How to solve non-linear equations with integrations?

Question

Yuting 2023-7-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2002262-how-to-solve-non-linear-equations-with-integrations

评论： Star Strider 2023-7-31

I want to solve a group of equations like this:

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);

in which the pl and pr are:

pl = -12*mu_L*rho_S*(int((Uxl*x/(deltal^3)),x,0,x)+Cl*int((1/(deltal^3)),x,0,x))/rho_L+P0;

pr = -12*mu_L*rho_S*(int((Uxr*x/(deltar^3)),x,0,x)+Cr*int((1/(deltar^3)),x,0,x))/rho_L+P0;

The solving prt of code like:

initial_guess = [0.5, 0.1, 2];

solve_equations = @(vars) double(subs([f1, f2, f3], {x1,y1,Uratio}, vars));

result = fsolve(solve_equations, initial_guess);

x1 = result(1);

y1 = result(2);

Uratio = result(3);

But the calculating process time was extremly long and couldn't get the solution, I thought the reason is: there are lots of integrations in the equations and the fsolve function cannot solve the int() part. Is this correct? How to solve it?

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Torsten 2023-7-31

编辑：Torsten 2023-7-31

在 MATLAB Online 中打开

Then follow @Star Strider to work with "matlabFunction", but compute pl and pr first since the result will also depend on x. This will influence int(pl*x,x,-L,0) etc. in the computation of f1, f2 and f3.

syms x L M d alph tilt G P_e delta rho_L Uxl Uxr rho_S mu_L P0 Cr Cl deltal deltar

pl = -12*mu_L*rho_S*(int((Uxl*x/(deltal^3)),x,0,x)+Cl*int((1/(deltal^3)),x,0,x))/rho_L+P0;

pr = -12*mu_L*rho_S*(int((Uxr*x/(deltar^3)),x,0,x)+Cr*int((1/(deltar^3)),x,0,x))/rho_L+P0;

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

f1 =

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)

f2 =

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)

f3 =

equations = matlabFunction(f1,f2,f3)

equations = function_handle with value:

@(Cl,Cr,G,L,M,P0,P_e,Uxl,Uxr,alph,d,deltal,deltar,mu_L,rho_L,rho_S,tilt)deal(M./d+(L.^2.*1.0./deltal.^3.*(L.^2.*Uxl.*mu_L.*rho_S.*3.0-Cl.*L.*mu_L.*rho_S.*8.0))./(rho_L.*2.0)-(L.^2.*1.0./deltar.^3.*(L.^2.*Uxr.*mu_L.*rho_S.*3.0+Cr.*L.*mu_L.*rho_S.*8.0))./(rho_L.*2.0),sin(alph+tilt).*(L.*P0-(L.*1.0./deltal.^3.*(L.^2.*Uxl.*mu_L.*rho_S.*2.0-Cl.*L.*mu_L.*rho_S.*6.0))./rho_L)-G./d+sin(alph-tilt).*(L.*P0-(L.*1.0./deltar.^3.*(L.^2.*Uxr.*mu_L.*rho_S.*2.0+Cr.*L.*mu_L.*rho_S.*6.0))./rho_L)-L.*P_e.*sin(alph).*cos(tilt).*2.0,cos(alph+tilt).*(L.*P0-(L.*1.0./deltal.^3.*(L.^2.*Uxl.*mu_L.*rho_S.*2.0-Cl.*L.*mu_L.*rho_S.*6.0))./rho_L)-cos(alph-tilt).*(L.*P0-(L.*1.0./deltar.^3.*(L.^2.*Uxr.*mu_L.*rho_S.*2.0+Cr.*L.*mu_L.*rho_S.*6.0))./rho_L)+L.*P_e.*sin(alph).*sin(tilt).*2.0)

Yuting 2023-7-31

@Torsten Thank you so much!!! Very important!!

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Answer 1

Star Strider 2023-7-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2002262-how-to-solve-non-linear-equations-with-integrations#answer_1280592

在 MATLAB Online 中打开

It is not possible to solve this numerically without substituting numerical values for the known variables.

That aside, use the matlabFunction function to create an anonymous function that fsolve can work with —

syms pl x pr L M d alph tilt G P_e delta rho_L Uxl Uxr rho_S mu_L P0 Cr Cl deltal deltar

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

f1 =

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)

f2 =

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)

f3 =

pls = -12*mu_L*rho_S*(int((Uxl*x/(deltal^3)),x,0,x)+Cl*int((1/(deltal^3)),x,0,x))/rho_L+P0;

prs = -12*mu_L*rho_S*(int((Uxr*x/(deltar^3)),x,0,x)+Cr*int((1/(deltar^3)),x,0,x))/rho_L+P0;

f1 = simplify(subs(f1, {pl,pr},{pls,prs}), 500)

f1 =

f2 = simplify(subs(f1, {pl,pr},{pls,prs}), 500)

f2 =

f3 = simplify(subs(f1, {pl,pr},{pls,prs}), 500)

f3 =

equations = matlabFunction(f1,f2,f3)

equations = function_handle with value:

@(Cl,Cr,L,M,Uxl,Uxr,d,deltal,deltar,mu_L,rho_L,rho_S,x)deal(M./d+(L.^2.*1.0./deltal.^3.*1.0./deltar.^3.*mu_L.*rho_S.*x.*(Cl.*deltar.^3.*2.0-Cr.*deltal.^3.*2.0+Uxl.*deltar.^3.*x-Uxr.*deltal.^3.*x).*3.0)./rho_L,M./d+(L.^2.*1.0./deltal.^3.*1.0./deltar.^3.*mu_L.*rho_S.*x.*(Cl.*deltar.^3.*2.0-Cr.*deltal.^3.*2.0+Uxl.*deltar.^3.*x-Uxr.*deltal.^3.*x).*3.0)./rho_L,M./d+(L.^2.*1.0./deltal.^3.*1.0./deltar.^3.*mu_L.*rho_S.*x.*(Cl.*deltar.^3.*2.0-Cr.*deltal.^3.*2.0+Uxl.*deltar.^3.*x-Uxr.*deltal.^3.*x).*3.0)./rho_L)

I leave the rest to you.

.

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Yuting 2023-7-31

Hello~ @Star Strider,

I have a question about the code f1 = simplify(subs(f1, {pl,pr},{pls,prs}), 500),

what does the "500" mean?

Star Strider 2023-7-31

My pleasure!

The ‘500’ tells simplify too keep simplifying until it cannot simplify the expressins further, or reaches the limit of 500 iterations. Usually 500 is enough, however if not specifically stated, simplify stops after one iteration. For complicated expressions, that is rarely enough.

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