2
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.