Let \(R[x,y,z]\) be the ring of the real polynomials in the variables x, y, and z. We say that \(h(x,y,z)\in R(x,y,z)\) is a Darboux polynomial of system (1.1) if it satisfies
$$ \frac{\partial h}{\partial x}P+\frac{\partial h}{\partial y}Q+\frac {\partial h}{\partial z}R=hL_{h} $$
(4.1)
for some polynomial \(L_{h}\), which is called the cofactor of \(h(x,y,z)\) [17–28]. If \(h(x,y,z)\) is a Darboux polynomial, then the surface \(h(x,y,z)=0\) is an invariant algebraic surface [18–21, 25, 28], which means that if an orbit of system (1.1) has a point on the surface \(h(x,y,z)=0\), then the whole orbit is contained in it. We assert categorically that the order k of \(L_{h}\) is at most to 1. This is because
$$\begin{aligned} &\operatorname{deg}(L_{h}) + \operatorname{deg}(h) \\ &\quad = \max\bigl\{ \operatorname{deg}(h)-1+ \operatorname{deg}(P),\operatorname{deg}(h)-1 + \operatorname{deg}(Q),\operatorname{deg}(h)-1 + \operatorname{deg}(R)\bigr\} \\ &\quad \leq \operatorname{deg}(h) + 1. \end{aligned}$$
(4.2)
Assume that the cofactor is
$$ L_{h}=b_{0}+b_{1}x+b_{2}y+b_{3}z, $$
then (4.1) becomes
$$ \frac{\partial h}{\partial x}P+\frac{\partial h}{\partial y}Q+\frac {\partial h}{\partial z}R=h(b_{0}+b_{1}x+b_{2}y+b_{3}z), $$
(4.3)
where \(b_{0},b_{1},b_{2},b_{3}\in R\).
Supposing that \(E =e^{\frac{g}{h}} \) is an exponential factor of (1.1). Thus, h, g both are Darboux polynomials of equations (1.1) and h, g are coprime [22–28]. If real-valued polynomial \(L_{e}\) is the cofactor of E, then
$$ \frac{\partial E}{\partial x}P+\frac{\partial E}{\partial y}Q+\frac {\partial E}{\partial z}R= EL_{e}. $$
(4.4)
If \(H(x,y,z)\) is a first integral [26, 27], then
$$ \frac{\partial H}{\partial x}P+\frac{\partial H}{\partial y}Q+\frac {\partial H}{\partial z}R= 0. $$
(4.5)
Theorem 4.1
When
\(r\neq0\), the following four conclusions are suitable for equations (1.1).
-
(1)
There are no Darboux polynomials with cofactors not equal to zero.
-
(2)
Their polynomial first integrals do not exist.
-
(3)
Their exponential factors cannot be found.
-
(4)
Their first integrals of Darboux type also do not exist.
Proof
Since there are four parts to the proof of Theorem 4.1, we consider two cases of \(p = 0\) and \(p \neq0\) for each of these four parts in turn.
Part (1). Let
$$ h(x,y,z)=\sum_{i=0}^{n}h_{i}, $$
(4.6)
where \(h_{i}=h_{i}(x,y,z)\) is a polynomial and its each item is of degree i. Assume \(h_{n}\neq0\), \(n\neq0\).
Collecting the terms of degree \(n+1\) in (4.3) yields
$$ xz\frac{\partial h_{n}}{\partial y}+g\bigl(mx^{2}-(1+m)xy\bigr)\frac{\partial h_{n}}{ \partial z}=h_{n}(b_{1}x+b_{2}y+b_{3}z). $$
(4.7)
Solving this equation for \(h_{n}\), we can get
$$\begin{aligned} \begin{aligned}[b] h_{n}(x,y,z)&=F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr) \\ &\quad {}\times\exp\biggl(b_{3}\frac{y}{x}-\frac{b_{1}}{\sqrt{g(1+m)}}k(x,y,z)-b_{2} \frac{z\sqrt{g(1+m)}+xgmk(x,y,z)}{\sqrt{g(1+m)}g(1+m)x}\biggr), \end{aligned} \end{aligned}$$
(4.8)
where F is a smooth function and
$$ k(x,y,z)=\arctan\biggl(\frac{\sqrt{g(1+m)}(mx-(1+m)y)}{(1+m)z}\biggr). $$
(4.9)
For the assumption of \(h_{n}(x,y,z)\), we have \(b_{1}=b_{2}=b_{3}=0\).
Let
$$ h_{n}(x,y,z)=x^{n-2p}\bigl(g(1+m)y^{2}-2gmxy+z^{2} \bigr)^{p}, $$
where the (nonnegative) integer p satisfies \(p\leq\frac{n}{2}\).
