Can the expansion of the Universe be explained by a giant gravitational wave with a wavelength bigger than observable Universe's diameter? In that case the expansion would be temporary, only happening while we are on the trough of the wave. After billions of years when the crest of the wave gets to us, we will see constriction instead.
Can this type of gravitational wave explain the accelerated expansion we observe?
According to general relativity gravitational waves are transverse. This means they only act in directions perpendicular to their direction of propagation. The way gravitational waves are polarized also means that as they stretch space in one direction, they squish it in a second.
For example if we define the direction of propagation of a wave to be the $\hat{z}$ direction, then the stretching/squishing would be happening in the $\hat{x}$ and $\hat{y}$ directions only. If we then define the direction of maximal stretching to be the $\hat{x}$ direction, then space will be maximally squished in the $\hat{y}$ direction.
The expansion of the universe is isotropic. This means it happens equally in all directions. If a giant gravitational wave were affecting the whole observable universe, the amount of stretching or squishing (or not at all) would be different in different directions.
The short answer is no, but let me give some more details. In this answer, I'm using units where $c=G=1$.
Let's write the metric for gravitational waves on an expanding background as \begin{equation} g_{\mu\nu} = g_{\rm \mu\nu}^{\rm FLRW} + h_{\mu\nu} \end{equation} where $g_{\rm \mu\nu}^{\rm FLRW}$ is the Friedmann-Lemaitre-Robertson-Walker metric, which we can express as \begin{equation} g_{\rm \mu\nu}^{\rm FLRW} {\rm d} x^\mu {\rm d} x^\nu = - {\rm d} t^2 + a^2(t) {\rm d} \vec{x}^2 \end{equation} where ${\rm d} \vec{x}^2$ is the line element for a spatially flat space and $a(t)$ is the scale factor. We can further choose a gauge where $h_{\mu\nu}$ is determined by its spatial components, $h_{ij}$, and is traceless and transverse, meaning $h^i_i=0$ and $\nabla^i h_{ij}=0$, where indices are raised and lowered with $g^{\rm FLRW}_{\mu\nu}$, and where $\nabla$ is a covariant derivative defined using $g^{\rm FLRW}_{\mu\nu}$,.
Gravitational waves in an expanding universe obey a modified version of the wave equation, which accounts for the effects of the expansion.
\begin{equation} \ddot{h}_{ij} + 3 H \dot{h}_{ij} - \nabla^2 h_{ij} = 0 \end{equation} If we consider a plane wave (and for convenience introduce a factor of $a$ to make the equation nicer), $h_{ij} = a v_{ij} e^{i (\vec{k} \cdot \vec{x} - \omega t)}$, then the equation becomes \begin{equation} \ddot{v}_{ij} + \left(k^2 - \frac{\ddot {a}}{a}\right) v_{ij} = 0 \end{equation} For small wavelengths, which means $k = 2\pi/\lambda$ is very large compared to $\ddot{a}/a$, we can ignore the $\ddot{a}/a$ term above and this equation reduces to the standard wave equation in flat spacetime \begin{equation} \ddot{v}_{ij} + k^2 v_{ij} = 0 \end{equation} describing propagating waves.
For large wavelengths, $k^2 \ll \ddot{a}/a$, the equation becomes \begin{equation} \ddot{v}_{ij} - \frac{\ddot{a}}{a} v_{ij}=0 \end{equation} The opposite sign here means that waves will decay instead of oscillate (actually this depends a bit on your choice of coordinates and variables; a better way to express this is that the waves are frozen and do not oscillate in the normal way a wave does). Crucially, this equation is independent of $k$, and so all perturbations with a wavelength much larger than $a/\ddot{a}$ will be frozen.
Just to get some intuition (the qualitative results of this section hold for other $a(t)$), since we approximately live at the moment in a Universe undergoing exponential expansion of the scale factor, let's suppose $a=e^{Ht}$. Then $\ddot{a}/a = H^2$. So the condition $k^2 \ll \ddot{a}/a$ means (ignoring factors of $2\pi$): \begin{equation} \lambda \gg \frac{1}{H} \end{equation} The scale $1/H$ is the "horizon size" of the Universe, the scale beyond which we can no longer see things because they are receding from us too rapidly.
What this condition expresses is that precisely in the regime you are interested -- a "giant" gravitational wave with a wavelength larger than the size of the observable Universe, the gravitational wave stops being a "wave" per se, and acts essentially like a constant part of the metric, which we can remove with a gauge transformation (change of coordinates).
A gravitational wave (or superposition of gravitational waves) itself can't cause a homogeneous and isotropic expansion of the Universe, because gravitational wave polarizations have a spin-2 character to them. If space is expanding in one direction, it contracts in another direction. One way to say this is that we can always choose a gauge where gravitational waves are traceless, but an isotropic expansion of the Universe means we can choose coordinates where each of the diagonal spatial components has the same value and is expanding at the same rate, so there is a non-zero trace.
Furthermore, cosmological observations depend on a very specific rate of growth of the scale factor (different power laws with time in the radiation and matter dominated eras, and an exponential expansion more recently), which you wouldn't expect from a gravitational wave.
The meaning of my original approach to the answer is that we know the Universe has a horizon, and that gravitational waves with a wavelength longer than this horizon aren't really waves.
No. This would be at odds with Hubble's Law, i.e. the observation that objects recede from us faster the further they are away.
