# Solutions to PDE $\langle\nabla\psi,\nabla\psi+\mathbf{f} \rangle= 0$

Consider the vector field $$\mathbf{f}(x_1,x_2)=\begin{bmatrix}f_1(x_1,x_2)\\ f_2(x_1,x_2)\end{bmatrix}=\begin{bmatrix}\frac{1}{1+x_2}+x_1\\ \frac{1}{1+x_1}+x_2\end{bmatrix}.$$

I'm interested in finding a scalar function $$\psi(x_1,x_2)$$ such that: $$\tag{\ast}\label{ast} \langle\nabla\psi,\nabla\psi+\mathbf{f} \rangle= \frac{\partial\psi(x_1,x_2)}{\partial x_1} \left(f_1(x_1,x_2)+\frac{\partial\psi(x_1,x_2)}{\partial x_1}\right)+\frac{\partial\psi(x_1,x_2)}{\partial x_2} \left(f_2(x_1,x_2)+\frac{\partial\psi(x_1,x_2)}{\partial x_2}\right)=0$$

My question. Does there exist a solution $$\psi(x_1,x_2)$$ to \eqref{ast}? If so, how to compute it?

I'm aware that my question could be a trivial or naive one, but since I'm new to this kind of problems I would really appreciate any help/comment/suggestion. Thank you.

On the one hand, the nullity of the scalar product $$\nabla\psi\cdot (f+\nabla\psi) = \psi_{,1} (f_1+ \psi_{,1} ) + \psi_{,2} (f_2+ \psi_{,2} ) = 0$$ leads to the orthogonality of $$\nabla\psi$$ and $$f+\nabla\psi$$. Thus, $$f+\nabla\psi$$ is proportional to the vector $$\nabla\psi^\perp = (\psi_{,1}, -\psi_{,2})^\top$$. There exists $$\alpha(x_1,x_2)$$ such that $$f+\nabla\psi = \alpha \nabla\psi^\perp$$, or equivalently $$(1-\alpha)\psi_{,1} = -f_1 \qquad\text{and}\qquad (1+\alpha)\psi_{,2} = -f_2 .$$ By eliminating successively $$\alpha$$ and unity of the previous system, we get to \begin{aligned} 2\psi_{,1}\psi_{,2} &= -f_2\psi_{,1} - f_1\psi_{,2} \\ 2\alpha\psi_{,1}\psi_{,2} &= -f_2\psi_{,1} + f_1\psi_{,2} . \end{aligned}
On the other hand, the nullity of the scalar product rewrites as $$(\psi_{,1} + \psi_{,2})^2 - 2\psi_{,1}\psi_{,2} + f_1\psi_{,1} + f_2\psi_{,2} = 0 ,$$ so that previous identities lead to $$(\psi_{,1} + \psi_{,2})^2 + (f_1+f_2)(\psi_{,1} + \psi_{,2}) = 0 .$$ Thus, the solutions satisfy $$\psi_{,1} + \psi_{,2} = -(f_1+f_2)$$ if they have nonzero divergence -- otherwise, they satisfy $$\psi_{,1} + \psi_{,2} = 0$$, which is a particular case of the previous non-homogeneous equation. Solutions to these linear first-order PDEs can be obtained by using the method of characteristics for the Lagrange-Charpit system $$\text d x_1 = \text d x_2 = {\text d \psi}/{r}$$ with $$r = -(f_1+f_2)$$ (see related posts on this site). Solutions are of the form $$\psi(x_1, x_2) = \int^{x_1} r(\xi,\xi + x_2-x_1)\, \text d \xi + F(x_2-x_1) ,$$ where $$F$$ is an arbitrary function. The latter is determined by injecting the previous expression in the orthogonality condition $$\nabla\psi\cdot (f+\nabla\psi) = 0$$.
If $$f$$ is constant, then linear solutions of the form \begin{aligned} \psi(x_1,x_2) &= -f_1x_1 - f_2x_2 + C \\ \text{or}\qquad \psi(x_1,x_2) &= -\tfrac12(f_1+f_2)(x_1+x_2)+ C \end{aligned} are obtained. In particular, one notes that the solutions are constant if $$f \equiv 0$$. In theory, such solutions can be obtained similarly in the case $$f_1(x_1,x_2) = x_1 + (1+x_2)^{-1}, \qquad f_2(x_1,x_2) = x_2 + (1+x_1)^{-1} ,$$ but computations are more involved.
• Once again, thanks! It seems to me that the problem now is to find the function $F(x_2-x_1)$ which is determined by the orthogonality condition. This looks like a problem as hard as the starting one, so I don't understand how this method could be helpful in computing the solution. But probably I'm missing something... Jan 25 '20 at 21:21
• For constant $\vec f$, $\vec\psi=-\vec r\cdot\vec f$ is a solution. You can knock out any part of the $\vec f$ vector that is the gradient of a potential this way leaving the case $\vec f=\vec\nabla\times\vec A$. Jan 25 '20 at 22:41