After this I want to integrate P1 from 0 to x (let's say this function to be P2) and then again integrating the function P2 from 0 to 1. I don't know how to proceed.

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AVINASH SAHU
AVINASH SAHU 2022-6-6
评论： AVINASH SAHU 2022-6-6
% For plane slider: H = Ho + a(1-x)
Ho = 1;
alpha = 0.1;
eps = 0.1;
a = 1.0;
lbar = 0.1;
sigma = 0.05;
H = @(x) Ho + a*(1 - x);
G1 = @(x) H(x).^3 + 3 .* H(x).^2 .* alpha + 3*sigma^2*alpha + alpha^3 - 12*lbar^3*(H(x) + alpha);
G2 = @(x) 24 * lbar^3 .* tanh(H(x)./(2*lbar));
G3 = @(x) (12*lbar^2*alpha - eps - alpha^3 - 3*sigma^2*alpha) .* (1 - (tanh(H(x)./(2*lbar))).^2);
G = @(x) G1(x) + G2(x) + G3(x);
Hm1 = @(x) H(x).* (1 ./ G(x));
Hm2 = @(x) (1 ./ G(x));
IntHm1 = integral(Hm1,0,1);
IntHm2 = integral(Hm2,0,1);
Hm = IntHm1 / IntHm2
P1 = @(x) 6 .* (1 ./ G(x)) .* (H(x) - Hm);

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采纳的回答

Torsten
Torsten 2022-6-6
Ran in:
% For plane slider: H = Ho + a(1-x)
Ho = 1;
alpha = 0.1;
eps = 0.1;
a = 1.0;
lbar = 0.1;
sigma = 0.05;
H = @(x) Ho + a*(1 - x);
G1 = @(x) H(x).^3 + 3 .* H(x).^2 .* alpha + 3*sigma^2*alpha + alpha^3 - 12*lbar^3*(H(x) + alpha);
G2 = @(x) 24 * lbar^3 .* tanh(H(x)./(2*lbar));
G3 = @(x) (12*lbar^2*alpha - eps - alpha^3 - 3*sigma^2*alpha) .* (1 - (tanh(H(x)./(2*lbar))).^2);
G = @(x) G1(x) + G2(x) + G3(x);
Hm1 = @(x) H(x).* (1 ./ G(x));
Hm2 = @(x) (1 ./ G(x));
IntHm1 = integral(Hm1,0,1);
IntHm2 = integral(Hm2,0,1);
Hm = IntHm1 / IntHm2;
P1 = @(x) 6 .* (1 ./ G(x)) .* (H(x) - Hm);
fun = @(x)integral(@(u)P1(u),0,x);
result = integral(fun,0,1,'ArrayValued',true)
result = 0.1310

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