How to solve these two equations for 'tau' and 'b'? All the other symbols are constants. Please help??

I've tried using the sysms command but not sure I'm getting the input formatting correct

[Stephan]

Did you miss to attach something?

[Anik Faisal]

Yeah I did. I think the image is attached now

[Stephan]

you should also show us what you have done so far by using syms...

[Anik Faisal]

syms m b nu R rc tau g b0 u0
eqn1=(m*b/2)*((2-nu)/(1-nu))*R*log(R/rc)-tau*pi*R^2+(g*pi^2*R^2/b0)*sin(2*pi*(u0+b)/b0)==0;
eqn2=(m*b^2/4)*((2-nu)/(1-nu))*(1+log(R/rc))-2*pi*tau*R*b-g*pi*R*sin(pi*b/b0)*sin(pi*(2*u0+b)/b0)==0;
eqns=[eqn1 eqn2];
vars=[b tau];
[solv,solu]=solve(eqns,vars)

Answer 1

[Stephan]

Hi,

i used the isolate function instead of solve and found:

sol_tau =

tau == -(2778046668940015*R*g*sin((2*pi*(b + u0))/b0) + 281474976710656*b*b0*m*log(R/rc) - 2778046668940015*R*g*nu*sin((2*pi*(b + u0))/b0) - 140737488355328*b*b0*m*nu*log(R/rc))/(281474976710656*R*b0*pi*(nu - 1))

sol_b =

b == 0

by using these lines of code:

syms m b nu R rc tau g b0 u0
eqn1=(m*b/2)*((2-nu)/(1-nu))*R*log(R/rc)-tau*pi*R^2+(g*pi^2*R^2/b0)*sin(2*pi*(u0+b)/b0)==0;
eqn2=(m*b^2/4)*((2-nu)/(1-nu))*(1+log(R/rc))-2*pi*tau*R*b-g*pi*R*sin(pi*b/b0)*sin(pi*(2*u0+b)/b0)==0;
sol_tau = simplifyFraction(isolate(eqn1,tau))
eqn2_new = simplifyFraction(subs(eqn2,tau,rhs(sol_tau)));
sol_b = simplifyFraction(isolate(eqn2_new,b))

I did not check if you have typos in your equations...

Best regards

Stephan

[Anik Faisal]

Hi Stephan,

I tried to check for consistency using the following lines of codes. I'm getting a different 'tau' expression also 'b' seems to be not isolated.

tauSol=simplifyFraction(isolate(eqn2,tau))
eqn1_test=simplifyFraction(subs(eqn1,tau,rhs(tauSol)));
bSol=simplifyFraction(isolate(eqn1_test,b))

[Stephan]

Do you have numeric values for all the other constants, so that subs can be used?

[Anik Faisal]

I do have numeric values for m(mu)=45e9; nu=0.33 & b0=0.2555e-9 and let's say rc=1. The goal is to find expressions for 'tau=f (R)' and 'b=f (R)'. Since I do not have specific values for g(gamma0) & 'u0', those can remain in the expressions for 'tau' & 'b'?

[Stephan]

I would try to set the both expressions for tau equal and check if it is consistent.

Is_equal_tau = sol_tau == tauSol

is expected to give a logical 1.

[Anik Faisal]

Do you mean these lines of code?

Is_equal_tau = sol_tau == tauSol 
a=logical(Is_equal_tau)

I'm getting a=0

[Stephan]

Thats a reason to think about

[Stephan]

Did you get it working?

[Anik Faisal]

no I have not. I don't exactly know what to do next with this

[Anik Faisal]

could you suggest any other technique for solving this?

[Walter Roberson]

Assuming that your initial code was correct in https://www.mathworks.com/matlabcentral/answers/413950-how-to-solve-these-two-equations-for-tau-and-b-all-the-other-symbols-are-constants-please-help#comment_597512 then, No, there is no closed form solution for this, aside from

tau = g*Pi*sin(2*Pi*u0/b0)/b0
b = 0

The other solutions involve the roots of something that is similar to a degree 4 polynomial but involving terms that are also trig.

Well, correction: there are also a couple of analytic solutions if the other parameters happen to have special relationships, such as if

-8*R*g*Pi^2*(nu-1)*sin(2*Pi*u0/b0)/(b0^2*m*(nu-2)*(3*ln(R/rc)-1))

happens to be an integer

[Anik Faisal]

Hi Walter,

I got the same

tau = g*Pi*sin(2*Pi*u0/b0)/b0
b = 0

I am curious about the technique for other analytical solutions you're speaking of.