function - 构造函数

Question

从这段代码开始：

    clc, clear all, close all
tic

k1 = 0.01:0.1:100;
k2 = 0.01:0.1:100;
k3 = 0.01:0.1:100;

k = sqrt(k1.^2 + k2.^2 + k3.^2);

c = 1.476;
gamma = 3.9;

colors = {'cyan'};
Ek = (1.453*k.^4)./((1 + k.^2).^(17/6));
E = @(k) (1.453*k.^4)./((1 + k.^2).^(17/6));
E_int = zeros(1,numel(k1));
E_int(1) = 1.5;

for i = 2:numel(k)
    E_int(i) = E_int(i-1) - integral(E,k(i-1),k(i));
end

beta = c*gamma./(k.*sqrt(E_int));


F_11 = zeros(1,numel(k1));
F_22 = zeros(1,numel(k1));
F_33 = zeros(1,numel(k1));

count = 0;
for i = 1:numel(k1)
    count = count + 1;
    phi_11 = @(k2,k3) phi_11_new(k1,k2,k3,beta,i);
    phi_22 = @(k2,k3) phi_22_new(k1,k2,k3,beta,i);
    phi_33 = @(k2,k3) phi_33_new(k1,k2,k3,beta,i);
    F_11(count) = integral2(phi_11,-100,100,-100,100);
    F_22(count) = integral2(phi_22,-100,100,-100,100);
    F_33(count) = integral2(phi_33,-100,100,-100,100);
end

figure
hold on
plot(k1,F_11,'b')
plot(k1,F_22,'cyan')
plot(k1,F_33,'magenta')
hold off

在哪里

function phi_11 = phi_11_new(k1,k2,k3,beta,i)
k = sqrt(k1(i).^2 + k2.^2 + k3.^2);
k30 = k3 + beta(i).*k1(i);
k0 = sqrt(k1(i).^2 + k2.^2 + k30.^2);
E_k0 = 1.453.*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta(i).*k1(i).^2).*(k1(i).^2 + k2.^2 - k3.*k30)./(k.^2.*(k1(i).^2 + k2.^2));
C2 = k2.*k0.^2./((k1(i).^2 + k2.^2).^(3/2)).*atan2((beta(i).*k1(i).*sqrt(k1(i).^2 + k2.^2)),(k0.^2 - k30.*k1(i).*beta(i)));
xhsi1 = C1 - k2./k1(i).*C2;
xhsi1_q = xhsi1.^2;
phi_11 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k1(i).^2 - 2.*k1(i).*k30.*xhsi1 + (k1(i).^2 + k2.^2).*xhsi1_q);
end

function phi_22 = phi_22_new(k1,k2,k3,beta,i)
k = sqrt(k1(i).^2 + k2.^2 + k3.^2);
k30 = k3 + beta(i).*k1(i);
k0 = sqrt(k1(i).^2 + k2.^2 + k30.^2);
E_k0 = 1.453.*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta(i).*k1(i).^2).*(k1(i).^2 + k2.^2 - k3.*k30)./(k.^2.*(k1(i).^2 + k2.^2));
C2 = k2.*k0.^2./((k1(i).^2 + k2.^2).^(3/2)).*atan2((beta(i).*k1(i).*sqrt(k1(i).^2 + k2.^2)),(k0.^2 - k30.*k1(i).*beta(i)));
xhsi2 = k2./k1(i).*C1 + C2;
xhsi2_q = xhsi2.^2;
phi_22 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k2.^2 - 2.*k2.*k30.*xhsi2 + (k1(i).^2 + k2.^2).*xhsi2_q);
end

function phi_33 = phi_33_new(k1,k2,k3,beta,i)
k = sqrt(k1(i).^2+k2.^2+k3.^2);
k30 = k3 + beta(i).*k1(i);
k0 = sqrt(k1(i).^2+k2.^2+k30.^2);
E_k0 = (1.453.*k0.^4./((1+k0.^2).^(17/6)));
phi_33 = (E_k0./(4*pi.*(k.^4))).*(k1(i).^2+k2.^2);
end

