4

I have a cubic equation: $X^3-UX^2-KX-L=0$ (1)

with $X=1-E+U$, $K=4(1-\gamma^2-\lambda^2)$, $L=4\gamma^2U$.

I solve Eq. (1) for the variable $E$ numerically for $U=2$ and different sets of parameter $\gamma =$ 0.1, 0.25, and 0.45 and plot real and imaginary parts of $E$ against $\lambda$. The real parts of the solutions for $X$ and $E$ differ by a constant shift for a fixed value of $U$ while the imaginary parts remain identical.

See the plots below. The plots plot all the roots or solutions. They are colored with three different colors by looking at the possible continuity of the roots or the solutions (here stressed on the continuity of the imaginary parts).

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

Here we see for the first case $\gamma = 0.1$, the continuity in the real parts of $E$ (Re $E$) break down while the imaginary parts (Im $E$) remain continuous for all parameter values of $\gamma$.

Since the solutions are found for discrete values of $\lambda$, we may think that extreme left and right of the cyan and blue curves of the first plot can be interchanged and hence made Re $E$ continuous all the way. However, that may lead Im $E$ to be discontinuous for other parameter values of $\gamma$ (I can add images if further clarity is needed).

How can we interpret or understand this? Or is there something fundamentally wrong?

The alternative coloring for the first two plots could be the following (in this sense all the curves appear continuous function of $\lambda$).

Real part:

enter image description here

Imaginary part:

enter image description here

Though the above fixes the continuity issue, it appears that all three curves do not smoothly evolve with parameter $\gamma$. Experts' opinions awaited.

hbaromega
  • 141
  • 1
    "we see for the first case ... the continuity in the real parts of E ... break down" Sorry, but that's not obvious to me, and the question is a bit hard to follow in general. You start with an equation in $X$ but what you solve and graph is a different cubic in $E$. It would help if you posted the actual equation you are solving, how exactly you solve it, and how you trace the individual roots across parameter changes where you suspect a discontinuity. One graph with one range and some annotations could suffice for that. – dxiv Feb 01 '22 at 22:49
  • It doesn't matter because $E$ is just a constant real shift to $X$ for fixed $U$. Continuity's typical definition: $f(x=a+)=f(x=a-)=f(a)$ which is not followed in the real part around $\lambda=\pm 1$ (for $\gamma=0.1$). A constant shift to this won't change the scenario. It would be great if you don't blame the question if an answer is not readily available. I'm not tracing, I'm just plotting all of them and trying to paint in one color following continuity of a curve. – hbaromega Feb 02 '22 at 11:21
  • I have edited the question a bit to clarify further. – hbaromega Feb 02 '22 at 12:06
  • Short answer to your question is that the roots are continuous functions of coefficients, and that covers both real and complex roots, as well as the real and imaginary parts of the latter. If this is not the kind of answer you are looking for, then you'll need to provide enough details for someone to duplicate the particular issue you (think you) are seeing - which means specifying a narrow range for $\lambda$, exact values for the other parameters etc. – dxiv Feb 02 '22 at 17:37
  • You don't say how you solve the equations, and numerical instability is possible around the points where a pair of real roots branch out into complex ones. You also don't say how you "trace" the roots, in other words how you decide which one is which when they are all packed tightly together. The first graph, for example, looks discontinuous only because of your assignment of colors i.e. such "tracing". – dxiv Feb 02 '22 at 17:37
  • @dxiv I already mentioned "They are colored with three different colors by looking at the possible continuity of the roots or the solutions (here stressed on the continuity of the imaginary parts)." I also mentioned we can change the colors to make the curves (e.g. blue and cyan at the two ends). But later when we change $\gamma$, imaginary parts become discontinuous. I can rephrase the question: Once we plot all the roots, can we color or paint three of them so that each appears a continuous function of $\lambda$ for various sets of parameter $\gamma$? – hbaromega Feb 02 '22 at 20:25
  • I'll see if I can simplify the problem further and re-edit the question. – hbaromega Feb 02 '22 at 20:26
  • Related: Continuous root map of the coefficients of a polynomial, in particular the "depend only on a real parameter" case mentioned in the answer. – dxiv Feb 02 '22 at 21:10
  • I have added two other alternative plots with comments. Let me know what you think. – hbaromega Feb 03 '22 at 12:08
  • The real and imaginary parts will vary continuously on the parameter, but not necessarily smoothly in general. Simplest case is to look at the roots of $x^2 = \alpha$ when $\alpha \in \mathbb R$ crosses $0$. That is what appears to happen in your example as well, and it is not unexpected. – dxiv Feb 04 '22 at 00:27
  • Sorry, you mean when $\alpha$ becomes negative? I think the real parts of the roots continuously approach zero. Do you have any example of two variables dependent function, say $f(x,t)$ where we plot $f$ vs $x$ while changing the parameter $t$? We want to see if the evolution of the plot $f(x)$ changes smoothly w.r.t. $t$ or not. $f$ can be the roots of an analytic equation as well. – hbaromega Feb 07 '22 at 22:23
  • The real part of the principal root of $,x^2=\alpha,$ is $,f(\alpha)=\begin{cases}\begin{align}\sqrt{\alpha} && \alpha \ge 0\ 0 && \alpha \lt 0\end{align}\end{cases},$. This $,f(\alpha),$ is continuous on $,\mathbb R,$ but not smooth at $,0,$ since $,f'(0),$ does not exist. – dxiv Feb 07 '22 at 22:36
  • Thanks. I understand that $\lim_{x\to 0-} \sqrt{x}$ doesn't exist. However, can this be noticed with numerics as well? – hbaromega Feb 08 '22 at 00:30
  • Not sure I follow, and there is nothing about $,\lim_{x \to 0-}\sqrt{x},$ in it. There is $,f(\alpha),$ above, which is defined and continuous on the entire $,\mathbb R,$. Then there is its derivative $,f'(\alpha)=\begin{cases}\begin{align}1 ,/,2\sqrt{\alpha} && \alpha \gt 0 \ 0 && \alpha < 0\end{align}\end{cases},$ which is only defined (and continuous) on $,\mathbb R \color{red}{\setminus {0}},$. Point of the example being that the real part of the root is a continuous function of the parameter, though not a "smooth" (differentiable) one at points where multiple roots occur. – dxiv Feb 08 '22 at 00:48
  • I was on the line of the definition of continuity of a function $f(x)$ at point $a$: $\lim_{x\to a-} f(x) = \lim_{x\to a+} f(x) = f(a)$ [equating left and right limits or small neighborhoods to the function's exact value at the specific point $a$]. In your example, $f(x)=\sqrt{x}$ and $a=0$. I used to know $f(x)=|x|$ is a classic example of a continuous but non-differentiable function. – hbaromega Feb 09 '22 at 22:46

0 Answers0