I am trying to solve a system of first-order linear PDEs. In the best case, I would want to solve it explicitly but proving that there exists a (unique) solution would also be helpful.

Let $\Omega \subset \mathbb{R}^{2}$ be a bounded and connected domain with a smooth boundary $\Gamma$. We assume that a smooth function $\phi:\Omega \rightarrow \mathbb{R}$ which is zero at the boundary is given. Then, I am looking for a function $u:\Omega \rightarrow \mathbb{R}$ such that \begin{align} \nabla u = \begin{bmatrix} \phi_{x}^{2} \\ \phi_{x}\phi_{y} \end{bmatrix} \qquad & \text{ in } \Omega, \\ u=g \qquad & \text{ on } \Gamma, \end{align} where $g$ is an arbitrary smooth and bounded function. Note we use the notation $\phi_{x} = \partial_{x}\phi=\frac{\partial \phi}{\partial x}$ and $\phi_{y} = \partial_{y}\phi=\frac{\partial \phi}{\partial y}$ here.

Now let me explain what I have done so far. I tried to solve the problem by integrating the individual equations, $$ \partial_{x}u=\phi_{x}^{2},$$ $$ \partial_{y}u=\phi_{x}\phi_{y}.$$ This way I get \begin{align*} u &= \int \phi_{x}^{2}(x,y)dx + C_{1}(y) \\ u &= \int \phi_{x}(x,y)\phi_{y}(x,y) dy + C_{2}(x). \end{align*}

Using integration by parts I get, \begin{align*} u &= \phi\phi_{x}-\int \phi(x,y)\phi_{xx}(x,y)dx, + C_{1}(y), \\ u &= \phi\phi_{x}-\int \phi(x,y)\phi_{xy}(x,y)dy + C_{2}(x). \end{align*}

Now we would try to find $C_{1}$ and $C_{1}$ based on the boundary conditions.

However, the main conclusion is that the PDE has a solution if for all $(x,y) \in \Omega$ holds $$ \phi\phi_{x}-\int \phi(x,y)\phi_{xx}(x,y)dx, + C_{1}(y) = u = \phi\phi_{x}-\int \phi(x,y)\phi_{xy}(x,y)dy + C_{2}(x)$,$$ or equivalently $$ -\int \phi(x,y)\phi_{xx}(x,y)dx, + C_{1}(y) = u = -\int \phi(x,y)\phi_{xy}(x,y)dy + C_{2}(x).$$

Am I correct that the equality of the two integrals is a necessary condition for the existence of a solution of the PDE?

EDIT: Based on the feedback from the comments I fixed some of my mistakes.

  • 1
    $\begingroup$ I haven't been through the whole calculation but how do you get $$\int \phi_{x}^{2} dx = - \int \phi \phi_{x} dx$$ using 'partial integration' (by which I assume you mean integration by parts)? The result should be $$\int \phi_{x}^{2} dx = \phi \phi_{x} - \int \phi \phi_{xx} dx$$ $\endgroup$ Nov 18, 2023 at 3:32
  • $\begingroup$ Yes, I mean integration by parts. I believe the name partial integration is only common in the German-speaking world. Sorry for that. You are right that's my mistake. It clearly should be $$ \int \phi_{x}^{2}dx=- \int \phi\phi_{xx}dx. $$ The boundary part finishes since $\phi$ is zero on the boundary by assumption. This should not change the conclusion that the only solution to the problem is trivial? I fixed the original post via an edit. $\endgroup$
    – SebastianP
    Nov 18, 2023 at 16:39
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    $\begingroup$ You are taking antiderivatives, not definite integrals over slices of $\Omega$, so the boundary terms don't drop out. To be precise, what you're really doing is computing $\int_a^x \phi_x^2(t,y) dt$, so you get boundary terms at $a$ and $x$. $\endgroup$ Nov 18, 2023 at 16:48
  • $\begingroup$ Thanks @kieransquared! I get your point. You are obviously right. I edited the original question to fix my mistakes. $\endgroup$
    – SebastianP
    Nov 18, 2023 at 22:41


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