| Sbornik: Mathematics |
|
|
|
|
|
Existence of polynomial solutions of degree 4 of the Monge-Ampère equation. Large deflections of thin plates Yu. A. Aminovab a B. Verkin Institute for Low Temperature Physics and Engineering of the National Academy of Sciences of Ukraine, Kharkiv, Ukraine b Brasília, Brazil
Bibliographic databases:
     § 1. IntroductionWe consider polynomial solutions $z(x,y)$ of the Monge-Ampère equation Given a polynomial $f(x,y)$, it is required to find a polynomial $z(x,y)$ of degree 4 solving this equation. In the case when $f(x,y)=1$ Jörgens [1] proved that if the solution of the Monge-Ampère equation is defined on the whole plane, then this solution is a quadratic polynomial. Calabi and Pogorelov extended this result to the graphs of convex multidimensional hypersurfaces. It is also natural to consider the general case when $f(x,y)$ is a polynomial of some degree, especially in view the fact that polynomial solutions are easily applicable. In this statement of the problem a solution can fail to exist. Necessary conditions for solvability were obtained by this author earlier in [2] and [3]. We set and First consider the homogeneous part of degree 4. The set of coefficients $b_{ij}$ defines the coordinates of the polynomial $f$ in the five-dimensional space, which we denote by $E^5(b)$. A point in this space is denoted by $b=(b_{40},\dots,b_{04})$. By $E^5(a)$, we denote the same space, but with coordinates $a_{40},\dots,a_{04}$. The following two invariants have been defined: and
$$
\begin{equation}
L(b)=\begin{vmatrix} \dfrac{b_{22}}{3}&\dfrac{b_{13}}{2}&2b_{04} \\ \dfrac{b_{31}}{2}&\dfrac{b_{22}}{3}&\dfrac{b_{13}}{2} \\ 2b_{40}&\dfrac{b_{31}}{2}&\dfrac{b_{22}}{3} \end{vmatrix}.
\end{equation}
\tag{4}
$$
The above quantities, which are invariant under rotations and shifts in the $(x,y)$-plane, appear in several important problems. Replacing the coefficients $b_{ij}$ by $a_{ij}$ we obtain the invariants $T(a)$ and $L(a)$ for the polynomial $z(x,y)$. It is noteworthy that these invariants are related by the functional relations:
$$
\begin{equation}
T(b)=12^23[T(a)]^2\quad\text{and} \quad L(b)= \sigma [L(a)]^2 +\mu [T(a)]^3,
\end{equation}
\tag{5}
$$
where $\sigma=-3^62^7$ and $\mu=3^32^6$. All these relations were obtained by this author in [3]. As a result, we have the first and second necessary conditions for solvability: if $L(a)\geqslant 0$, then
$$
\begin{equation}
1) \ T(b)\geqslant 0, \qquad 2) \ \biggl(\frac{T(b)}{3}\biggr)^{3/2}-L(b)=3^62^7[L(a)]^2\geqslant 0,
\end{equation}
\tag{6}
$$
and if $L(a)\leqslant0$, then
$$
\begin{equation}
1) \ T(b)\geqslant 0, \qquad 2) \ -\biggl(\frac{T(b)}{3}\biggr)^{3/2}-L(b)=3^62^7[L(a)]^2\geqslant 0.
\end{equation}
\tag{7}
$$
Geometrically, the solvability domain is bounded by the quadric $T(b)=0$, that is, a generalized cone, and by the image of the cubic $L(a)=0$ under the mapping of the space $E^5(a)$ onto $E^5(b)$ by means of the Monge-Ampère operator. The first condition means that the solvability domain lies in the complement of the generalized cone in the five-dimensional space and is a subdomain of this complement. The solvability domain can have several connected components. In this problem multidimensional geometry works for two-dimensional. Comparing the coefficients of like powers on both sides of the Monge-Ampère equation (provided that $f(x,y)$ is a homogeneous polynomial of degree 4) we obtain the system of five equations These equations define a mapping $\rho$ of the space $E^5(a)$ with coordinates $a_{40}, a_{31},\dots,a_{04}$ to the space $E^5(b)$ with coordinates $b_{40},\dots,b_{04}$. This author showed earlier that the Jacobian of this mapping can be expressed in terms of the above invariants as follows:
$$
\begin{equation}
J\biggl(\frac{b_{40},\dots,b_{04}}{a_{40},\dots,a_{04}}\biggr)= -2^{12}3^5T(a)L(a).
