vpasolve did not work with some special values of variables

Does anyone please help me to clarify what the problem might be. I use vpasolve to solve Manning's equation. This is the code:

%%Variable

c1=1.486; Q=50; b=20; m=2; n=0.02; S=0.01; g=32.2;

syms yo

Sol_yo = vpasolve((c1/n)*(b+m*yo)*yo*(S^0.5) *(((b+m*yo)*yo)^(2/3)) - ((b+2*yo*sqrt(1+m*m))^(2/3)*Q)==0,yo);

disp('Normal depth: ');     vpa(Sol_yo)

Problem is: with m = 2 or some other values around 2. The function gave me a complex number, which is not correct. With other value of m, the function work well.

If I use while loop:

yo=3;

Sol_yo = (c1/n)*(b+m*yo)*yo*(S^0.5) *(((b+m*yo)*yo)^(2/3)) - ((b+2*yo*sqrt(1+m*m))^(2/3)*Q);

while Sol_yo > 0.000001

      yo=yo-0.001

      Sol_yo = (c1/n)*(b+m*yo)*yo*(S^0.5) *(((b+m*yo)*yo)^(2/3)) - ((b+2*yo*sqrt(1+m*m))^(2/3)*Q);

end

then everything is just fine.

My question is: how can I get vpasolve worked with any values of variable m? and why it did not work in the first place?

Thank you very much for your time and help!

Best,

Duc.

Answer 1

[Walter Roberson]

vpasolve() is defined to find any one solution that satisfies all the criteria it has been told about. vpasolve() will not be restricted to real-valued solutions unless you have added constraints to the variables in order to disallow such solutions.

[Walter Roberson]

Note: the formula has multiple positive solutions for some m, such as m = -1

[Duc Tran]

Thank you for answering, Walter.

Before ending up using vpasolve I had tried "slove" with (ReturnConditions, true) and "subs" to limit down the solutions but sometime it worked, sometime not. I do not know why.

Would you mind suggesting how I can add constraints to get the correct answer? Given that all the variables and solutions are positive and real number.

Thank you very much,

Duc.

[Walter Roberson]

You can specify a search range of [realmin, inf] for vpasolve()

Checking the formula further, for positive m, y0 will be in the range 0 to .5310799190444193; larger y0 require negative m.

Also, for negative m, there will be two positive roots.

[Duc Tran]

Thanks, I will try it.

Cheers,