   -modulo young
w=2;%MPa                                        -pressione int/ext
K=0.5;%                                         -incasto incastro


%equazioni derivanti dall'equilibrio radiale:

E1=E/(1-(ni^2));%                     -modulo young dilataz laterale impedita
I=(2*pi*R*(s^3))/12;%                 -momento di inerzia cilindro
k=2*pi*((E*s)/R);%                    -rigidezza molla fondazione
Pcr=sqrt(4*k*E1*I)%                   -carico critico
Pcre=((pi)^2*E*I)/(K*((L/1000)^2))%   -carico critico secondo Eulero


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    
f = @(z,u) [u(2); u(3); u(4); (-1/(E1*I))*((P*u(3))+(k*u(1)))]; %equazione differenziale w=0

bc = @(ua, ub) [ua(1); ua(2)-0.00000001; ub(1); ub(2)+0.00000001];

zmesh = linspace(0, L, 100000);%crea vettore equispaziato da 0->lunghezza cilindro
iguess= @(z) [0; 0; 0; 0];%initial guess @soluzione dalla quale inizia l'iterazione della function bvp4c

solinit = bvpinit(zmesh, iguess); %reference @doc bvpinit (form initial guess value)

opts = bvpset('RelTol',0.1,'AbsTol',0.1,'Stats','on'); %setting per ottenere le statistiche della function bvp4c
sol= bvp4c(f, bc, solinit,opts)% sol in x, y values 

figure
hold on
grid on
plot(sol.x(1,:),sol.y(1,:),'.') %I need the y values from this plot for x from 50 to 180

legend('w=0');
xlabel('z');
ylabel('solution u');

Answer 1

[Ameer Hamza]

You can use logical indexing. Add these lines at the end of your code

x = sol.x;
y = sol.y;
idx = (x > 50) & (x < 180);

figure
hold on
grid on
plot(x(idx), y(1, idx))

[federico midei]

Is it also possibile to get the max of the y values from this new range?

[Ameer Hamza]

Yes,

x = sol.x;
y = sol.y;
idx = (x > 50) & (x < 180);
x_new = x(idx);
y_new = y(1, idx);

figure
hold on
grid on
plot(x_new, y_new)

y_new_max = max(y_new);

[federico midei]

I'm sorry for bothering you but I have another problem.

I need to run the code written in the first question for the variable "P" running from 400000 to 900000, then consider only the max( sol.y) value considered in the sol.x>50 and sol.x<180 range for every "P(i)" solution.

i've tried to fit the code in a for loop i=1:length(n) allocating a P(i) value to the P variable where n is a "

linspace" vector but I don't know if it is correct.

Cuold you help me please?

this is the for loop that should solve the differential equation f for every P value:

n=linspace(400000,900000);
for i=1:length(n)
f = @(z,u) [u(2); u(3); u(4); (-1/(E1*I))*((P(i)*u(3))+(k*u(1)))];

bc = @(ua, ub) [ua(1); ua(2)-0.00000001; ub(1); ub(2)+0.00000001];
zmesh = linspace(0, L, 100000);
iguess= @(z) [0; 0; 0; 0];
solinit = bvpinit(zmesh, iguess);
Sol= bvp4c(f, bc, solinit);
end

[Ameer Hamza]

Instead of n, assign the linspace values to P.

P = linspace(400000,900000);
y_max_values = zeros(size(P));
for i=1:length(P)
    f = @(z,u) [u(2); u(3); u(4); (-1/(E1*I))*((P(i)*u(3))+(k*u(1)))];
    bc = @(ua, ub) [ua(1); ua(2)-0.00000001; ub(1); ub(2)+0.00000001];
    zmesh = linspace(0, L, 100000);
    iguess= @(z) [0; 0; 0; 0];
    solinit = bvpinit(zmesh, iguess);
    Sol= bvp4c(f, bc, solinit);
    
    x = Sol.x;
    y = Sol.y;
    idx = (x > 50) & (x < 180);
    x_new = x(idx);
    y_new = y(1, idx);
    
    y_max_values(i) = max(y_new);
end

[federico midei]

running this I get this error :

Unable to resolve the name sol.x.

Error in Untitled (line 36)

% x = sol.x;

[Ameer Hamza]

Please check the code in the updated comment. You were using Sol as a variable name. I mistakenly typed sol.