这个程序导致我的结果与来自研究的其他一些结果不匹配。我应该匹配的结果发布在下面：

而我的看起来像这些

很容易估计只有comp w 与理论结果相匹配；因此，我认为该缺陷可能存在于函数 phi_11_new（和 phi_22_new）之外的 beta 定义中。

你们中的任何人都可以建议如何在 phi_11_new（和 phi_22_new）中计算 beta，而不是我目前的方式吗？

我提前感谢大家的支持。

最好的问候， fpe

Answer 1

我改进了插值，使其不再因小值而崩溃。它还返回更正确的值，因为它现在内插值的对数。

这是代码，就像现在一样。

function test15()

[k1,k2,k3] = deal(0.01:0.1:400);

k = sqrt(k1.^2 + k2.^2 + k3.^2);

c = 1.476;
gamma = 3.9;

Ek = (1.453*k.^4)./((1 + k.^2).^(17/6));
E_int = 1.5-cumtrapz(k,Ek);
beta = c*gamma./(k.*sqrt(E_int));

[F_11,F_22,F_33] = deal(zeros(1,numel(k1)));

k_vec = k;
beta_vec = beta;

kLim = 100;

for ii = 1:numel(k1)
    phi_11 = @(k2,k3) phi_11_new(k1(ii),k2,k3,k_vec,beta_vec);
    phi_22 = @(k2,k3) phi_22_new(k1(ii),k2,k3,k_vec,beta_vec);
    phi_33 = @(k2,k3) phi_33_new(k1(ii),k2,k3,k_vec,beta_vec);
    F_11(ii) = quad2d(phi_11,-kLim,kLim,-kLim,kLim);
    F_22(ii) = quad2d(phi_22,-kLim,kLim,-kLim,kLim);
    F_33(ii) = quad2d(phi_33,-kLim,kLim,-kLim,kLim);
end

figure
loglog(k1,F_11,'b')
hold on
loglog(k1,F_22,'cyan')
loglog(k1,F_33,'magenta')
hold off
grid on

end

function phi_11 = phi_11_new(k1,k2,k3,k_vec,beta_vec)
k = sqrt(k1^2 + k2.^2 + k3.^2);

log_beta_vec = interp1(log(k_vec),log(beta_vec),log(k(:)),'linear','extrap');
log_beta = reshape(log_beta_vec,size(k));
beta = exp(log_beta);

k30 = k3 + beta*k1;
k0 = sqrt(k1^2 + k2.^2 + k30.^2);
E_k0 = 1.453*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta*k1^2).*(k1^2 + k2.^2 - k3.*k30)./(k.^2.*(k1^2 + k2.^2));
C2 = k2.*k0.^2./((k1^2 + k2.^2).^(3/2)).*atan2((beta*k1.*sqrt(k1^2 + k2.^2)),(k0.^2 - k30*k1.*beta));
xhsi1 = C1 - (k2/k1).*C2;
xhsi1_q = xhsi1.^2;
phi_11 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k1^2 - 2*k1*k30.*xhsi1 + (k1^2 + k2.^2).*xhsi1_q);
end

function phi_22 = phi_22_new(k1,k2,k3,k_vec,beta_vec)
k = sqrt(k1^2 + k2.^2 + k3.^2);

log_beta_vec = interp1(log(k_vec),log(beta_vec),log(k(:)),'linear','extrap');
log_beta = reshape(log_beta_vec,size(k));
beta = exp(log_beta);

