I have an equation as follows:
$$a \Delta \mathbf{u} + \mathbf{\nabla}(\mathbf{\nabla} \cdot \mathbf{u}) = 0$$
in which $a$ is a constant, $\mathbf{u}$ is a vector, $\Delta$ is the Laplacian operator, $\mathbf{\nabla}$ is the gradient operator and $\mathbf{\nabla} \cdot$ is the divergence operator.
If I take a Laplacian of the above equation, the answer is supposed to be:
$$\Delta \Delta \mathbf{u} = 0.$$
But how could I reach this?
(Actually, these two equations are eq.(7.4) and eq.(7.7) in the book of Theory of Elasticity – 1970 written by L. Landau)
$\bf{Edit.1}$ – A possible solution
Since we know: $\mathbf{\nabla}(\mathbf{\nabla} \cdot \mathbf{u}) = \Delta \mathbf{u} + \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{u})$, if we insert this equation to the first equation and then apply Laplacian operator on it, we will have a following equation:
$$
a \Delta \Delta \mathbf{u} + \Delta \Delta \mathbf{u} + \mathbf{\nabla} \cdot ( \mathbf{\nabla} ( \mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{u}) ) ) = 0
$$
where I used the fact that $\Delta \mathbf{v} = \mathbf{\nabla} \cdot (\mathbf{\nabla} \mathbf{v} )$, in which $\mathbf{v}$ is another vector.
Now, if we assume that we can commute $\mathbf{\nabla}$ with $\mathbf{\nabla} \times $, the above equation would become:
$$
a \Delta \Delta \mathbf{u} + \Delta \Delta \mathbf{u} + \mathbf{\nabla} \cdot ( \mathbf{\nabla} \times ( \mathbf{\nabla} (\mathbf{\nabla} \times \mathbf{u}) ) ) = 0
$$
Since it is known that $\mathbf{\nabla} \cdot ( \mathbf{\nabla} \times \mathbf{v}) = 0$, finally we will have:
$$
(a + 1) \Delta \Delta \mathbf{u} = 0
$$
Then, either we say $a = -1$, or we say $\Delta \Delta \mathbf{u} = 0$.
It turns out that $a = 1- 2 \sigma$, in which $\sigma$ represents the Poisson's ratio. For a stable, isotropic and linear elastic material, $\sigma$ must be between -1.0 and 0.5. Thus, the possibility of $a = -1$ can be ruled out, and we have $\Delta \Delta \mathbf{u} = 0$ at the end.
Anyway, the possibility of the interchange of the gradient and curl would be critical for my approach.
$\bf{Edit.2}$ There is a problem in Edit.1
The curl of a gradient is 0, but it is not necessary that the gradient of a curl is also 0. Thus, the assumption that I made in Edit.1, which is that we can commute the curl and gradient, is wrong!
Best Answer
The problem can proved by this way.
If we take a divergence of the first equation, we get
$$ \Delta (\mathbf{\nabla} \cdot \mathbf{u}) = 0$$
Now, if we take the Laplacian of the first equation and take into account that a Laplacian and a gradient is commutable, we get
$$ a \Delta \Delta \mathbf{u} + \mathbf{\nabla}[\Delta (\mathbf{\nabla} \cdot \mathbf{u})] = 0 $$
then it is easy to see
$$\Delta \Delta \mathbf{u} = 0$$