以圆盘或圆形形式模拟波浪

Question

在为波浪模拟选择圆盘视图或圆形视图选项时运行此代码时出现错误。附上代码和错误。我认为这部分代码通常在 fzero 函数中存在一些问题。任何帮助都会很棒。

代码：

function z = bjzeros(n,k)
% BJZEROS  Zeros of the Bessel function.
% z = bjzeros(n,k) is the first k zeros of besselj(n,x)
% delta must be chosen so that the linear search can take
% steps as large as possible 

delta = .99*pi;
Jsubn = inline('besselj(n,x)''x','n');
a = n+1;
fa = besselj(n,a);
z = zeros(1,k);
j = 0;
while j < k
   b = a + delta;
   fb = besselj(n,b);
   if sign(fb) ~= sign(fa)
      j = j+1;
      z(j) = fzerotx(Jsubn,[a b],n);
   end
   a = b;
   fa = fb;
end

错误：

<块引用>

未定义函数“fzerotx”，用于“inline”类型的输入参数。

waves>bjzeros 中的错误（第 292 行）

 z(j) = fzerotx(Jsubn,[a b],n);

waves 中的错误（第 137 行）

       mu = [bjzeros(0,2) bjzeros(1,2)];

Answer 1

函数声明和语法

不得声明 fzerotx() 函数。您可以按照以下文件结构在同一目录中创建所需的 M 文件/函数。另一个小错误可能是缺少逗号引起的，我通过更改行摆脱了错误：

Jsubn = inline('besselj(n,x)''x','n');

到

Jsubn = inline('besselj(n,x)','x','n');

文件 1：主文件/函数调用 → [main.m]

mu = [bjzeros(0,2) bjzeros(1,2)];

---

文件 2：bjzeros() 函数 → [bjzeros.m]

function z = bjzeros(n,k)
% BJZEROS  Zeros of the Bessel function.
% z = bjzeros(n,k) is the first k zeros of besselj(n,x)
% delta must be chosen so that the linear search can take
% steps as large as possible 

delta = .99*pi;
Jsubn = inline('besselj(n,x)','x','n');
a = n+1;
fa = besselj(n,a);
z = zeros(1,k);
j = 0;
while j < k
   b = a + delta;
   fb = besselj(n,b);
   if sign(fb) ~= sign(fa)
      j = j+1;
      z(j) = fzerotx(Jsubn,[a b],n);
   end
   a = b;
   fa = fb;
end
end

---

文件 3：fzerotx() 函数 → [fzerotx.m]

函数参考：MATLAB: Textbook version of FZERO

function b = fzerotx(F,ab,varargin)
%FZEROTX  Textbook version of FZERO.
%   x = fzerotx(F,[a,b]) tries to find a zero of F(x) between a and b.
%   F(a) and F(b) must have opposite signs.  fzerotx returns one 
%   end point of a small subinterval of [a,b] where F changes sign.
%   Arguments beyond the first two, fzerotx(F,[a,b],p1,p2,...),
%   are passed on, F(x,p1,p2,..).
%
%   Examples:
%      fzerotx(@sin,[1,4])
%      F = @(x) sin(x); fzerotx(F,[1,4])

%   Copyright 2014 Cleve Moler
%   Copyright 2014 The MathWorks, Inc.

% Initialize.
a = ab(1);
b = ab(2);
fa = F(a,varargin{:});
fb = F(b,varargin{:});
if sign(fa) == sign(fb)
   error('Function must change sign on the interval')
end
c = a;
fc = fa;
d = b - c;
e = d;

% Main loop, exit from middle of the loop
while fb ~= 0

   % The three current points, a, b, and c, satisfy:
   %    f(x) changes sign between a and b.
   %    abs(f(b)) <= abs(f(a)).
   %    c = previous b, so c might = a.
   % The next point is chosen from
   %    Bisection point, (a+b)/2.
   %    Secant point determined by b and c.
   %    Inverse quadratic interpolation point determined
   %    by a, b, and c if they are distinct.

   if sign(fa) == sign(fb)
      a = c;  fa = fc;
      d = b - c;  e = d;
   end
   if abs(fa) < abs(fb)
      c = b;    b = a;    a = c;
      fc = fb;  fb = fa;  fa = fc;
   end
   
   % Convergence test and possible exit
   m = 0.5*(a - b);
   tol = 2.0*eps*max(abs(b),1.0);
   if (abs(m) <= tol) | (fb == 0.0)
      break
   end
   
   % Choose bisection or interpolation
   if (abs(e) < tol) | (abs(fc) <= abs(fb))
      % Bisection
      d = m;
      e = m;
   else
      % Interpolation
      s = fb/fc;
      if (a == c)
         % Linear interpolation (secant)
         p = 2.0*m*s;
         q = 1.0 - s;
      else
         % Inverse quadratic interpolation
         q = fc/fa;
         r = fb/fa;
         p = s*(2.0*m*q*(q - r) - (b - c)*(r - 1.0));
         q = (q - 1.0)*(r - 1.0)*(s - 1.0);
      end;
      if p > 0, q = -q; else p = -p; end;
      % Is interpolated point acceptable
      if (2.0*p < 3.0*m*q - abs(tol*q)) & (p < abs(0.5*e*q))
         e = d;
         d = p/q;
      else
         d = m;
         e = m;
      end;
   end
   
   % Next point
   c = b;
   fc = fb;
   if abs(d) > tol
      b = b + d;
   else
      b = b - sign(b-a)*tol;
   end
   fb = F(b,varargin{:});
end

使用 MATLAB R2019b 运行