k30 = k3 + beta*k1;
k0 = sqrt(k1^2 + k2.^2 + k30.^2);
E_k0 = 1.453*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta*k1^2).*(k1^2 + k2.^2 - k3.*k30)./(k.^2.*(k1^2 + k2.^2));
C2 = k2.*k0.^2./((k1^2 + k2.^2).^(3/2)).*atan2((beta*k1.*sqrt(k1^2 + k2.^2)),(k0.^2 - k30.*k1.*beta));
xhsi2 = (k2/k1).*C1 + C2;
xhsi2_q = xhsi2.^2;
phi_22 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k2.^2 - 2.*k2.*k30.*xhsi2 + (k1^2 + k2.^2).*xhsi2_q);
end

function phi_33 = phi_33_new(k1,k2,k3,k_vec,beta_vec)
k = sqrt(k1^2+k2.^2+k3.^2);

log_beta_vec = interp1(log(k_vec),log(beta_vec),log(k(:)),'linear','extrap');
log_beta = reshape(log_beta_vec,size(k));
beta = exp(log_beta);

k30 = k3 + beta*k1;
k0 = sqrt(k1^2+k2.^2+k30.^2);
E_k0 = (1.453*k0.^4./((1+k0.^2).^(17/6)));
phi_33 = (E_k0./(4*pi*(k.^4))).*(k1^2+k2.^2);
end

该图似乎与原始结果非常吻合。即使还是有一些不同。

旁注：由于在模拟中将 k 值 100 设置为上限，因此图中大于此值的值是不正确的。它们的计算不使用完整 (k2,k3)-“圆”中的所有值。我们还可以看到这些值的偏差。

Answer 2

好的，这就是我目前得到的。我想听听您对此的看法——它还不完美。我无法访问这些函数integral，或者integral2如果您可以重新插入它们（而不是我quad2d的例子）并测试代码，您可能会得到比我现在更好的结果。

我的第一个想法是beta在一个 for 循环中计算 -functions[k1,k2,k3]中的每个三元组phi。结果证明这非常慢，所以我改用了一个k-values 向量，并像以前一样计算了相应的向量beta。然后将这两个向量传递到phi插值函数 ( interp1) 中使用这些值的位置，以找到beta特定值的值k。

function myFunction()

[k1,k2,k3] = deal(0.01:0.1:400);

k = sqrt(k1.^2 + k2.^2 + k3.^2);

c = 1.476;
gamma = 3.9;

Ek = (1.453*k.^4)./((1 + k.^2).^(17/6));
E_int = 1.5-cumtrapz(k,Ek);
beta = c*gamma./(k.*sqrt(E_int));

[F_11,F_22,F_33] = deal(zeros(1,numel(k1)));

k_vec = k;
beta_vec = beta;

for ii = 1:numel(k1)
    phi_11 = @(k2,k3) phi_11_new(k1(ii),k2,k3,k_vec,beta_vec);
    phi_22 = @(k2,k3) phi_22_new(k1(ii),k2,k3,k_vec,beta_vec);
    phi_33 = @(k2,k3) phi_33_new(k1(ii),k2,k3,k_vec,beta_vec);
    F_11(ii) = quad2d(phi_11,-100,100,-100,100);
    F_22(ii) = quad2d(phi_22,-100,100,-100,100);
    F_33(ii) = quad2d(phi_33,-100,100,-100,100);
end

figure
loglog(k1,F_11,'b')
hold on
loglog(k1,F_22,'cyan')
loglog(k1,F_33,'magenta')
hold off
grid on

end

function phi_11 = phi_11_new(k1,k2,k3,k_vec,beta_vec)
k = sqrt(k1^2 + k2.^2 + k3.^2);

beta = reshape(interp1(k_vec,beta_vec,k(:)),size(k));

