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Let $u_0$ and $u_1$ to be smooth functions defined on $\Omega\subset\mathbb{R}^n$, consider the following system of equations

$$\triangle u_1 = C_1(\partial_{ij}u_0),$$ $$\triangle u_0 = C_0u_1,$$

where $\partial_{ij} = \frac{\partial^2}{\partial x_i\partial x_j}$ and $C_0,C_1$ are constants. The boundary of $\Omega$ is assumed to be smooth, and $u_0|_{\partial\Omega} = 0$ and $u_1|_{\partial\Omega} = 0$. My question is that if we can solve the system (in the sense of nonzero strong solutions) with the given boundary data?

First of all, it is very easy to see that this system can be solved if $C_0=0$. Maybe it somehow suggests that we should assume $C_0$ small? In addition, we can also assume $\Omega$ is anything you want in order to see what happens first. In addition, one can put any conditions on $C_0$ and $C_1$, except for them to be 0.

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  • $\begingroup$ Are you picking $i$ and $j$ or is that supposed to be the determinant of the Hessian? $\endgroup$
    – AHusain
    Commented Jun 13, 2016 at 1:48
  • $\begingroup$ @AHusain Let's fix $i,j$ for the moment. But it is ok to assume that it is the determinant of Hessian if that helps. $\endgroup$
    – M.G. Tsai
    Commented Jun 13, 2016 at 2:15
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    $\begingroup$ Just a note that inserting the determinant of the Hessian in the first equation would make the system fully non-linear making it MUCH more complicated than the linear system that you get by selecting some particular values of $i$ and $j$ or even taking linear combinations of such choices. So you would better have a very good reason if you want to involve the determinant of the Hessian here. $\endgroup$
    – Igor Khavkine
    Commented Jun 13, 2016 at 16:16
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    $\begingroup$ Let me just make precise Michael Renardy's remark: Consider the case where $n = 1$, $\Omega = [0,1]$. If as you wrote the equations are linear, we can reduce the problem to $\triangle u_1 = C_0 C_1 u_1$. With the boundary conditions you get non-zero solutions for a discrete set of negative values for the produce $C_0 C_1$. As written your system depends highly on the properties of the coefficients; please pin them down first. $\endgroup$
    – Willie Wong
    Commented Jun 13, 2016 at 18:13
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    $\begingroup$ "The constant $C_0$ is positive but it can be taken arbitrarily small." But that's the problem. If $C = C_0 C_1 > -\pi^2$ (in particular when $C_0$ arbitrarily small) there are no non-trivial solutions. But if $C = C_0 C_1 = -\pi^2$ you have $u_1 = \sin (\pi x)$. The existence of non-trivial solutions to your system in general would depend sensitively on the values of $C_0$ and $C_1$, and how they relate to the eigenvalues of the Laplacian on $\Omega$. So you need to be precise about what the constants should be and what values they can take. $\endgroup$
    – Willie Wong
    Commented Jun 13, 2016 at 20:02

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