What's the mechanism of the reaction of (E)-methyl 3-(furan-2-yl)acrylate in acidic methanol?
Unfortunately I cannot agree with the currently accepted answer. This 1,5-hydride shift that was invoked seems to have a very strained transition state (the cyclic, planar system means that the "1" and "5" atoms are quite far apart) and with all due respect, I am quite skeptical as to whether it would really work. Furthermore furan does not enjoy a particularly large aromatic stabilisation and "rearomatisation" as a driving force is not too convincing. Furan, in all its reactions, has a very pronounced tendency to lose its aromaticity. For example even nitration with the mild reagent acetyl nitrate ($\ce{AcONO2}$) leads to a 2,5-addition product and you actually need to add a second reagent (pyridine) to eliminate $\ce{AcOH}$ and rearomatise. For all intents and purposes furan reactivity is much closer to that of an enol rather than that of benzene.

Instead of a 1,5-hydride shift, proton transfers should suffice, especially since there are carbonyl groups all over the molecule (i.e. the proton transfer is simply a tautomerism and doesn't go through any particularly unstable intermediates).

One way of thinking about this question would be to try and do it in reverse. 1,4-Dicarbonyl compounds are known to cyclise to form furans in acidic solution, so this should simply be the microscopic reverse of that reaction. The key cyclisation step involves the attack of an enol oxygen onto a protonated carbonyl group. So, in this reaction, the fragmentation step should be the reverse: the ejection of an enol oxygen from a (hemi)acetal.

I would propose this mechanism instead.



So far, this is the same as in the currently accepted answer. However the usage of the furan lone pair to push electrons saves time in drawing all the resonance forms (this is what was meant by K_P's comment also).

Protonation of the furan oxygen (not drawn) brings us to the start of the next step:



As described previously this fragmentation is the reverse of the cyclisation reaction used to generate furans. Following tautomerisation of two of the enols, most of the molecule is already in place and the remaining question is how to generate the second ester functionality. The issue is that the carbon that should become the ester has an extra hydrogen on it. Instead of having a hydride shift it makes much more sense to move it around by tautomerisation!



The ketene acetal can be protonated and attacked by methanol at the carbonyl carbon, but that would be unproductive as there is no way of getting rid of one of the methyl groups on oxygen (unless we invoke a $\mathrm{S_N2}$ reaction on the methyl group with methanol acting as the nucleophile). So, it probably needs to wait until water is added during the reaction workup, at which point it is hydrolysed to give the desired product.