Let $B(x,y) \geq 0$ be a function defined for $x, y \geq 0$ such that $B(x,0)=B(0,y)=0$ and $B''_{xx}\leq 0, B''_{yy}\leq 0$ (i.e. it is bicocncave function). I am looking for the solutions among of such functions of the following differential equation: $$ 2B''_{xx}B''_{yy}-(B''_{xy})^{2}=0 $$

Besides of the trivial solutions, I know that these type of functions $x^{\alpha}y^{\beta}$ solve the problem (for appropriate choice of $\alpha, \beta$). I am wondering if there are other solutions and if somebody can describe all possible solutions.

I know how to solve homogeneous Monge–Ampère equation, and unfortunately any technique used in that case does not apply here (as far as I understand).


Update 1, Q2:

Describe all nonnegative functions $B(x_{1},x_{2},x_{3})$, defined for $x_{1},x_{2},x_{3} \geq 0$, such that $B$ is concave function with respect to each variable i.e. $B_{x_{j}x_{j}}\leq 0$ and $B$ satisfies the following PDE:

\begin{align*} \det \left( {\begin{array}{ccc} 2B_{x_{1}x_{1}} & B_{x_{1}x_{2}} &B_{x_{1}x_{3}} \\ B_{x_{2}x_{1}} & 2B_{x_{2}x_{2}} &-B_{x_{2}x_{3}}\\ B_{x_{3}x_{1}} & -B_{x_{3}x_{2}} &2B_{x_{3}x_{3}}\\ \end{array} } \right)=0 \end{align*}

Note that besides the trivial solutions, the function $B(x_{1},x_{2},x_{3})=Cx_{1}^{2/3}x_{2}^{2/3}x_{3}^{2/3}$ satisfies the above conditions.

  • $\begingroup$ What does the notation $B''_{xx}$ mean? Is it just $B_{xx}$? $\endgroup$ May 17, 2014 at 9:30
  • $\begingroup$ yes, it is just second derivative $B''_{xx}=\frac{\partial^{2} B}{\partial x^{2}}$ $\endgroup$ May 17, 2014 at 9:40

1 Answer 1


I had a little time on a flight today to think about your problem, and so I applied the standard integration method to see whether or not your equation could be explicitly integrated (in the sense that the Monge-Ampère equation $u_{xx}u_{yy}-{u_{xy}}^2=1$ can be integrated by transforming it to Laplace's equation). The answer is that it cannot be integrated that way. However, one can make the problem equivalent to a linear one, at least locally, that has an explicit solution in series, and this may or may not help you. I'll record the results here, just in case you find it useful.

Suppose that we have a solution $B(x,y)$ on some simply connected domain $D$ in the $xy$-plane and suppose that it satisfies $B_{xx} = -2p^2 < 0$, $B_{yy} = -q^2 < 0$, and $B_{xy} = 2pq$ for some positive functions $p$ and $q$. (One can also treat the case when one of $p$ or $q$ is negative, but that's a minor variation that I'll leave to you.) Then, we have $$ \mathrm{d} B_x = -2p^2\,\mathrm{d} x + 2pq\,\mathrm{d} y \quad\text{and}\quad \mathrm{d} B_y = 2pq\,\mathrm{d} x - q^2\,\mathrm{d} y $$ Thus, the functions $p$ and $q$ must satisfy $$ \mathrm{d}\bigl(-2p^2\,\mathrm{d} x + 2pq\,\mathrm{d} y\bigr) =\mathrm{d}\bigl(2pq\,\mathrm{d} x - q^2\,\mathrm{d} y\bigr) = 0, $$ and, conversely, if $p$ and $q$ satisfy these two conditions (which are first order PDE), then the above equations determine $B_x$ and $B_y$ (up to an additive constant) and then $\mathrm{d}B = B_x\,\mathrm{d}x + B_x\,\mathrm{d}y$ determines $B$ up to an additive constant, so the two PDE on $p$ and $q$ are essentially equivalent to the original equations.

