Phase plot for quadratic damping oscillator

22-Apr-2021

As an appendix promised for the article How fast helium balloon floats…, and also appendix for the follow up article Hint for solving quadratic drag force on a damping oscillator, I’m going to show the phase plot or phase diagram of the oscillation of quadratically damped floating balloon.

Originally, I want to make it interactive, but I realized the computation is complicated, it might make your browser hangs. I am using GeoGebra instead to make some screenshots of it.

Recalling the equation of motions

Our original equation of motions is this:

x¨=ω2xdx˙x˙\ddot x=-\omega^2 x - d \dot x \left| \dot x \right|

In the previous article we call dd as D2D_2 the damping coefficient for quadratic drag force. Because we are very clear now that this article only consider quadratic case, let’s just call the coefficient dd.

The equation of motion above have interesting property. It will converge to a stable position, x=0x=0 and x˙=0\dot x=0 and x¨=0\ddot x=0, due to losing energy. That means, the center of the coordinate system is the position where the object converges and stop moving.

Any oscillation with linear restoring force ωx-\omega x and quadratic damping drag force dx˙x˙- d \dot x \left| \dot x \right|, can be reduced into a generic equation above. If the equation have constant force CC, we shall find C=ωAC=\omega A where A is the amplitude value that can be used to do a coordinate shift from the original reference frame to a reference frame where position 0 is where the object will be at rest.

So, it’s suffice for us to find the behaviour of above equation, because any similar motions can be reduced to our equation above.

The term in the drag force dx˙x˙- d \dot x \left| \dot x \right|, the absolute value of x˙\dot x, is just a convenient way for us to say that the direction of the drag force is always opposing the direction of the velocity. With the magnitude of drag force proportional to x˙2\dot x^2.

So, our equation above can be broken down into positive (positive x) direction and negative (negative x) direction.

x˙>0;x¨=ω2xdx˙2x˙<0;x¨=ω2x+dx˙2\dot x >0; \ddot x=-\omega^2 x - d \dot x^2 \\ \dot x <0; \ddot x=-\omega^2 x + d \dot x^2

We solved the phase space equation in the previous article. We also defined more variables that will make our life simpler.

X=2dxV2=2d2ω2x˙2α=2dω2x¨K+=k+ω2K=kω2X=2dx \\ V^2=\frac{2d^2}{\omega^2}\dot{x}^2 \\ \alpha =\frac{2d}{\omega^2}\ddot{x} \\ K_+=\frac{k_+}{\omega^2} \\ K_-=\frac{k_-}{\omega^2}

The phase space relations conveniently becomes:

V>0;K+=eX(1+α)V<0;K=eX(1α)V>0;K+=eX(1XV2)V<0;K=eX(1+XV2)V > 0; K_+=e^X(1+\alpha) \\ V < 0; K_-=e^{-X}(1-\alpha) \\ V > 0; K_+=e^X(1-X-V^2) \\ V < 0; K_-=e^{-X}(1+X-V^2)

Some interesting insight from this. When we normalized the phase space variables XX and VV, it seems that the phase is deterministic. The constant K+K_+ and KK_- is entirely defined by the substitution of XX and α\alpha. That means, for any such system with this behavior, we can transform the initial condition to fit such phase plot. Then find the solution and behavior, then transform back from this standard plot.

Now, let’s define our standard initial condition.

First, we define that our standard coordinate system x=0x=0 is a position when the motion has converges.

Second, we define that our initial condition start when the object is displaced from x=0x=0, with maximum displacement. We can pick either in the left side or right side of origin. But I prefer to start from the left side. By starting from the left side, we will use the positive equation of motion (where velocity direction is positive).

If the distance between initial position and x=0x=0 is AA (the maximum amplitude). Then the initial position x0=Ax_0=-A. The initial acceleration becomes the maximum acceleration possible, and we can easily calculate it using Newton’s First Law in the initial condition. Or, shortly a0=ω2Aa_0=\omega^2 A

After defining the initial condition above, we have completely mapped a phase from our standard phase space XX and VV into the actual phase space of the object xx and x˙\dot x.

