how to separate two graphe in one figure

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how to separate two graphe in one figure like this picture in matlab

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Subhajyoti
Subhajyoti 2024-11-26
编辑:Subhajyoti 2024-11-26
It is my understanding that you want to draw multiple plots on the same axes. You can use 'hold on' to retains the plots in the current axes so that new plots added to the axes do not delete existing plots.
Here, in the following implementation, I have added a surface plot and a 3D line plot on the same same using the same.
% Define the grid
[x, y] = meshgrid(linspace(-1, 1, 100), linspace(-1, 1, 100));
% Define the surface function
z1 = sin(pi*x).*cos(pi*y);
% Define the second graph (e.g., a circle with noise)
theta = linspace(0, 2*pi, 100);
r = 0.5 + 0.1*rand(1, 100); % Add some noise
x2 = r .* cos(theta);
y2 = r .* sin(theta);
z2 = -2 * ones(size(x2)); % Place it below the surface
% Create the figure
figure;
% Plot the surface
surf(x, y, z1, 'EdgeColor', 'none');
hold on;
% Plot the second graph
plot3(x2, y2, z2, 'b-', 'LineWidth', 2);
plot3(x2, y2, z2, 'r.', 'MarkerSize', 10);
% Customize the view
colormap('jet');
colorbar;
xlabel('x');
ylabel('y');
zlabel('z');
grid on;
hold off;
Refer to the following MathWorks Documentation to know more about 'hold' and '3D Plots' in MATLAB Graphics:
  2 个评论
Belaid
Belaid 2024-11-29
how can plot the curve gare with equation 1+(1/10)*tanh(10*sin(25t)) in the boundary and analytical function u(x,y)=exp(2.*x+y)+sin(pi.*x).*cosh(y); to find the graphe like picture
% Define the grid
[x, y] = meshgrid(linspace(0, 2*pi, 1000), linspace(0, 2*pi, 1000));
%r=cos(2.*x).^2.*exp(cos(x)) + sin(2.*y).^2.*exp(sin(y));
% r=sqrt(cos(2.*x)+sqrt(1.1-sin(2.*x).^2));
r=1+1/10*tanh(10*sin(25*x));
r1=1+1/10*tanh(10*sin(25*y));
x1 = r .* cos(x);
y1 = r .* sin(y);
% Define the surface function
z1=exp(2.*x1+y1)+sin(pi.*x1).*cosh(y1);
% Define the second graph (e.g., a circle with noise)
theta = linspace(0, 2*pi, 100);
%r = 0.5 + 0.1*rand(1, 100);
% r=sqrt(cos(2.*theta)+sqrt(1.1-sin(2.*theta).^2));
%r=cos(2.*theta).^2.*exp(cos(theta)) + sin(2.*theta).^2.*exp(sin(theta));
r=1+1/10*tanh(10*sin(25*theta));
% Add some noise
x2 = r .* cos(theta);
y2 = r .* sin(theta);
z2 = -2 * ones(size(x2)); % Place it below the surface
% Create the figure
figure;
% Plot the surface
surf(x1, y1, z1, 'EdgeColor', 'none');
hold on;
% Plot the second graph
plot3(x2, y2, z2, 'b-', 'LineWidth', 2);
% plot3(x2, y2, z2, 'r.', 'MarkerSize', 10);
% Customize the view
colormap('jet');
colorbar;
xlabel('x');
ylabel('y');
zlabel('z');
grid on;
hold off;

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更多回答(1 个)

Image Analyst
Image Analyst 2024-11-26
If you want one plot or surface to be "lifted" above the other, you can add an offset to the one you want shifted. Be aware that the axes units will not align with the original units, but with the shifted units.
  4 个评论
Belaid
Belaid 2024-11-29
编辑:Image Analyst 2024-11-29
How can plot the curve gare with equation 1+(1/10)*tanh(10*sin(25t)) in the boundary and analytical function u(x,y)=exp(2.*x+y)+sin(pi.*x).*cosh(y); to find the graph like picture?
% Define the grid
[x, y] = meshgrid(linspace(0, 2*pi, 1000), linspace(0, 2*pi, 1000));
%r=cos(2.*x).^2.*exp(cos(x)) + sin(2.*y).^2.*exp(sin(y));
% r=sqrt(cos(2.*x)+sqrt(1.1-sin(2.*x).^2));
r=1+1/10*tanh(10*sin(25*x));
r1=1+1/10*tanh(10*sin(25*y));
x1 = r .* cos(x);
y1 = r .* sin(y);
% Define the surface function
z1=exp(2.*x1+y1)+sin(pi.*x1).*cosh(y1);
% Define the second graph (e.g., a circle with noise)
theta = linspace(0, 2*pi, 100);
%r = 0.5 + 0.1*rand(1, 100);
% r=sqrt(cos(2.*theta)+sqrt(1.1-sin(2.*theta).^2));
%r=cos(2.*theta).^2.*exp(cos(theta)) + sin(2.*theta).^2.*exp(sin(theta));
r=1+1/10*tanh(10*sin(25*theta));
% Add some noise
x2 = r .* cos(theta);
y2 = r .* sin(theta);
z2 = -2 * ones(size(x2)); % Place it below the surface
% Create the figure
figure;
% Plot the surface
surf(x1, y1, z1, 'EdgeColor', 'none');
hold on;
% Plot the second graph
plot3(x2, y2, z2, 'b-', 'LineWidth', 2);
% plot3(x2, y2, z2, 'r.', 'MarkerSize', 10);
% Customize the view
colormap('jet');
colorbar;
xlabel('x');
ylabel('y');
zlabel('z');
grid on;
hold off;
Image Analyst
Image Analyst 2024-11-29
I don't understand the question. What is t? Is it x or y? And if it's one of those, what value would the other x or y have? Maybe you should start your own question rather than continue here in @Belaid Boutqlmount's question (for which he'll get emails about).

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