Solving a integral with a unknown value with known limits

Hello guys,

I am trying the solve the following equations in matlab to find Bg. All other variables are known.

Answer 1

[Walter Roberson]

The below is Maple notation, easily change to MATLAB notation with the Symbolic Toolbox

simplify([solve](LHS = 2*B*int(-B^2*((1 - xi)^2 - 2*(1 - rho__v/rho__1)*xi - 1), xi = 0 .. 1), B));

[                                           (1/3)  
[  (1/3) /                                2\       
[12      \LHS rho__1 (5 rho__1 - 3 rho__v) /       
[------------------------------------------------, 
[              10 rho__1 - 6 rho__v                

                                             (1/3)                   
    (1/3) /                                2\      /   (1/2)    \    
  12      \LHS rho__1 (5 rho__1 - 3 rho__v) /      \I 3      - 1/    
  ---------------------------------------------------------------, - 
                       20 rho__1 - 12 rho__v                         

                                             (1/3)               
    (1/3) /                                2\      /   (1/2)    \
  12      \LHS rho__1 (5 rho__1 - 3 rho__v) /      \I 3      + 1/
  ---------------------------------------------------------------
                       20 rho__1 - 12 rho__v                     

  ]
  ]
  ]
  ]
  ]

The first is certain to be real-valued if the coefficients are real-valued. The other two might be real-valued for particular combinations of values, if you use the definition that x^(1/3) is exp(log(x)/3) and x is negative, then the complex part can vanish.

[Sam Joseph]

Okay I will test this out to see if it will work.

Thanks for the quick reply.

[Sam Joseph]

Apologies I do not have access to symbolic toolbox. Is there a way to convert this to matlab code more easily?

[Walter Roberson]

You can use the results:

[ 12^(1/3)*(LHS*rho__1*(5*rho__1-3*rho__v)^2)^(1/3)/(10*rho__1-6*rho__v), ...
  12^(1/3)*(LHS*rho__1*(5*rho__1-3*rho__v)^2)^(1/3)*(1i*3^(1/2)-1)/(20*rho__1-12*rho__v), ...
 -12^(1/3)*(LHS*rho__1*(5*rho__1-3*rho__v)^2)^(1/3)*(1i*3^(1/2)+1)/(20*rho__1-12*rho__v) ]

where LHS is the expression that appears to the left of your =

[Sam Joseph]

What is "i" refering to?

[John D'Errico]

sqrt(-1)

Answer 2

[Steven Lord]

Subtract the left side of your equation from both sides to give an equation of the form 0 = someFunctionOf(B_g). Then use fzero to find a root of someFunctionOf.

[Sam Joseph]

Will the integral and exponential term work with someFunctionOf?

[Steven Lord]

Inside the someFunctionOf function you write you can certainly call exp and integral.