Drawing the plot of first iteration

To illustrate, we are going to do a transformation X=2dxX=2dx. this will map x0=Ax_0=-A into X0=2dAX_0=-2dA. From the initial condition, we got the constant K+K_+

K+=(1X0)eX0K_+=(1-X_0)e^{X_0}

The constant for our standard phase is determined by it’s original position.

If we use Lambert W function to find XX at V=0V=0, we should find two values of XX. One at X0X_0 when the object began to move and another one at X1X_1, which is when the object stop and change it’s velocity direction.

X=1+W(K+e)X=1+W(-\frac{K_+}{e})

By the way, we got the above expression using Lambert W substitution: wew=zw e^w = z into w=W(z)w=W(z), with w=X1w=X-1.

Most of numerical WW function only implements the principal branch. But the equation above has two solution, which means it will use two branch of WW function, the principal and W1W_{-1} branch. Luckily for us, the WW function returns the other XX value, X1X_1 and not X0X_0 again. This is probably because the absolute value of X0X_0 has to be bigger than X1X_1 due to the damping.

With just this information we are able to plot the phase. The equation is:

V=1XK+eXV=\sqrt{1-X-K_+e^{-X}}

It looks like this (plotted using GeoGebra):

As you can see that the maximum of VV is slanted a little bit left. We can calculate this position using the acceleration relation above, setting α=0\alpha=0.

Xpeak=ln(K+)X_{peak}=\ln(K_+)

Drawing the plot of the second iteration

The second iteration of the phase is when the velocity is in the negative direction. So, we use the second set of the equation:

V=1+XKeXV=-\sqrt{1+X-K_-e^{X}}

The initial XX value is the value that we find before: X1X_1. From there, we calculate KK_-

K=(1+X1)eX1K_-=(1+X_1)e^{-X_1}

Then, we can calculate X2X_2.

X=1W(Ke)X=-1-W(\frac{K_-}{e})

The peak:

Xpeak=ln(K)X_{peak}=-\ln({K_-})

The plot looks like this:

The plot for the next n iteration

By this point, you probably have found some patterns.

Our equation basically tells us the same shape of phase, but it is rotated, and scaled, due to the Lambert W function behaviour.

Let’s do a small recap. Our most basic relation tells us this:

V2=1XKeXV^2=1-X-Ke^{-X}

It will have two zeroes, X0X_0 will be negative and it’s absolute value is bigger than the positive X1X_1

The peak is only determined by KK, which is X=ln(K)X=\ln(K).

KK is determined only by it’s initial position.

K=(1X0)eX0K=(1-X_0) e^{X_0}

Or the other way around is also valid. It’s starting initial position is determined by KK.

The equation when VV is negative is actually the same equation but with a transform X=XX'=-X and KK'. After that, the equation just alternates it’s sign with different value of KK.

We can reformulate which equation/constants to use based on which iteration number.

Note, that if we define K0=0K_0=0, there is only one XX value possible, which is 1. We can imagine that the original amplitude X0X_0 is so far away at -\infty that it technically doesn’t cross XX axis. Let’s use this as a base.

X0=K0=0X1=1X_0=-\infty \\ K_0=0 \\ X_1=1

We got the next KK, K1K_1 by flipping the equation, so the amplitude now is X=X1=1X=-X_1=-1. I’m sure you got the pattern if you continue through multiple iteration. We got this relation:

Xn=(1)n1(1+W(Kn1e)Kn=(1+(1)n1Xn)e(1n1)XnVn=(1)n1+(1)n1XKne(1n1)XXpeakn=(1)nln(Kn)X_n=(-1)^{n-1}(1+W(\frac{K_{n-1}}{e}) \\ K_n=(1+(-1)^{n-1} X_n)e^{(-1^{n-1}) X_n} \\ V_n=(-1)^n \sqrt{1+(-1)^{n-1} X- K_n e^{(-1^{n-1}) X}} \\ X_{peak-n}=(-1)^n\ln(K_n)

Choice of index nn is completely arbitrary. With this kind of choice, that means, when nn is even, velocity is positive. When nn is odd, velocity is negative.

From this recursive relation, you can make a simple program that calculate all the necessary points of interests.