\end{equation}
\tag{13}
$$
In addition, the characteristic polynomial of the derived mapping $\rho'$ has coefficients expressed as follows in terms of these invariants:
$$
\begin{equation}
|\rho'-\lambda E|=-\lambda^5+\lambda^3\mu_3T(a)+\lambda^2\mu_2L(a)+\lambda \mu_1T^2(a)+\mu_0T(a)L(a)=0,
\end{equation}
\tag{14}
$$
where $\mu_3=2^43^77$, $\mu_2=2^73^6$, $\mu_1=-2^{10}3^2$ and $\mu_0=-2^{12}3^5$. We omit the proof of this result, since this equation is not required in what follows. If $f(x,y)$ is a complete polynomial (that is, all of its 15 coefficients are nonzero), then there is an additional system of equations for the coefficients $a_{ij}$. We consider this system below. In the following theorem we assume that $f(x,y)$ is a complete polynomial. Theorem 1. If the coefficients of the polynomial $f(x,y)$ satisfy If $b_{10}=c_1$, $b_{01}=c_2$ and $b_{00}=c_0$, then $z$ is a complete solution of the Monge-Ampère equation. It can be said that the result of the action of the Monge-Ampère operator on a function $z(x,y)$ presents the polynomial $f(x,y)$ up to linear terms. A constructive solution can be obtained for the whole system of equations. The system for the $a_{ij}$ is solved successively: first, a solution of (8)–(12) is found (the coefficients $a_{ij}$ of power 4 terms are found, then the coefficients of power 3 are evaluated, and so on). As the sum of indices $i+j$ corresponds to these powers, for simplicity we will say that $a_{ij}$ is a coefficient of degree $i+j$. The result of our analysis will be used in § 7 where we deal with the system of equations describing large deflection of a thin plate under internal stresses and an external load (see Landau and Lifschitz [4]). We hope that further advances in this direction will allow one to apply this theory to the description of Earth’s terrain or seismic events — in these problems Earth’s crust is regarded as a relatively thin plate in comparison with the size of the whole Earth. Note that degree 4 of polynomials corresponds to the following exceptional case. If $z(x,y) $ is a polynomial of degree $n$, then in general the result of the application of the Monge-Ampère operator is a polynomial of degree $2n-4$. However, none of the attempts to find similar invariants for the polynomials $T(b)$ and $L(b)$ for $n>4$ has come to fruition, For $ n=5$ the action of the Monge-Ampère operator on polynomials of degree 5 produces a hypersurface in $E^7$ defined parametrically by the coefficients $a_{ij}$; for $n>5$ this hypersurface is a submanifold of codimension $n-4$. The study of such submanifolds is of certain interest. § 2. Constructing a solution of the basic system (8)–(12)This system contains five nonlinear equations and five unknowns. We use the method of systematic elimination. First we eliminate $a_{31} $ and $a_{13}$. We set
$$
\begin{equation*}
\tau=\frac{b_{22}}{3}\quad\text{and} \quad t=\sqrt{\frac{T(b)}{3}}.
\end{equation*}
\notag
$$
The relation between $T(b)$ and $T(a)$ can be written as
$$
\begin{equation}
a_{13}a_{31}=4a_{40}a_{04}+\frac{a_{22}^2}{3}\pm\frac{t}{12}.
\end{equation}
\tag{20}
$$
Substituting this expression into the left-hand side of equation (10) we obtain where $\kappa=\tau\pm{t}/{2}$. Consider the subsystem of equations (9) and (11) for $a_{31}$ and $a_{13}$. The determinant of this system is We have
$$
\begin{equation}
a_{31}=\frac{6a_{40}b_{13}+a_{22}b_{31} }{6\kappa}\quad\text{and} \quad a_{13}=\frac{6a_{04}b_{31}+a_{22}b_{13}}{6\kappa}.
\end{equation}
\tag{23}
$$
The case $\kappa=0$ is considered separately. From (8) and (12), for the squares we obtain
$$
\begin{equation}
a_{31}^2=\frac{24a_{40}a_{22}-b_{40}}{9}, \qquad a_{13}^2=\frac{24a_{04}a_{22}-b_{04}}{9}.
\end{equation}
\tag{24}
$$
We express $a_{22}^2$ and $a_{31}a_{13}$ in terms of $a_{40}a_{04}$. From (21) we obtain Substituting this expression into (20) we have where $\overline\kappa=\frac{1}{6}(\tau+t)$. Multiplying $a_{31}$ and $a_{13}$ and using (24) we get that
$$
\begin{equation}
36a_{40}a_{04}(b_{31}b_{13}-16\kappa^2)+a_{22}^2b_{31}b_{13}+6 a_{22}(a_{40}b_{13}^2+a_{04}b_{31}^2)=-36\overline\kappa\kappa^2.
\end{equation}
\tag{27}
$$
Substituting (25) for $a_{22}^2$ into this equation we obtain
$$
\begin{equation}
72a_{40}a_{04}(b_{31}b_{13}-8\kappa^2) +6a_{22}(a_{40}b_{13}^2+a_{04}b_{31}^2)=\frac{\kappa b_{31}b_{13}}{2}-36\overline\kappa\kappa^2.
\end{equation}
\tag{28}
$$
Recalling the expressions in (23) and (24) for $a_{31}$, $a_{13}$ and their squares we have the two further equations
$$
\begin{equation}
36a_{04}^2b_{31}^2+12 a_{04}a_{22}b_{31}b_{13}+a_{22}b_{13}^2=(24a_{04}a_{22}-b_{04})4\kappa^2
\end{equation}
\tag{29}
$$
and
$$
\begin{equation}
36a_{40}^2b_{13}^2+12a_{40}a_{22}b_{31}b_{13}+a_{22}^2b_{31}^2 =(24a_{40}a_{22}-b_{40})4\kappa^2.
\end{equation}
\tag{30}
$$
Multiplying both parts of (29) by $a_{40}$, multiplying (30) by $a_{04}$, and subtracting the second equation from the first we obtain
$$
\begin{equation}
(36a_{40}a_{04}-a_{22}^2)(a_{04}b_{31}^2-a_{40}b_{13}^2) =-(b_{04}a_{40}-b_{40}a_{04})4\kappa^2.
