How to solve $\nabla(u\cdot u)=\lambda u$

Question

Suppose $C^2(\mathbb R^n)$ vector field $u(x)$, how to solve this pde? $$\nabla(u\cdot u)=\lambda u$$ for some constant $\lambda$.

The physical intuition is that: The left hand side of the equation is advection acceleration of irrotational flow, while the right hand side is something like viscosity force.

I get two solutions: one is $u=0$ and the other is $u=\frac12\lambda x$. I have no idea how to get other solutions, neither know the existence of other solutions. Can anyone help?

Answer 1

Here is my attempt.

\begin{align} u:\mathbb R^n&\to\mathbb R, & \nabla\,\big(\,u\cdot u\,\big)&=\lambda \,u &\iff && \begin{cases} \dfrac{\partial}{\partial x_1} \Big(u_1^2 + \dots + u_n^2 \Big) = \lambda u_1 \\ \qquad\qquad\vdots \\ \dfrac{\partial}{\partial x_n} \Big(u_1^2 + \dots + u_n^2 \Big) = \lambda u_n \\ \end{cases} \end{align}

Denote $\,F := u\cdot u = u_1^2 + \dots + u_n^2 \,$, so that the original equation will take form

\begin{align} \nabla F &= \lambda \,u\,& \iff & & \dfrac{\partial F}{\partial x_i} & = \lambda\,u_i & \forall \; i = 1,\,\dots,\, n \end{align}

But then

\begin{align} F = \sum_{i=1}^n u_i^2 = \dfrac{1}{\lambda^2}\sum_{i=1}^n (\dfrac{\partial F}{\partial x_i})^2 \implies \lambda^2\,F = \big(\,\nabla F\cdot\nabla F\,\big)= \big\| \nabla F\big\|^2 \end{align}

Therefore all you need to do is to solve equation \begin{align} \bbox[4pt, border:2pt solid #FF0000]{\left\| \nabla F\right\|^2_{\phantom{y}} = \lambda^2\,F} \end{align} and the solution of the original problem will be $\;u=\dfrac{1}{\lambda}\,\nabla F.\,$

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In particular, consider, for example, solution \begin{align} F &= a\,\left\|x\right\|^2 = x_1^2 + \dots + x_n^2 &\implies && \nabla F &= 2\,a\,x &\implies&& a&= \dfrac{\lambda^2}{4} \\ &&\implies && u & = \dfrac{1}{2}\,\lambda\,x \end{align}

Answer 2

Here are some nontrivial solutions to the problem. Note that the value of $\lambda$ has no intrinsic meaning.

Consider a smooth closed convex hypersurface $S\subset{\mathbb R}^n$ (an egg), and let $\Omega$ be the exterior of $S$. For ${\bf x}\in\Omega$ define $$f({\bf x}):=d({\bf x},S)=\inf_{{\bf y}\in S}|{\bf x}-{\bf y}|\ .$$ For a fixed ${ \bf p}\in\Omega$ there is a unique ${\bf z}\in S$ with $f({\bf p})=|{\bf p}-{\bf z}|$, and one has $$f\bigl({\bf p}+t({\bf p}-{\bf z})\bigr)=f({\bf p})+t|{\bf p}-{\bf z}|\qquad(t\geq0)\ .$$ It follows that $${\bf E}:=\nabla f$$ is a unit vector field on $\Omega$. Now put $${\bf U}({\bf x}):=f({\bf x})\>{\bf E}({\bf x})\qquad({\bf x}\in\Omega)\ .$$ Then $$\nabla({\bf U}\cdot{\bf U})=\nabla( f^2)=2 f\>\nabla f=2f\>{\bf E}=2{\bf U}\ .$$