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For each $t$, letA weak energy solution of a variable-coefficient wave equation is a function $u\in H^1(U_T)$ with zero lateral trace, equivalently $u\in L^2(0,T;H_0^1(U))$, whose time trace satisfies $u(0)={\psi }_0$ in $L^2(U)$ and for whichfor every $v\in H^1(U_T)$ with zero lateral trace and $v(T)=0$. Here $u\in H^1(0,T;L^2(U))$ has a continuous $L^2$ representative, so the displacement trace is well defined.

The velocity condition is encoded by the boundary term in time; an arbitrary space-time $H^1$ function need not have an $L^2$ trace of $u_t$. The equation implies $u_{tt}=f-L(t)u\in L^2(0,T;H^{-1}(U))$, hence $u_t$ has a continuous $H^{-1}$ representative and $u_t(0)={\psi }_1$ in that sense. This is the weak formulation obtained by integration by parts once in time and once in space.

For the weak energy solution of a variable-coefficient wave equation, uniqueness can be proved without using $u_t$ as a spatial $H_0^1$ test function. For a difference with zero data, fix $s$ and test by $v(t)={\int }_t^{s}u(r)dr$ for $t<s$, extended by zero afterwards. Put $z(t)={\int }_0^{t}u(r)dr$, so $v(t)=z(s)-z(t)$. Integration by parts in time for the symmetric principal form, and in space for the first-order terms, giveFor short enough intervals the last term is absorbed. The Gronwall inequality forces $u=0$, and repetition proves uniqueness on the full interval. This antiderivative test is valid at the stated space-time $H^1$ regularity, unlike an unqualified direct energy test by $u_t$.