How to find unknown variable in the below equation?

I wish to solve the equation as shown below:

So, I made a new variable "A1" as coded below with i/p parameters.

r = 3.88;
Vau = 43;
theta_u = 71.8;
gamma = 5/3;
Vsu = 47.45;
syms Unu
A1 = double(solve(((Unu^2 - (r * Vau^2 *(cosd(theta_u))^2))^2) * (Unu^2 - ((2 * r * Vsu^2)/(r+1-gamma*(r-1)))) - ((r * (sind(theta_u)^2) * Unu^2 * Vau^2) *((2 * r - gamma * (r-1))*(Unu^2)/(r+1-gamma(r-1)) - r * Vau^2 * (cosd(theta_u))^2))))

A1 shows error like:

Subscript indices must either be real positive integers or logicals.

Error in IPS_P2 (line 333)
A1 = double(solve(((Unu^2 - (r * Vau^2 *(cosd(theta_u))^2))^2) * (Unu^2 - ((2 * r * Vsu^2)/(r+1-gamma*(r-1)))) - ((r * (sind(theta_u)^2) * Unu^2 * Vau^2) *((2 * r - gamma * (r-1))*(Unu^2)/(r+1-gamma(r-1)) - r * Vau^2 * (cosd(theta_u))^2))))

Can anyone please help me, if there is any error in coding equation?

Answer 1

[Alan Stevens]

The equation is a cubic in Unu^2, so the following uses roots to find solutions:

r = 3.88;
Vau = 43;
theta_u = 71.8;
gamma = 5/3;
Vsu = 47.45;

A = r*Vau^2*cos(theta_u)^2;
B = 2*r*Vsu^2/(r+1-gamma*(r-1));
C = r*Vau^2*sin(theta_u)^2;
D = (2*r - gamma*(r-1))/(r+1 - gamma*(r - 1));

% Let x = Unu^2
% (x - A)^2*(x - B) - C*x*(D*x - A) = 0
% (x/A - 1)^2*(x/A - B/A) - C/A*(x/A)*(D*x/A - 1) = 0
% Let y = x/A
% (y^2 - 2y + 1)*(y - B/A) - D*C/A*y^2 + C/A*y = 0
% Expand and collect like terms to get:
% y^3 -(2 + B/A + D*C/A)y^2 + (1 + 2*B/A + C/A)y - B/A = 0 
  
p = [1; -(2 + B/A + D*C/A); (1 + 2*B/A + C/A); -B/A];

y = roots(p);
x = A*y;       % reconstruct x
Unu = sqrt(x); % reconstruct Unu
disp(y)                
disp(x)
disp(Unu)

% Check (f should be zero)
f = polyval(p,y);
disp(f)

[John D'Errico]

As is often the case, Alan has shown that just because you don't know the value of a variable, this does not mean it must be made symbolic. All the symbolic parameter does is make the solution happen considerably slower.

A symbolic solution here would be useful only if you truly wanted to see an algebraic form for the solution. Since this is a cubic polynomial at heart, that is doable. But roots can handle a cubic far faster and more efficiently.

[Megha]

@ Alan Stevens : thank you so much spending your valuable time on this. However, I see that the answer of Unu is 678 (in my text refrences). I dont know how to find this answer from the above equation using above input parameters. I am unsure if solving this equation would lead me to this answer. Could you please help?

[Alan Stevens]

Ah! The problem seems to be that i used the constants exactly as you specified them. This means I used theta_u = 71.8. However, If you change theta_u to 71.8*pi/180 (i.e. turn degrees into radians), Matlab produces Unu = 677.25.

(Alternatively, you could keep theta_u as 71.8 and use cosd and sind instead of cos and sin).

[Megha]

@ Alan Stevens, Great!! That was such a great great help. That woked. So happy.

Could you please tell what difference does that make if i use "Solve" and "Roots"?

[Alan Stevens]

"solve" produces a symbolic solution, but not all equations have symbolic solutions. "roots" provides numerical solutions only. As John said (above) "roots" is faster and more efficient.

[Megha]

Great. Thank you so much