Krdytkyu

I have a parameterized system of equations describing the crossed ladders problem.

• 
  $(x, y)$ are the $2$ horizontal distances respectively on the left/right of the junction of the ladders


• 
  $(a, b, c)$ are parameters for, respectively; the length of the first/second ladder and the distance between the ground and the junction.

Here is my system:

$$ (x + y)^2 (x^2 + c^2) - a^2 x^2 = 0$$
$$ (x + y)^2 (y^2 + c^2) - b^2 y^2 = 0$$

I've obtained the solution $(x, y) approx$ $(0.8572, 1.7461)$ for $(a, b, c) = (4, 3, 1)$.

Is there any way to find a function $f(a, b, c)$ that would give me a good starting candidate for my system?

Thank you.

You could make at least life simpler writing $$(x + y)^2 (x^2 + c^2) = a^2 x^2tag 1$$
$$(x + y)^2 (y^2 + c^2) = b^2 y^2tag 2$$ Make the ratio
$$frac{x^2+c^2}{y^2+c^2}=frac{a^2 x^2 }{b^2 y^2 }tag 3$$ This gives two solutions
$$y_pm=pmfrac{a, c ,x}{sqrt{(b^2-a^2), x^2+b^2, c^2}}tag 4$$
Suppose that we use $y_+$ that we plug in $(1)$ to get as equation
$$left(a^2 c^4+b^2 c^4-a^2 b^2 c^2right)+x^2 left(a^4-a^2 b^2+2 b^2
c^2right)+x^4 left(b^2-a^2right)+2 a c left(c^2+x^2right) sqrt{(b^2-a^2)x^2 +b^2c^2}=0$$ and you can notice that we can solved for $x^2$ instead of $x$.

If $a >b$, this is limiting the search domain since we need $(b^2-a^2)x^2 +b^2c^2 geq 0$ that is to say (admitting that we look for $x>0$) $$0 leq x leq frac{b c}{sqrt{a^2-b^2}}$$ Next, what I should do is a Taylor expansion around $x=0$ which, assuming that $a,b,c$ are all positive to get
$$left(c^4 (a+b)^2-a^2 b^2 c^2right)+frac{ (a+b) left(a^2 b (a-b)-c^2 (a-2 b)
(a+b)right)}{b}x^2+Oleft(x^4right)$$ to get an estimate
$$x_0=sqrt{frac{a^2 b^3 c^2-b c^4 (a+b)^2}{(a+b) left(a^2 b (a-b)-c^2 (a-2 b) (a+b)right)}}$$

Applied to your case $(a, b, c) = (4, 3, 1)$, the upper bound would be $frac{3}{sqrt{7}}approx 1.134$ and $x_0=sqrt{frac{285}{434}}approx 0.810$.

Now, Newton method will generate the following iterates
$$ left(
begin{array}{cc}
n & x_n \
0 & 0.81035920 \
1 & 0.85758341 \
2 & 0.85721270 \
3 & 0.85721269
end{array}
right)$$ and, back to $(4)$, $y=1.74607506$.

Edit

We can even do better if, instead of using Taylor, we build the $[2,2]$ Padé approximant. Solving for $0$ th numerator, we get
$$x_0=sqrt{frac A {(a+b)B}}$$ where
$$A=4 b^2 c^2 left(a^4 b^3 (a-b)-a^2 b c^2 (a+b) left(a^2+a b-3 b^2right)+c^4 (a-2 b) (a+b)^3right)$$
$$B=4 a^4 b^3 (a-b)^2+c^4 (a+b)^2 left(a^3+2 a^2 b-11 a b^2+12 b^3right)+3 a^2 b^2
c^2 left(-3 a^3+6 a^2 b+a b^2-4 b^3right)$$ Applied to the example, this would give $x_0=sqrt{frac{26505}{37037}}approx 0.846$.

Newton iterates would then be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.84595255 \
1 & 0.85722574 \
2 & 0.85721269
end{array}
right)$$

For the worked example, the function we look for the zero of is
$$f(x)=-7 x^4+2 left(4 sqrt{9-7 x^2}+65right) x^2+8 sqrt{9-7 x^2}-119$$ and its $[2,2]$ Padé approximant is
$$g(x)=frac{37037 x^2-26505}{35 x^2+279}$$ Just for curiosity, plot the two functions on the same graph. You would be amazed (I hope !).

We could still do better building the $[2,4]$ Padé approximant; the expression of $x_0$ would be really nasty but for the example we would have
$$h(x)=frac{74568309 x^2-54285660}{20650 x^4+85245 x^2+571428}$$ giving $x_0=sqrt{frac{18095220}{24856103} } approx 0.853$.

Update

We could make the problem simpler defining first $X=x^2$ and then $X=frac{b^2 c^2-T^2}{a^2-b^2}$. This makes the equation to be
$$-a^4 c^4-2 a^3 c^3T+ a^2left(a^2- b^2right)T^2+2 a c T^3+T^4=0tag5$$ which can be solved analytically for $T$. This quartic has two real roots since
$$Delta=-16 a^{12} c^4 (a-b)^2 (a+b)^2 left(left(a^2-b^2right)^2+27 c^4right) < 0qquad qquadforall a,b,c$$ (see here).

I suppose that the roots have opposite signs and that the one to keep is the positive root (very ugly expressions with awful radicals).

For the worked example, we should then have to solve
$$T^4+8 T^3+112 T^2-128 T-256=0$$ for which the real roots are $T=-1.06797$ and $T= 1.96375$

So, there is an explicit solution to the problem.