Next, make a change of variables: Set $s = px$ and $t = qy - px$ and then $p= e^u$ and $q = e^v$. Then set $w = s + it$ and $z = -\tfrac12(v-iu)$. Then the above two PDE relating $(x,y,p,q)$ simply become the real and imaginary parts of the linear elliptic complex equation $$ \frac{\partial w}{\partial \bar z} = \bar w $$ Now, this equation is known not to be integrable by the method of Darboux, so, in particular, you cannot write down its general solution in terms of a single holomorphic function of $z$. However, all the $C^1$ solutions are real analytic (because the equation is elliptic), and there is an explicit representation of the analytic solution in terms of power series: $$ w(z)=\sum_{k=0}^\infty c_k\ f^{(k)}(z\bar z) z^k+\overline{c_k}\ f^{(k+1)}(z\bar z) \bar z^{k+1} $$ where $f^{(k)}$ is the $k$-th derivative of the modified Bessel $I$-function whose series representation is $$ f(r) = 1 + \sum_{j=1}^\infty \frac{r^j}{(j!)^2} $$ Using this representation, you can trace back through and integrate by parts to get series representations of $B(x,y)$, $x$, and $y$ in terms of $u$ and $v$, which gives you a graphical representation of the local solutions in this case.

Whether this will help you with the global solutions, I don't know. However, it does give you the 'general' solution.

  • $\begingroup$ Thanks. We used your idea of linearization this Monge--Ampere type equation and eventually succeeded to solve two dimensional case. Now we are wondering how to solve three dimensional case (which was initial problem) and unfortunately this trick of linearization does not work anymore. We hope that maybe you can give some idea how to linearize the differential equation. So the question is the following: we need to solve the following equation $\endgroup$ May 27, 2014 at 11:14
  • $\begingroup$ Here is the equation: \begin{align*} \det \left( {\begin{array}{ccc} 2B_{x_{1}x_{1}} & B_{x_{1}x_{2}} &B_{x_{1}x_{3}} \\ B_{x_{1}x_{2}} & 2B_{x_{2}x_{2}} &-B_{x_{2}x_{3}}\\ B_{x_{3}x_{1}} & -B_{x_{3}x_{2}} &2B_{x_{3}x_{3}}\\ \end{array} } \right) =0 \end{align*} where $B(x_{1},x_{2},x_{3})$ is concave function with respect to each variable and it is given for $x_{1},x_{2},x_{3} \geq 0$. Assume $B_{x_{i}x_{j}}\geq 0$ for $i\neq j$ and $B=0$ if one of the coordinate is zero. One can see that the function $B(x_{1},x_{2},x_{3})=Cx_{1}^{2/3}x_{2}^{2/3}x_{3}^{2/3}$ works. $\endgroup$ May 27, 2014 at 11:49
  • $\begingroup$ I'm glad the 2D information was useful; I may think about the 3D case. There are methods that give something useful in higher dimensions, but, as you say, the simple linearization idea that worked in the 2D case does not. It would help if you could give some motivation for wanting to solve this problem. Not only does motivation often suggest a good approach, it makes the problem more interesting (as opposed to being some random PDE to solve). N.B.: Instead of adding to the problem in the comments, you really should append it to the question, which helps keep things together better. $\endgroup$ May 27, 2014 at 15:40
  • $\begingroup$ This is about Brascamp-Lieb inequality which is better than Holder's inequality. Right now we are trying to prove that in the case $n=3$ there is only one (or maybe not, any answer is interesting) Brascamp-Lieb inequality which corresponds to the function $B(x,y,z)=x^{1/p_{1}}y^{1/p_{2}}z^{1/p_{3}}$ where $\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}} = 2$ and $1 \leq p_{j} \leq \infty$. I will add my second question to my first question. We also have homogeneity for the function $B$, it is homogeneous of degree 2. $\endgroup$ May 27, 2014 at 17:34

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