非线性积分方程组中含​变限积分fsolve​数值求解报错: Limits of integration must be double or single scalars

问题描述：

待求方程组{f1=0, f2=0, f3=0}中包含变量pl以及pr，pl和pr的表达式中包含上限是变量的积分（代码中加粗部分）。在通过matlabFunction处理为匿名函数后，int被integral替代，而integral不支持变量作为上下限，故报错：

Error using integral

Limits of integration must be double or single scalars.

可能有用的参考：

https://www.zhihu.com/question/46915508

这是我找到的一个相近的问题，但是我看不懂。。。

问题代码：

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% 参数定义部分省略

beta = atan(x1/y1);

xL = sqrt(x1^2+y1^2)*cos(pi-alph-beta);

DL = sqrt(x1^2+y1^2)*sin(pi-alph-beta);

xR = sqrt(x1^2+y1^2)*cos(beta-alph);

DR = sqrt(x1^2+y1^2)*sin(beta-alph);

LL = DL^2+(xL-x)^2;

RR = DR^2+(xR-x)^2;

pl = -12*mu_L*rho_S^4*h_m^3*Uratio^3*(Uratio*int(LL^2*x,0,x)+C*int(LL^1.5, 0,x))/(rho_L*T0^3*K_L^3)+P0;

pr = -12*mu_L*rho_S^4*h_m^3*Uratio^3*(Uratio*int(RR^2*x,0,x)+C*int(RR^1.5, 0,x))/(rho_L*T0^3*K_L^3)+P0;

% 对pl和pr进行定积分，得到目标方程组

f1 = int(pl*x,x,-L,0)+int(pr*x,x,0,L)+M/d;

f2 = int(pl,x,-L,0)*sin(alph+tilt)+int(pr,x,0,L)*sin(alph-tilt)-G/d-2*L*P_e*sin(alph)*cos(tilt);

f3 = int(pl,x,-L,0)*cos(alph+tilt)-int(pr,x,0,L)*cos(alph-tilt)+2*L*P_e*sin(alph)*sin(tilt);

% 进行求解

% 初始值猜测

initial_guess = [0.5, 0.1, 2];

% 定义求解函数

equations = matlabFunction(f1,f2,f3, 'Vars', {x1, y1, Uratio});

% 使用fsolve求解方程组

result = fsolve(@(vars) equations(vars(1), vars(2), vars(3)), initial_guess);

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Answer 1

[Ayush]

Hi @Yuting,

此邮件使用英语，旨在为您提供最快速的答复。如果您希望得到中文回复，请告诉我

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As mentioned in the documentation the `integral` function does support symbolic expressions and numbers as limits of integration. This means you can use symbolic variables as limits without encountering the error mentioned earlier.

% Define the symbolic variables and expressions
syms x pl pr
pl_expr = -12*mu_L*rho_S^4*h_m^3*Uratio^3*(Uratio*integral(LL^2*x,0,x)+C*integral(LL^1.5, 0,x))/( rho_L*T0^3*K_L^3)+P0;
pr_expr = -12*mu_L*rho_S^4*h_m^3*Uratio^3*(Uratio*integral(RR^2*x,0,x)+C*integral(RR^1.5, 0,x))/( rho_L*T0^3*K_L^3)+P0;

Hence, you can use symbolic expressions as limits of integration in the `integral` function without encountering any errors.