how do i get a proper time plot for this qudratic equation?

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Pranay
Pranay 2024-9-28
编辑： William Rose 2024-9-29
Ran in:
the script seems to be not behaving properly,as it is supposed to be ?
% Define the range for x
x = linspace(-5, 5, 10); % 10 points between -5 and 5
dx1 = 2*x.^2 - 8; % Compute the derivative (velocity field)
% Create a figure
figure;
hold on;
% Plot the real line (x-axis)
plot(x, zeros(size(x)), 'k'); % The x-axis (real line)
% Plot the vector field as arrows
for i = 1:length(x)
if dx1(i) > 0
% Arrow pointing to the right (positive dx)
quiver(x(i), 0, 0.5, 0, 'r', 'MaxHeadSize', 0.5);
elseif dx1(i) < 0
% Arrow pointing to the left (negative dx)
quiver(x(i), 0, -0.5, 0, 'b', 'MaxHeadSize', 0.5);
end
end
% Plot equilibria points
plot([-2, 2], [0, 0], 'bo', 'MarkerFaceColor', 'b', 'MarkerSize', 8); % Equilibria at x = -2 and x = 2
% Labels and title
xlabel('x');
ylabel('x-dot');
title('1D Vector Field for x-dot = 2x^2 - 8');
ylim([-1 1]); % Set y-axis limits to keep focus on x-axis
xlim([-4 4]); % Set x-axis limits
grid on; % Add grid
% Add text annotations for equilibrium points
text(-2, 0.2, 'Stable Equilibrium: x = -2', 'HorizontalAlignment', 'center');
text(2, 0.2, 'Unstable Equilibrium: x = 2', 'HorizontalAlignment', 'center');
hold off;
t = [-5 5];
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
figure;
plot(time, x, 'LineWidth',0.5);
title('Solution of dx/dt = 2x^2 - 8 vs Time');
xlabel('Time t');
ylabel('x(t)');
xlim([-t(2) t(2)]); % Adjust x-axis limits for clarity
ylim([-3 3]); % Set y-axis limits to see behavior around -2 and 2
grid on;
the first part works perfectly fine, but the 2nd part where i have to plot wrt time, is where i am having trouble? its supposed to be going from x-2 being stable , and 2 being unstable, the plot should depict that.
but I am unable to get that done properly
  1 个评论
Walter Roberson
Walter Roberson 2024-9-28
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
Your ode_function is written in a way that appears to be single input and single output. But you are passing in vector x0, so the input x to ode_function will be a 2 x 1 vector. Your 2*x.^2-8 is going to be calculated with that 2 x 1 vector, returning a 2 x 1 vector. We must ask whether this is really what you want.

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回答（1 个）

William Rose
William Rose 2024-9-29
编辑：William Rose 2024-9-29
Ran in:
@Pranay,
Let us try different initial values for x: x0=[-2,...,2]. Use a different initial value on each pass. Store results in a cell array, because t,x will have different lengths on different passes, so an array would not work.
t = [-5 5];
x0 = [2,1.99,1.95,1.8,1.5,0,-2]; % initial values
N=length(x0);
C={1,N}; % allocate cell array for results
legstr={1,N}; % allocate cell array for plot legend strings
ode_function = @(t,x) 2*x.^2-8;
for i=1:N
[time, x] = ode45(ode_function, t, x0(i));
C{i}=[time,x];
legstr{i}=num2str(x0(i));
end
% Plot results
figure
hold on
linespec={'-r.','-g.','-b.','-c.','-m.','-y.','-k.'};
for i=1:N
data=C{i};
time=data(:,1); x=data(:,2);
plot(time, x, linespec{i});
end
title('dx/dt = 2x^2 - 8, at different initial values');
xlabel('Time t'); ylabel('x(t)');
legend(legstr); grid on
OK

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