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I have a question concerning 2nd order evolution equation of the form $u''(t)+A(t)u(t) = f(t)$ in $L^2(0,T;V^*)$, where $f\in\ L^2(0,T;H)$ holds. Under what assumptions is it possible, to guarantee a unique solution for a more general $f\in L^2(0,T;V^*)$. I wrote down an exact setting following Wloka in his book partial differential equations:

Be $t\in [0,T]$ with $0<T<\infty$ and let the bilinear form $a(t;u,v)$ be symmetric and continuous for all $t$, i.e.,

$$|a(t;u,v)|\le \text{const.} \|u\|_V\|\ v\|_V,\quad \forall\ u,v\in V.$$

We assume for all $u,v\in V$ the mapping $t\mapsto a(t;u,v)$ to be continuously differentiable and V-coercive, i.e.,

$$\exists\ k_1,k_2>0:\ \forall\ t\in [0,T],\ v\in V:\ a(t;v,v) + k_1|v|_H^2 \ge k_2\|v\|_V^2$$

Suppose

$$V:=W^{1,2}(\Omega),\ H:=L^{2}(\Omega),\ f\in L^2(0,T;H)$$

and

$$u_0\in V,\ u_1\in H$$

then the second order evolution equation

$$\langle u''(t),v\rangle +a(t;u(t),v)=\langle f(t),v\rangle\quad in\ V^*\quad f.a.a.\ t\in (0,T)$$

with initial values

$$u(0)=u_0\in V,\ u'(0)=u_1\in H$$

possesses a unique solution (cf. Zeidler: Nonlinear functional analysis and its applications IIa, Wloka (as above))

$$u\in L^2(0,T;V),\ u'\in L^2(0,T;H),\ u''\in L^2(0,T;V^*).$$

I found a possibility in Dautray and Lions: Mathematical analysis and numerical methods for science and technology volume 5, but there an additional term on the left-hand side is necessary, with a coercive Operator acting on $u'$. These and comparable results give me the impression, that my question has to be answered with no. But since I am fairly new to this topic, I am thankfull for every suggestion for regularity theory on hyperbolic equations and general ideas to my problem.

This is my first post in this forum, so please do not hesitate to comment on any possible improvements for any future post regarding form, e.g. Thanks in advance.

Best regards, Simon.

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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.

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  • $\begingroup$ thank you very much for your response. That is an option, I did not think of. $\endgroup$
    – sgr
    Jan 3, 2015 at 19:14

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