This is my first question here, so hopefully I will be able to explain my problem in a coherent way.

Phoebe Xiong

Phoebe Xiong

Answered question

2022-02-23

This is my first question here, so hopefully I will be able to explain my problem in a coherent way. My ultimate question is: do you see any way I can simplify the following system so I can have a more intuitive solution to it? Let me explain in details what I mean:
I want to characterize x,y,z that solve the following system of non-linear equations:
Δx=θΔzΔyΔyθΔz     (EQ1)
Δx+Δy+Δz=Y     (EQ2)
θ(Δz)2+(Δy)2=U     (EQ3)
where
Δx=xx
Δy=yy
Δz=zz
and 0<θ<1,Y,U,x,y and z are parameters.
The way I've approached this was to parameterize (EQ2):
Δy+Δz=α
such that α=YΔx
For θ=0.8,x=y=z=0.5 and Y=0.5 and U=0.2 this parametric system would look like the following (I apologize I don't have reputation to embed the photo here):
The parametric system
I can characterize Δy and Δz as a function of α:
Δy=θα±U(1+θ)θα21+θ
Δz=αU(1+θ)θα21+θ
My problem arrives in the characterization of α using (EQ1). Although we can show such an α exists, it is far from an intuitive closed-form characterization, as α solves:
±(Yα)(1θ)U(1+θ)θα2θα±U(1+θ)θα2=θ(αU(1+θ)θα2)
Am I using the correct approach to this problem? Or do you see anyway I could simplify this? Thank you!

Answer & Explanation

Kathryn Duggan

Kathryn Duggan

Beginner2022-02-24Added 7 answers

I will reason in terms of the Δ rather than the non-centered x,y,z.
The first equation, which can be written
ΔxΔy=θΔz(Δx+Δy)
is that of a cone with apex at the origin. The second is a plane and the third and elliptic cylindre of vertical axis. The latter two define an ellipse, which you can represent by the parametric equations
Δy=U1u21+u2
Δz=Uθ2u1+u2
Δx=YΔyΔz.
(The rational expressions can be obtained from the trigonometric representation.)
Plugging these expressions in the first equation, you will get a quartic equation in u, which can have zero, two or four solutions (intersection points of the ellipse and the cone).
It is unlikely that there will be a simple expression for these solutions, even though there are only three independent parameters and the quartic equation is not fully general.
Elena Ray

Elena Ray

Beginner2022-02-25Added 4 answers

a=θbcbθc
a+b+c=12
θc2+b2=15
a=x12
b=y12
c=z12

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