Positive Semi-Definite and Sum of Squares Biquadratic Polynomials

Abstract

Hilbert proved in 1888 that a positive semi-definite (PSD) homogeneous quartic polynomial of three variables always can be expressed as the sum of squares (SOS) of three quadratic polynomials, and a psd homogeneous quartic polynomial of four variables may not be sos. Only after 87 years, in 1975, Choi gave the explicit expression of such a psd-not-sos (PNS) homogeneous quartic polynomial of four variables. An $m\times n$ biquadratic polynomial is a homogeneous quartic polynomial of $m+n$ variables. In this paper, we show that an $m\times n$ biquadratic polynomial can be expressed as a tripartite homogeneous quartic polynomial of $m+n-1$ variables. Therefore, by Hilbert’s theorem, a $2\times 2$ PSD biquadratic polynomial can be expressed as the sum of squares of three quadratic polynomials. This improves the result of Calderón in 1973, who proved that a $2\times 2$ biquadratic polynomial can be expressed as the sum of squares of nine quadratic polynomials. Furthermore, we present a necessary and sufficient condition for an $m\times n$ psd biquadratic polynomial to be sos, and show that if such a polynomial is sos, then its sos rank is at most $mn$. Then we give a constructive proof of the sos form of a $2\times 2$ psd biquadratic polynomial in three cases.

1. Introduction

In 1888, when he was 26, David Hilbert [1] pointed out that for homogeneous polynomials, only in the following three cases, a positive semi-definite (PSD) polynomial definite is a sum of squares (SOS) polynomial: (i) $N=2$; (ii) $M=2$; (iii) $M=4$; and $N=3$, where M is the degree of the polynomial and N is the number of variables. Hilbert proved that in all the other possible combinations of $M=2K$ and N, there are PSD non-SOS (PNS) homogeneous polynomials. Furthermore, Hilbert established that every PSD homogeneous quartic polynomial in three variables admits a representation as an SOS of three quadratic forms. His original proof of this result, demonstrating the equivalence between PSD and SOS for ternary quartics, combines striking conciseness with pioneering topological methods that were remarkably advanced for their era, while mathematically elegant, the proof’s innovative approach proved difficult to fully comprehend, and even modern readers must carefully work through several non-obvious details. Recent literature has witnessed both comprehensive reconstructions of Hilbert’s original proof and the emergence of novel alternative proofs. For instance, in 2012, Pfister and Scheiderer [2] presented a completely new approach that uses only elementary techniques.

While Hilbert established in 1888 that PSD polynomials coincide with SOS in only three specific cases, the first explicit example of a PNS homogeneous polynomial was not constructed until 77 years later, in 1965, when Motzkin [3] gave the first PNS homogeneous polynomial:

$$f_M\left(\mathbf{x}\right)=x_3^6+x_1^2{x}_2^4+x_1^4{x}_2^2-3x_1^2{x}_2^2{x}_3^2,$$

where $M=6$ and $N=3$. Thus, the remaining case with small M and N is that $M=N=4$. In 1977, Choi and Lam [4] gave such a PNS polynomial:

$$f_{CL}\left(\mathbf{x}\right)=x_1^2{x}_2^2+x_1^2{x}_3^2+x_2^2{x}_3^2-4x_1{x}_2{x}_3{x}_4+x_4^4.$$

(1)

Also, see [5,6] for PNS homogeneous polynomials. In 2017, Goel et al. [7] showed that a symmetric PSD polynomial with $M=N=4$ is SOS. This raises a natural question: Do analogous results hold for nonsymmetric quartic forms under additional structural constraints?

In this paper, we study the PSD-SOS problem of a special class of homogeneous quartic polynomials: biquadratic polynomials. In general, an $m\times n$ biquadratic polynomial can be expressed as follows:

$$f(\widehat{\mathbf{x}},\widehat{\mathbf{y}})=\sum _{i,j=1}^m\sum _{k,l=1}^n{a}_{ijkl}{x}_i{x}_j{y}_k{y}_l,$$

(2)

where $\widehat{\mathbf{x}}={(x_1,\dots ,x_m)}^{\top }$ and $\widehat{\mathbf{y}}={(y_1,\dots ,y_n)}^{\top }$. Biquadratic polynomials and biquadratic tensors arise in solid mechanics, statistics, quantum physics, spectral graph theory, and polynomial theory [8]. Biquadratic SOS decompositions enable efficient solutions to non-convex optimization problems via SDP relaxations. In the analysis of nonlinear dynamical systems, stability can be rigorously verified through the construction of Lyapunov functions that admit an SOS decomposition.

Without loss of generality, we may assume that $m\ge n\ge 2$. Thus, an $m\times n$ biquadratic polynomial is a quartic homogeneous polynomial of $m+n$ variables, i.e., $M=4$ and $N=m+n$. However, we will show that for the PSD-SOS problem, this value of $m+n$ can be reduced by one. In Section 2, we show that for an $m\times n$ biquadratic polynomial f, there exists a special quartic homogeneous polynomial h, which we call a tripartite quartic polynomial, with $N=m+n-1$ variables, such that f is PSD if and only if h is PSD, and f is SOS if and only if h is SOS. Then, according to Hilbert’s result, we immediately conclude that a $2\times 2$ PSD biquadratic polynomial can be expressed as the sum of squares of three quadratic polynomials. This greatly improves the result of Calderón [9] in 1973, who proved that a $2\times 2$ biquadratic polynomial can be expressed as the sum of squares of nine quadratic polynomials. Furthermore, we present a necessary condition for a tripartite quartic polynomial to be PSD, and a necessary and sufficient condition for it to be PSD under the condition that the coefficient of the quartic power term of the flexible variable is zero.

In Section 3, we give a sufficient and necessary condition for an $m\times n$ PSD biquadratic polynomial to be SOS. We show that if an $m\times n$ PSD biquadratic polynomial is SOS, than its SOS rank is at most $mn$.

For a $2\times 2$ PSD biquadratic polynomial, we may explicitly construct its SOS representation in some cases. A $2\times 2$ biquadratic polynomial f has four half-cross terms and one full-cross term. In Section 4, we show that for a $2\times 2$ PSD biquadratic polynomial f, there is another $2\times 2$ PSD biquadratic polynomial h, which has at most two half-cross terms, such that if h has an SOS form, then the SOS form of f can be explicitly constructed. Then we give a constructive proof of the SOS form of a $2\times 2$ PSD biquadratic polynomial f, if it has either at most one half-cross term, or two half-cross terms but no full-cross term.

Some final remarks and open questions are made and raised in Section 5.

2. Biquadratic Polynomials and Tripartite Quartic Polynomials

The following theorem is one main result of this paper.

Theorem 1. 

Suppose that we have an $m\times n$ biquadratic polynomial f, expressed by (2), with $m\ge n\ge 2$. Then there is a tripartite homogeneous quartic polynomial $h(\mathbf{x},\mathbf{y},z)$ of $m+n-1$ variables, where $\mathbf{x}={(x_1,\dots ,x_{m-1})}^{\top }$ and $\mathbf{y}={(y_1,\dots ,y_{n-1})}^{\top }$, such that f is PSD if and only if h is PSD, and f is SOS if and only if h is SOS. In particular, a $2\times 2$ PSD biquadratic polynomial is always SOS of three quadratic polynomials.

Proof. 

Let $x_m=1$ and $y_n=1$ for $j=1,\dots ,n-1$. Then we have a non-homogeneous quartic polynomial $g(\mathbf{x},\mathbf{y})$, of $m+n-2$ variables $x_1,\dots ,x_{m-1}$, $y_1,\dots ,y_{n-1}$, such that f is PSD if and only if g is PSD, and f is SOS if and only if g is SOS.

Now, we may convert g to a homogeneous quartic polynomial h of $m+n-1$ variables, by adding an additional z in the position of $x_m$ and $y_n$, i.e., h is obtained by replacing $x_m$ and $y_n$ by z. Then g is PSD if and only if h is PSD, and g is SOS if and only if h is SOS. The final conclusion now follows from the 1888 Hilbert result [1]. The proof is complete. □

We call $h(\mathbf{x},\mathbf{y},z)$ a tripartite homogeneous quartic polynomial, as it has the following property: in any term of h, the total degree of x-variables is at most 2, and the total degree of y-variables is also at most 2. We call z the flexible variable. Only the flexible variable can have degree four in h.

A tripartite quartic homogeneous polynomial $h(\mathbf{x},\mathbf{y},z)$ can be written as follows:

$$h(\mathbf{x},\mathbf{y},z)=h_0{z}^4+h_1(\mathbf{x},\mathbf{y})z^3+h_2(\mathbf{x},\mathbf{y})z^2+h_3(\mathbf{x},\mathbf{y})z+h_4(\mathbf{x},\mathbf{y}),$$

(3)

where $h_0$ is a constant coefficient; $h_4(\mathbf{x},\mathbf{y})$ is an $(m-1)\times (n-1)$ biquadratic polynomial; $h_1(\mathbf{x},\mathbf{y})$ is a linear homogeneous polynomial of $(\mathbf{x},\mathbf{y})$; $h_2(\mathbf{x},\mathbf{y})$ is the sum of a quadratic homogeneous polynomial of $\mathbf{x}$, a quadratic homogeneous polynomial of $\mathbf{y}$, and a bilinear polynomial of $(\mathbf{x},\mathbf{y})$; and $h_3(\mathbf{x},\mathbf{y})$ is the sum of a linear quadratic polynomial and a quadratic-linear polynomial of $(\mathbf{x},\mathbf{y})$. We have the following theorem.