Calculating the terms of order n in (4.3), one can obtain
$$\begin{aligned}& xz\frac{\partial h_{n-1}}{\partial y}+gx\bigl(m(x-y)-y\bigr)\frac{\partial h_{n-1}}{ \partial z}+r(y-x) \frac{\partial h_{n}}{\partial x}+\bigl(m(x-y)-y\bigr) \frac{\partial h_{n}}{\partial y}-\frac{\partial h_{n}}{\partial z} fz \\& \quad =b_{0}h_{n}. \end{aligned}$$
(4.10)
Case 1. \(p=0\). Solving equation (4.10), one can obtain
$$\begin{aligned} h_{n-1}(x,y,z) =&F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr)-\frac{x^{n-1}k(x,y,z)}{ \sqrt{g(1+m)}}b_{0} \\ &{}-\frac{nrx^{n-2}(gxk(x,y,z)-\sqrt{g(1+m)}z)}{\sqrt{g(1+m)}g(1+m)}, \end{aligned}$$
where F is a smooth function. Considering the definition of polynomial, we must have \(n=0\), \(b_{0}=0\), which is impossible.
Case 2. \(p>0\). Solving equation (4.10), one can obtain
$$\begin{aligned} h_{n-1}(x,y,z) =&F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr)-b_{0}\frac{k(x,y,z)}{ x\sqrt{g(1+m)}}h_{n} \\ &{}-\frac{zy}{(1+m)^{2}}k_{1}k_{2}-k\frac{(1+m)rn+pk_{3}h_{n}+k_{1}k_{4}m^{2}x^{3}g}{ x\sqrt{gm+m}(1+m)^{2}} \\ &{}+\frac{z}{g(1+m)^{2}x^{2}}\bigl((m+1) (nr-2pr)h_{n}+k_{1}k_{5}gmx^{3} \bigr), \end{aligned}$$
where F is a smooth function and
$$\begin{aligned}& k_{1}(x,y,z) =px^{n-2p-1}\bigl(-2gmxy+(gm+g)y^{2}+z^{2} \bigr)^{p-1}, \\& k_{2}(x,y,z) =m^{3}+(-f+r+3)m^{2}+(-2f+r+3)m-f+1, \\& k_{3}(x,y,z) =m^{3}+(f+r+3)m^{2}+(2f-r+3)m+f-2r+1, \\& k_{4}(x,y,x) =m^{2}+(f+r+2)m+(f-2r+1), \\& k_{5}(x,y,z) =m^{2}+(-f-r+2)m+(-f+2r+1). \end{aligned}$$
Since \(h_{n-1}(x, y, z)\) is a polynomial, we obtain \(p=0\), \(n=0\), \(b_{0}=0\). And the case is again not possible.
Part (2). Let \(b_{0}=b_{1}=b_{2}=b_{3}=0\) in (4.3). Then h satisfies
$$ \frac{\partial h}{\partial x}P+\frac{\partial h}{\partial y}Q+ \frac{\partial h}{\partial z}R=0. $$
(4.11)
We still assume
$$ h=\sum_{i=0}^{n}h_{i}(x,y,z), $$
(4.12)
where \(h_{i}=h_{i}(x,y,z)\) is a polynomial and its each item is of degree i. Assume that \(h_{n}\neq0\) and \(n\neq0\).
Collecting terms of degree \(n+1\) in (4.11) yields
$$ xz\frac{\partial h_{n}}{\partial y}+g\bigl(mx^{2}-(1+m)xy\bigr)\frac{\partial h_{n}}{\partial z}=0. $$
(4.13)
Solving this equation for \(h_{n}\) gives
$$ h_{n}(x,y,z)=F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr), $$
where F is a smooth function.
Assume
$$ h_{n}(x,y,z)=x^{n-2p}\bigl(g(1+m)y^{2}-2gmxy+z^{2} \bigr)^{p}, $$
where p is a nonnegative integer and \(p\leq\frac{n}{2}\).
Terms of degree n in (4.11) are
$$ xz\frac{\partial h_{n-1}}{\partial y}+gx\bigl(m(x-y)-y\bigr)\frac{\partial h_{n-1}}{ \partial z}+r(y-x) \frac{\partial h_{n}}{\partial x}+\bigl(m(x-y)-y\bigr) \frac{\partial h_{n}}{\partial y}-\frac{\partial h_{n}}{\partial z}fz=0. $$
(4.14)
Case 1. Solving equation (4.14) for \(p=0\), we can give
$$\begin{aligned} h_{n-1}(x,y,z) =&F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr) \\ &{}-\frac{nrx^{n-2}(gxk(x,y,z)-\sqrt{g(1+m)}z)}{\sqrt{g(1+m)}g(1+m)}, \end{aligned}$$
where F is a smooth function. Since the expression of \(h_{n-1}(x, y, z)\), it forces \(n=0\), which is illogicality, because one has made the assumption that \(n\neq0\). This hypothesis therefore does not hold.