It’s also worth noting that subsequent values of KnK_n is only determined by the initial values in the range of <X01-\infty < X_0 \leq -1. If X0X_0 were to happen in the range 1<X0<0-1 < X_0 < 0 it is probably not the first iteration, and you can determine the the original XX that is capable of generating that value.

Simply speaking, if you make a phase plot with X0=1X_0=-1 then you will get the same phase plot if you draw the same plot, but starting at X2X_2 that is generated from X0=1X_0=-1. So, hypothetically speaking, if we have a canonical phase plot and the other phase plot can be transformed into this canonical phase plot, that means it is possible to define a standard function that will cover the whole dynamics.

Other insights from the above recursive relations is the maxima/minima of each cycle, from the value of XpeaknX_{peak-n}. We can infer that eventually KnK_n will converge to 1. In which the XpeaknX_{peak-n} will converge to 0.

This is a plot of graph with X0=1X_0=-1 with 5 iterations:

Same plot with but now with X0=X2X'_0=X_2, also with 5 iterations as you can see, same plot is constructed:

Deriving interesting identities

We can go a little bit further deep. There is an interesting relation that we don’t have if the plot phase is a simple harmonic cycles. If we choose two arbitrary index of a different parity KK, with the same value of V2V^2. We have an interesting relationship between positive VV and negative VV. Obviously in the different half cycle, it will corresponds with a different XX value that maps to the same V2V^2. With simple harmonics, this is not interesting because the XX value will be the same, but with this plot phase, we will have extra identity. Let’s suppose that XaX_a is the position where velocity is positive, meanwhile XbX_b is the position where velocity is negative.

1XaKaeXa=1+XbKbeXbKbebXKaeXa=Xb+XaHsinh(ϕ)=Xb+Xa21-X_a-K_ae^{-X_a}=1+X_b-K_be^{X_b} \\ K_be^X_b-K_ae^-X_a=X_b+X_a \\ H \sinh(\phi)=\frac{X_b+X_a}{2}

We just introduced a new variable HH and hyperbolic angle ϕ\phi. For every same V2V^2 between consecutive half cycles, this relation emerges. The definition of the new variables:

H=KaKbeXbXa2ϕ=ln(KbKa)+Xb+Xa2H=\sqrt{K_a K_b}e^\frac{X_b-X_a}{2} \\ \phi=\ln(\sqrt{\frac{K_b}{K_a}})+\frac{X_b+X_a}{2}

We can do lots of things with this relation. For example, one of the things I immediately recognize is the term Xb+Xa2\frac{X_b+X_a}{2} shows up twice in the identity above and the definition of ϕ\phi. That means we can use the identity to approximate something using recursion.

We know that in the final iteration, the value of K=1K=1, because the peak has to be in the middle. We can then inspect every XX where V2=0V^2=0. By setting either KaK_a or KbK_b to 1, then, HH quantifies how far the current cycle is with the final cycles. This is set unambigously because either Odd XaX_a or even XbX_b will corresponds to a positive value of XbXaX_b-X_a. For example if X=1X=1 then it must be XbX_b due to the recursion definition above. If X=0.59X=-0.59 then it must be XaX_a, causing XbXa=0.59X_b-X_a=0.59.

If we compare with X0X_0, then HH\approx \infty which is the maximum value. When X=1X=1, H=2H=\sqrt{2}. We can immediately know which cycle is earlier or later. We can even compare different HH to get relative cycle between the two. The idea is very powerful, that we can think about HH as the absolute cycle lengths. Then HaH_a and HbH_b can be divided to get a relative cycles. We can even know, if 2<H\sqrt{2} < H then that means it is the very first initial cycle.

Now, we are going to try to interpret the hyperbolic angle ϕ\phi. We can immediately recognize that it is possible to set the right hand side to zero. But the cycle length HH can not be zero. So the zero must comes from sinh(ϕ)\sinh(\phi). The reason why it happens, maybe because of these:

First, Xa=XbX_a=-X_b whatever that is. But in order for that to be valid. It also must satisfy

ln(KbKa)+Xb+Xa2=0Kb=Ka\ln(\sqrt{\frac{K_b}{K_a}})+\frac{X_b+X_a}{2}=0 \\ K_b=K_a

Because we assumed that KbK_b and KaK_a from a different parity of nn, that means this can only happen if K=1K=1 and it is in the final cycle. We actually proved otherwise. The ϕ\phi then cannot be zero if we are inspecting two cycles from a different parity.