k30 = k3 + beta*k1;
k0 = sqrt(k1^2 + k2.^2 + k30.^2);
E_k0 = 1.453*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta*k1^2).*(k1^2 + k2.^2 - k3.*k30)./(k.^2.*(k1^2 + k2.^2));
C2 = k2.*k0.^2./((k1^2 + k2.^2).^(3/2)).*atan2((beta*k1.*sqrt(k1^2 + k2.^2)),(k0.^2 - k30*k1.*beta));
xhsi1 = C1 - (k2/k1).*C2;
xhsi1_q = xhsi1.^2;
phi_11 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k1^2 - 2*k1*k30.*xhsi1 + (k1^2 + k2.^2).*xhsi1_q);
end

function phi_22 = phi_22_new(k1,k2,k3,k_vec,beta_vec)
k = sqrt(k1^2 + k2.^2 + k3.^2);

beta = reshape(interp1(k_vec,beta_vec,k(:)),size(k));

k30 = k3 + beta*k1;
k0 = sqrt(k1^2 + k2.^2 + k30.^2);
E_k0 = 1.453*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta*k1^2).*(k1^2 + k2.^2 - k3.*k30)./(k.^2.*(k1^2 + k2.^2));
C2 = k2.*k0.^2./((k1^2 + k2.^2).^(3/2)).*atan2((beta*k1.*sqrt(k1^2 + k2.^2)),(k0.^2 - k30.*k1.*beta));
xhsi2 = (k2/k1).*C1 + C2;
xhsi2_q = xhsi2.^2;
phi_22 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k2.^2 - 2.*k2.*k30.*xhsi2 + (k1^2 + k2.^2).*xhsi2_q);
end

function phi_33 = phi_33_new(k1,k2,k3,k_vec,beta_vec)
k = sqrt(k1^2+k2.^2+k3.^2);

beta = reshape(interp1(k_vec,beta_vec,k(:)),size(k));

k30 = k3 + beta*k1;
k0 = sqrt(k1^2+k2.^2+k30.^2);
E_k0 = (1.453*k0.^4./((1+k0.^2).^(17/6)));
phi_33 = (E_k0./(4*pi*(k.^4))).*(k1^2+k2.^2);
end

这将产生下图。请注意，对于 的最小值，积分不会成功k1。

编辑 - 关于在 phi 函数中计算 beta 的评论

由于您基本上尝试了与我最初所做的相同的事情，因此我添加了一个示例，说明如何在-functions 中计算beta矩阵。phi请注意，这段代码太慢了，我从来没有真正运行过它。

function phi_11 = phi_11_new(k1,k2,k3)
k = sqrt(k1^2 + k2.^2 + k3.^2);

c = 1.476;
gamma = 3.9;
beta = zeros(size(k));
E = @(x) (1.453*x.^4)./((1 + x.^2).^(17/6));
for ii = 1:size(k,1)
    for jj = 1:size(k,2)
        E_int = 1.5-quad(E,0.001,k(ii,jj));
        beta(ii,jj) = c*gamma/(k(ii,jj)*sqrt(E_int));
    end
end


k30 = k3 + beta*k1;
k0 = sqrt(k1^2 + k2.^2 + k30.^2);
E_k0 = 1.453*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta*k1^2).*(k1^2 + k2.^2 - k3.*k30)./(k.^2.*(k1^2 + k2.^2));
C2 = k2.*k0.^2./((k1^2 + k2.^2).^(3/2)).*atan2((beta*k1.*sqrt(k1^2 + k2.^2)),(k0.^2 - k30*k1.*beta));
xhsi1 = C1 - (k2/k1).*C2;
xhsi1_q = xhsi1.^2;
phi_11 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k1^2 - 2*k1*k30.*xhsi1 + (k1^2 + k2.^2).*xhsi1_q);
end

Answer 3

目前我正在运行上述代码的不同版本。

它如下

clc, clear all, close all
tic



k1 = (0.01:0.1:100);

c = 1.476;
gamma = 3.9;


