Dgdrxrt

I'm working on an analytical solution to the crossed ladders problem.

The solution is almost done and already useable (see my answer to Crossed Ladders Problem for details).

However I'm left with a problem in the end: depending on which ladder I choose using to calculate $w$, the expression of $w$ seems not to be the same.

Let $left(j,kright)inlbraceleft(1,2right),left(2,1right)rbrace$, $minlbrace1,2,3,4rbrace$, and $ninlbrace-,+rbrace$.

Any quantity with a $j$ or $k$ index below seems to depend on the choice of the ladder while, in all numerical applications I tried, I always have $w_jmn=w_kmn$ with $w_jmn=nsqrtl_k^2-h_jm^2$ and $w_kmn=nsqrtl_j^2-h_km^2$.

This leads to:

beginequation
beginsplit
w_j1+&=fracsqrtR_9+R_7+c_F6+textifracc_jLsqrtR_9-R_7-c_F6\
w_j1-&=frac-sqrtR_9+R_7+c_F6+textifrac-c_jLsqrtR_9-R_7-c_F6\
w_j2+&=fracsqrtR_9+R_7+c_F6+textifracc_jMsqrtR_9-R_7-c_F6\
w_j2-&=frac-sqrtR_9+R_7+c_F6+textifrac-c_jMsqrtR_9-R_7-c_F6\
w_j3+&=fracc_jNsqrtleft6+textifracc_jOsqrtleft6\
w_j3-&=frac-c_jNsqrtleft6+textifrac-c_jOsqrtleft6\
w_j4+&=fracc_jPsqrt6+textifracc_jQsqrt6\
w_j4-&=frac-c_jPsqrt6+textifrac-c_jQsqrt6\
endsplit
notag
endequation

With (again see the answer to Crossed Ladders Problem for all details):

beginequation
beginsplit
R_j8&=6 textsgnleft(3h+R_j2right)sqrthright\
c_jL&=1-left(-1right)^j textsgnleft(3h-R_j2right)-left| textsgnleft(3h-R_j2right)right|\
c_jM&=1+left(-1right)^j textsgnleft(3h-R_j2right)-left| textsgnleft(3h-R_j2right)right|\
c_jN&=frac1+ textsgnleft(2c_F-2R_7+R_j8right)2\
c_jO&=frac1- textsgnleft(2c_F-2R_7+R_j8right)2\
c_jP&=frac1+ textsgnleft(2c_F-2R_7-R_j8right)2\
c_jQ&=frac1- textsgnleft(2c_F-2R_7-R_j8right)2\
endsplit
notag
endequation

Thus I wonder if it is possible to prove that:

beginequation
beginsplit
R_k8&=R_j8\
c_kL&=c_jL\
c_kM&=c_jM\
endsplit
notag
endequation

$R_k8=R_j8$ would imply $c_kN=c_jN$, $c_kO=c_jO$, $c_kP=c_jP$, and $c_kQ=c_jQ$.

Proving $ textsgnleft(3h+R_k2right)= textsgnleft(3h+R_j2right)$ would be enough to prove all these relations.

For $c_kL=c_jL$ and $c_kM=c_jM$, proving $ textsgnleft(3h-R_k2right)=- textsgnleft(3h-R_j2right)$ would be enough.

In my attempts, and when working on the general solution, I came across the following equations, which I write in case they are helpful:

beginequation
beginsplit
R_j1^3&=27c_jA^2R_j1+54c_jC\
R_j2R_j3&=54left(hc_jA+h^3right)\
R_j3&=2sqrtR_j1^2+c_jBR_j1+9c_jA^2-108h^2c_jA+81h^4\
R_6&=fracR_j1^2+6c_jAR_j1-18c_jA^29\
R_7&=frac6hR_j2+R_j32\
R_j8&=2R_j4left(3h+R_j2right)\
R_9&=sqrtleft(c_F+R_7right)^2+R_j5^2left(3h-R_j2right)^2\
R_j1&geq6c_jA-9h^2\
R_k1&=-R_j1\
textsgnleft(R_j1right)&= textsgnleft(c_jAright)\
hc_jA+h^3=0&implies R_j1-c_jB=0\
R_j3&geqleft|R_j1+12c_jA-18h^2right|\
R_j3&geq-6hR_j2\
R_j5=0&implies3h-R_j2=0
endsplit
notag
endequation

Thanks in advance for your answers.

Here is a proof by counterexample, so there might be a better one.

The system $left(1right)$ of the answer to Crossed Ladders Problem gives:

beginequation
beginsplit
0&=hleft(h_j+h_kright)-h_jh_k\
0&=h_j^2-h_k^2+c_jA\
endsplit
tag8
endequation

Let the real numbers:

beginequation
beginsplit
left(S_j4,S_j5right)&inlbraceleft(-1,0right),left(0,-1right),left(0,1right),left(1,0right)rbrace\
left(S_k4,S_k5right)&inlbraceleft(-1,0right),left(0,-1right),left(0,1right),left(1,0right)rbrace\
endsplit
notag
endequation

According to $left(7right)$ of the answer to Crossed Ladders Problem, the general expressions of $h_j$ and $h_k$ are:

beginequation
beginsplit
h_j&=frac3h+left(S_j4^2-S_j5^2right)R_j2+S_j4R_j46+textifracS_j5R_j56\
h_k&=frac3h+left(S_k4^2-S_k5^2right)R_k2+S_k4R_k46+textifracS_k5R_k56\
endsplit
tag9
endequation

Combining $left(8right)$ and $left(9right)$:

beginequation
beginsplit
&S_j4R_j4left(3h-left(S_k4^2-S_k5^2right)R_k2right)+S_k4R_k4left(3h-left(S_j4^2-S_j5^2right)R_j2right)\
&=S_j4S_k4R_j4R_k4-S_j5S_k5R_j5R_k5+left(S_j4^2-S_j5^2right)left(S_k4^2-S_k5^2right)R_j2R_k2\
&-left(S_j4^2-S_j5^2right)3hR_j2-left(S_k4^2-S_k5^2right)3hR_k2-27h^2\
&S_j5R_j5left(3h-left(S_k4^2-S_k5^2right)R_k2right)+S_k5R_k5left(3h-left(S_j4^2-S_j5^2right)R_j2right)\
&=S_j4S_k5R_j4R_k5+S_k4S_j5R_k4R_j5\
&S_j4R_j4left(3h+R_j2right)-S_k4R_k4left(3h+R_k2right)=left(S_k4^2-S_j4^2+S_j5^2-S_k5^2right)R_7\
&S_j5R_j5left(3h-R_j2right)-S_k5R_k5left(3h-R_k2right)=0\
endsplit
notag
endequation

Many counterexamples can be found for these equations if $S_k4neq S_j4$ or $S_k5neq S_j5$.

The one given in Crossed Ladders Problem, $left(h,l_j,l_kright)=left(3,8,10right)$, is one of them.

Thus:

beginequation
beginsplit
S_k4&=S_j4\
S_k5&=S_j5\
endsplit
notag
endequation

Thus each $h_km$ corresponds to each $h_jm$, and thus:

beginequation
beginsplit
R_k8=R_j8&=R_8\
c_kL=c_jL&=c_L\
c_kM=c_jM&=c_M\
w_kmn=w_jmn&=w_mn\
endsplit
notag
endequation