Let $(M,g)$ be a closed (compact and without boundary) and oriented Riemannian manifold and let us consider the Poisson equation for a smooth function $\varphi$:

$\Delta \phi = f$,

where $f$ is a fixed smooth function on $M$. It is then well-known that the previous Poisson equation has a solution if and only if

$\int_M f = 0$,

and in that case, the solution is unique up to the addition of a constant. I was wondering if this result generalizes to a system of Poisson equations for smooth functions $\phi_1$ and $\phi_2$:

$\Delta \phi_1 = f_1 \phi_1 +f_2 \phi_2\, , \qquad \Delta \phi_2 = g_1 \phi_1 + g_2 \phi_2$

for fixed functions $f_1$, $g_1$, $f_2$ and $g_2$. Clearly, a necessary condition for $\phi_1$ and $\phi_2$ to be solutions is:

$ \int_M (f_1 \phi_1 +f_2 \phi_2) = 0\, , \qquad \int_M (g_1 \phi_1 + g_2 \phi_2) = 0$

I wonder if the converse is also true. I am sure this problem has been extensively studied, but I have not found a reference about it.


  • $\begingroup$ $\Delta\phi=f$ is inhomogeneous, while $\Delta\phi^i=f^i{}_j\phi^j$ is homogeneous. The second problem is slightly different. You can tell this from the fact that your "necessary conditions" involve $\phi^i$, so they are not integrability conditions on $f^i{}_j$. $\endgroup$ – AccidentalFourierTransform Mar 30 '18 at 17:56
  • 1
    $\begingroup$ The problem with your "necessary conditions" is that they depend on the unknown solution. If you are looking for something more useful, I do not think you can say much beyond what you get from Fredholm theory. For generic choices of the $f_i$ and $g_i$ you can expect a unique solution, but you could also have, for instance, $f_2=g_1=0$ and $f_1$ and $g_2$ equal to eigenvalues of the Laplacian. $\endgroup$ – Michael Renardy Mar 30 '18 at 17:59
  • $\begingroup$ @MichaelRenardy Thanks. Could you elaborate a bit on "beyond what you get from Fredholm theory"? Indeed the "necessary conditions" depend on the solution, but even so, I thought they could be sufficient, in the sense that if $\phi_1$ and $\phi_2$ satisfy the necessary conditions then they are actually solutions. $\endgroup$ – Bilateral Mar 30 '18 at 18:34
  • $\begingroup$ No way that could be sufficient. Consider the case where there exists some open $\Omega \subset M$ on which $f_1 = f_2 = g_1 = g_2 = 0$. Then any $\phi_1, \phi_2\in C^\infty_c(\Omega)$ satisfies the "necessary condition", but only one of them is the true solution. $\endgroup$ – Willie Wong Mar 30 '18 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.