Solution of hyperbolic equations with $V^*$ data

Question

Let $V\subset H\subset V^*$ a Hilbert triple and consider a 2nd order evolution equation of the form $$u''(t)+Au(t) = f(t)\quad \text{ in }\ L^2(0,T;V^*),$$ where $f\in\ L^2(0,T;H)$.

Can we let $f\in L^2(0,T;V^*)$?

This question is a special case ($A(t)=A$) of Regularity of solution to a hyperbolic pde. There, the answer says

If you want $f$ to take values in $V^*$ rather than $H$, you can do this if you assume more temporal regularity on $f$. Basically, the idea is to integrate by parts in the term $\int_0^t \langle u',f \rangle$ in the energy estimate. You will have no trouble finding results of this type in the literature.

I think by integration by parts it is meant $$\int_0^t \big\langle f(s),v'(s)\big\rangle_{V^*,V}\,\mathrm{d}s=\big(f(t),v(t)\big)_H-\big(f(0),v(0)\big)_H-\int_0^t \big\langle v(s),f'(s)\big\rangle_{V^*,V} \,\mathrm{d}s,$$ but the right hand side does not make sense unless $f'(s)\in V$, and I am confused.

Where can I find this type of result?

Answer 1

Chapter 9 in Volume 1 of Lions/Magenes [1] treats this case, even for nonautonomous operators. One essentially gets (somewhat as expected?) a regularity shift just in the spatial components, so the solution $u$ will satisfy $(u,u') \in C([0,T];H \times V^*)$. (This is provided the initial values for $(u,u')$ are in $H \times V^*$, of course.) For nonautonomous operators $A$ you will however have to assume additional time regularity in order to recover uniqueness of solutions. This circumvents the lack of an energy equality which one would normally use, I guess.

[1] Lions, J. L.; Magenes, E., Non-homogeneous boundary value problems and applications. Vol. I. Translated from the French by P. Kenneth, Die Grundlehren der mathematischen Wissenschaften. Band 181. Berlin-Heidelberg-New York: Springer-Verlag. XVI,357 p. (1972). ZBL0223.35039.

Answer 2

You need to use the integration by parts formula

$$\int_0^t \langle f(s), v'(s)\rangle ds = \langle f(t), v(t)\rangle - \langle f(0), v(0)\rangle - \int_0^t \langle f'(s), v(s)\rangle ds$$

where each of the duality pairing is the one in $V^*\times V$. After getting rid of $v(0)$ through the initial conditions the right hand side is well defined for almost all $t\in(0,T)$ for $v\in L^2(0, T; V)$ and $f\in H^1(0,T; V^*)$.

This is what the answer in the linked question hinted at: You can get away with $f(t)\in V^*$ as long as $f$ has more regularity in time.