Theorem 2. 

Suppose that we have a tripartite quartic polynomial $h(\mathbf{x},\mathbf{y},z)$, expressed by (3), where $\mathbf{x}\in {\mathbb{R}}^{m-1},\mathbf{y}\in {\mathbb{R}}^{n-1}$, $z\in \mathbb{R}$ and $m\ge n\ge 2$. Then there is an $m\times n$ biquadratic polynomial $f(\widehat{\mathbf{x}},\widehat{\mathbf{y}})$, where $\widehat{\mathbf{x}}={(x_1,\dots ,x_m)}^{\top }$ and $\widehat{\mathbf{y}}={(y_1,\dots ,y_n)}^{\top }$, such that h is PSD if and only if f is PSD, and h is SOS if and only if f is SOS.

Proof. 

In (3), let $z=1$. Then we have a non-homogeneous quartic polynomial $g(\mathbf{x},\mathbf{y})$, of $m+n-2$ variables $x_1,\dots ,x_{m-1},y_1,\dots ,y_{n-1}$, such that h is PSD if and only if g is PSD, and h is SOS if and only if g is SOS.

Now, we may convert g to an $m\times n$ biquadratic polynomial $f(\widehat{\mathbf{x}},\widehat{\mathbf{y}})$, by adding an additional variable $x_m$ or $x_m^2$, if needed, to terms of g to make the total degrees of x-variables of all terms of g as 2, and an additional variable $y_n$ or $y_n^2$, if needed, to terms of g to make the total degrees of y-variables of all terms of g as 2. Then g is PSD if and only if f is PSD, and g is SOS if and only if f is SOS. The proof is complete. □

Thus, the PSD-SOS problem of $m\times n$ biquadratic polynomials is equivalent to the PSD-SOS problem of tripartite quartic polynomials of $m+n-1$ variables.

To consider the PSD-SOS problem of tripartite quartic polynomials, we have the following theorem.

Theorem 3. 

Suppose that $h(\mathbf{x},\mathbf{y},z)$ is a tripartite quartic polynomial, expressed by (3). If $h(\mathbf{x},\mathbf{y},z)$ is PSD, then $h_0\ge 0$ and $h_4(\mathbf{x},\mathbf{y})$ is PSD.

Furthermore, if $h_0=0$, then the tripartite quartic polynomial $h(\mathbf{x},\mathbf{y},z)$ is PSD if and only if (i) $h_1(\mathbf{x},\mathbf{y})\equiv 0$, (ii) $h_2(\mathbf{x},\mathbf{y})$, $h_4(\mathbf{x},\mathbf{y})$ and $4h_2(\mathbf{x},\mathbf{y})h_4(\mathbf{x},\mathbf{y})-h_3{(\mathbf{x},\mathbf{y})}^2$ are PSD.

Proof. 

Assume that h is PSD. Let x-variables and y-variables be zero and $z=1$. Since h is PSD, we have $h_0\ge 0$. Let $z=0$. Since h is PSD, $h_4(\mathbf{x},\mathbf{y})$ should be PSD.

Suppose now that $h_0=0$.

First, we assume that h is PSD. Since $h_1(\mathbf{x},\mathbf{y})$ is a linear homogeneous polynomial of $(\mathbf{x},\mathbf{y})$, if $h_1(\mathbf{x},\mathbf{y})\not\equiv 0$, we may choose adequate values of $(\mathbf{x},\mathbf{y})$ such that $h_1$ takes a negative value. Let z goes to infinity. Then, we have $h(\mathbf{x},\mathbf{y},z)<0$, contradicting with the assumption that f is PSD. Thus, $h_1(\mathbf{x},\mathbf{y})\equiv 0$. Similarly, if $h_0=0$, we can show that $h_2(\mathbf{x},\mathbf{y})$ is PSD. Furthermore, if $4h_2(\mathbf{x},\mathbf{y})h_4(\mathbf{x},\mathbf{y})-h_3{(\mathbf{x},\mathbf{y})}^2$ is not PSD, then there are some values $({\mathbf{x}}^{*},{\mathbf{y}}^{*})$ such that the function $4h_2({\mathbf{x}}^{*},{\mathbf{y}}^{*})h_4({\mathbf{x}}^{*},{\mathbf{y}}^{*})-h_3{({\mathbf{x}}^{*},{\mathbf{y}}^{*})}^2<0$. Then, $h({\mathbf{x}}^{*},{\mathbf{y}}^{*},z)$ is an indeterminate quadratic polynomial of z. There is a value $z^{*}$ such that $h({\mathbf{x}}^{*},{\mathbf{y}}^{*},z^{*})<0$, contradicting to the assumption that h is PSD. Thus, $4h_2(\mathbf{x},\mathbf{y})h_4(\mathbf{x},\mathbf{y})-h_3{(\mathbf{x},\mathbf{y})}^2$ must be PSD.

On the other hand, assume that conditions (i) and (ii) hold. Then for any value $({\mathbf{x}}^{*},{\mathbf{y}}^{*})$, $h({\mathbf{x}}^{*},{\mathbf{y}}^{*},z)$ is a PSD quadratic polynomial of z. Then we always have $h(\mathbf{x},\mathbf{y},z)\ge 0$, i.e., h is PSD. The proof is complete. □

Thus, to consider the PSD-SOS problem of tripartite quartic polynomials, we only need to investigate two types of tripartite quartic polynomials. One type is that $h_0=1$ and $h_4(\mathbf{x},\mathbf{y})$ is PSD in (3). We call such a tripartite quartic polynomial a nondegenerated tripartite quartic polynomial. Another type is that $h_0=0,h_1(\mathbf{x},\mathbf{y})\equiv 0$, $h_2(\mathbf{x},\mathbf{y})$, $h_4(\mathbf{x},\mathbf{y})$ and $4h_2(\mathbf{x},\mathbf{y})h_4(\mathbf{x},\mathbf{y})-h_3{(\mathbf{x},\mathbf{y})}^2$ are PSD in (3). We call such a tripartite quartic polynomial a degenerated tripartite quartic polynomial.

In fact, $h_0=0$ holds if and only if $a_{mnmn}=0$. Moreover, in the construction of the function g in Theorem 1, the choice of indices $i\in \{1,\dots ,m\}$ and $j\in \{1,\dots ,n\}$ is arbitrary. Consequently, h is degenerated whenever there exist i and j such that $a_{ijij}=0$.

Note that a degenerated tripartite quartic polynomial is PSD. A natural question is as follows: Is a degenerated tripartite quartic polynomial always SOS?

3. SOS Biquadratic Polynomials

Let $\mathcal{A}=\left(a_{ijkl}\right)$ be an $m\times n$ symmetric biquadratic tensor [10] that satisfies $a_{ijkl}=a_{klij}$. We can rewrite the $m\times n$ biquadratic polynomial as follows:

$$\begin{array}{c}\hfill f(\mathbf{x},\mathbf{y})=\sum _{i,k=1}^m\sum _{j,l=1}^n{a}_{ijkl}{x}_i{y}_j{x}_k{y}_l={\mathbf{z}}^T(\begin{array}{c}\hfill B\end{array}+P\left(\mathrm{\Gamma }\right))\mathbf{z}:={\mathbf{z}}^{T}M\left(\mathrm{\Gamma }\right)\mathbf{z},\end{array}$$

(4)

where $\mathbf{z}=\mathbf{x}\otimes \mathbf{y}\in {\mathbb{R}}^{mn}$, the matrix $B=\left(b_{st}\right)\in {\mathbb{R}}^{mn\times mn}$ is obtained by flattening the fourth-order tensor $\mathcal{A}$ into a two-dimensional representation, where the first two dimensions of the tensor are arranged as matrix rows and the last two dimensions as matrix columns. Namely,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill B\end{array}\end{array}=\left[\begin{array}{cccc}{\begin{array}{c}\hfill B\end{array}}_{11}& {\begin{array}{c}\hfill B\end{array}}_{12}& \dots & {\begin{array}{c}\hfill B\end{array}}_{1m}\\ {\begin{array}{c}\hfill B\end{array}}_{21}& {\begin{array}{c}\hfill B\end{array}}_{22}& \dots & {\begin{array}{c}\hfill B\end{array}}_{2m}\\ \dots & \dots & \dots & \dots \\ {\begin{array}{c}\hfill B\end{array}}_{m1}& {\begin{array}{c}\hfill B\end{array}}_{m2}& \dots & {\begin{array}{c}\hfill B\end{array}}_{mm}\end{array}\right],\begin{array}{c}\hfill B_{ik}\end{array}=\left[\begin{array}{cccc}{a}_{i1k1}& a_{i1k2}& \dots & a_{i1kn}\\ a_{i2k1}& a_{i2k2}& \dots & a_{i2kn}\\ \dots & \dots & \dots & \dots \\ a_{ink1}& a_{ink2}& \dots & a_{inkn}\end{array}\right].$$

Here, each element is given by $b_{st}=a_{ijkl}=a_{klij}$, where $s=(i-1)n+j$ and $t=(k-1)n+l$.

