Solving a system of five polynomials

Question

I am trying to solve the following system of equations for tuple $\left(a,b,c,d,t\right) \in \mathbb{R}^{4} \times [0,1]$, with parameter $\ell\in\mathbb{R}$.

$$ \begin{eqnarray} a\frac{t^{2}}{2} - bt + 1 = 0 \qquad (1)\\ a \frac{t^{3}}{6} - b \frac{t^{2}}{2} + t - \ell = 0 \qquad (2)\\ c \frac{t^{2}}{2} - dt + \left(d - \frac{c}{2} - 1\right) = 0 \qquad (3)\\ c \frac{t^{3}}{6} - d \frac{t^{2}}{2} + \left(d - \frac{c}{2} - 1\right)t + \left(\frac{c}{3} - \frac{d}{2} + 1 - \ell\right) = 0 \qquad (4)\\ at - b - ct + d = 0 \qquad (5) \end{eqnarray} $$

This arose while solving an optimal control problem, whose solution is known. That known solution tells that if $\frac{1}{6} \leq \ell \leq \frac{1}{4}$, then the above system of equations has unique solution given by $a = 24(1-4\ell)$, $b = 8(1-3\ell)$, $c = -24(1-4\ell)$, $d = 8(1-3\ell) - 24(1-4\ell)$, $t = \frac{1}{2}$. Otherwise, (when $\ell$ does not satisfy the stated inequality) the above system has no solution. While this seems like an easy exercise, I am not able to recover these results. I feel that these facts are not as difficult to establish as I am making it ... so any help or insight in proving them would be appreciated.

So far I have tried these (other than verifying the claimed solution):

1. Brute force: from (5), substitute $t = \frac{b-d}{a-c}$ in the first four equations and try to solve the resulting four polynomials for $(a,b,c,d)$. As you can see, this soon leads to complications, specially in establishing the bound for $\ell$, and in using the fact that $0 \leq \frac{b-d}{a-c} \leq 1$.

2. Let us call the LHS of (2), a cubic polynomial in $t$, as $p(t)$, and the LHS of (4), another cubic, as $q(t)$. Then (1) and (2) say that $p(t) = p^{\prime}(t) = 0$, meaning $p(t)$ has (at least) a double real root $t$. This, using (1), tells that the quadratic discriminant $= b^{2} - 2 a \geq 0$, and using (2), tells that the cubic discriminant $ = \frac{3a^{2}}{4}\ell^{2} + \frac{1}{2}\left(3ab + b^{3}\right)\ell + \left(\frac{2a}{3} - \frac{b^{2}}{4}\right) = 0$. Similarly, (3) and (4) tells that $q(t) = q^{\prime}(t) = 0$, meaning $q(t)$ has (at least) a double real root $t$, which in turn, yields two similar conditions on the quadratic discriminant of (3) and on the cubic discriminant of (4). It is not clear to me how to use the resulting relations together with (5).

3. Consider the quadratic (in $t$) equation obtained by subtracting (1) from (3), the cubic (in $t$) equation obtained by subtracting (2) from (4), and the original equation (5), which is linear in $t$. If we call the cubic (in $t$) LHS obtained by subtracting (2) from (4), as $r(t)$, then the resulting three equations are $r(t) = r^{\prime}(t) = r^{\prime\prime}(t) = 0$, which tells that the cubic $r(t)$ has a real root with multiplicity 3, with the root being $\frac{b-d}{a-c}$. Again, not sure how to go from here to recover the desired results.

Answer 1

Puting $t=0$ in Eq(1) and $t=1$ in Eq(3) shows us that both $0$ and $1$ are not possible values for t. So, dividing by $t$ and $t-1$ should be fine.

Solving Eq(1) and Eq(2) for $a$ and $b$ in terms of $l$ and $t$ we get

\begin{eqnarray*} a &=& \frac{6(t-2l)}{t^3} \\ b &=& \frac{2(2t-3l)}{t^2} \\ at - b &=& \frac{6t-12l-4t+6l}{t^2} = \frac{2(t-3l)}{t^2} \end{eqnarray*}

Solving Eq(3) and Eq(4) for $c$ and $d$ in terms of $l$ and $t$ we get

\begin{eqnarray*} c &=& \frac{-6\left(t-1+2l\right)}{(t-1)^3} \\ d &=& \frac{-2((2t+1)(t-1)+3l(t+1))}{(t-1)^3} \\ ct-d&=& -2\frac{\left[3t^2-3t+6lt-2t^2+t+1-3lt-3l)\right]}{(t-1)^3} \\ &=& -2\frac{t^2-2t+3lt-3l+1}{(t-1)^3} \\ &=& -2\frac{t-1+3l}{(t-1)^2} \end{eqnarray*}