If we define, the absolute cycle length of cycle n as HnH_n:

Hn=KnexH_n=K_ne^x

Then our previous HH is actually a square root/geometric average between two absolute cycles:

H=Haverage=HaHbH=H_{average}=\sqrt{H_a \cdot H_b}

While ϕ\phi can be defined as half the difference between two absolute cycles, and can be interpreted as hyperbolic angle difference:

ϕn=12ln(Hn)=arctanh(Hn1Hn+1)ϕ=Δϕ=ϕbϕa=12ln(HbHa)\phi_n=\frac{1}{2}\ln(H_n)=arctanh(\frac{H_n-1}{H_n+1}) \\ \phi=\Delta \phi=\phi_b-\phi_a=\frac{1}{2}\ln(\frac{H_b}{H_a})

The quantity ϕ\phi can becomes a natural choice to understand which cycle is currently early or late if we compare them. We can retrieve the cycle’s hyperbolic angle from any absolute cycle length HnH_n, either by calculating it directly with the natural logarithm, or by using arctanharctanh Like what we understand before. But now, we can conveniently give it a name. It arises intuitively that when ϕ\phi is small, that means the cycles is at it’s end and very similar.

From these results and understanding, we can then try to generalize these identities.

It follows that it turns out this identity is valid for any time that we observe two different cycles with the same V2V^2.

Havgsinh(Δϕ)=XavgH_{avg} sinh(\Delta \phi)=X_{avg}

However, remember that the sign alternates whenever if it’s not the same parity. Currently, the above identity holds if bb odd and aa even.

It is a direct result, that if we compare two different XX in the same even cycle (positive velocity) with Xb>XaX_b > X_a:

K+eXbXa2sinh(XbXa2)=XbXa2K_+ e^\frac{{-X_b-X_a}}{2} sinh(\frac{X_b-X_a}{2})=\frac{X_b-X_a}{2}

In the odd cycle (negative velocity)

KeXb+Xa2sinh(XbXa2)=XbXa2K_- e^\frac{{X_b+X_a}}{2} sinh(\frac{X_b-X_a}{2})=\frac{X_b-X_a}{2}

From both equation, we can’t determine if we set Xa=XbX_a=X_b, which is make sense because the only time it happens is when X=XpeaknX=X_{peak-n}.

If both XX are zeroes of the equation, then XbXa\left|X_b-X_a\right| is the full range/length of XX of that cycle. We can call this the span S=XbXaS=\left|X_b-X_a\right|. Meanwhile the quantity XbXa2\frac{{-X_b-X_a}}{2} and Xb+Xa2\frac{{X_b+X_a}}{2} turns out to be always positive due to the slanted shape of the phase plot. It is the average of absolute value of left and right amplitude difference. Let’s call this ΔA\Delta A.

KeΔA2sinh(S2)=S2K e^{\frac{\Delta A}{2}} sinh(\frac{S}{2})=\frac{S}{2}

Regardless of if it is an odd or even cycle, ΔA\Delta A is actually a good quantity to compare against difference cycles. If ΔA\Delta A is small or near zero, that means it is closer to final cycle. If we remember our definition of absolute cycle length HnH_n. Then the value KeΔA2K e^{\frac{\Delta A}{2}} is a relative cycle length between the same span of cycle HavgH_{avg}.

With this property, we can easily scale any phase plot with a given span SS into another span. If we treat span as a continous variables (we don’t have any reason now, but for the sake of arguments, let’s pretend), we can take partial derivative. A span is invariant under origin translation. That means, in any given cycle, if we know the span and the motion constant KK, we can determine where the canonical origin should be from the ΔA\Delta A value.

Next, we are going to observe the quantity ϕ\phi.