While the coefficients of diagonal (where $i=k$ and $j=l$) and half-cross (where $i=k$ or $j=l$) terms are uniquely determined by the symmetry of B, the full-cross terms (where $i\ne k$ and $j\ne l$) exhibit non-uniqueness, as evidenced by the identity:

$$\begin{array}{ccc}& & \left(a_{ijkl}+a_{klij}+a_{kjil}+a_{ijkl}\right)x_i{y}_j{x}_k{y}_l\hfill \\ & =& 2\left(a_{ijkl}+\gamma \right)x_i{y}_j{x}_k{y}_l+2\left(a_{kjil}-\gamma \right)x_i{y}_j{x}_k{y}_l\hfill \end{array}$$

for any $\gamma \in \mathbb{R}$. Specifically, let us define the index mappings $s_1=(i-1)n+j$, $t_1=(k-1)n+l$, and $s_2=(k-1)n+j$, $t_2=(i-1)n+l$. The quadratic form remains invariant under the transformation $b_{s_1{t}_1}\to b_{s_1{t}_1}+\gamma$ and $b_{s_2{t}_2}\to b_{s_2{t}_2}-\gamma$. Therefore, we use $\mathrm{\Gamma }=\left({\gamma }_{\begin{array}{c}\hfill t_i{t}_j\end{array}}\right)\in {\mathbb{R}}^{\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2}}\right)\times \left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}$ to represent a free parameter that captures the degrees of freedom in expressing the full cross terms, and construct $P\left(\mathrm{\Gamma }\right)=\left(P_{ij}\left(\mathrm{\Gamma }\right)\right)\in {\mathbb{R}}^{mn\times mn}$ as follows:

$$\begin{array}{c}\hfill P_{ij}\left(\mathrm{\Gamma }\right)=\left\{\begin{array}{cc}{\gamma }_{\begin{array}{c}\hfill t_i{t}_j\end{array}},\hfill & \mathrm{if}\exists i_1<i_2,j_1<j_2\mathrm{s}.\mathrm{t}.i=n_{i_1{j}_1},j=n_{i_2{j}_2},\begin{array}{c}\hfill t_i\end{array}=t_{i_1{i}_2},\begin{array}{c}\hfill t_j\end{array}=t_{j_1{j}_2};\hfill \\ {\gamma }_{\begin{array}{c}\hfill t_i{t}_j\end{array}},\hfill & \mathrm{if}\exists i_1>i_2,j_1>j_2\mathrm{s}.\mathrm{t}.i=n_{i_1{j}_1},j=n_{i_2{j}_2},\begin{array}{c}\hfill t_i\end{array}=t_{i_1{i}_2},\begin{array}{c}\hfill t_j\end{array}=t_{j_1{j}_2};\hfill \\ -{\gamma }_{\begin{array}{c}\hfill t_i{t}_j\end{array}},\hfill & \mathrm{if}\exists i_1<i_2,j_1>j_2\mathrm{s}.\mathrm{t}.i=n_{i_1{j}_1},j=n_{i_2{j}_2},\begin{array}{c}\hfill t_i\end{array}=t_{i_1{i}_2},\begin{array}{c}\hfill t_j\end{array}=t_{j_1{j}_2};\hfill \\ -{\gamma }_{\begin{array}{c}\hfill t_i{t}_j\end{array}},\hfill & \mathrm{if}\exists i_1>i_2,j_1<j_2\mathrm{s}.\mathrm{t}.i=n_{i_1{j}_1},j=n_{i_2{j}_2},\begin{array}{c}\hfill t_i\end{array}=t_{i_1{i}_2},\begin{array}{c}\hfill t_j\end{array}=t_{j_1{j}_2};\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

Here, $\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2}}\right)={\displaystyle \frac{m(m-1)}{2}}$, $n_{i_1{j}_1}=(i_1-1)n+j_1$, $t_{i_1{i}_2}={\displaystyle \frac{(i_1-1)(i_1-2)}{2}}+i_2$ if $i_1>i_2$ and $t_{i_1{i}_2}={\displaystyle \frac{(i_2-1)(i_2-2)}{2}}+i_1$ if $i_1<i_2$.

For instance, when $m=n=2$, it holds that $\mathbf{z}={[x_1{y}_1,x_1{y}_2,x_2{y}_1,x_2{y}_2]}^{\top }$,

$$B=\left[\begin{array}{cccc}{a}_{1111}& a_{1112}& a_{1121}& a_{1122}\\ a_{1211}& a_{1212}& a_{1221}& a_{1222}\\ a_{2111}& a_{2112}& a_{2121}& a_{2122}\\ a_{2211}& a_{2212}& a_{2221}& a_{2222}\end{array}\right]\mathrm{and}P\left(\gamma \right)=\left[\begin{array}{cccc}& & & \gamma \\ & & -\gamma & \\ & -\gamma & & \\ \gamma & & \end{array}\right].$$

Then we have the following theorem.

Theorem 4. 

Suppose that the $m\times n$ biquadratic polynomial f expressed by (4) is PSD. Then the following three statements are equivalent.

(i) 

f is SOS, with its SOS rank at most $mn$, and the coefficients can be orthogonal to each other;

(ii) 

there exists $\mathrm{\Gamma }=\left({\gamma }_{\begin{array}{c}\hfill t_i{t}_j\end{array}}\right)\in {\mathbb{R}}^{\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2}}\right)\times \left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}$ such that $M\left(\mathrm{\Gamma }\right)$ is PSD;

(iii) 

the following condition holds,

$$\underset{\mathrm{\Gamma }\in {\mathbb{R}}^{\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2}}\right)\times \left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}}{max}\underset{{\mathbf{z}}^{\top }\mathbf{z}=1}{min}{\mathbf{z}}^{\top }M\left(\mathrm{\Gamma }\right)\mathbf{z}\ge 0.$$

(5)

Proof.$\left(i\right)⟹\left(ii\right)$ If f is SOS, then there exists $T>0$ such that $f={\sum }_{t=1}^T{f}_t^2(\mathbf{x},\mathbf{y})$. By Theorem 1 in [6], we obtain that $\mathrm{supp}\left(f_t\right)\subseteq {\displaystyle \frac{1}{2}}\mathrm{New}\left(f\right)$, $\forall t=1,\dots ,T$. Here, New $\left(f\right)$ is the Newton polytope of f. Therefore, we deduce that each $f_t$ must be of the linear form $f_t={\mathbf{c}}_t^{\top }\mathbf{z}$, where ${\mathbf{c}}_t$ is a coefficient vector and $\mathbf{z}=\mathbf{x}\otimes \mathbf{y}$. Let $C=[{\mathbf{c}}_1,\dots ,{\mathbf{c}}_T]\in {\mathbb{R}}^{mn\times T}$ denote the coefficient matrix. This immediately gives the quadratic representation $f={\mathbf{z}}^{\top }C{C}^{\top }\mathbf{z}$. Consequently, there exists $\mathrm{\Gamma }$ such that $M\left(\mathrm{\Gamma }\right)=C{C}^{\top }$ is PSD.

$\left(ii\right)⟹\left(i\right)$ Since $M\left(\mathrm{\Gamma }\right)$ is PSD, it admits a Cholesky decomposition. Specifically, there exists a matrix $C\in {\mathbb{R}}^{mn\times mn}$ such that $M\left(\mathrm{\Gamma }\right)=C{C}^{\top }$. Therefore, $f={\sum }_{t=1}^{mn}{\left({\mathbf{c}}_t^{\top }\mathbf{z}\right)}^2$ is an SOS of tetranomials, with SOS rank at most $mn$, and the coefficients are orthogonal to each other.

$\left(ii\right)⟺\left(iii\right)$ Since $M\left(\mathrm{\Gamma }\right)$ is symmetric, its positive semidefiniteness is equivalent to the nonnegativity of all its eigenvalues. In particular, $M\left(\mathrm{\Gamma }\right)$ is PSD if and only if its smallest eigenvalue ${\lambda }_{min}={min}_{{\mathbf{z}}^{\top }\mathbf{z}=1}{\mathbf{z}}^{\top }M\left(\mathrm{\Gamma }\right)\mathbf{z}$ is nonnegative.

This completes the proof. □

Next, we use the $2\times 2$ example in page 358 of Qi et al. [10] to demonstrate that B is not PSD, yet there exists $\gamma$ such that $f(\mathbf{x},\mathbf{y})$ is SOS.

Example 1. 

Given a $2\times 2$ biquadratic tensor $\mathcal{A}$ with the following components: $a_{1111}=1$, $a_{1112}=2$, $a_{1122}=4$, $a_{1212}=12$, $a_{2121}=12$, $a_{1222}=1$, $a_{1121}=2$, $a_{2122}=1$, and $a_{2222}=2$. The associated biquadratic polynomial can be expressed as follows:

$$\begin{array}{ccc}\hfill f(\mathbf{x},\mathbf{y})& =& x_1^2{y}_1^2+4x_1^2{y}_1{y}_2+12x_1^2{y}_2^2+4x_1{x}_2{y}_1^2+16x_1{x}_2{y}_1{y}_2\hfill \\ & & +2x_1{x}_2{y}_2^2+12x_2^2{y}_1^2+2x_2^2{y}_1{y}_2+2x_2^2{y}_2^2.\hfill \end{array}$$

Then we can obtain

$$B=\left[\begin{array}{cccc}{a}_{1111}& a_{1112}& a_{1121}& a_{1122}\\ a_{1211}& a_{1212}& a_{1221}& a_{1222}\\ a_{2111}& a_{2112}& a_{212}& a_{2122}\\ a_{2211}& a_{2212}& a_{2221}& a_{2222}\end{array}\right]=\left[\begin{array}{cccc}1& 2& 2& 4\\ 2& 12& 4& 1\\ 2& 4& 12& 1\\ 4& 1& 1& 2\end{array}\right].$$