\end{equation}
\tag{31}
$$
Using (22) to replace the first bracket in the left-hand side and cancelling by $\kappa$ we obtain
$$
\begin{equation}
a_{04}b_{31}^2-a_{40}b_{13}^2+(b_{04}a_{40}-b_{40}a_{04})8\kappa=0,
\end{equation}
\tag{32}
$$
which relates $a_{40}$, $a_{04}$ and $b_{ij}$ only. In what follows, to keep the transformation symmetric, we introduce the new variables $\xi$ and $ \eta$ by
$$
\begin{equation}
a_{40}=\frac{\xi+\eta}{2}\quad\text{and} \quad a_{04}=\frac{\xi-\eta}{2}.
\end{equation}
\tag{33}
$$
By (32), $\xi$ and $\eta$ are related by
$$
\begin{equation}
\eta=\xi\frac{b_{31}^2-b_{13}^2-(b_{40}-b_{04})8\kappa} {b_{31}^2+b_{13}^2-(b_{40}+b_{04})8\kappa}.
\end{equation}
\tag{34}
$$
Assume that either the numerator or the denominator of the fraction on the right is nonzero. (The case of special interest when they both vanish, that is, $b_{31}^2=8\kappa b_{40}$ and $b_{13}^2=8\kappa b_{04}$, is considered below.) We turn back to (29) and (30). Multiplying these equations by $a_{40}$ and $a_{04}$, respectively, and summing we have
$$
\begin{equation}
\begin{aligned} \, \notag &(36a_{40}a_{04}+a_{22}^2)(a_{04}b_{31}^2+a_{04}b_{13}^2) +24a_{40}a_{04}a_{22}b_{31}b_{13} \\ &\qquad =[48a_{40}a_{04}a_{22}-(b_{04}a_{40}+b_{40}a_{04})]4\kappa^2. \end{aligned}
\end{equation}
\tag{35}
$$
Collecting the coefficients of $a_{22}$ we have
$$
\begin{equation}
\begin{aligned} \, &(36a_{04}a_{04}+a_{22}^2)(a_{04}b_{31}^2+a_{04}b_{13}^2) +24a_{40}a_{04}a_{22}(b_{31}b_{13}-8\kappa^2) \nonumber \\ &\qquad=-(b_{04}a_{40}+b_{40}a_{04})4\kappa^2. \end{aligned}
\end{equation}
\tag{36}
$$
Now using equation (28) we replace $a_{40}a_{04}(b_{31}b_{13}-8\kappa^2)$ in the second factor on the left in (36). This results in the expression $(36a_{40}a_{04}-a_{22}^2)$ in the first term. Using (22) for this expression, we obtain
$$
\begin{equation}
\frac{\kappa}{2}(a_{04}b_{31}^2+a_{40}b_{13}^2) +a_{22}\Bigl(\frac{\kappa}{6}b_{31}b_{13}-12\overline\kappa\kappa^2\Bigr) =-(b_{04}a_{04}+b_{40}a_{04})4\kappa^2 .
\end{equation}
\tag{37}
$$
Cancelling the factor of $\kappa$ and making change to $\xi$ and $\eta$ we have
$$
\begin{equation}
\begin{aligned} \, &3\xi(b_{31}^2+b_{13}^2+(b_{40}+b_{04})8\kappa) +3\eta(b_{13}^2-b_{31}^2+(b_{04}-b_{40})8\kappa) \nonumber \\ &\qquad +2a_{22}(b_{31}b_{13}-72\overline\kappa\kappa)=0. \end{aligned}
\end{equation}
\tag{38}
$$
From this equation we can express $a_{22}$ in terms of $\xi$ and $\eta$, and then use (34) to express $a_{22}$ in terms of $\xi$ alone. Setting $b_{31}b_{13}=\beta$ and $b_{31}^2+b_{13}^2=\alpha$ we have
$$
\begin{equation}
a_{40}a_{04}=\frac{\xi^2-\eta^2}{4} =\xi^2\frac{\beta^2-8\kappa(b_{31}^2b_{04}+b_{13}^2b_{40})+4b_{40}b_{04}8 \kappa^2}{(\alpha-(b_{04}+b_{40})8\kappa)^2}.
\end{equation}
\tag{39}
$$
Substituting the expressions for $a_{40}a_{04}$ and $a_{22}^2$ into equation (20) we obtain the required equation for $\xi^2$:
$$
\begin{equation}
\begin{aligned} \, \notag & 36\xi^2[(\beta-72\overline\kappa\kappa)^2 (\beta^2-8\kappa(b_{31}^2b_{04}+b_{13}^2b_{40})+8^2\kappa^2b_{40}b_{04})- (\beta^2-8^2\kappa^2b_{40}b_{04})^2] \\ &\qquad =\frac{\kappa}{2}(\beta-72\overline\kappa\kappa)^2 (\alpha-(b_{40}+b_{04})8\kappa)^2. \end{aligned}
\end{equation}
\tag{40}
$$
This equation has a solution if the coefficient $\xi^2$ has the same sign as $\kappa$. This should be secured by the second sufficient condition in (15). The coefficient being quite involved, the general case is fairly difficult to deal with. But this property does indeed hold. To prove this we need an additional construction in the next section. Note that all coefficients $a_{ij}$ can be found from equation (40). § 3. Transformation of the polynomial $f(x,y)$The coefficient on the left in (40) can be handled more easily if some of them vanish. In particular, the case when $\beta=b_{31}b_{13}=0$ can be treated conveniently. The general situation can be reduced to this case by a shift in the $(x,y)$-plane. Suppose that $b_{40}^2+b_{04}^2\ne 0$. For example, let $b_{40}\ne 0$. Consider the univariate polynomial Changing the variable $x=u-\sigma$, where $\sigma=b_{31}/(4b_{40})$, and substituting we obtain the polynomial of $u$ with coefficients $c_{ij}$, where
$$
\begin{equation*}
\begin{gathered} \, c_{40}=b_{40}, \qquad c_{31}=-4\sigma b_{40}+b_{31}=0, \\ c_{22}=6\sigma^2b_{40}-3\sigma b_{31}+b_{22}, \\ c_{13}=-4\sigma^3b_{40}+3\sigma^2b_{31}-2\sigma b_{22}+b_{13} \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
c_{04}=\sigma^4b_{40}-\sigma^3b_{31}+\sigma^2b_{22}-\sigma b_{13}+b_{04}.
