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Denis Serre
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Consider a linear homogeneous 2nd-order PDE in a convex planar domain $\Omega$ : $$a(x,y)\frac{\partial^2u}{\partial x^2}+2b(x,y)\frac{\partial^2u}{\partial x\partial y}+c(x,y)\frac{\partial^2u}{\partial y^2}=0.$$ The coefficients $a,b,c$ are smooth functions. Let me assume a non-degeneracy condition : $\Delta(x,y):=b^2-ac$ does not vanish. It has therefore a constant sign.

When the equation is elliptic ($\Delta<0$), the only convex solutions $u$ are affine. This is because if $A,B$ are positive definite symmetric matrices, then ${\rm Tr}(AB)>0$.

I therefore assume that the equation is hyperbolic : $\Delta>0$. It is not so much difficult (solve a Goursat problem) to prove that given a point $p\in\Omega$, the equation admits solutions $u$ that are strongly convex at $p$, meaning that ${\rm D}^2u(p)$ is positive definite. Such a solution is convex in a small neighbourhood of $p$.

Question: in general, does there exist a solution $u$ which is strongly convex over the whole domain $\Omega$ ? If not, are there known obstructions ?

Somehow, this is a question about $h$-principle, because we do have global maps $(x,y)\mapsto S$ with values in $2\times 2$ symmetric positive definite matrices, which satisfy $$as_{11}+2bs_{12}+cs_{22}\equiv0$$ and we may ask whether such sections are homotopic, within positive definite sections, to the Hessian of the solution.

Consider a linear homogeneous 2nd-order PDE in a convex planar domain $\Omega$ : $$a(x,y)\frac{\partial^2u}{\partial x^2}+2b(x,y)\frac{\partial^2u}{\partial x\partial y}+c(x,y)\frac{\partial^2u}{\partial y^2}=0.$$ The coefficients $a,b,c$ are smooth functions. Let me assume a non-degeneracy condition : $\Delta(x,y):=b^2-ac$ does not vanish. It has therefore a constant sign.

When the equation is elliptic ($\Delta<0$), the only convex solutions $u$ are affine. This is because if $A,B$ are positive definite symmetric matrices, then ${\rm Tr}(AB)>0$.

I therefore assume that the equation is hyperbolic : $\Delta>0$. It is not so much difficult (solve a Goursat problem) to prove that given a point $p\in\Omega$, the equation admits solutions $u$ that are strongly convex at $p$, meaning that ${\rm D}^2u(p)$ is positive definite. Such a solution is convex in a small neighbourhood of $p$.

Question: in general, does there exist a solution $u$ which is strongly convex over the whole domain $\Omega$ ? If not, are there known obstructions ?

Consider a linear homogeneous 2nd-order PDE in a convex planar domain $\Omega$ : $$a(x,y)\frac{\partial^2u}{\partial x^2}+2b(x,y)\frac{\partial^2u}{\partial x\partial y}+c(x,y)\frac{\partial^2u}{\partial y^2}=0.$$ The coefficients $a,b,c$ are smooth functions. Let me assume a non-degeneracy condition : $\Delta(x,y):=b^2-ac$ does not vanish. It has therefore a constant sign.

When the equation is elliptic ($\Delta<0$), the only convex solutions $u$ are affine. This is because if $A,B$ are positive definite symmetric matrices, then ${\rm Tr}(AB)>0$.

I therefore assume that the equation is hyperbolic : $\Delta>0$. It is not so much difficult (solve a Goursat problem) to prove that given a point $p\in\Omega$, the equation admits solutions $u$ that are strongly convex at $p$, meaning that ${\rm D}^2u(p)$ is positive definite. Such a solution is convex in a small neighbourhood of $p$.

Question: in general, does there exist a solution $u$ which is strongly convex over the whole domain $\Omega$ ? If not, are there known obstructions ?

Somehow, this is a question about $h$-principle, because we do have global maps $(x,y)\mapsto S$ with values in $2\times 2$ symmetric positive definite matrices, which satisfy $$as_{11}+2bs_{12}+cs_{22}\equiv0$$ and we may ask whether such sections are homotopic, within positive definite sections, to the Hessian of the solution.

added top-level tag; https://meta.mathoverflow.net/questions/1457/why-are-mo-tags-formatted-as-they-are
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Martin Sleziak
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Denis Serre
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Convex solutions of linear hyperbolic PDEs in a planar domain

Consider a linear homogeneous 2nd-order PDE in a convex planar domain $\Omega$ : $$a(x,y)\frac{\partial^2u}{\partial x^2}+2b(x,y)\frac{\partial^2u}{\partial x\partial y}+c(x,y)\frac{\partial^2u}{\partial y^2}=0.$$ The coefficients $a,b,c$ are smooth functions. Let me assume a non-degeneracy condition : $\Delta(x,y):=b^2-ac$ does not vanish. It has therefore a constant sign.

When the equation is elliptic ($\Delta<0$), the only convex solutions $u$ are affine. This is because if $A,B$ are positive definite symmetric matrices, then ${\rm Tr}(AB)>0$.

I therefore assume that the equation is hyperbolic : $\Delta>0$. It is not so much difficult (solve a Goursat problem) to prove that given a point $p\in\Omega$, the equation admits solutions $u$ that are strongly convex at $p$, meaning that ${\rm D}^2u(p)$ is positive definite. Such a solution is convex in a small neighbourhood of $p$.

Question: in general, does there exist a solution $u$ which is strongly convex over the whole domain $\Omega$ ? If not, are there known obstructions ?