The matrix B fails to be PSD, as evidenced by its smallest eigenvalue of $-2.6110$. However, by selecting $\gamma =-3$, we obtain a modified matrix $M\left(\gamma \right)$ that is PSD, with its smallest eigenvalue now being $M\left(\gamma \right)$ is $0.2069$. This positive semidefiniteness is further confirmed by the existence of the following Cholesky decomposition

$$M\left(\gamma \right)=\left[\begin{array}{cccc}1& 2& 2& 1\\ 2& 12& 7& 1\\ 2& 7& 12& 1\\ 1& 1& 1& 2\end{array}\right]=C{C}^{\top },\mathit{with}C=\left[\begin{array}{cccc}1& 0& 0& 0\\ 2& 2.8284& 0& 0\\ 2& 1.0607& 2.6220& 0\\ 1& -0.3536& -0.2384& 0.9045\end{array}\right]$$

Consequently, $f(\mathbf{x},\mathbf{y})$ admits the following SOS expression

$$\begin{array}{ccc}\hfill f(\mathbf{x},vy)& =& {(x_1{y}_1+2x_1{y}_2+2x_2{y}_1+x_2{y}_2)}^2+{(2.8284x_1{y}_2+1.0607x_2{y}_1-0.3536x_2{y}_2)}^2\hfill \\ & & +{(2.6220x_2{y}_1-0.2384x_2{y}_2)}^2+0.9045x_2{y}_2^2.\hfill \end{array}$$

Theorem 4 shows that if an $m\times n$ PSD biquadratic polynomial is SOS, then its SOS rank is at most $mn$, and the coefficients can be orthogonal to each other.

It may be insightful to examine f in relation to the PSDness of $M\left(\mathrm{\Gamma }\right)$. Define

$$S_{mn}=\{\mathbf{z}:\mathbf{z}=\mathbf{x}\otimes \mathbf{y},\mathbf{x}\in {\mathbb{R}}^m,\mathbf{y}\in {\mathbb{R}}^n\}.$$

(6)

Suppose that f is PSD, then there exists $\mathrm{\Gamma }$ such that ${\mathbf{z}}^{\top }M\left(\mathrm{\Gamma }\right)\mathbf{z}\ge 0$ for all $\mathbf{z}\in S_{mn}$. Denote ${\mathbf{e}}_i\left(mn\right)$ as the i-th standard basis in ${\mathbb{R}}^{mn}$. It follows from ${\mathbf{e}}_1\left(mn\right),\dots ,{\mathbf{e}}_{mn}\left(mn\right)\in S_{mn}$ that $\dim \left(S_{mn}\right)=mn$. However, $S_{mn}$ is not a convex set. Consequently, the PSD property of f does not necessarily guarantee that f is SOS, in general.

4. $\mathbf{2}\times \mathbf{2}$ Biquadratic Polynomials

Given a $2\times 2$ biquadratic polynomials as follows:

$$\begin{array}{cc}\hfill f(\mathbf{x},\mathbf{y})=& a_{11}{x}_1^2{y}_1^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2+a_{22}{x}_2^2{y}_2^2+b{x}_1{x}_2{y}_1{y}_2\hfill \\ \hfill & +c_{x1}{x}_1^2{y}_1{y}_2+c_{x2}{x}_2^2{y}_1{y}_2+c_{y1}{x}_1{x}_2{y}_1^2+c_{y2}{x}_1{x}_2{y}_2^2,\hfill \end{array}$$

(7)

where the first four terms are called diagonal terms, the fifth term is called the full-cross term, and the last four terms are called half-cross terms. In Section 2, we have shown that a $2\times 2$ PSD biquadratic polynomial is always SOS. In this section, we present some SOS forms of $2\times 2$ PSD biquadratic polynomials.

We first show that a $2\times 2$ PSD biquadratic polynomial f expressed by (7) can be reformulated equivalently to another PSD $2\times 2$ biquadratic polynomial with two neighbor half-cross terms and one full-cross terms.

Proposition 1. 

Suppose that the $2\times 2$ biquadratic polynomial f expressed by (7) is PSD. Then $a_{11},a_{12},a_{21},a_{22}\ge 0$, and

$$4a_{11}{a}_{12}\ge c_{x1}^2,4a_{11}{a}_{21}\ge c_{y1}^2,4a_{12}{a}_{22}\ge c_{y2}^2,4a_{21}{a}_{22}\ge c_{x2}^2.$$

Furthermore, there is a $2\times 2$ PSD biquadratic polynomial g with only three half-cross terms:

$$\begin{array}{cc}\hfill g(\mathbf{x},\mathbf{y})=& {\overline{a}}_{11}{x}_1^2{y}_1^2+{\overline{a}}_{12}{x}_1^2{y}_2^2+{\overline{a}}_{21}{x}_2^2{y}_1^2+{\overline{a}}_{22}{x}_2^2{y}_2^2+\overline{b}{x}_1{x}_2{y}_1{y}_2\hfill \\ \hfill & +{\overline{c}}_{x1}{x}_1^2{y}_1{y}_2+{\overline{c}}_{y1}{x}_1{x}_2{y}_1^2+{\overline{c}}_{y2}{x}_1{x}_2{y}_2^2,\hfill \end{array}$$

(8)

such that if g has an SOS form, then the SOS form of f can be resulted.

Proof. 

Let $x_1=y_1=1$ and $x_2=y_2=0$. Then we obtain that $a_{11}\ge 0$. Similarly, we derive that $a_{12},a_{21},a_{22}\ge 0$. Next, setting $x_1=1$ and $x_2=0$ leads to the inequality $4a_{11}{a}_{12}\ge c_{x1}^2$. Similar arguments establish the remaining conditions: $4a_{11}{a}_{21}\ge c_{y1}^2$, $4a_{12}{a}_{22}\ge c_{y2}^2$ and $4a_{21}{a}_{22}\ge c_{x2}^2$.

Suppose $a_{21}=0$. It follows from $4a_{21}{a}_{22}\ge c_{x2}^2$ that $c_{x2}=0$. Then f has only three half-cross terms itself.

Now suppose that $a_{21}>0$. Let ${\overline{x}}_1=x_1$, ${\overline{x}}_2=x_2$, ${\overline{y}}_2=y_2$,

$${\overline{y}}_1=y_1+{\displaystyle \frac{c_{x2}}{2a_{21}}}{y}_2$$

and $g(\overline{\mathbf{x}},\overline{\mathbf{y}})\equiv f(\mathbf{x},\mathbf{y})$. Then g has only three half-cross terms as follows:

$$\begin{array}{cc}\hfill g(\overline{\mathbf{x}},\overline{\mathbf{y}})=& f(\mathbf{x},\mathbf{y})\hfill \\ \hfill =& a_{11}{\overline{x}}_1^2{\overline{y}}_1^2+\left({\displaystyle \frac{a_{11}{c}_{x2}^2}{4a_{21}^2}}+a_{12}-{\displaystyle \frac{c_{x1}{c}_{x2}}{2a_{21}}}\right){\overline{x}}_1^2{\overline{y}}_2^2+a_{21}{\overline{x}}_2^2{\overline{y}}_1^2+\left(a_{22}-{\displaystyle \frac{c_{x2}^2}{4a_{21}}}\right){\overline{x}}_2^2{\overline{y}}_2^2\hfill \\ \hfill & +\left(b-{\displaystyle \frac{c_{y1}{c}_{x2}}{a_{21}}}\right){\overline{x}}_1{\overline{x}}_2{\overline{y}}_1{\overline{y}}_2+\left(c_{x1}-{\displaystyle \frac{a_{11}{c}_{x2}}{2a_{21}}}\right){\overline{x}}_1^2{\overline{y}}_1\overline{y_2}\hfill \\ \hfill & +c_{y1}{\overline{x}}_1{\overline{x}}_2{\overline{y}}_1^2+\left(c_{y2}+{\displaystyle \frac{c_{y1}{c}_{x2}^2}{4a_{21}^2}}-{\displaystyle \frac{b{c}_{x2}}{2a_{21}}}\right){\overline{x}}_1{\overline{x}}_2{\overline{y}}_2^2,\hfill \end{array}$$

which is in the form of (8). This completes the proof. □

Proposition 2. 

Suppose that the $2\times 2$ biquadratic polynomial g expressed by (8) is PSD. Then, there is a $2\times 2$ PSD biquadratic polynomial h containing only two half-cross terms are in neighbor positions as follows:

$$\begin{array}{cc}\hfill h(\mathbf{x},\mathbf{y})=& {\widehat{a}}_{11}{x}_1^2{y}_1^2+{\widehat{a}}_{12}{x}_1^2{y}_2^2+{\widehat{a}}_{21}{x}_2^2{y}_1^2+{\widehat{a}}_{22}{x}_2^2{y}_2^2+\widehat{b}{x}_1{x}_2{y}_1{y}_2\hfill \\ \hfill & +{\widehat{c}}_{x1}{x}_1^2{y}_1{y}_2+{\widehat{c}}_{y1}{x}_1{x}_2{y}_1^2,\hfill \end{array}$$

(9)

such that if h has an SOS form, then the SOS form of g can be resulted.

Proof. 

Suppose ${\overline{a}}_{22}=0$. It follows from $4{\overline{a}}_{12}{\overline{a}}_{22}\ge {{\overline{c}}_{y2}}^2$ that ${\overline{c}}_{y2}=0$. Then g itself has only two half-cross terms, which are in neighbor positions.

