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I deal with two-dimensional Kirchhoff equation with $L^\infty$ coefficient and distributional right hand side: $$ \Delta\Delta w+u(x,y)\left(\alpha^2\frac{\partial w}{\partial t}+\beta^2w\right)+\gamma^2\frac{\partial^2 w}{\partial t^2}=P\left[\theta(t)-\theta(t-\tau)\right]\delta(x-x_0(t))\delta(y-y_0(t)),~~ (x,y,t)\in(-l_1,l_1)\times(-l_2,l_2)\times(0,T), $$ subject to $$ w=\frac{\partial^2 w}{\partial x^2}=0,~~ x=-l_1; l_1,~~ {\rm for~ all}~ (y,t)\in[-l_2,l_2]\times[0,T], $$ $$ w=\frac{\partial^2 w}{\partial y^2}=0,~~ y=-l_2; l_2,~~ {\rm for~ all}~ (x,t)\in[-l_1,l_1]\times[0,T], $$ and $$ w(x,y,0)=w_0(x,y),~~ \frac{\partial w}{\partial t}\bigg|_{t=0}=w_0^1(x,y),~~ {\rm for~ all}~ (x,y)\in[-l_1,l_1]\times[-l_2,l_2]. $$

Here $\Delta$ is the Laplacian, $\theta(t)$ is the Heaviside`s function, $\delta(t)$ is that of Dirac, $u\in L^\infty[-l_1,l_1]\times[-l_2,l_2]$, $x_0$, $y_0$ are continuous, $w_0$ and $w_0^1$ are, at least, piecewise continuous and in consistency with boundary conditions, $\tau<T$.

Before starting to find the coefficient $u$, a question arose: in which Sobolev space the problem is well posed?

The energy of the system without damping term is $$ E(t)=\int_{[-l_1,l_1]\times[-l_2,l_2]}\left[\left(\Delta w\right)^2+\gamma^2\left(\frac{\partial w}{\partial t}\right)^2\right]dxdy, $$ so the starting point is that $w\in H^2\left([-l_1,l_1]\times[-l_2,l_2];W^{2,1}[0,T]\right)$, but only intuitively.

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For an evolution (time-dependent) problem with a Cauchy data, you must consider the pair $W:=(w,\partial_tw)$, which satisfies a first-order equation in the time variable. According to your a priori estimate, a convenient space is $$ L^2(0,T;H^2(\Omega))\times L^2(0,T;L^2(\Omega)), $$ with $\Omega=(-\ell_1,\ell_1)\times(-\ell_2,\ell_2)$ your spatial domain. However, because of your Dirichlet boundary condition, it is better to consider $$ L^2(0,T;(H^2\cap H^1_0)(\Omega))\times L^2(0,T;L^2(\Omega)). $$ If you are willing to employ the Hille-Yosida Theorem in semi-group theory, the second-order boundary condition will come in the domain of the operator.

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The right hand side belongs to $\mathfrak{D}'(\Omega\times(0,T))$, therefore it has a distributional solution $w\in\mathfrak{D}'(\Omega\times(0,T))$. See, for example, Vladimirov: Methods of mathematical physics (in Russian), or Melnikova, Filinkov: Abstract Cauchy problems.

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