The ϕn\phi_n is an equivalent of HnH_n but it is in the form of hyperbolic angle. Since we have :

ϕn(X)=12ln(Hn(X))\phi_n(X) = \frac{1}{2}\ln(H_n(X))

It’s value may signify some importance with the hyperbolic geometry. The range of values of ϕn(X)\phi_n(X), depends on it’s span and amplitude difference in a given cycle nn. The minimum value ϕn,min=S+ΔA4+12ln(Kn)\phi_{n,min}=-\frac{S+\Delta A}{4}+\frac{1}{2}\ln(K_n), the maximum value ϕn,max=SΔA4+12ln(Kn)\phi_{n,max}=\frac{S-\Delta A}{4}+\frac{1}{2}\ln(K_n). It’s very similar with the phase of circular function like, sin(θ)\sin(\theta). I’m very suspicious. The midpoint of ϕn\phi_n is a very special value where ϕn,mid=Xpeakn=ln(Kn)\phi_{n,mid}=X_{peak-n}=\ln(K_n).

From, that we have an interesting hyperbolic identity, where any hyperbolic angle ϕn(X)\phi_n(X) can be described with relative to the midpoint. Then, we have a quantity to for us to actually which phase the oscillation is currently in with arbitrary XX value.

Φn(X)=ϕnϕmid=12ln(eXKn)Φn,min=S+ΔA412ln(Kn)Φn,max=SΔA412ln(Kn)\Phi_n(X) = \phi_n-\phi_{mid}=\frac{1}{2}\ln(\frac{e^X}{K_n}) \\ \Phi_{n,min}=-\frac{S+\Delta A}{4}-\frac{1}{2}\ln(K_n) \\ \Phi_{n,max}=\frac{S-\Delta A}{4}-\frac{1}{2}\ln(K_n)

The new quantity Φn(X)\Phi_n(X) is the relative hyperbolic angle from the midpoint angle. From this form, we can intuitively understand that, the Φn(X)\Phi_n(X) corresponds to how big the value of eXe^X is compared with the current KnK_n. If Φn(X)\Phi_n(X) is negative, that mean eXe^X is less than KnK_n is magnitude, (also the other way around by inference).

You can check the graph of Φ(X)\Phi(X) and ϕ(x)\phi(x). It is very convenient that Φ(X)\Phi(X) can be used to show the phase angle of the cycle. Check that the zero value of Φ(X)\Phi(X) does match with the maximum of V2V^2.

Now, because we had a new definition of Φn(X)\Phi_n(X), we can add a new definition, analoguous with the relation ϕn(X)\phi_n(X) and Hn(X)H_n(X). Since Φn(X)\Phi_n(X) is a normalized value, then we shall define a normalized absolute cycles Ln(X)L_n(X), with the definition:

Ln(X)=e2Φ(X)=eXKn=Hn(X)Kn2Φ(X)=12ln(Ln(X))L_n(X)=e^{2\Phi(X)}=\frac{e^X}{K_n}=\frac{H_n(X)}{K_n^2} \\ \Phi(X)=\frac{1}{2}\ln(L_n(X))

Here’s the graph of both Hn(X)H_n(X) and Ln(X)L_n(X).

Because of how Hn(X)H_n(X) and Ln(X)L_n(X) cramped together, I only included the first three cycles. Exponential line above V2=1V^2=1 corresponds to Ln(X)L_n(X). Exponential line below V2=1V^2=1 corresponds to Hn(X)H_n(X). Each color of the line corresponds to the phase cycle plot directly below.

From what I observed in the graph. I understand that:

  1. The closer Hn(X)H_n(X) to cross (X,V)=(0,1)(X,V)=(0,1), that means the closer the cycle to the final state. Cycle with K=0K=0 is defined with H(X)=H(X)= everywhere (the furthest from final state)
  2. The closer Ln(X)L_n(X) to cross (X,V)=(0,1)(X,V)=(0,1), that means the closer the cycle to the final state. Cycle with K=0K=0 is defined have undefined normalized cycle length. (because division by zero). However we can intuitively says that if the value Ln(X)L_n(X) is bigger in the same phase, that means it is furthest from the final state.

