I'm just wondering about the following definition of the stress tensor $\mathbb{T}$:
$\mathbb{T}(u):= 2\mu \mathbb{D}(u) + \lambda \mathbb{I}\text{div}u$.
Here $\lambda$ and $\mu$ are the viscosity coefficients and $\mathbb{D}(u):= \frac{1}{2}(\nabla u + \nabla u^t)$. Here I'm not sure about the lower case $t$, does this just mean $\mathbb{D}(u)$ is the symmetric part $\frac{1}{2}(\nabla u + (\nabla u)^T)$?
And I'm also not sure about the notation $\mathbb{I}$. Since this has to be a matrix, i would just assume it to be the identity, but the lecture notes don't say anything about it. Is this correct?
Thank you for any answers.
You are correct, $\nabla u^t \equiv (\nabla u)^T$, and $\mathbb{I}$ is the identity matrix. There's only a small typo in your expression, it should be
$$ \mathbb{T}(u) = 2\mu \mathbb{D}(u) + \lambda \mathbb{I}(\color{red}{\nabla \cdot u}) $$
That is, the last term includes the divergence of the velocity field, not the gradient.