Question on why a solution of this PDE is of class $\;C^4\;$

Question

Let $\;u:\mathbb R^2 \to \mathbb R^2\;$ ,$\;W \in C^3(\mathbb R^2;\mathbb R)\;$ and consider the following PDE:

$\;-\Delta u+W_u(u)=0\;$ where $\;W_u(u)=(\frac{\partial W}{\partial u_1}(u),\frac{\partial W}{\partial u_2}(u))^{T}\;$

I was reading a paper consisting of the above pde and in the first lines I saw this: "Since $\;W\in C^3\;$ then $\;W_u \in C^2\;$ and hence a solution of the above pde should be of class $\;C^4\;$"

I was wondering why is this true? Is it related to the elliptic estimates and the fact that if we have in general the pde $\;\Delta u=f\;$ and $\;f \in C^{\alpha} \Rightarrow u \in C^{\alpha+2}\;$?

But $\;f\in C^{\alpha}\;$ is a Holder estimate and here I don't have such an estimate for $\;W\;$...

I'm having a really hard time getting my head around this so any help would be valuable! Thanks in advance

Answer 1

$\bigtriangleup u$ should be of class $C^2$, since $\bigtriangleup u=W_u(u)$ and $W_u$ is $C^2$. But when you compute Laplacian, you have to differentiate twice, so you should "loose" 2 degrees of regularity.