Recap of the definition and identities

From the phase and cycle plot, we summarized the following proposed definition and identities to understand the cycles.

Recursive relations

Standard base:

X0=K0=0X1=1X_0=-\infty \\ K_0=0 \\ X_1=1

Recursive relation, starting from n=1n=1

Xn=(1)n1(1+W(Kn1e)Kn=(1+(1)n1Xn)e(1n1)XnVn=(1)n1+(1)n1XKne(1n1)XXpeakn=(1)nln(Kn)X_n=(-1)^{n-1}(1+W(\frac{K_{n-1}}{e}) \\ K_n=(1+(-1)^{n-1} X_n)e^{(-1^{n-1}) X_n} \\ V_n=(-1)^n \sqrt{1+(-1)^{n-1} X- K_n e^{(-1^{n-1}) X}} \\ X_{peak-n}=(-1)^n\ln(K_n)

Qualitative definition

  • The span SnS_n is the length from Xn,minX_{n,min} to Xn,maxX_{n,max} in a given cycle n.
  • The amplitude AnA_n is the maximum displacement from the origin point to a point where the velocity is zero in a given cycle.
  • The absolute amplitude difference ΔA\Delta A is the difference between the left amplitude and right amplitude of the given cycle, relative to the origin point.

Quantitative definition

The absolute cycle length Hn(X)H_n(X) quantifies how far the current cycle nn into the final cycle state n=n=\infty.

Hn(X)=KneXH_n(X)=K_n e^X

Absolute cycle length preserves ordering. Whenever b>ab > a, Hb(X)>Ha(X)H_b(X) > H_a(X) if compared in the same value of X. Maximum value is Hn(X)=eXH_n(X)=e^X.

Relative cycle length is a quantity to measure the ordering of the cycle Hba(X)=Hb(X)Ha(X)H_{ba}(X)=\frac{H_b(X)}{H_a(X)}, if compared in the same value of X.

Cycle average is a quantity that measures the average of two cycles. Havg=HaHbH_{avg}=\sqrt{H_a \cdot H_b}

Hyperbolic phase angle ϕn(X)\phi_n(X) quantifies the phase of motion in the current cycle n. It can be used for ordering in the same cycle (but with some care about the parity).

ϕn(X)=X2+12ln(Kn)\phi_n(X)=\frac{X}{2}+\frac{1}{2}\ln(K_n)

The mid of the half cycle phase corresponds to when the acceleration is zero.

ϕn,mid=Xpeakn=ln(Kn)\phi_{n,mid}=X_{peak-n}=\ln(K_n)

The minima and maxima of the phase in a given cycle nn is completely determined from the current span SnS_n and amplitude difference ΔA\Delta A.

ϕn,min=Sn+ΔAn4+12ln(Kn)ϕn,max=SnΔAn4+12ln(Kn)\phi_{n,min}=-\frac{S_n+\Delta A_n}{4}+\frac{1}{2}\ln(K_n) \\ \phi_{n,max}=\frac{S_n-\Delta A_n}{4}+\frac{1}{2}\ln(K_n)

The phase angle difference is the difference between two phases. Δϕ=ϕbϕa\Delta \phi=\phi_b-\phi_a

The normalized phase angle Φn(X)\Phi_n(X) is a translated phase where the origin is ϕn,mid\phi_{n,mid}. The normalized phase angle has a special property that the sign can also determined the sign of the equation of motions (combination of sign of acceleration, and velocity).

Φn(X)=ϕn(X)ϕmid=12ln(eXKn)Φn,mid(X)=0Φn,min=S+ΔA412ln(Kn)Φn,max=SΔA412ln(Kn)\Phi_n(X) = \phi_n(X)-\phi_{mid}=\frac{1}{2}\ln(\frac{e^X}{K_n}) \\ \Phi_{n,mid}(X)=0 \\ \Phi_{n,min}=-\frac{S+\Delta A}{4}-\frac{1}{2}\ln(K_n) \\ \Phi_{n,max}=\frac{S-\Delta A}{4}-\frac{1}{2}\ln(K_n)

The normalized phase angle difference is the same phase angle difference, ΔΦ=Δϕ\Delta \Phi=\Delta \phi.

