We say $v \in C^2(\bar{U})$ is subharmonic if $-\Delta v \le 0$ in $U \subset \mathbb{R^n}$.
Prove that $v := |Du|^2$ is subharmonic, whenever $u$ is harmonic.
This is Exercise 5, part d, in PDE Evans, 2nd edition. Here is the work I have tried so far:
- $Du = \sum_{i=1}^n \frac{\partial u}{\partial x_i}=\sum_{i=1}^n u_{x_i}$
- $\Delta u = \sum_{i=1}^n u_{x_i x_i}$
- (I am working with $x=(x_1,\cdots,x_n)\in\mathbb{R}^n$.)
Since $u$ is harmonic, $\Delta u =0$. We have as follows (apologies for algebraic mistakes along the way):
\begin{align} \Delta v &= \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2} |Du|^2 \\ &= \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2} \left(\sum_{i=1}^nu_{x_i} \right)^2 \\ &= \sum_{i=1}^n \frac{\partial}{\partial x_i} \left(2\sum_{i=1}^nu_{x_i} \cdot u_{x_i x_i} \right) \\ &= \sum_{i=1}^n \left(2\sum_{i=1}^n u_{x_i x_i} \cdot u_{x_i x_i} + \sum_{i=1}^n u_{x_i} \cdot u_{x_i x_i x_i} \right) \\ &= \sum_{i=1}^n \left(\Delta u \cdot u_{x_i x_i} + |Du| \cdot u_{x_i x_i x_i} \right) \\ &= ??? \end{align}
Basically, I am lost at a sea of differentiation rules for $\mathbb{R}^n$. My algebra is probably incorrect as well. How can I fix this and/or proceed? I am trying to establish that $\Delta v \ge 0$, which means $-\Delta v \le 0$, so that by definition $v := |Du|^2$ is subharmonic as required.
Don't overthink this. It is easier to take things one derivative at a time, then sum over the indices.
Let $\nu = |Du|^2 = \sum_{i=1}^n (u_{x_i})^2$. Then $$ \frac{\partial}{\partial x_j}\nu = 2\sum_{i=1}^n u_{x_i}u_{x_ix_j} $$ and $$ \frac{\partial^2}{\partial x_j^2}\nu = 2\left[ \sum_{i=1}^n u_{x_i}u_{x_ix_jx_j} + \sum_{i=1}^n (u_{x_ix_j})^2 \right]. $$ Consider the first term in the bracket, and sum over $j=1,\ldots, n$. We obtain $$ \sum_{j=1}^n\sum_{i=1}^n u_{x_i}u_{x_ix_jx_j} = \sum_{i=1}^n\sum_{j=1}^n u_{x_i}u_{x_ix_jx_j} = \sum_{i=1}^n u_{x_i}(\Delta u)_{x_i}. $$ Now we're essentially done; you should be able to fill in the rest of the proof.