# Properties of nonlinear equation system

Considering the very simple electrical dc power flow problem with two nodes, with voltage V1 and V2 respectively, connected via a resistance R, we end up with a system of two nonlinear equations:

$P_1=\frac{V_1(V_1-V_2)}{R}$

$P_2=\frac{V_2(V_2-V_1)}{R}$

We have two equations and four unknowns. I would expect then that, defining two variables (let's say $P_1$ and $P_2$), the system would be solvable for the other two.

However, when using $P_1 = P_2 = 0$, I get infinite solutions where $V_1 = V_2$.

More specifically, when solving those kind of systems, how do I know the conditions where there will be infinite solutions instead of one or two?

• Also, if $P_1 =0$ then $V_1(V_1-V_2)=0$ so either $V_1=0$ or $V_1=V_2$.
– Surb
Commented Mar 9, 2017 at 23:00
• Yes, thanks for pointing out! Commented Mar 9, 2017 at 23:01
• If $V_1 = 0$, then $P_2=0$ implies $(V_2)^2=0$ and so $V_2=0$ implying $V_1=V_2=0$.
– Surb
Commented Mar 9, 2017 at 23:02
• So, it is normal that $P_1=P_2=0$ implies $V_1=V_2$
– Surb
Commented Mar 9, 2017 at 23:02
• On the other hand, if you assume $P_2\neq 0$, then $V_2\neq 0$ (or it would lead to a contradiction) and it holds $\frac{P_1}{P_2}=\frac{V_1}{V_2}$
– Surb
Commented Mar 9, 2017 at 23:04

Short answer. There is a unique solution if $P_1 + P_2 \not= 0$. If $P_1 + P_2 = 0$, but $P_1 \not= 0$ (and hence $P_2\not= 0$), then there are no solutions. Finally if $P_1 = P_2 = 0$, there are infinitely many solutions.

Indeed, your system is equivalent to

(1) $RP_1= V_1(V_1-V_2)$ and

(2) $RP_2= -V_2(V_1-V_2)$.

Adding each side of (1) and (2) yields

(3) $R(P_1 + P_2)= (V_1-V_2)^2$

Squaring (1) and (2) and using (3) gives respectively

(4) $R^2P_1^2= V_1^2(V_1-V_2)^2 = V_1^2R(P_1 + P_2)$

and

(5) $R^2P_2^2 = V_2^2(V_1-V_2)^2 = V_2^2R(P_1 + P_2)$

whence, if $P_1 + P_2 \not= 0$, $$V_1^2 = \frac{RP_1^2}{P_1 + P_2} \quad \text{and}\quad V_2^2 = \frac{RP_2^2}{P_1 + P_2}$$ If now $P_1 + P_2 = 0$, then you get from (3) $V_1 = V_2$. It now follows from (1) and (2) that $P_1 = P_2 = 0$.

• Thank you for the answer! For linear equations, usually infinite solutions is related to an underdetermined system, especially when we have two dependent equations. In this case, for $P_1=P_2=0$, can we say that our nonlinear equations form also an undetermined system? If true, does that mean that $V_1(V_1-V_2)$ and $V_2(V_2-V_1)$ are in fact the same equation? Commented Mar 10, 2017 at 13:15
• If $P_1 = P_2 = 0$, there are infinitely many solutions, thus the terminology "undetermined system" probably makes sense in this case. In this case, (1) becomes $V_1(V_1-V_2) = 0$ and (2) becomes $V_2(V_1-V_2) = 0$. However, these two equations are not equivalent. If you just keep (1), the solutions are ($V_1 = 0$ and $V_2$ arbitrary) or ($V_1 = V_2$). If you just keep (2), the solutions are ($V_2 = 0$ and $V_1$ arbitrary) or ($V_1 = V_2$). If you consider both equations, the solutions are ($V_1 = V_2$). Commented Mar 10, 2017 at 14:51
• Ok! Geometrically then, the equation $V_1(V_1-V_2)$, in a plane ($V_2$ as x,$V_1$ as y), would be constructed as the union of a horizontal line $V_1$=0 and a diagonal line $V_1$=$V_2$. Analogously, the other equation $V_2(V_2-V_1)$ would be, in the same plane as before, the union of a vertical line $V_2=0$ and the same diagonal $V_1=V_2$. Then the intersection of the two curves would end up being just the diagonal line (considering that the crossing of the vertical and horizontal line, the origin, is also comprised in the diagonal), crossing each other at infinite points. Is that right? Commented Mar 10, 2017 at 15:10
• Absolutely, this is a perfect description of the solutions. Commented Mar 10, 2017 at 15:37
• Thank you! For linear equations we could write them in a matrix form and study the rank of the matrices to deduce the properties of the system without recurring to any further algebraic manipulations. Is there anything analogous for nonlinear systems? Commented Mar 10, 2017 at 15:40