From the normalized phase angle, we can construct a normalized cycle length Ln(X)L_n(X). The contour of Ln(X)L_n(X) preserves ordering between cycles.

Ln(X)=eXKnL_n(X)=\frac{e^X}{K_n}

Relation and identities

For any XaX_a and XbX_b, with b>ab>a, with the same squared velocity V2V^2, it follows:

Havgsinh(Δϕ)=ΔX2H_{avg} sinh(\Delta \phi) = \frac{\Delta X}{2} \\

Expanded as:

HaHbsinh(ϕbϕa2)=XbXa2\sqrt{H_a \cdot H_b} sinh(\frac{\phi_b-\phi_a}{2}) = \frac{X_b-X_a}{2}

Above equation assumes both odd parity cycle nn. If one of the arguments is even cycle, switch the sign of XX. If both XX is even cycle, switch both sign.

If expanded, odd b and a:

KbKaeXb+Xa2sinh(ln(KbKa)+XbXa2)=XbXa2\sqrt{K_b K_a} e^{\frac{X_b+X_a}{2}} sinh(\ln(\sqrt{\frac{K_b}{K_a}})+\frac{X_b-X_a}{2})=\frac{X_b-X_a}{2}

odd b, even a:

KbKaeXbXa2sinh(ln(KbKa)+Xb+Xa2)=Xb+Xa2\sqrt{K_b K_a} e^{\frac{X_b-X_a}{2}} sinh(\ln(\sqrt{\frac{K_b}{K_a}})+\frac{X_b+X_a}{2})=\frac{X_b+X_a}{2}

even b, even a:

KbKaeXbXa2sinh(ln(KbKa)+XbXa2)=XbXa2\sqrt{K_b K_a} e^{\frac{-X_b-X_a}{2}} sinh(\ln(\sqrt{\frac{K_b}{K_a}})+\frac{X_b-X_a}{2})=\frac{X_b-X_a}{2}

even b, odd a:

KbKaeXb+Xa2sinh(ln(KbKa)+Xb+Xa2)=Xb+Xa2\sqrt{K_b K_a} e^{\frac{-X_b+X_a}{2}} sinh(\ln(\sqrt{\frac{K_b}{K_a}})+\frac{X_b+X_a}{2})=\frac{X_b+X_a}{2}

If it is the same cycle, relation reduced to:

if odd:

KeXb+Xa2sinh(XbXa2)=XbXa2Ke^{\frac{X_b+X_a}{2}} sinh(\frac{X_b-X_a}{2})=\frac{X_b-X_a}{2}

if even:

KeXbXa2sinh(XbXa2)=XbXa2Ke^{\frac{-X_b-X_a}{2}} sinh(\frac{X_b-X_a}{2})=\frac{X_b-X_a}{2}

If both XX are zeroes in the same cycle, relation is simplified into:

KeΔA2sinh(S2)=S2Ke^{\frac{\Delta A}{2}}sinh(\frac{S}{2})=\frac{S}{2}

Some defition have connecting relation:

The relation below happens in the same cycle n:

ΔΦn=Δϕn\Delta \Phi_n = \Delta \phi_n

The relation below connects between ϕn(X)\phi_n(X), Hn(X)H_n(X), Ln(X)L_n(X) and Φn(X)\Phi_n(X) in the same cycle n:

ϕn(X)=12ln(Hn(X))Φn(X)=12ln(Hn(X)Kn2)=12ln(Ln(X))Ln(X)=Hn(X)Kn2Hn(X)=e2ϕn(X)Ln(X)=e2Φn(X)\phi_n(X)=\frac{1}{2}\ln(H_n(X)) \\ \Phi_n(X)=\frac{1}{2}\ln(\frac{H_n(X)}{K_n^2})=\frac{1}{2}\ln(L_n(X)) \\ L_n(X)=\frac{H_n(X)}{K_n^2} \\ H_n(X)=e^{2\phi_n(X)} \\ L_n(X)=e^{2\Phi_n(X)}


Rizky Maulana Nugraha

Written by Rizky Maulana Nugraha
Software Developer. Currently remotely working from Indonesia.
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