\end{equation*}
\notag
$$
We have
$$
\begin{equation*}
\begin{aligned} \, &T(c)=4c_{40}c_{04}+\frac{c_{22}^2}{3}- c_{31}c_{13}=4b_{40}b_{04}+\frac{b_{22}^2}{3}-b_{31}b_{13} \\ &\quad +\sigma^4\biggl(4b_{40}^2+\frac{36}{3}b_{40}^2-16b_{40}^2\biggr) -\sigma^3(-4b_{40}b_{31}-12b_{40}b_{31}+4b_{40}b_{31}+12b_{40}b_{31}) \\ &\quad +\sigma^2(\dots)+\sigma(\dots)=T(b). \end{aligned}
\end{equation*}
\notag
$$
Here the coefficient of each power of $\sigma$ is equal to zero. Let us check that $L(b)$ is invariant under the shift by $\sigma=b_{31}/(4b_{40})$. We have
$$
\begin{equation*}
L(c)=\biggl(\frac{c_{22}}{3}\biggr)^3-\frac{c_{22}}{3} \biggl(4c_{40}c_{04} +\frac{c_{31}c_{13}}{2}\biggr) +\frac{c_{31}^2c_{04}+c_{13}^2c_{40}}{2}.
\end{equation*}
\notag
$$
Substituting in the above expressions for the coefficients $c_{ij}$ and using the condition $c_{31}=0$ we find that
$$
\begin{equation*}
\begin{aligned} \, L(c)7&=\biggl(-\frac{b_{31}^2}{8b_{40}}+\frac{b_{22}}{3}\biggr)^3 -\biggl(-\frac{b_{31}^2}{8b_{40}}+\frac{b_{22}}{3}\biggr) \biggl[4b_{40}\biggl(-\frac{3b_{31}^4}{4^4b_{40}^3}+ \frac{b_{31}^2b_{22}}{(4b_{40})^2} \\ &\qquad\qquad -\frac{b_{31}b_{13}}{4b_{40}}+b_{04}\biggr)\biggr] +\biggl[\frac{b_{31}^3}{8b_{40}^2}-\frac{b_{31}b_{22}}{2b_{40}} +b_{13}\biggr]^2\frac{b_{40}}{2} \\ &=\frac{b_{31}^6}{b_{40}^3}\biggl[-\frac{1}{8^3} -\frac{3}{4^42}+\frac{1}{8^22}\biggr] +\frac{b_{22}^2b_{31}^2}{b_{40}} \biggl[-\frac{1}{24}-\frac{1}{12}+\frac{1}{8}\biggr] \\ &\qquad+\frac{b_{22}b_{31}^4}{b_{40}^2} \biggl[\frac{1}{8^2}+\frac{1}{4^22}+\frac{1}{4^3}-\frac{1}{16}\biggr] -\frac{b_{22}b_{31}b_{13}}{6} +\biggl(\frac{ b_{22}}{3}\biggr)^3 \\ &\qquad-\frac{b_{22}4b_{40}b_{04}}{3}+ \frac{1}{2}(b_{31}^2b_{04}+b_{13}^2b_{40})=L(b). \end{aligned}
\end{equation*}
\notag
$$
Note that each expression in square brackets vanishes. Therefore, $L(c)=L(b)$. So the polynomials $T(b)$ and $L(b)$ are shift invariant. It can also be shown easily that these polynomials are invariant under rotations of the $(x,y)$-plane. Let us rewrite equation (40) using the fact that $b_{13}b_{31}=0$.
§ 4. Compatibility conditionsAssume, for example, that $b_{31}=0$. In this case equation (40) has the form
$$
\begin{equation}
36\xi^2[(72\overline\kappa\kappa)^28\kappa b_{40}(8\kappa b_{04}-b_{13}^2)-(8^2\kappa^2b_{40}b_{04})^2] =\frac{1}{2}\kappa(72\overline\kappa\kappa)^2 (\alpha-(b_{40}+b_{04})8\kappa)^2.
\end{equation}
\tag{41}
$$
Note that both parts of equation (41) involve the factor $\kappa^3$. Let $A$ be the expression in square brackets after dividing by $\kappa^3$. Using the above notation for $\tau$ and $t$ we obtain
$$
\begin{equation}
A=8^3\biggl[b_{40}b_{04}(9(\tau+t)^2-4b_{40}b_{04})2 \biggl(\tau+\frac{t}{2}\biggr)-\frac{9}{4}(\tau+t)^2b_{13}^2b_{40}\biggr].
\end{equation}
\tag{42}
$$
We have that is, $3(t^2-\tau^2)=4b_{40}b_{04}$. Substituting this expression into (42), after some elaborate transformations we have
$$
\begin{equation*}
A=8^3(\tau+t)^2\frac{9}{2}[(t-\tau)(2\tau+t)^2-b_{13}^2b_{40}].
