I will appreciate any suggestion on how I could have a solution to this.

 SX=1000*[1 2 3];
 SY=2000*[1.5 2 3];
 SXY = 1258[1 2 3];
 a = [0.3 0.6 0.9];
syms rb
 for j=1:1:3
     if pwmid(j)<=pwc(j)
         SRR(j)=0.5*(SX(j)+SY(j)).*(1-(a(j).^2)/rb^2)+0.5*(SX(j)-SY(j)).*(1+(3*a(j).^4/rb^4)-(4*a(j).^2/rb^2))*cos(2*thbkso(j))...
             +SXY(j).*(1+(3*a(j).^4/rb^4)-(4*a(j).^2/rb^2))*sin(2*thbkso(j))+(a(j).^2/rb^2).*pwmid(j);
         STT(j)=0.5*(SX(j)+SY(j)).*(1+(a(j).^2)/rb^2)-0.5*(SX(j)-SY(j)).*(1+(3*a(j).^4/rb^4))*cos(2*thbkso(j))...
             -SXY(j).*(1+(3*a(j).^4/rb^4))*sin(2*thbkso(j))-(a(j).^2/rb^2).*pwmid(j);
         SRT(j)=(0.5*(SX(j)-SY(j)).*sin(2*thbkso(j))+SXY(j).*cos(2*thbkso(j))).*(1-(3*a(j).^4/rb^4)+(2*a(j).^2/rb^2));
         SIGMA1A(j)=0.5*(STT(j)+SRR(j))+0.5*((STT(j)-SRR(j)).^2+4*SRT(j).^2).^0.5;
         SIGMA3A(j)=0.5*(STT(j)+SRR(j))-0.5*((STT(j)-SRR(j)).^2+4*SRT(j).^2).^0.5;
         C0FUN(j)=SIGMA1A(j)-SIGMA3A(j);
         rbsoln{j}=double(vpasolve(C0FUN(j)==C0(j),rb));
         cell(rbsoln);
         rw(j)=rbsoln{j}(1);

         rbkt_art(j) = rbsoln{j}(1)-a(j);
        else
          rw(j)=a(j); 
          rbkt_art(j)=rbkt_int(j);
     end 
 end

[Isaac]

pwmid=[1000 2000 3000]; pwc = [500 1000 3500];

[Isaac]

thbkso is a matrix with known values...any values can be used; but my concern is the use of solve or vpasolve not giving me favorable results

[Isaac]

the unknown is rb

[Walter Roberson]

I do not understand what your question is? vpasolve() is going to give you a solution if it can find one. If it is not the "right" solution then you need some way of characterizing the "right" solution. For example is there a range of values that the "right" solution will always be in?

[Walter Roberson]

Should

SXY = 1258[1 2 3];

read

SXY = 1258*[1 2 3];

[Walter Roberson]

C0(j) has not been assigned a meaning.

[Isaac]

Sorry

SXY = 125*[1 2 3]; C0 = 100*[1 2 3];

Thanks

[Isaac]

thbkso = pi/2*[1 1 1];

Answer 1

[Isaac]

Sorry

SXY = 125*[1 2 3]; C0 = 100*[1 2 3];

Thanks

[Isaac]

thbkso = pi/2*[1 1 1];

Answer 2

[Walter Roberson]

All solutions to those equations are strictly imaginary for the parameters you give.

For example, for j = 1, the solutions are

(3/10)*sqrt(5)*sqrt(roots([+8105,-9500,+4790,-1164,+117]))

and the negatives of those.

[Walter Roberson]

Please explain what you mean when you said you were concerned about solve or vpasolve "not giving favorable results" ?

If you want all of the results, then you may have to use solve() instead of vpasolve(), and you might have to double() the result of solve() to get numeric values. I do not have the Symbolic Toolbox so I cannot check exactly what would be returned.

[Walter Roberson]

I could have made a mistake along the way, but if I got it right then:

for j = 1 : 3
  A = 32 * SXY(j) * sin(thbkso(j)) * cos(thbkso(j))^3 * SX(j) - 32 * SXY(j) * sin(thbkso(j)) * cos(thbkso(j))^3 * SY(j) - 16 * SXY(j) * sin(thbkso(j)) * cos(thbkso(j)) * SX(j) + 16 * SXY(j) * sin(thbkso(j)) * cos(thbkso(j)) * SY(j) - C0(j)^2 + SX(j)^2 - 2 * SX(j) * SY(j) + 4 * SXY(j)^2 + SY(j)^2;

