Symbolically solving in mathcad, bound the solution output

[jroth]

Hello,

I'm wanting to plot some stream functions of which I have the functions for. I know what my stream function is in terms of u & v, but now I want to express the solution in terms of x & y, of which I have. Since I know that the stream function is a constant, any choice for u or v has a solution.

Currently psi = -2*C*sinh(u)*cos(v) = constant, and C is an arbitrary constant.

To plot out the points on a continuous line, I let v for instance = 0, phi/4...2*phi. I can then get the values for u, which are unique, and can then through a programming function, get the converted values of these two vectors for x(u,v) and y(u,v), therby giving me a line. I can add other lines by changing the value of psi = constant.

However, when I want to get the stream lines for say, letting u=0,0.5,3 , there are multiple solutions to the value of v. As I am symbolically solving the equations, I'm wanting to know if there is a way to 'bound' the symbolic solution to v between say 0 & phi, which would allow a vector output, of which I can then use to plot.

I have attached (the second version) of the spreadsheet. The first one crashed when I tried solve, v, 0, 2phi.

I would also like to know if I can make the stream functions solve, some variable output be equal to a vector iteself.

Thank you for any help in advance =).

[Werner_E · Accepted solution]

No idea from your description and sheet what you need.

Look if the attached helps.

You can try to constrain symbolic solve by adding an "assume ..."

[jroth]

Werner,

I have gone back and put the items in as functions. I am attaching a new sheet that will detail the problem abit more closely. It is when I solve for the velocity potential that mathcad provides two answers for.

I redefined the velocity potential so that the solving for u wouldn't fail, so I renamed it. The velocity potential 'should' be orthogonal to the stream function. I'm trying to plot it to confirm my results.

Any help would be much appreciated.

[Werner_E]

Its not that difficult to simply select the first of the two solutions for q(), but to be able to plot a function at least one single argument in the plotrange must yield a real result.

You are plotting acosh(B/sin(v)), so you must chose B and the v-range in a way that at least for one of the v-values B/sin(v) is greater than 1. This means the plot fails for B=0 and it also fails for B>0 and pi<v<2pi.