\end{equation*}
\notag
$$
Note that $\overline\kappa=\frac{1}{6}(\tau+t)$. Both parts of equation (41) involve the factor $(72\overline\kappa)^2$. After cancellation, we obtain
$$
\begin{equation}
2^73^2\xi^2\biggl[(t-\tau)(2\tau+t)^2-\frac{b_{13}^2b_{40}}{2}\biggr] =(b_{31}^2 +b_{13}^2-(b_{40}+b_{04})8\kappa)^2.
\end{equation}
\tag{43}
$$
Let us show that in this equation the sign of the coefficient of $\xi^2$ is controlled by the second expression in the sufficient conditions of Theorem 1, that is, by inequality (15). In the case when $b_{31}=0$ this inequality has the form
$$
\begin{equation}
\biggl(\frac{T(b)}3\biggr)^{3/2}-L(b)=t^3-\tau^3+\tau(4b_{40}b_{04}) -\frac{b_{13}^2b_{40}}{2}>0.
\end{equation}
\tag{44}
$$
Substituting in $4b_{40}b_{04}=3(t^2-\tau^2)$, on the right-hand side of inequality (44) we have
$$
\begin{equation}
(t-\tau)[t^2+t\tau+\tau^2+3\tau(t+\tau)]-\frac{b_{13}^2b_{40}}{2} =(t-\tau)(t+2\tau)^2-\frac{b_{13}^2b_{40}}{2}.
\end{equation}
\tag{45}
$$
This is exactly the expression in square brackets in (43). By the hypotheses of Theorem 1 this expression is positive. Hence (43) is solvable with respect to $\xi$. Having the solution for the simplified polynomial at our disposal, we can transform this solution using the inverse transformation, thereby obtaining a solution for the original polynomial. But we have found an equation for $\xi^2$ for the general polynomial without simplification. Using this equation we can obtain in succession the coefficients of the required polynomial $z(x,y)$ under conditions (15). A transition to a simpler polynomial is used only for the proof of the solvability of equation (40). So we do not need this transformation in delivering the required solution.
§ 5. Special casesWe have several special cases to consider. Let us show that, in each case, either there is no solution, or conditions (15) are not met. 1. Let $\kappa=0$. Then $t=-2b_{22}/3$. Assume that $b_{13}^2+b_{13}^2\ne 0$. Using the above notation we have From (22) we obtain $36a_{40}a_{04}=a_{22}^2$. Hence $a_{40}$ and $a_{04}$ have the same sign. Let these coefficients be positive, and let $a_{22}=6\sqrt{a_{40}a_{04}}$. From equations (9) and (11) we obtain
$$
\begin{equation*}
72\sqrt{a_{40}}(\sqrt{a_{40}}a_{13}-\sqrt{a_{04}}a_{31})=b_{31}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
72\sqrt{a_{04}}(\sqrt{a_{04}}a_{31}-\sqrt{a_{40}}a_{13})=b_{13}.
\end{equation*}
\notag
$$
Therefore, We have
$$
\begin{equation}
a_{40}=-\frac{a_{22}b_{31}}{6b_{13}}\quad\text{and} \quad a_{04}=-\frac{a_{22}b_{13}}{6b_{31}}.
\end{equation}
\tag{47}
$$
Equations (8) and (12) can be written as and Using (24), we can express $(a_{31}a_{13})^2$ in terms of $a_{22}$ and $b_{ij}$. Next, from (10) we obtain An appeal to (46)–(48) produces the final equation
$$
\begin{equation}
16a_{22}^2\bigl[b_{22}b_{31}b_{13}-(b_{31}^2b_{04}+b_{13}^2b_{40})\bigr] =(b_{31}b_{13})^2.
\end{equation}
\tag{49}
$$
Note that in the derivation of this equation we used that fact that the terms involving $a_{22}^4$ annihilate. Let $F$ denote the expression in square brackets. That this the expression is positive follows from condition (15). In fact, in view of (4) we can write the second condition in (15) as
$$
\begin{equation*}
\begin{aligned} \, &t^3-\tau^3+\tau\biggl(4b_{40}b_{04}+\frac{b_{31}b_{13}}{2}\biggr) -\frac{b_{31}^2b_{04}+b_{13}^2b_{40}}{2} \\ &\qquad =-8\tau^3 - \tau^3+\tau\biggl(9\frac{b_{22}^2}{9}+b_{31}b_{13}+\frac{b_{31}b_{13}}{2}\biggr) - \frac{b_{31}^2b_{04}+b_{13}^2b_{40}}{2}=\frac{1}{2}F>0. \end{aligned}
\end{equation*}
\notag
$$
Hence equation (49) has a solution. Using (46), (47) and (49) we obtain the following explicit expressions for $a_{31}$ and $a_{13} $:
$$
\begin{equation*}
a_{31}=\pm\frac{b_{31}b_{22}-2b_{13}b_{40}}{6\sqrt F}\quad\text{and} \quad a_{13}=\pm\frac{b_{13}b_{22}-2b_{31}b_{04}}{6\sqrt F}.