    B =  - 32 * cos(thbkso(j))^4 * SX(j)^2 * a(j)^2 + 64 * cos(thbkso(j))^4 * SX(j) * SY(j) * a(j)^2 + 128 * cos(thbkso(j))^4 * SXY(j)^2 * a(j)^2 - 32 * cos(thbkso(j))^4 * SY(j)^2 * a(j)^2 - 8 * sin(thbkso(j)) * cos(thbkso(j)) * SX(j) * SXY(j) * a(j)^2 - 8 * sin(thbkso(j)) * cos(thbkso(j)) * SXY(j) * SY(j) * a(j)^2 + 16 * sin(thbkso(j)) * cos(thbkso(j)) * SXY(j) * a(j)^2 * pwmid(j) + 28 * cos(thbkso(j))^2 * SX(j)^2 * a(j)^2 - 64 * cos(thbkso(j))^2 * SX(j) * SY(j) * a(j)^2 + 8 * cos(thbkso(j))^2 * SX(j) * a(j)^2 * pwmid(j) - 128 * cos(thbkso(j))^2 * SXY(j)^2 * a(j)^2 + 36 * cos(thbkso(j))^2 * SY(j)^2 * a(j)^2 - 8 * cos(thbkso(j))^2 * SY(j) * a(j)^2 * pwmid(j) - 2 * SX(j)^2 * a(j)^2 + 8 * SX(j) * SY(j) * a(j)^2 - 4 * SX(j) * a(j)^2 * pwmid(j) + 16 * SXY(j)^2 * a(j)^2 - 6 * SY(j)^2 * a(j)^2 + 4 * SY(j) * a(j)^2 * pwmid(j);

    C = 128 * sin(thbkso(j)) * cos(thbkso(j))^3 * SX(j) * SXY(j) * a(j)^4 - 128 * sin(thbkso(j)) * cos(thbkso(j))^3 * SXY(j) * SY(j) * a(j)^4 + 48 * cos(thbkso(j))^4 * SX(j)^2 * a(j)^4 - 96 * cos(thbkso(j))^4 * SX(j) * SY(j) * a(j)^4 - 192 * cos(thbkso(j))^4 * SXY(j)^2 * a(j)^4 + 48 * cos(thbkso(j))^4 * SY(j)^2 * a(j)^4 - 48 * sin(thbkso(j)) * cos(thbkso(j)) * SX(j) * SXY(j) * a(j)^4 + 80 * sin(thbkso(j)) * cos(thbkso(j)) * SXY(j) * SY(j) * a(j)^4 - 32 * sin(thbkso(j)) * cos(thbkso(j)) * SXY(j) * a(j)^4 * pwmid(j) - 40 * cos(thbkso(j))^2 * SX(j)^2 * a(j)^4 + 96 * cos(thbkso(j))^2 * SX(j) * SY(j) * a(j)^4 - 16 * cos(thbkso(j))^2 * SX(j) * a(j)^4 * pwmid(j) + 192 * cos(thbkso(j))^2 * SXY(j)^2 * a(j)^4 - 56 * cos(thbkso(j))^2 * SY(j)^2 * a(j)^4 + 16 * cos(thbkso(j))^2 * SY(j) * a(j)^4 * pwmid(j) + 7 * SX(j)^2 * a(j)^4 - 18 * SX(j) * SY(j) * a(j)^4 + 4 * SX(j) * a(j)^4 * pwmid(j) - 8 * SXY(j)^2 * a(j)^4 + 15 * SY(j)^2 * a(j)^4 - 12 * SY(j) * a(j)^4 * pwmid(j) + 4 * a(j)^4 * pwmid(j)^2;

    E = 288 * SXY(j) * a(j)^8 * sin(thbkso(j)) * cos(thbkso(j))^3 * SX(j) - 288 * SXY(j) * a(j)^8 * sin(thbkso(j)) * cos(thbkso(j))^3 * SY(j) - 144 * SXY(j) * a(j)^8 * sin(thbkso(j)) * cos(thbkso(j)) * SX(j) + 144 * SXY(j) * a(j)^8 * sin(thbkso(j)) * cos(thbkso(j)) * SY(j) + 9 * SX(j)^2 * a(j)^8 - 18 * SY(j) * a(j)^8 * SX(j) + 36 * SXY(j)^2 * a(j)^8 + 9 * SY(j)^2 * a(j)^8;

    sols_plus = sqrt( roots([A, B, C, 0, E]) );
    sols{j} = [sols_plus; -sols_plus];
  end

I am not certain of these coefficients; I am concerned that the previous solution did not have a 0 in the x^1 position but this does.

Answer 3

[Isaac]

Thanks Walter...yes, the solutions are all imaginary with the current inputs