Now suppose that $a_{22}>0$. Let ${\widehat{x}}_1={\overline{x}}_1$,

$${\widehat{x}}_2={\overline{x}}_2+{\displaystyle \frac{{\overline{c}}_{y2}}{2{\overline{a}}_{22}}}{\overline{x}}_1,$$

${\widehat{y}}_1={\overline{y}}_1$, ${\widehat{y}}_2={\overline{y}}_2$ and $h(\widehat{\mathbf{x}},\widehat{\mathbf{y}})\equiv g(\overline{\mathbf{x}},\overline{\mathbf{y}})$. Then h has only two half-cross terms, which are in neighbor positions, i.e.,

$$\begin{array}{cc}\hfill h(\widehat{\mathbf{x}},\widehat{\mathbf{y}})=& g(\overline{\mathbf{x}},\overline{\mathbf{y}})\hfill \\ \hfill =& \left({\overline{a}}_{11}+{\displaystyle \frac{{\overline{a}}_{21}{\overline{c}}_{y2}^2}{4{\overline{a}}_{22}^2}}-{\displaystyle \frac{{\overline{c}}_{y1}{\overline{c}}_{y2}}{2{\overline{a}}_{22}}}\right){\widehat{x}}_1^2{\widehat{y}}_1^2+\left({\overline{a}}_{12}-{\displaystyle \frac{{\overline{c}}_{y2}^2}{4{\overline{a}}_{22}}}\right){\widehat{x}}_1^2{\widehat{y}}_2^2+{\overline{a}}_{21}{\widehat{x}}_2^2{\widehat{y}}_1^2+{\overline{a}}_{22}{\widehat{x}}_2^2{\widehat{y}}_2^2\hfill \\ \hfill & +\overline{b}{\widehat{x}}_1{\widehat{x}}_2{\widehat{y}}_1{\widehat{y}}_2+\left({\overline{c}}_{x1}-{\displaystyle \frac{b{\overline{c}}_{y2}}{2{\overline{a}}_{22}}}\right){\widehat{x}}_1^2{\widehat{y}}_1{\widehat{y}}_2+\left({\overline{c}}_{y1}-{\displaystyle \frac{{\overline{a}}_{21}{\overline{c}}_{y2}}{{\overline{a}}_{22}}}\right){\widehat{x}}_1{\widehat{x}}_2{\widehat{y}}_1^2,\hfill \end{array}$$

which is in the form of (9). This completes the proof. □

Combining Propositions 1 and 2, we obtain the following theorem.

Theorem 5. 

Suppose that the $2\times 2$ biquadratic polynomial f expressed by (7) is PSD. Then $a_{11},a_{12},a_{21},a_{22}\ge 0$, and

$$4a_{11}{a}_{12}\ge c_{x1}^2,4a_{11}{a}_{21}\ge c_{y1}^2,4a_{12}{a}_{22}\ge c_{y2}^2,4a_{21}{a}_{22}\ge c_{x2}^2.$$

(10)

Furthermore, there is a $2\times 2$ PSD biquadratic polynomial h with only two half-cross terms as expressed by (9), such that if h has an SOS form, then the SOS form of f can be resulted.

By Theorem 1, any $2\times 2$ PSD biquadratic polynomial admits an SOS decomposition. Thus, in the remaining part of this section, we consider a $2\times 2$ PSD biquadratic polynomial f, with the form

$$\begin{array}{cc}\hfill f(\mathbf{x},\mathbf{y})=& a_{11}{x}_1^2{y}_1^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2+a_{22}{x}_2^2{y}_2^2+b{x}_1{x}_2{y}_1{y}_2\hfill \\ \hfill & +c_{x1}{x}_1^2{y}_1{y}_2+c_{y1}{x}_1{x}_2{y}_1^2,\hfill \end{array}$$

(11)

to identify its concrete SOS form. In Section 4.1, Section 4.2 and Section 4.3, we give constructive proofs for the SOS form of a $2\times 2$ PSD biquadratic polynomials, in three cases: I. f has no half-cross terms; II. f has only one half-cross term; III. f has two neighbor half-cross terms and no full-cross terms. The case that the $2\times 2$ bipartite polynomial with one full-cross term along with two neighbor half-cross terms is somewhat complicated. We do not consider this case here.

4.1. Case I: No Half-Cross Terms

If f has no cross terms at all, the case is trivial. In this subsection, we assume that f has one full-cross term but no half-cross terms.

Lemma 1. 

Let

$$f(\mathbf{x},\mathbf{y})=a_{11}{x}_1^2{y}_1^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2+a_{22}{x}_2^2{y}_2^2-4x_1{x}_2{y}_1{y}_2.$$

Then the following three statements are equivalent:

(a) 

$f(\mathbf{x},\mathbf{y})$ is PSD.

(b) 

$f(\mathbf{x},\mathbf{y})$ is SOS.

(c) 

The parameters $a_{11},a_{12},a_{21},a_{22}$ satisfy

$$a_{11},a_{12},a_{21},a_{22}\ge 0\mathrm{and}\sqrt{a_{11}{a}_{22}}+\sqrt{a_{12}{a}_{21}}\ge 2.$$

Proof. $\left(a\right)\Rightarrow \left(c\right)$ Since f is PSD when $x_1=y_1=1$ and $x_2=y_2=0$, we obtain $a_{11}\ge 0$. Similarly, $a_{12},a_{21},a_{22}\ge 0$. If $\sqrt{a_{11}{a}_{22}}\ge 2$, then

$$\begin{array}{c}\hfill f(\mathbf{x},\mathbf{y})=\left(a_{11}-{\displaystyle \frac{4}{a_{22}}}\right)x_1^2{y}_1^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2+{\left({\displaystyle \frac{2}{\sqrt{a_{22}}}}{x}_1{y}_1-\sqrt{a_{22}}{x}_2{y}_2\right)}^2.\end{array}$$

By analogous reasoning, if $\sqrt{a_{12}{a}_{21}}\ge 2$, then

$$\begin{array}{c}\hfill f(\mathbf{x},\mathbf{y})=a_{11}{x}_1^2{y}_1^2+\left(a_{12}-{\displaystyle \frac{4}{a_{21}}}\right)x_1^2{y}_2^2+{\left({\displaystyle \frac{2}{\sqrt{a_{21}}}}{x}_2{y}_1-\sqrt{a_{21}}{x}_2{y}_1\right)}^2+a_{22}{x}_2^2{y}_2^2.\end{array}$$

Now we assume that $\sqrt{a_{11}{a}_{22}}<2$ and $\sqrt{a_{12}{a}_{21}}<2$. It follows from f is PSD when $x_2=y_2=1$, $\sqrt{a_{12}}{x}_1{y}_2=\sqrt{a_{21}}{x}_2{y}_1$, and $x_1,y_1\to \infty$ that $a_{11}>0$. Following similar methodology, we obtain that $a_{12},a_{21},a_{22}>0$. By direct computation, we derive that

$$\begin{array}{ccc}\hfill f(\mathbf{x},\mathbf{y})& =& {\left(\sqrt{a_{11}}{x}_1{y}_2-\sqrt{a_{22}}{x}_2{y}_1\right)}^2+{\left({\displaystyle \frac{2-\sqrt{a_{11}{a}_{22}}}{\sqrt{a_{21}}}}{x}_1{y}_2-\sqrt{a_{21}}{x}_2{y}_1\right)}^2\hfill \\ & & +\left(a_{12}-{\displaystyle \frac{{(2-\sqrt{a_{11}{a}_{22}})}^2}{a_{21}}}\right)x_1^2{y}_2^2.\hfill \end{array}$$

When the first two terms vanish, the PSD condition of f implies $\sqrt{a_{11}{a}_{22}}+\sqrt{a_{12}{a}_{21}}\ge 2$.

$\left(c\right)\Rightarrow \left(b\right)$ follows directly from the above three reformulations of f.

$\left(b\right)\Rightarrow \left(a\right)$ is obvious. □

Building upon Lemma 1, we now establish the following theorem.

Theorem 6. 

Let

$$f(\mathbf{x},\mathbf{y})=a_{11}{x}_1^2{y}_1^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2+a_{22}{x}_2^2{y}_2^2-b{x}_1{x}_2{y}_1{y}_2.$$

Then, the following three statements are equivalent:

(a) 

$f(\mathbf{x},\mathbf{y})$ is PSD.

(b) 

$f(\mathbf{x},\mathbf{y})$ is SOS.

(c) 

The parameters $a_{11},a_{12},a_{21},a_{22}$ satisfy

$$a_{11},a_{12},a_{21},a_{22}\ge 0,\mathrm{and}\sqrt{a_{11}{a}_{22}}+\sqrt{a_{12}{a}_{21}}\ge {\displaystyle \frac{\left|b\right|}{2}}.$$

(12)

4.2. Case II: Only One Half-Cross Term

In this subsection, we assume that f has one half-cross term and one full-cross term.

Lemma 2. 

Suppose that

$$f(\mathbf{x},\mathbf{y})=a_{11}{x}_1^2{y}_1^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2+a_{22}{x}_2^2{y}_2^2-4x_1{y}_1{x}_2{y}_2-c_{y1}{x}_1{x}_2{y}_1^2.$$

Then, the following three statements are equivalent:

(a) 

$f(\mathbf{x},\mathbf{y})$ is PSD.

(b) 

$f(\mathbf{x},\mathbf{y})$ is SOS.

(c) 

There exist $a_{111},a_{211}\ge 0$ and $a_{112},a_{212}>0$ such that $a_{11}=a_{111}+a_{112}$, $a_{21}=a_{211}+a_{212}$, $4a_{112}{a}_{212}=c_{y1}^2$, and $\sqrt{a_{111}{a}_{22}}+\sqrt{a_{12}{a}_{211}}\ge 2$.