Using Eq(5) and above results \begin{eqnarray*} \frac{(t-3l)}{t^2}&=&-\frac{t-1+3l}{(t-1)^2} \\ 3l(t^2-(t-1)^2) + t(t-1)(2t-1)&=& 0 \\ (3l-t(1-t))(2t-1) &=& 0 \end{eqnarray*}

If $t \ne \frac{1}{2}$ then we can see there are multiple solutions for the system but given $t\in(0,1)-{\frac{1}{2}}$, $l\in(0,\frac{1}{12})$ as $3l=t(1-t)$. But given the input condition that $l\in[\frac{1}{6},\frac{1}{4}]$, $3l$ cant be equal to $t(1-t)$ and hence $t=\frac{1}{2}$. The rest of the solution follows by substituting $t=\frac{1}{2}$ in the above equations.

Answer 2

Seems like a perfect opportunity to use computer algebra (such as Mathematica), where one imposes the range of values of $l$ explicitly:

Assuming[1/6 <= l <= 1/4, 
 Solve[{a t^2/2 - b t + 1 == 0, a t^3/6 - b t^2/2 + t - l == 0, 
   c t^2/2 - d t + (d - c/2 - 1) == 0, 
   c t^3/6 - d t^2/2 + (d - c/2 - 1) t + (c/3 - d/2 + 1 - l) == 0, 
   a t - b - c t + d == 0}, {a, b, c, d, t}]
 ]

$\left\{\left\{b\to \frac{1}{3} \left(-\frac{\sqrt{1-12 l}}{l}+\frac{1}{l}+6\right),a\to \frac{-6 \sqrt{1-12 l} l-\sqrt{1-12 l}+1}{9 l^2},d\to \frac{1}{9} \left(-\frac{\sqrt{1-12 l}}{l^2}-\frac{1}{l^2}-\frac{3 \sqrt{1-12 l}}{l}+\frac{3}{l}+18\right),c\to \frac{-6 \sqrt{1-12 l} l-\sqrt{1-12 l}-1}{9 l^2},t\to \frac{1}{2} \left(\sqrt{1-12 l}+1\right)\right\},\left\{b\to \frac{\sqrt{1-12 l}}{3 l}+\frac{1}{3 l}+2,a\to \frac{6 \sqrt{1-12 l} l+\sqrt{1-12 l}+1}{9 l^2},d\to \frac{\sqrt{1-12 l}}{9 l^2}-\frac{1}{9 l^2}+\frac{\sqrt{1-12 l}}{3 l}+\frac{1}{3 l}+2,c\to \frac{6 \sqrt{1-12 l} l+\sqrt{1-12 l}-1}{9 l^2},t\to \frac{1}{2} \left(1-\sqrt{1-12 l}\right)\right\},\left\{b\to -8 (3 l-1),a\to -24 (4 l-1),d\to 8 (9 l-2),c\to 24 (4 l-1),t\to \frac{1}{2}\right\}\right\}$

Answer 3

The Maple code

sol := solve({(1/2)*a*t^2-b*t+1 = 0, (1/6)*a*t^3-(1/2)*b*t^2+t-l = 0,
 (1/2)*c*t^2-d*t+d-(1/2)*c-1 = 0,
 (1/6)*c*t^3-(1/2)*d*t^2+(d-(1/2)*c-1)*t+(1/3)*c-(1/2)*d+1-l = 0, a*t-c*t-b+d = 0}, {a, b, c, d, t}, parametric = full):
    allvalues(sol);

does the job. See its huge nested output in this Maple 2015 file. The one can be viewed by free MaplePlayer.

Answer 4

Your statement, that l has to be between 1/6 and 1/4 for a real solution with t between 0 and 1, does not seem to be correct. As you wrote, let

t = 1/2, a = -96*l + 24, b = 8 - 24*l,  c = 96*l - 24,  d = 72*l - 16

Then this is a solution no matter what l is.

Furthermore, if t is arbitrary then this is a solution:

l = (t - t^2)/3, a = (4*t + 2) / t^2 , b = (2*t + 2) / t,
c = (4*t - 6)/(t^2 - 2*t + 1), d = (2*t^2 - 2*t - 2)/(t^2 - 2*t + 1)

For t between 0 and 1, l is between 0 and 1/12.