我想知道是否有任何方法繞過呼叫與嵌套trapz
循環。我會用一些更詳細的討論我的問題:目前,我執行二重積分這樣計算:嵌套trapz雙重集成
clc, clear all, close all
load E_integral.mat
c = 1.476;
gamma = 3.0;
beta_int = (c*gamma)./(k_int.*sqrt(E_integral));
figure, loglog(k_int,beta_int,'r','LineWidth',2.0), grid on;
k1 = (.01:.1:100);
k2 = .01:.1:100;
k3 = -100:.1:100;
int_k3 = zeros(size(k2));
int_k3k2 = zeros(size(k1));
tic
for ii = 1:numel(k1)
phi11 = @(k2,k3) PHI11(k1(ii),k2,k3,k_int,beta_int);
int_k3(ii) = 2*quad2d(phi11,-100,100,-100,100);
end
toc
其中PHI11
被定義爲
function phi11 = PHI11(k1,k2,k3,k_int,beta_int)
k = sqrt(k1.^2 + k2.^2 + k3.^2);
ksq = k.^2;
k1sq = k1.^2;
fourpi = 4.*pi;
beta = exp(interp1(log(k_int),log(beta_int),log(k),'linear'));
k30 = k3 + beta.*k1;
k0 = sqrt(k1.^2 + k2.^2 + k30.^2);
k0sq = k0.^2;
k04sq = k0.^4;
Ek0 = (1.453.*k04sq)./((1 + k0sq).^(17/6));
C1 = (beta.*k1sq.*(k0sq - 2.*(k30.^2) + beta.*k1.*k30))./(ksq.*(k1.^2 + k2.^2));
C2 = ((k2.*k0sq)./((k1.^2 + k2.^2).^(3/2))).*atan2((beta.*k1.*sqrt(k1.^2 + k2.^2)),(k0sq - k30.*k1.*beta));
xhsi1 = C1 - (k2./k1).*C2;
xhsi1_sq = xhsi1.^2;
phi11 = (Ek0./(fourpi.*k04sq)).*(k0sq - k1sq - 2.*k1.*k30.*xhsi1 + (k1.^2 + k2.^2).*xhsi1_sq);
end
和E_integral.mat
可以通過這種方式獲得:
clc,clear all,close all
k_int = .001:.01:1000;
Ek = (1.453.*k_int.^4)./((1 + k_int.^2).^(17/6));
E = @(k_int) (1.453.*k_int.^4)./((1 + k_int.^2).^(17/6));
E_integral = zeros(size(k_int));
for ii = 1:numel(k_int)
E_integral(ii) = integral(E,k_int(ii),Inf);
end
save('E_integral','k_int','E_integral')
現在的問題是:有可能忽略quad2d
和最愛的handle function
或更實際的方法,通過使用嵌套的trapz
函數?
到目前爲止,我已經試過下面的代碼段,這還沒有產生預期的結果:
int_k33 = zeros(size(k2));
S_11 = zeros(size(k1));
tic
for ii = 1:1
for jj = 1:numel(k2)
int_k33(jj) = trapz(k3,PHI11(k1(ii),k2(jj),k3,k_int,beta_int));
end
S_11(ii) = 4*trapz(k2,int_k33);
end
toc
任何有趣的想法? – fpe 2013-05-07 16:34:52
請解釋一下「實用方法」是什麼意思?我只給出了這種前瞻性的外觀,我當然可以看到改進代碼的地方。這就是說,一般來說,如果你可以寫'f(x),那麼你只能將一個雙積分∫∫f(x,y)dA'分解成兩個獨立的積分∫g(x)dx *∫h(y)dy' (x,y)'作爲兩個函數g(x)* h(y)'的乘積。你能發表實際的方程式和他們的上下文嗎? (因子1 /(4pi)使我想到與磁場有關的事情。) – 2013-05-13 21:45:47
我*假設*你想這樣做,因爲你給了實驗數據?否則,選擇這樣一種簡單的方法而不是更精確的方法是沒有意義的......你能詳細說明*爲什麼*你想這樣做? – 2013-05-16 11:35:52