Proof. 

If $c_{y1}=0$, then the conclusion follows from Lemma 1. Thus, we may assume that $c_{y1}\ne 0$.

(a) ⇒ (c) Because f is PSD when $y_2=0$, we obtain that $4a_{11}{a}_{21}\ge c_{y1}^2$. Therefore, there exist $a_{111},a_{211}\ge 0$ and $a_{112},a_{212}>0$ such that $a_{11}=a_{111}+a_{112}$, $a_{21}=a_{211}+a_{212}$, and $4a_{112}{a}_{212}=c_{y1}^2$. In the following, we show $\sqrt{a_{111}{a}_{22}}+\sqrt{a_{12}{a}_{211}}\ge 2$. Let $x_1=\sqrt{a_{212}}>0$ and $x_2=\mathrm{sgn}\left(c_{y1}\right)\sqrt{a_{112}}\ne 0$. Then,

$$f(\mathbf{x},\mathbf{y})=\left(a_{111}{a}_{212}+a_{211}{a}_{112}\right)y_1^2+\left(a_{22}{a}_{112}+a_{12}{a}_{212}\right)y_2^2-4\mathrm{sgn}\left(c_{y1}\right)\sqrt{a_{112}{a}_{212}}{y}_1{y}_2.$$

Since f is PSD, we see that $a_{111}$ and $a_{211}$ cannot be both zero, and $a_{22}$ and $a_{12}$ cannot be both zero. We consequently classify three scenarios based on the parameters $a_{22}$ and $a_{12}$:

Case (i) $a_{22}=0$ and $a_{12}>0$. It follows from f is PSD when $x_1=\sqrt{a_{21}}$, $x_2=\sqrt{a_{11}}$, $y_2=1$, and $y_1\to \infty$ that $a_{21}>{\displaystyle \frac{c_{y1}^2}{4a_{11}}}$. By selecting the parameters as $a_{112}=a_{11}$, $a_{212}={\displaystyle \frac{c_{y1}^2}{4a_{11}}}$, $a_{211}=a_{21}-a_{212}$, we obtain $a_{211}>0$ and

$$\begin{array}{ccc}\hfill f(\mathbf{x},\mathbf{y})& =& \left(a_{12}-{\displaystyle \frac{4}{a_{211}}}\right)x_1^2{y}_2^2+{\left({\displaystyle \frac{2}{\sqrt{a_{211}}}}{x}_1{y}_2-\sqrt{a_{211}}{x}_2{y}_1\right)}^2\hfill \\ & & +{\left(\sqrt{a_{11}}{x}_1-\mathrm{sgn}\left(c_{y1}\right)\sqrt{a_{212}}{x}_2\right)}^2{y}_1^2.\hfill \end{array}$$

Let $x_1=\sqrt{a_{212}}$, $x_2=\mathrm{sgn}\left(c_{y1}\right)\sqrt{a_{112}}$, $y_2=\sqrt{a_{211}}{x}_2$ and $y_1=\sqrt{a_{12}}{x}_1$. Then $f(\mathbf{x},\mathbf{y})=\left(a_{12}-{\displaystyle \frac{4}{a_{211}}}\right)x_1^2{y}_2^2$. It follows from f is PSD that $\sqrt{a_{12}{a}_{211}}\ge 2$. Consequently, we conclude that there exist parameters $a_{111},a_{211},a_{112},a_{212}$ satisfying condition (c).

Case (ii) $a_{12}=0$ and $a_{22}>0$. This can be proved similarly as Case (i).

Case (iii) $a_{22}>0$ and $a_{12}>0$. Note that $a_{111}$ and $a_{211}$ cannot both be zero. Suppose $a_{111}\ne 0$. If $\sqrt{a_{111}{a}_{22}}>2$, condition (c) hold directly. Suppose that $\sqrt{a_{111}{a}_{22}}<2$. Then it follows from f is PSD that $a_{211}>0$ and

$$\begin{array}{cc}\hfill f(\mathbf{x},\mathbf{y})=& {(\sqrt{a_{111}}{x}_1{y}_1-\sqrt{a_{22}}{x}_2{y}_2)}^2+\left(a_{12}-{\displaystyle \frac{{(2-\sqrt{a_{111}{a}_{22}})}^2}{a_{211}}}\right)x_1^2{y}_2^2\hfill \\ \hfill & +{\left({\displaystyle \frac{2-\sqrt{a_{111}{a}_{22}}}{\sqrt{a_{211}}}}{x}_1{y}_2-\sqrt{a_{211}}{x}_2{y}_1\right)}^2\hfill \\ \hfill & +{\left(\sqrt{a_{112}}{x}_1-\mathrm{sgn}\left(c_{y1}\right)\sqrt{a_{212}}{x}_2\right)}^2{y}_1^2.\hfill \end{array}$$

Let $x_1=\sqrt{a_{212}}$, $x_2=\mathrm{sgn}\left(c_{y1}\right)\sqrt{a_{112}}$, $y_1=\sqrt{a_{22}}{x}_2$ and $y_2=\sqrt{a_{111}}{x}_1$. Then the first two terms are zeros. Furthermore, define

$$\varphi \left(a_{112}\right)=\sqrt{a_{111}}(2-\sqrt{a_{111}{a}_{22}})c_{y1}^2-4a_{211}{a}_{112}^2\sqrt{a_{22}}.$$

It follows from $a_{111}=a_{11}-a_{112}$, $a_{211}=a_{21}-a_{212}$, and $a_{212}={\displaystyle \frac{c_{y1}^2}{4a_{112}}}$ that $\varphi$ is a function of $a_{112}$. Since $a_{212}={\displaystyle \frac{c_{y1}^2}{4a_{112}}}\le a_{21}$, it yields that $a_{112}\in \left[{\displaystyle \frac{c_{y1}^2}{4a_{21}}},a_{11}\right]$. When $a_{112}={\displaystyle \frac{c_{y1}^2}{4a_{21}}}$, we derive that $a_{211}=0$ and $\varphi \left({\displaystyle \frac{c_{y1}^2}{4a_{21}}}\right)\ge 0$. When $a_{112}=a_{11}$, we obtain that $a_{111}=0$ and $\varphi \left(a_{11}\right)\le 0$. Thus, there is ${\eta }_0\in \left[{\displaystyle \frac{c_{y1}^2}{4a_{21}}},a_{11}\right]$ such that $\varphi \left(a_{112}\right)=0$. Then by setting $a_{112}={\eta }_0$, $a_{212}={\displaystyle \frac{c_{y1}^2}{4{\eta }_0}}$, we obtain that $f(\mathbf{x},\mathbf{y})=\left(a_{12}-{\displaystyle \frac{{(2-\sqrt{a_{111}{a}_{22}})}^2}{a_{211}}}\right)x_1^2{y}_2^2$. It follows from f is PSD that $\sqrt{a_{111}{a}_{22}}+\sqrt{a_{12}{a}_{211}}\ge 2$. Consequently, we derive that there exist parameters $a_{111},a_{211},a_{112},a_{212}$, specifically chosen as $a_{112}={\eta }_0$, $a_{212}={\displaystyle \frac{c_{y1}^2}{4{\eta }_0}}$, that satisfy (c).

The case when $a_{211}\ne 0$ follows by an analogous argument with $a_{111}\ne 0$.

(c) ⇒ (b) follows directly from the above reformulations of f.

(b) ⇒ (a) follows directly from the definitions.

This completes the proof. □

Furthermore, we establish an equivalent condition based on Lemma 2 that depends solely on the coefficients of f.

Theorem 7. 

Suppose that

$$f(\mathbf{x},\mathbf{y})=a_{11}{x}_1^2{y}_1^2+a_{22}{x}_2^2{y}_2^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2-b{x}_1{y}_1{x}_2{y}_2-c_{y1}{x}_1{x}_2{y}_1^2.$$

Let

$$g\left(a_{112}\right)=\sqrt{\left(a_{11}-a_{112}\right)a_{22}}+\sqrt{a_{12}\left(a_{21}-{\displaystyle \frac{c_{y1}^2}{4a_{112}}}\right)}.$$

and $a_{112}^{*}\in \left({\displaystyle \frac{c_{y1}^2}{4a_{21}}},a_{11}\right)$ be the unique parameter such that the derivative $g^{\prime }\left(a_{112}^{*}\right)=0$. Then the following three statements are equivalent:

(a) 

$f(\mathbf{x},\mathbf{y})$ is PSD.

(b) 

$f(\mathbf{x},\mathbf{y})$ is SOS.

(c) 

It holds that $g\left(a_{112}^{*}\right)\ge {\displaystyle \frac{\left|b\right|}{2}}$.

Proof. 