F_11 = zeros(1,numel(k1));
F_22 = zeros(1,numel(k1));
F_33 = zeros(1,numel(k1));

count = 0;
for i = 1:numel(k1)
    count = count + 1;
    phi_11 = @(k2,k3) phi_11_new(k1,k2,k3,gamma,i);
    phi_22 = @(k2,k3) phi_22_new(k1,k2,k3,gamma,i);
    phi_33 = @(k2,k3) phi_33_new(k1,k2,k3,gamma,i);
    F_11(count) = integral2(phi_11,-100,100,-100,100);
    F_22(count) = integral2(phi_22,-100,100,-100,100);
    F_33(count) = integral2(phi_33,-100,100,-100,100);
end

这次 phi_11、phi_22 和 phi_33 分别为

function phi_11 = phi_11_new(k1,k2,k3,gamma,i)
k = sqrt(k1(i).^2 + k2.^2 + k3.^2);
beta = gamma./((k.^(2/3)).*sqrt(hypergeom([1/3,17/6],4/3,-k.^(-2))));  
k30 = k3 + beta.*k1(i);
k0 = sqrt(k1(i).^2 + k2.^2 + k30.^2);
E_k0 = 1.453.*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta(i).*k1(i).^2).*(k0.^2 - 2.*k30.^2 + beta.*k30.*k1(i))./(k.^2.*(k1(i).^2 + k2.^2));
C2 = k2.*k0.^2./((k1(i).^2 + k2.^2).^(3/2)).*atan2((beta.*k1(i).*sqrt(k1(i).^2 + k2.^2)),(k0.^2 - k30.*k1(i).*beta));
xhsi1 = C1 - k2./k1(i).*C2;
xhsi1_q = xhsi1.^2;
phi_11 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k1(i).^2 - 2.*k1(i).*k30.*xhsi1 + (k1(i).^2 + k2.^2).*xhsi1_q);
end

function phi_22 = phi_22_new(k1,k2,k3,gamma,i)
k = sqrt(k1(i).^2 + k2.^2 + k3.^2);
beta = gamma./((k.^(2/3)).*sqrt(hypergeom([1/3,17/6],4/3,-k.^(-2)))); 
k30 = k3 + beta.*k1(i);
k0 = sqrt(k1(i).^2 + k2.^2 + k30.^2);
E_k0 = 1.453.*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta.*k1(i).^2).*(k0.^2 - 2.*k30.^2 + beta.*k1(i).*k30)./(k.^2.*(k1(i).^2 + k2.^2));
C2 = k2.*k0.^2./((k1(i).^2 + k2.^2).^(3/2)).*atan2((beta.*k1(i).*sqrt(k1(i).^2 + k2.^2)),(k0.^2 - k30.*k1(i).*beta));
xhsi2 = (k2./k1(i)).*C1 + C2;
xhsi2_q = xhsi2.^2;
phi_22 = E_k0./(4.*pi.*k0.^4).*(k0.^2 - k2.^2 - 2.*k2.*k30.*xhsi2 + (k1(i).^2 + k2.^2).*xhsi2_q);
end

function phi_33 = phi_33_new(k1,k2,k3,gamma,i)
k = sqrt(k1(i).^2+k2.^2+k3.^2);
beta = gamma./((k.^(2/3)).*sqrt(hypergeom([1/3,17/6],4/3,-k.^(-2)))); 
k30 = k3 + beta.*k1(i);
k0 = sqrt(k1(i).^2+k2.^2+k30.^2);
E_k0 = (1.453.*k0.^4./((1+k0.^2).^(17/6)));
phi_33 = (E_k0./(4*pi.*(k.^4))).*(k1(i).^2+k2.^2);
end

请注意，现在 beta 是在 phi 函数中计算的。此外，我正在使用 beta 的等效表达式。欲了解更多详情，请查看下面的图片

因此，在更新的模型中，我调用了 hypergeom，它的性能相当慢。这就是为什么我想通过能谱积分来计算第一个代码中的 beta 的主要原因。

顺便说一句，目前我不知道如何成功地执行此操作。

最好的问候， fpe