表示山洞表面的方程

Equation to represent mountain-with-hollow-like surface

我想使用函数,其中 x 和 y 是形成圆的坐标,z(x, y) 从边缘的 0 增长到 R/2 的最大值,然后回到中心的 0 而没有急剧变化.我坚持

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

for i = 1:361
  for j = 1:361
     z(i,j) = exp( sqrt((x(i)).^2 + (y(j)).^2));
  end
end
[u, v] = meshgrid(x, y);
mesh(u, v, z), grid on;

我应该怎么实现z把这个drop加到中间呢?任何建议表示赞赏!

也许减去两个具有不同标准偏差的 2D 高斯分布?

% the area
x = linspace(-5,5,1E2);
y = linspace(-5,5,1E2);

sig1=1;
sig2=2;

%2D gaussian
efac = 1/(2*sig1);
X   = exp(-efac*x.^2);
Y   = exp(-efac*y.^2)';
z1    = Y*X;
z1=z1./max(z1(:));

%2D gaussian
efac = 1/(2*sig2);
X   = exp(-efac*x.^2);
Y   = exp(-efac*y.^2)';
z2    = Y*X;
z2=z2./max(z2(:));

[u, v] = meshgrid(x, y);
mesh(u, v, z2-z1), grid on;

在 R 之外具有零并且更符合您自己的代码的替代方案:

x = linspace(-2*pi,2*pi,1E2);
y = linspace(-2*pi,2*pi,1E2);
[u, v] = meshgrid(x, y);
r = sqrt(u.^2+v.^2);
z = sin(r);
z(r>pi)=0;

mesh(u, v, z), grid on;

或者让它底部不那么尖锐:

z = sin(r).^2;