Based on Lemma 2, we obtain that $a_{11}=a_{111}+a_{112}$, $a_{21}=a_{211}+a_{212}$, $4a_{112}{a}_{212}=c_{y1}^2$, and

$$\begin{array}{c}\hfill g\left(a_{112}\right)=\sqrt{\left(a_{11}-a_{112}\right)a_{22}}+\sqrt{a_{12}\left(a_{21}-{\displaystyle \frac{c_{y1}^2}{4a_{112}}}\right)}=\sqrt{a_{111}{a}_{22}}+\sqrt{a_{12}{a}_{211}}.\end{array}$$

For any $a_{112}\in \left({\displaystyle \frac{c_{y1}^2}{4a_{21}}},a_{11}\right)$, we may derive the derivative of g as follows:

$$g^{\prime }\left(a_{112}\right)=-{\displaystyle \frac{\sqrt{a_{22}}}{2\sqrt{a_{11}-a_{112}}}}+{\displaystyle \frac{\sqrt{a_{12}}\frac{c_{y1}^2}{4a_{112}^2}}{2\sqrt{a_{21}-\frac{c_{y1}^2}{4a_{112}}}}}.$$

Here, the first term ${\displaystyle \frac{\sqrt{a_{22}}}{2\sqrt{a_{11}-a_{112}}}}$ is monotonously increasing to infinity and the second term ${\displaystyle \frac{\sqrt{a_{12}}\frac{c_{y1}^2}{4a_{112}^2}}{2\sqrt{a_{21}-\frac{c_{y1}^2}{4a_{112}}}}}$ is monotonously decreasing from infinity. Therefore, there is one unique parameter $a_{112}^{*}\in \left({\displaystyle \frac{c_{y1}^2}{4a_{21}}},a_{11}\right)$ such that $g^{\prime }\left(a_{112}^{*}\right)=0$. Furthermore, $g\left(a_{112}\right)$ is monotonously increasing in $a_{112}\in \left({\displaystyle \frac{c_{y1}^2}{4a_{21}}},a_{112}^{*}\right)$ and then monotonously decreasing in $a_{112}\in \left(a_{112}^{*},a_{11}\right)$. Consequently, $g\left(a_{112}^{*}\right)={max}_{a_{112}\in \left[{\displaystyle \frac{c_{y1}^2}{4a_{21}}},a_{11}\right]}g\left(a_{112}\right)$. Therefore, if f is PSD, then it holds that $g\left(a_{112}^{*}\right)\ge {\displaystyle \frac{\left|b\right|}{2}}$. □

4.3. Case III: Two Neighbor Half-Cross Terms and No Full-Cross Terms

In this subsection, we study the case of $2\times 2$ biquadratic polynomials with two half-cross terms in the neighbor sides and no full-cross term. The case that the two half-cross terms in the opposite sides and no full-cross term is somewhat trivial. Consider

$$f(\mathbf{x},\mathbf{y})=a_{11}{x}_1^2{y}_1^2+a_{12}{x}_1^2{y}_2^2+a_{21}{x}_2^2{y}_1^2+a_{22}{x}_2^2{y}_2^2+2c_x{x}_1^2{y}_1{y}_2+2c_y{x}_1{x}_2{y}_1^2.$$

(13)

Here, we denote $c_x={\displaystyle \frac{c_{x1}}{2}}$ and $c_y={\displaystyle \frac{c_{y1}}{2}}$ to simplify our notation. Suppose that $c_x\ne 0$ and $c_y\ne 0$, and f is PSD. Then we derive that

$$a_{11}{a}_{12}\ge c_x^2>0,a_{11}{a}_{21}\ge c_y^2>0,$$

$a_{11},a_{12},a_{21}>0$ and $a_{22}\ge 0$.

For simplicity, we may replace $x_2$ and $y_2$ by ${\displaystyle \frac{{\overline{x}}_2}{\sqrt{a_{21}}}}$ and ${\displaystyle \frac{{\overline{y}}_2}{\sqrt{a_{12}}}}$, respectively, and assume that $a_{12}=a_{21}=1$, i.e.,

$$f(\mathbf{x},\mathbf{y})=a_{11}{x}_1^2{y}_1^2+x_1^2{y}_2^2+x_2^2{y}_1^2+a_{22}{x}_2^2{y}_2^2+2c_x{x}_1^2{y}_1{y}_2+2c_y{x}_1{x}_2{y}_1^2.$$

(14)

We also may assume $c_x>0$ (resp. $c_y>0$), since otherwise we could replace $y_1$ with $-y_1$ (resp. $x_1$ with $-x_1$). Without loss of generality, assume that $c_x\ge c_y>0$.

If $a_{11}\ge c_x^2+c_y^2$, then

$$\begin{array}{cc}\hfill f(\mathbf{x},\mathbf{y})=& \left(a_{11}-c_x^2-c_y^2\right)x_1^2{y}_1^2+x_1^2{\left(c_x{y}_1+y_2\right)}^2+{\left(c_y{x}_1+x_2\right)}^2{y}_1^2+a_{22}{x}_2^2{y}_2^2,\hfill \end{array}$$

which gives the SOS form of f explicitly.

When $a_{11}=c_x^2$, $f(\mathbf{x},\mathbf{y})$ is not PSD for any $a_{22}>0$. This can be seen by letting $y_1=1$, $y_2=-c_x$, $x_1=-c_y$ and

$$0<x_2<{\displaystyle \frac{2c_y^2}{1+a_{22}{c}_x^2}}.$$

Theorem 8. 

Suppose that the $2\times 2$ biquadratic polynomial f is expressed by (14), where $c_x^2<a_{11}<c_x^2+c_y^2$, $a_{22}>0$, and $c_x\ge c_y>0$. Then the following three statements are equivalent:

(a) 

$f(\mathbf{x},\mathbf{y})$ is PSD.

(b) 

$f(\mathbf{x},\mathbf{y})$ is SOS.

(c) 

It holds that

$$a_{22}\ge \left\{\begin{array}{cc}{\displaystyle \frac{8c_y}{{\left(3c_y-\sqrt{9c_y^2-4a_{11}}\right)}^3}}-{\displaystyle \frac{4}{{\left(3c_y-\sqrt{9c_y^2-4a_{11}}\right)}^2}},\hfill & \mathit{if}{c}_x=c_y\mathit{and}{\displaystyle \frac{5}{4}}{c}_y^2\le a_{11}\le 2c_y^2,\hfill \\ {\displaystyle \frac{1}{a_{11}-c_x^2}}-{\displaystyle \frac{4}{{\left(3c_y-\sqrt{9c_y^2-4a_{11}}\right)}^2}},\hfill & \mathit{if}{c}_x=c_y\mathit{and}{c}_y^2\le a_{11}\le {\displaystyle \frac{5}{4}}{c}_y^2,\hfill \\ {\displaystyle \frac{{\left({\gamma }_1^{*}\right)}^2(1-{\gamma }_1^{*}){(2-{\gamma }_1^{*})}^2}{(({\gamma }_1^{*}-1)c_x+c_y)(c_x+({\gamma }_1^{*}-1)c_y)}},\hfill & \mathit{if}{c}_x>c_y,\hfill \end{array}\right.$$

where ${\gamma }_1^{*}\in \left({\displaystyle \frac{c_x-c_y}{c_x}},1\right)$ is the root of function $\Omega$ defined by the following equation

$$\mathrm{\Omega }\left({\gamma }_1\right):=a_{11}{\gamma }_1^2{(2-{\gamma }_1)}^2-3{(1-{\gamma }_1)}^3{c}_x{c}_y+2{(1-{\gamma }_1)}^2(c_x^2+c_y^2)+(1-{\gamma }_1)c_x{c}_y-c_x^2-c_y^2=0.$$

(15)

Proof. 

We assume that f has the following SOS decomposition:

$$\begin{array}{cc}\hfill f(\mathbf{x},\mathbf{y})=& {\gamma }_1{(\alpha x_1+x_2)}^2{y}_1^2+{\gamma }_1{x}_1^2{(\beta y_1+y_2)}^2+{\gamma }_2{\left[(\alpha +\beta )x_1{y}_1+x_1{y}_2+x_2{y}_1\right]}^2\hfill \\ \hfill & +{\gamma }_3{(\alpha \beta x_1{y}_1-x_2{y}_2)}^2+\left(a_{22}-{\gamma }_3\right)x_2^2{y}_2^2,\hfill \end{array}$$

(16)

where parameters ${\gamma }_1,{\gamma }_2\ge 0$ and $0\le {\gamma }_3\le a_{22}$. By comparing (14) and (16), we observe that

$$\begin{array}{cc}\hfill {\gamma }_1({\alpha }^2+{\beta }^2)+{\gamma }_2{(\alpha +\beta )}^2+{\gamma }_3{\alpha }^2{\beta }^2& =a_{11},\hfill \end{array}$$

(17a)

$$\begin{array}{cc}\hfill {\gamma }_1+{\gamma }_2& =1,\hfill \end{array}$$

(17b)

$$\begin{array}{cc}\hfill {\gamma }_1\beta +{\gamma }_2(\alpha +\beta )& =c_x,\hfill \end{array}$$

(17c)

$$\begin{array}{cc}\hfill {\gamma }_1\alpha +{\gamma }_2(\alpha +\beta )& =c_y,\hfill \end{array}$$

(17d)

$$\begin{array}{cc}\hfill {\gamma }_2-{\gamma }_3\alpha \beta & =0.\hfill \end{array}$$

(17e)

The polynomial f admits an SOS decomposition if the system of Equation (17) admits solutions $\alpha ,\beta ,{\gamma }_1,{\gamma }_2,{\gamma }_3$ satisfying ${\gamma }_1,{\gamma }_2\ge 0$ and $0\le {\gamma }_3\le a_{22}$. Combining Equations (17c) and (17d) yields $(\alpha -\beta ){\gamma }_1=c_y-c_x$. In the case where $c_y=c_x$, either $\alpha =\beta$ or ${\gamma }_1=0$. We therefore proceed by analyzing two distinct cases separately: (i) $c_y=c_x$ and (ii) $c_y\ne c_x$. The former case is subsequently divided into subcases (i1) and (i2) for detailed examination.

