如何实现带中心凹陷的类山形曲面z(x,y)的MATLAB建模？

Hey there! Let's fix up your code to get that centrally depressed surface you're after—where the value hits 0 at the center and edge, peaks at R/2, and stays smooth with no abrupt changes.

First, let's clean up your original code: nested loops are unnecessary here, and we can calculate the radial distance r directly using meshgrid for efficiency.

Key Requirements Recap #

We need a function z(r) (where r = sqrt(x² + y²) is the distance from the center) that satisfies:

• z(0) = 0 (center is depressed to 0)
• z(R/2) = maximum value (peak at half the radius)
• z(R) = 0 (edge is 0)
• Smooth, no abrupt changes (continuous first/second derivatives across the entire circle)

Option 1: Smooth Polynomial Function #

A quartic (4th-degree) polynomial works perfectly here—it's smooth, symmetric, and meets all our constraints. The formula is:

z = 16 * (r.^2) .* ((R - r).^2) ./ (R^4);

• At r=0 and r=R, z=0
• At r=R/2, z=1 (you can scale this by multiplying by a constant if you want a larger peak)
• First and second derivatives are continuous everywhere, so no sharp edges or abrupt changes.

Option 2: Trigonometric Function #

If you prefer a trig-based approach, a squared sine function also fits the bill:

z = sin(pi * r / R).^2;

• This is also smooth, with z=0 at r=0 and r=R, and z=1 at r=R/2
• The shape is slightly different than the polynomial (more "rounded" near the peak), but still meets all your requirements.

Modified Full MATLAB Code #

Here's how to implement either option (we'll use the polynomial as an example, but you can swap in the trig function easily):

R = 5;
theta = -pi:pi/180:pi;
x = R * cos(theta);
y = R * sin(theta);

% Create meshgrid for x and y
[u, v] = meshgrid(x, y);

% Calculate radial distance from center
r = sqrt(u.^2 + v.^2);

% Option 1: Quartic polynomial
z = 16 * (r.^2) .* ((R - r).^2) ./ (R^4);

% Uncomment below for Option 2: Trigonometric function
% z = sin(pi * r / R).^2;

% Plot the surface
mesh(u, v, z);
grid on;
xlabel('X');
ylabel('Y');
zlabel('Z');
title('Centrally Depressed Circular Surface');

Why This Works #

• We eliminate the nested loops by vectorizing calculations with meshgrid, which is way faster and more readable in MATLAB.
• Both functions are symmetric around r=R/2, smooth, and exactly hit your required values at the center, peak, and edge.

内容的提问来源于stack exchange，提问作者Serge S