Case 2. Solving equation (4.14) for \(p>0\), we can give
$$\begin{aligned} h_{n-1}(x,y,z) =&F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr)-\frac{k_{7}(x,y,z)yz}{ (1+m)^{2}} \\ &{}+\frac{(m+1)(n-2p)rz}{gx^{2}(1+m)^{2}}h_{n}+\frac {mxzg_{5}k_{5}(x,y,z)}{(1+m)^{2}} \\ &{}-\frac{k(x,y,z)(((m+1)rn+k_{6}(x,y,z)p)h_{n}+gm^{2}x^{3}k_{4}(x,y,z)g_{5})}{ x\sqrt{gm+g}(1+m)^{2}} \end{aligned}$$
for a smooth function F. Also
$$\begin{aligned}& k_{6}(x,y,z)=m^{3}+1+m^{2}r-mr-2r+(1+m)^{2}f+3m(m+1), \\& k_{7}(x,y,z)=m^{3}+(-f+r+3)m^{2}+(-2f+r+3)m-f+1. \end{aligned}$$
Thus, \(p=0\), \(n=0\), there is no possibility, because in the above, one has assumed that \(n\neq0\). So this hypothesis again does not hold.
Part (3). By Theorem 4.1(1), one can make an assumption that equations (1.1) have an exponential factor \(E=x^{u}\) with \(u\in R[x,y,z]\). Suppose that the cofactor \(L=b_{0}+b_{1}x+b_{2}y+b_{3}z\) with \(b_{i}\in R\) for \(i=0,1,2,3\).
Obviously, by (4.4), u is the solution of equation (4.15), that is to say,
$$ \frac{\partial u}{\partial x}P+\frac{\partial u}{\partial y}Q+\frac {\partial u}{\partial z}R=b_{0}+b_{1}x+b_{2}y+b_{3}z. $$
(4.15)
We can write u as
$$ u=\sum_{i=0}^{n}u_{i}(x,y,z), $$
(4.16)
where each \(u_{i}=u_{i}(x,y,z)\) is homogeneous of degree i. Supposing that \(u_{n}\neq0\) and \(n\neq0\) again, calculating terms of order \(n+1\) in (4.15), we can have
$$ xz\frac{\partial u_{n}}{\partial y}+g\bigl(mx^{2}-(1+m)xy\bigr)\frac{\partial u_{n}}{ \partial z}=0. $$
(4.17)
Solving this equation for \(u_{n}\) gives
$$ u_{n}(x,y,z)=F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr) $$
for a smooth function F.
Assume
$$ u_{n}(x,y,z)=x^{n-2p}\bigl(g(1+m)y^{2}-2gmxy+z^{2} \bigr)^{p}, $$
where p is a nonnegative integer and \(p\leq\frac{n}{2}\).
Calculating terms of order n in (4.15) gets
$$ xz\frac{\partial u_{n-1}}{\partial y}+gx\bigl(m(x-y)-y\bigr)\frac{\partial u_{n-1}}{ \partial z}+r(y-x) \frac{\partial u_{n}}{\partial x}+\bigl(m(x-y)-y\bigr) \frac{\partial u_{n}}{\partial y}-\frac{\partial u_{n}}{\partial z}fz=0. $$
(4.18)
Case 1. When \(p=0\), equation (4.18) yields
$$\begin{aligned} u_{n-1}(x,y,z) =&F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr) \\ &{}-\frac{nrx^{n-2}(gxk(x,y,z)-\sqrt{g(1+m)}z)}{\sqrt{g(1+m)}g(1+m)} \end{aligned}$$
for a smooth function F. Thus, one can obtain that \(n=0\), which is also impossible for \(n\neq0\) in the assumption. This hypothesis cannot hold.
Case 2. Solving equation (4.18) for \(p > 0\) yields
$$\begin{aligned} u_{n-1}(x,y,z) =&F\bigl(x,-2gmxy+g(m+1)y^{2}+z^{2} \bigr)-\frac {k_{7}(x,y,z)yz}{(1+m)^{2}} \\ &{}+\frac{(m+1)(n-2p)rz}{gx^{2}(1+m)^{2}}u_{n}+\frac {mxzg_{5}k_{5}(x,y,z)}{(1+m)^{2}} \\ &{}-\frac{k(x,y,z)(((m+1)rn+k_{6}(x,y,z)p)u_{n}+gm^{2}x^{3}k_{4}(x,y,z)g_{5})}{ x\sqrt{gm+g}(1+m)^{2}}, \end{aligned}$$
where F is a smooth function. Since \(u_{n-1}(x,y,z)\) is a polynomial, we get \(p=0\), \(n=0\), contradicting the assumption that \(n\neq0\). This hypothesis again does not hold.
Part (4). By the arguments of Part (3) above, together with the definition of the exponential factor, Part (4) follows. □