# Closed form solution of tricky nonlinear equation system

I am developing a solution for a Chemical Engineering course, and I came across an interesting (tricky) equation system, for which I am curious if it is possible to solve it explicitly (closed form solutions). I am not interested in the solution "per-se" as I could just use MATLAB or any other program to get such solutions.

I have the following system of 2 nonlinear equations in the 2 unknowns $$\xi$$ and $$\eta$$: $$\begin{cases} A(1-\xi)(B+\xi)=(C+\eta)(D+\eta)\\[3pt] E(1-\xi)(B+\xi)=(F+\xi-\eta)(G+\xi-\eta) \end{cases}$$ where $$A,B,C,D,E,F,G$$ are all strictly positive reals. The conditions of existence of $$\xi$$ and $$\eta$$ are: $$0<\eta<\xi\le1$$ Is it possible to rearrange/transform, or more generally exploit the form of the two equations to obtain $$\xi$$ and $$\eta$$ in a closed form? I have tried to do so, but to no avail.

The only thing I noticed is that the first equation is in the form $$f(\xi)=g(\eta)$$, therefore the only condition is that both functions are equal to a constant. However, I have not been able to move past that point.

• Eliminating one variable leaves a quartic in the other. There's an explicit "quartic formula", so your system does have a closed-form solution ... but it's super-ugly.
– Blue
Mar 28, 2020 at 19:33
• Alternatively, you can use the algorithm described here for computing the intersection of two conics, but again, there’s not likely to be a “nice” general closed-form solution.
– amd
Mar 28, 2020 at 20:06
• @Blue Checked and confirmed, unfortunately. Mar 29, 2020 at 20:44
• @amd Unfortunately I confirm that the closed-form solution will be ugly. I will include your comments in my answer. Mar 29, 2020 at 20:45

A proposed solution method exploits the property of the first equation $$f(\xi)=g(\eta)$$, since two functions of two separate variables are equated, both must be equal to a constant unique value $$K$$.
The first equation (RHS) becomes: $$(C+\eta)(D+\eta)=K$$ which yields (since $$\eta>0$$) $$\eta(K)=\frac{-(C+D)+\sqrt{(C+D)^2-4(CD-K)}}{2} \$$ The first equation (LHS) becomes: $$(1-\xi)(B+\xi)=K/A$$ which yields (since $$\xi>0$$) $$\xi(K)=\frac{-(B-1)+\sqrt{(B-1)^2-4(K/A-B)}}{2} \$$ The second equation becomes: $$(F+[\xi-\eta])(G+[\xi-\eta])=EK/A$$ which yields (since $$\xi-\eta>0$$) $$[\xi-\eta](K)=\frac{-(F+G)+\sqrt{(F+G)^2-4(FG-EK/A)}}{2} \$$ Then, the unique constant $$K$$ must satisfy $$[\xi-\eta](K)=\xi(K)-\eta(K)\,,$$ thus by substituting the respective functional form of the above functions, the unique constant $$K$$ is obtained through an irrational equation.
In the special case when $$C=D=F=G=0$$, the solutions become: $$\eta(K)=\sqrt{K}$$ and $$[\xi-\eta](K)=\sqrt{\frac{EK}{A}} \$$ Thus, by combining the above solutions, $$\xi$$ is obtained. Since $$\xi(K)=\eta(K)\left(1+\sqrt{\frac{E}{A}}\right)=\sqrt{K}\left(1+\sqrt{\frac{E}{A}}\right) \,,$$ then by using the previous definition of $$\xi(K)$$, that is $$\xi(K)=\sqrt{K}\left(1+\sqrt{\frac{E}{A}}\right)=\frac{-(B-1)+\sqrt{(B-1)^2-4(K/A-B)}}{2} \,,$$ becomes an irrational equation with closed form solution, being: $$\sqrt{K}=\sqrt{A}\frac{-\alpha\beta+\sqrt{(\alpha\beta)^2+4B(\alpha^2+4)}}{\alpha^2+4}$$ where $$\alpha=2(\sqrt{A}+\sqrt{E})$$ and $$\beta=B-1$$. In this special case a quartic has been reduced to two decoupled quadratics.