Case (i1): $c_y=c_x$ and ${\displaystyle \frac{5}{4}}{c}_y^2\le a_{11}\le 2c_y^2$. By setting $\alpha =\beta \ne 0$ and ${\gamma }_2=1-{\gamma }_1$, we can simplify Equations (17a)–(17e) to

$$\begin{array}{cc}\hfill 2{\gamma }_1{\alpha }^2+4(1-{\gamma }_1){\alpha }^2+{\gamma }_3{\alpha }^4& =a_{11},\hfill \end{array}$$

(18a)

$$\begin{array}{cc}\hfill {\gamma }_1\alpha +2(1-{\gamma }_1)\alpha & =c_y,\hfill \end{array}$$

(18b)

$$\begin{array}{cc}\hfill 1-{\gamma }_1-{\gamma }_3{\alpha }^2& =0.\hfill \end{array}$$

(18c)

Combining Equations (18a) and (18c) yields

$$\begin{array}{c}\hfill {\alpha }^2(5-3{\gamma }_1)=a_{11}\mathrm{and}(2-{\gamma }_1)\alpha =c_y.\end{array}$$

Therefore, we obtain that ${\gamma }_1=2-{\displaystyle \frac{c_y}{\alpha }}$ and ${\alpha }^2-3c_y\alpha +a_{11}=0$. It follows from $a_{11}\le 2c_y^2$ that the quadratic equation in $\alpha$ admits two distinct real roots $\alpha ={\displaystyle \frac{3c_y\pm \sqrt{9c_y^2-4a_{11}}}{2}}$. By the fact that $0\le {\gamma }_1=2-{\displaystyle \frac{c_y}{\alpha }}\le 1$, only the root ${\alpha }^{*}={\displaystyle \frac{3c_y-\sqrt{9c_y^2-4a_{11}}}{2}}$ is feasible. Consequently, in Case (i1), if

$$a_{22}\ge {\gamma }_3^{*}={\displaystyle \frac{c_y}{{\left({\alpha }^{*}\right)}^4}}-{\displaystyle \frac{1}{{\left({\alpha }^{*}\right)}^2}},$$

then f is SOS.

Case (i2): $c_y=c_x$ and $c_y^2<a_{11}\le {\displaystyle \frac{5}{4}}{c}_y^2$. By setting ${\gamma }_1=0$ and ${\gamma }_2=1$, we can simplify (17a)–(17e) to

$$\begin{array}{c}\hfill {(\alpha +\beta )}^2+\alpha \beta =a_{11},\alpha +\beta =c_y,\mathrm{and}{\gamma }_3={\displaystyle \frac{1}{\alpha \beta }}.\end{array}$$

(19)

By direct computation, we obtain $\beta =c_y-\alpha$, $c_y^2+\alpha \beta =a_{11}$, and ${\gamma }_3={\displaystyle \frac{1}{a_{11}-c_y^2}}$. Consequently, we derive the following quadratic equation with respective to $\alpha$:

$$\begin{array}{c}\hfill {\alpha }^2-c_y\alpha +a_{11}-c_y^2=0.\end{array}$$

(20)

It follows from $c_y^2<a_{11}\le {\displaystyle \frac{5}{4}}{c}_y^2$ that (20) admits two real solutions, which derives that (19) is feasible. Hence, in Case (i2), if

$$a_{22}\ge {\gamma }_3^{*}={\displaystyle \frac{1}{a_{11}-c_x^2}},$$

then f is SOS.

Case (ii): $c_x>c_y$ and $c_x^2<a_{11}<c_x^2+c_y^2$. Then we obtain that $\alpha \ne \beta$, and ${\gamma }_1\ne 0$. It follows from (17b)–(17e) that

$$\begin{array}{c}\hfill {\gamma }_2=1-{\gamma }_1,{\gamma }_3={\displaystyle \frac{1-{\gamma }_1}{\alpha \beta }},\alpha ={\displaystyle \frac{({\gamma }_1-1)c_x+c_y}{{\gamma }_1(2-{\gamma }_1)}},\beta ={\displaystyle \frac{c_x+({\gamma }_1-1)c_y}{{\gamma }_1(2-{\gamma }_1)}}.\end{array}$$

(21)

Consequently, Equation (15) is derived directly from (17a). By the inequalities $\mathrm{\Omega }\left(1\right)=a_{11}-c_x^2-c_y^2<0$ and $\mathrm{\Omega }\left({\displaystyle \frac{c_x-c_y}{c_x}}\right)={\displaystyle \frac{{(c_x^2-c_y^2)}^2}{c_x^4}}(a_{11}-c_x^2)>0$, we conclude that there exists a solution ${\gamma }_1^{*}\in \left({\displaystyle \frac{c_x-c_y}{c_x}},1\right)$ such that $\mathrm{\Omega }\left({\gamma }_1^{*}\right)=0$. Subsequently, we obtain the corresponding values of ${\gamma }_2^{*},{\gamma }_3^{*},{\alpha }^{*}$ and ${\beta }^{*}$ by (21). Furthermore, it follows directly from $c_x>c_y$ and ${\gamma }_1^{*}\in (0,1)$ that ${\beta }^{*}>0$, and from ${\gamma }_1^{*}>{\displaystyle \frac{c_x-c_y}{c_x}}$ that ${\alpha }^{*}>0$. Consequently, ${\gamma }_3^{*}={\displaystyle \frac{1-{\gamma }_1^{*}}{{\alpha }^{*}{\beta }^{*}}}>0$. Thus, if

$$a_{22}\ge {\gamma }_3^{*}={\displaystyle \frac{{\left({\gamma }_1^{*}\right)}^2(1-{\gamma }_1^{*}){(2-{\gamma }_1^{*})}^2}{(({\gamma }_1^{*}-1)c_x+c_y)(c_x+({\gamma }_1^{*}-1)c_y)}}>0,$$

then f is SOS.

In both Cases (i) and (ii), if $a_{22}<{\gamma }_3^{*}$, then f is not PSD. This can be seen by letting $x_2=-\alpha x_1=1$ and $y_2=-\beta y_1=1$ in (16). This completes the proof. □

Example 2. 

Let $a_{11}=1.2$, $a_{22}=6$, $c_x=c_y=-1$. The corresponding biquadratic polynomial is given by:

$$\begin{array}{c}\hfill f(\mathbf{x},\mathbf{y})=1.2x_1^2{y}_1^2+x_1^2{y}_2^2+x_2^2{y}_1^2+6x_2^2{y}_2^2-2x_1^2{y}_1{y}_2-2x_1{x}_2{y}_1^2.\end{array}$$

We could verify that $c_y^2\le a_{11}\le {\displaystyle \frac{5}{4}}{c}_y^2$ and $a_{22}=6>{\displaystyle \frac{1}{a_{11}-c_x^2}}-{\displaystyle \frac{4}{{\left(3c_y-\sqrt{9c_y^2-4a_{11}}\right)}^2}}=4.8431$. According to Theorem 8, these conditions guarantee that f is an SOS. By YALMIP and SDPT4 within the MATLAB 2024a environment, we obtain an explicit SOS decomposition as follows:

$$\begin{array}{ccc}\hfill f(\mathbf{x},\mathbf{y})& =& {(-0.4876x_1{y}_1+0.1140x_1{y}_2+0.1140x_2{y}_1+2.4326x_2{y}_2)}^2\hfill \\ & & +{(0.9781x_1{y}_1-0.9685x_1{y}_2-0.9685x_2{y}_1+0.2869x_2{y}_2)}^2\hfill \\ & & +{(0.2181x_1{y}_2-0.2181x_2{y}_1)}^2\hfill \\ & & +{(0.0739x_1{y}_1+0.0390x_1{y}_2+0.0390x_2{y}_1+0.0112x_2{y}_2)}^2.\hfill \end{array}$$

5. Conclusions and Open Questions

In his seminal 1888 work [1], Hilbert demonstrated the existence of PSD homogeneous quartic polynomials in four variables that cannot be expressed as SOS. Nearly a century later, Choi and Lam (1977) [4] constructed an explicit example of such a PNS quartic polynomial. In the present work, we establish that despite being four-variable quartic polynomials, all $2\times 2$ PSD biquadratic polynomials admit an SOS representation. The key technique is that an $m\times n$ biquadratic polynomial can be expressed as a tripartite homogeneous quartic polynomial of $m+n-1$ variables. Furthermore, we established a necessary and sufficient condition for an $m\times n$ PSD biquadratic polynomial to admit an SOS representation, and demonstrated that the SOS rank of such polynomials was bounded above by $mn$. Subsequently, we provided a constructive proof for obtaining the SOS decomposition of $2\times 2$ PSD biquadratic polynomials in three specific cases.

In general, an $m\times n$ biquadratic polynomial can be written as

$$f(\mathbf{x},\mathbf{y})=\sum _{i,j=1}^m\sum _{k,l=1}^n{a}_{ijkl}{x}_i{x}_j{y}_k{y}_l.$$

In 1973, Calderón [9] proved that an $m\times 2$ PSD biquadratic polynomial can be expressed as the sum of squares of ${\displaystyle \frac{3m(m+1)}{2}}$ quadratic polynomials. In this paper, we proved that a $2\times 2$ PSD biquadratic polynomial can be expressed as the sum of squares of three quadratic polynomials. An open question is what is the SOS rank for an $m\times 2$ PSD biquaratic polynomial for $m\ge 3$. In 1975, Choi [11] gave a $3\times 3$ PNS biquadratic polynomial. Another open question is: What is the maximum SOS rank of an $m\times n$ SOS biquadratic polynomial for $m,n\ge 3$. An even more challenging question is to determine a given $m\times n$ PSD biquadratic polynomial is SOS or not numerically.