# Solution to simple system of PDEs - Did I prove that only trivial solutions exist?

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.

• 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 $$\int \phi_{x}^{2} dx = \phi \phi_{x} - \int \phi \phi_{xx} dx$$ Nov 18, 2023 at 3:32
• 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. Nov 18, 2023 at 16:39
• 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 $x$. Nov 18, 2023 at 16:48
• Thanks @kieransquared! I get your point. You are obviously right. I edited the original question to fix my mistakes. Nov 18, 2023 at 22:41