I'm working on part (c) of this problem:
The only argument I can come up with is that in the $x$ direction, there is $0$ velocity at intervals of $n\pi$ and the same can be said about the $y$ direction because of the $\sin(x)$ and $\sin(y)$ terms.
Is there a more robust mathematical way of saying this?
Also, I'm unsure what part ii) in (c) means for me to show because $\mu$ doesn't change the behaviour of the velocity components.
At any given time, a streamline is a level curve of the streamfunction in $\mathbb{R}^2$. This is a set of points $(x,y)$ such that $\psi(x,y) = C(t)$ where $C(t)$ is a constant when $t$ is fixed.
On the boundary of the square $S=\{(x,y): 0 \leqslant x \leqslant \pi, 0 \leqslant y \leqslant \pi\}$ we have $x = 0 \text{ or } \pi$ for $0 \leqslant y \leqslant \pi$ or $y = 0 \text{ or } \pi$ for $0 \leqslant x \leqslant \pi.$ Since,
$$\psi(0,y) = \psi(\pi,y) = \psi(x,0) = \psi(x,\pi) = g(t)= C(t),$$
it follows that the boundary is a streamline.
Equivalently, the velocity field is tangential to every point on a streamline. This is because $u_x = \frac{\partial \psi}{\partial y}$ and $u_y = -\frac{\partial \psi}{\partial x}$ relates the velocity field and the streamfunction. At any point $(x,y)$ on a streamline,the vector $\nabla\psi(x,y)$ is normal, and the vector $(u_x(x,y),u_y(x,y))$ is tangential since
$$(u_x(x,y),u_y(x,y))\cdot \nabla \psi(x,y) = \left(\frac{\partial \psi}{\partial y},-\frac{\partial \psi}{\partial x}\right)\cdot \left(\frac{\partial \psi}{\partial x},\frac{\partial \psi}{\partial y}\right)= \frac{\partial \psi}{\partial y}\frac{\partial \psi}{\partial x} - \frac{\partial \psi}{\partial x}\frac{\partial \psi}{\partial y} = 0$$
As you began to show, this is true for the segments that constitute the boundary of $S$. Individually, then, these segments are on streamlines. To show they are on the same streamline we revert back to the first argument where it was shown that $\psi(x,y) = C(t)$ for the same constant, i.e, $C(t) = g(t)$, for each of these segments.
For the last part, in viscous flow where $\mu \neq 0$, we have the no-slip condition where both components of the velocity field must vanish on a solid boundary. In this case, the boundary of the square could not be a solid surface. For example, when $x = 0$ and $0 < y < \pi$, we have
$$u_y = e^{-2\mu t/\rho}\sin y \neq 0$$
In inviscid flow, where $\mu = 0$, the only physical requirement is that the velocity field is tangential at a solid surface. This has already been shown for the boundary of the square.