\end{equation*}
\notag
$$
2. The case $b_{40}=b_{04}=b_{22}=0$. If the coefficients $b_{31}$ and $b_{13}$ are of the same sign, then no solution exists. Indeed, in this case $T(b)=-b_{31}b_{13}<0$. But if the polynomial $b$ is the result of the action of the Monge-Ampère operator on a polynomial $a$, then $T(b)>0$. A solution exists if the coefficients $b_{31}$ and $b_{13}$ are of different sign in our case. Let us show this. From equations (8) and (12) we obtain
$$
\begin{equation*}
a_{31}=\pm\sqrt{\frac{8}{3}a_{40}a_{22}}\quad\text{and} \quad a_{13}=\pm\sqrt{\frac{8}{3}a_{04}a_{22}}.
\end{equation*}
\notag
$$
Therefore, $a_{31}a_{13}=\pm\frac{8}{3}\sqrt{a_{40}a_{04}}a_{22}$. For example, consider the minus sign. Substituting this expression into (10) we obtain
$$
\begin{equation*}
a_{22}^2+4\sqrt{a_{40}a_{04}}a_{22}- 12a_{40}a_{04}=0.
\end{equation*}
\notag
$$
This gives us two values,
$$
\begin{equation*}
(a_{22})_1= 2\sqrt{a_{40}a_{04}}\quad\text{and} \quad (a_{22})_2=-6\sqrt{a_{40}a_{04}}.
\end{equation*}
\notag
$$
It suffices to consider the first root. Using the relation between $T(a)$ and $T(b)$ in (20) we find that
$$
\begin{equation*}
(a_{40}a_{04})_1=\frac{t}{128}\quad\text{and} \quad (a_{22})_1=\frac{\sqrt t}{4\sqrt 2}.
\end{equation*}
\notag
$$
Next, from (9) and (11) we obtain
$$
\begin{equation*}
\begin{gathered} \, (a_{40})_1=\frac{b_{31}^2}{24t^{3/2}},\qquad (a_{04})_1=\frac{b_{13}^2}{24t^{3/2}},\qquad (a_{22})_1=\frac{\sqrt t}{4\sqrt 2}, \\ a_{31}=\frac{b_{31}}{3\sqrt{8t}} \quad\text{and} \quad a_{13}=\frac{b_{13}}{3\sqrt{8t}}. \end{gathered}
\end{equation*}
\notag
$$
For example, if
$$
\begin{equation*}
z=\frac{1}{8\sqrt2}\, x^4+\frac{1}{\sqrt{24}}\, x^3y+\frac{1}{4\sqrt 2}\, x^2y^2 -\frac{1}{\sqrt{24}}\, xy^3+\frac{1}{8\sqrt2}\, y^4,
\end{equation*}
\notag
$$
then the result of the action of the Monge-Ampère operator is as follows: In this case $t=1$, $b_{31}=-\sqrt3$ and $b_{13}=\sqrt3$. The case $b_{22}\ne 0$ is dealt with similarly. 3. Now consider the case when $b_{31}^2=8\kappa b_{40}$ and $b_{13}^2=8\kappa b_{04}$. In this case the link between $\xi$ and $\eta$ disappears, but the second sufficient condition is not met. Indeed, consider the coefficient of $\xi^2$ in formula (40). Note that
$$
\begin{equation*}
\beta^2-8\kappa(b_{31}^2b_{04}+b_{13}^2b_{40}) +8^2\kappa^2b_{40}b_{04}=(b_{31}^2-8\kappa b_{40})(b_{13}^2-8\kappa b_{04})=0
\end{equation*}
\notag
$$
and Hence the coefficient of $\xi^2$ in (40) is zero. On the other hand, we have an expression for this coefficient in the form (44), that is, in this case which contradicts the fact that the expression on the left-hand side is positive.
§ 6. Existence of a solution of the Monge-Ampère equation for the complete polynomial $f(x,y)$It turns out that the sufficient conditions for the solvability of (15) that we used above when dealing with a homogeneous polynomial of degree 4 are also sufficient for the solvability of the general polynomial $f(x,y)$, which also involves polynomials of degrees 3, 2, $\dots$, with 15 coefficients in total. But, as mentioned in Theorem 1, this implies solvability up to linear terms of the polynomial $f(x,y)$. This author obtained in [5] a system of equations (0), (I), $\dots$, (IV) which expresses the coefficients $b_{ij}$ in terms of the $a_{ij}$. Assume that the coefficients $a_{40}$, $a_{31}$, $a_{22}$, $a_{13}$ and $a_{04}$ of degree 4 are already known. The subsystem of equations (III) for the coefficients $a_{30}$, $a_{21}$, $a_{12}$ and $a_{03}$ of degree 3 is as follows:
$$
\begin{equation}
Ba=b, \qquad a=\begin{pmatrix} a_{30}\\ a_{21}\\ a_{12}\\ a_{03} \end{pmatrix}, \qquad b=\begin{pmatrix} b_{30}\\ b_{21}\\ b_{12}\\ b_{03} \end{pmatrix}.
\end{equation}
\tag{III}
$$
The determinant of the matrix $B$ is
$$
\begin{equation*}
\begin{aligned} \, | B| &=\begin{vmatrix} 12a_{22}&-12a_{31}&24a_{40}&0\\ 36a_{13}&-12a_{22}&0&72a_{40}\\ 72a_{04}&0&-12a_{22}&36a_{31}\\ 0&24a_{04}&-12a_{13}&12a_{22} \end{vmatrix}=12^4\begin{vmatrix} a_{22}&-a_{31}&2a_{40}&0\\ 3a_{13}&-a_{22}&0&6a_{40}\\ 6a_{04}&0&-a_{22}&3a_{31}\\ 0&2a_{04}&-a_{13}&a_{22} \end{vmatrix} \\ &=12^49\biggl(4a_{40}a_{04}+\frac{a_{22}^2}{3}-a_{31}a_{13}\biggr)^2 =12^23T(b)>0. \end{aligned}
\end{equation*}
\notag
$$
Therefore, the system of equations for $a_{30},\dots,a_{03}$ has a solution. Now consider the system of equations for the coefficients $a_{20}$, $a_{11}$ and $a_{02}$ of second-power terms. We set
$$
\begin{equation*}
\begin{gathered} \, t_1=b_{20}-4(3a_{30}a_{12}-a_{21}^2), \\ t_2=b_{11}-4(9a_{30}a_{03}-a_{21}a_{12}) \end{gathered}
\end{equation*}
\notag
$$
and Now the system of equations (II) for $a_{20}$, $a_{11}$ and $a_{02}$ assumes the form
$$
\begin{equation}
C\begin{pmatrix} a_{20}\\ a_{11}\\ a_{02} \end{pmatrix} =\begin{pmatrix} t_1\\ t_2\\ t_3 \end{pmatrix}.
\end{equation}
\tag{II}
$$
The determinant of the matrix $C$ is
$$
\begin{equation*}
| C|=\begin{vmatrix} 4a_{22}&-6a_{31}&24a_{40} \\ 12a_{13}&-8a_{22}&12a_{31} \\ 24a_{04}&-6a_{13}&4a_{22} \end{vmatrix} =-2^53\begin{vmatrix} \dfrac{a_{22}}{3}&\dfrac{a_{31}}{2}&2a_{40} \\ \dfrac{a_{13}}{2}&\dfrac{a_{22}}{3}&\dfrac{a_{31}}{2} \\ 2a_{04}&\dfrac{a_{13}}{2}&\dfrac{a_{22}}{3} \end{vmatrix}=-2^53L(a)\ne 0.
\end{equation*}
\notag
$$
The last inequality here is secured by condition (15). Recall that $(T(b)/3)^{3/2}-L(b)=2^73^6L(a)^2>0$. Hence system (II) has a solution. This proves Theorem 1.
§ 7. Large deflections of thin platesLet us see how the above results can be applied to elasticity. In [4] Landau and Lifschitz derived (with a reference to Föppl, 1907) the following complete system of equations for large deflections of thin plates (see (14.6) and (14.7) in [7]):
$$
\begin{equation}
D\Delta^2\xi-h\biggl(\frac{\partial^2\chi}{\partial y^2}\,\frac{\partial^2\xi}{\partial x^2}-2\frac{\partial^2\chi}{\partial x\,\partial y}\,\frac{\partial^2\xi}{\partial x\,\partial y}+\frac{\partial^2\chi}{\partial x^2}\,\frac{\partial^2\xi}{\partial y^2}\biggr)=P,
\end{equation}
\tag{50}
$$
$$
\begin{equation}
\Delta^2\chi+E\biggl[\frac{\partial^2\xi}{\partial x^2}\frac{\partial^2\xi}{\partial y^2} -\biggl(\frac{\partial^2\xi}{\partial x\,\partial y}\biggr)^2\biggr]=0.
\end{equation}
\tag{51}
$$
Here $\chi$ is the Airy stress function, $\xi$ is the deflection of the plate, $P$ is the volume load per unit area of the plate, $D$, $h$ and $E$ are some positive constants, and $\Delta^2$ is the biharmonic operator:
$$
\begin{equation*}
\Delta^2=\frac{\partial^4}{\partial x^4}+2\frac{\partial^4}{\partial x^2\,\partial y^2}+\frac{\partial^4}{\partial y^4}.
\end{equation*}
\notag
$$
According to Landau and Lifschitz: “These equations are very complicated, and cannot be solved exactly, even in very simple cases. It should be noticed that they are nonlinear”. Consider these equations for polynomial functions. 1. Assume that the stress function $\chi$ is a polynomial of degree 4: The biharmonic operator takes a constant value at $\chi$: If this constant $c=-a^2$ is negative, then from equation (51) we have
$$
\begin{equation*}
\frac{\partial^2\xi}{\partial x^2}\,\frac{\partial^2\xi}{\partial y^2}-\biggl(\frac{\partial^2\xi}{\partial x\,\partial y}\biggr)^2=\frac{a^2}{E}.
\end{equation*}
\notag
$$
If the system of equations (50)–(51) is considered on the whole plane, then by Jörgens’s theorem $\xi$ is a quadratic polynomial. Let $a_{ij}$ be the coefficients of this polynomial. Consider equation (50). The value of the biharmonic operator at $\xi$ is zero, and the second derivatives of $\xi$ are constants. Hence the expression on the left-hand side of this equation is a polynomial of degree 2. Equation (51) assumes the form
$$
\begin{equation}
\begin{aligned} \, \notag &2(12c_{40}a_{02}-3c_{31}a_{11}+2c_{22}a_{20})x^2 +2(6c_{30}a_{02}-2c_{21}a_{11}+2c_{12}a_{20})x \\ \notag &\qquad +2(6c_{31}a_{02}-4c_{22}a_{11}+6c_{13}a_{20})xy +2(2c_{21}a_{02}-2c_{12}a_{11}+6c_{03}a_{20})y \\ &\qquad +2(2c_{22}a_{02}-3c_{13}a_{11}+12c_{04}a_{20})y^2 +2(2c_{20}a_{02}-c_{11}a_{11}+2c_{02}a_{20})= -\frac Ph. \end{aligned}
\end{equation}
\tag{53}
$$
Let us find out when the expression on the left-hand side is a constant number — this corresponds to the natural case of constant external load $P$. But this is only a particular case of external load. In some cases the load is known to vary along the plate. Note that this inconsistency between the external load, the internal stresses and the deflection of the pate may result in fractures of the plate, singular points, or some loss of regularity of the surface. In the case under consideration all the coefficients of powers of $x$ and $y$ vanish. Consider the system of equations obtained by equating to zero the coefficients of the second powers of $x$ and $ y$, where the $a_{ij}$ are the required coefficients. The determinant of this system is
$$
\begin{equation}
\begin{vmatrix} 12c_{40}&-3c_{31}&2c_{22} \\ 6c_{31}&-4c_{22}&6c_{13} \\ 2c_{22}&-3c_{13}&12c_{04} \end{vmatrix}=4^23^3L(c).
\end{equation}
\tag{54}
$$
A nonzero solution $a_{02}$, $a_{11}$, $a_{20}$ exists if and only if $L(c)=0$. In addition, we have the system of two more equations
$$
\begin{equation}
\begin{gathered} \, \notag 6c_{30}a_{02}-2c_{21}a_{11}+2c_{12}a_{20}=0, \\ 2c_{21}a_{02}-2c_{12}a_{11}+6c_{03}a_{20}=0. \end{gathered}
\end{equation}
\tag{55}
$$
So the external load is the same at all points in the plate if and only if the five vectors composed of the coefficients $c_{ij}$ lie in the same plane. The vector $\mathbf a=(a_{02},a_{11},a_{20})$ must be orthogonal to this plane. Equation (51) can be satisfied by choosing appropriately the length of the vector $\mathbf a$. Note that the last term on the left in (53) is the constant $-{P}/{h}$. So there can exist solutions of system (50)–(51) in which $\chi$ is a polynomial of degree 4, $\xi$ is a quadratic polynomial and $P$ is a constant number. 2. Consider the case when the constant in equation (51) is $c=a^2$. For example, let $a^2=E$. We have the equation In this case, a solution of equation (56) on the whole of the $(x$, $y)$-plane cannot be a quadratic polynomial. An example of such a solution was given by Kantor:
$$
\begin{equation*}
z(x,y)=xy+x\ln\bigl(x+\sqrt{x^2+e^{-2y}}\bigr)-\sqrt{x^2+e^{-2y}}.
\end{equation*}
\notag
$$
This example was constructed in connection with Efimov’s theorem (see [6]) stating that the normal image of a complete surface of negative curvature the Jacobian of whose normal mapping is negative and bounded away from zero should cover the whole plane, a half-plane, or a strip between parallel straight lines, In Kantor’s example, the normal image covers a half-plane (see [7]). Efimov’s theorem deals with mappings from the full $(x,y)$-plane to the $(p,q)$-plane (or a subdomain of it) under the conditions that the Jacobian of the mapping is negative and its absolute value is greater than some constant plus the absolute value of the curl times a positive coefficient. It is still unknown whether the plane can be mapped to a strip under the Efimov conditions. For some interesting results in this direction, see [7] and [8]. Note that the general solution of equation (56) was given by Goursat in [9]. A simple example of a solution of the equation $rt-s^2=-1$ was proposed in [1].3. If $\chi$ is a polynomial of degree 8, then $\Delta^2\chi$ is a polynomial of degree 4. The coefficients $d_{ij}$ of this polynomial in terms of the coefficients $c_{ij}$ of degree 8 can be easily written down:
$$
\begin{equation*}
\begin{gathered} \, d_{40}=\lambda_1c_{80}+\lambda_2c_{62}+\lambda_3c_{44}, \\ d_{31}=\mu_1c_{71}+\mu_2c_{53}+\mu_3c_{35}, \\ d_{22}=\nu_1c_{62}+\nu_2c_{44}+\nu_1c_{26}, \\ d_{13}=\mu_3c_{53}+\mu_2c_{35}+\mu_1c_{17}, \\ d_{04}=\lambda_3c_{44}+\lambda_2c_{26}+\lambda_1c_{08}. \end{gathered}
\end{equation*}
\notag
$$
Here
$$
\begin{equation*}
\begin{gathered} \, \lambda_1=\frac{8!}{4!}, \qquad \lambda_2=\frac{6!}{3!}, \qquad \lambda_3=4!, \\ \mu_1=\frac{7!}{3!}, \qquad \mu_2=5!\,2, \qquad \mu_3=\frac{5!}{2}, \\ \qquad \nu_1=\frac{6!}{2}\quad\text{and} \quad \nu_2=4^23^22. \end{gathered}
\end{equation*}
\notag
$$
The invariants of $T(d)$ and $L(d)$ can be given in terms of these expressions. In this case one can assume that the unknown function $\xi$ in (51) is a polynomial of degree 4. A solution exists under the sufficient conditions (15) expressed in terms of $c_{ij}$. In equation (50), $\Delta^2\xi$ is a constant, and in the general case the second term on the left-hand side is a polynomial of degree 8. Hence the external load $ P$ in this case should be a polynomial of degree 8. For completeness, it is worth pointing out that if $c=0$, then the Hessian of the deflection function is zero, and its graph is a developable surface. 
Citation in format AMSBIB
\Bibitem{Ami23} |
|
|
Contact us: |
Terms of Use
|
Registration to the website |
Logotypes |
|