Given two second-order tensors, \[\mathbf{U} = U_{ij}\mathbf{e}_i\mathbf{e}_j, \quad \mathbf{V} = V_{ij}\mathbf{e}_i\mathbf{e}_j.\]
Tensor Product (Outer Product): \[\mathbf{U}\otimes\mathbf{V} = U_{ij}V_{kl}\mathbf{e}_i\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l.\]
Singal Dot Product (Contraction Operator): \[\mathbf{U}\cdot\mathbf{V} = \mathbf{U}\mathbf{V} = (U_{ij}\mathbf{e}_i\mathbf{e}_j)\cdot(V_{kl}\mathbf{e}_k\mathbf{e}_l) = U_{ij}\mathbf{e}_i\delta_{ij}V_{kl}\mathbf{e}_l = U_{ij}V_{jl}\mathbf{e}_i\mathbf{e}_l.\]
Double Dot Product (Scalar or Inner Product): \[\mathbf{U}: \mathbf{V} = (U_{ij}\mathbf{e}_i\mathbf{e}_j):(V_{kl}\mathbf{e}_k\mathbf{e}_l) = (U_{ij}\mathbf{e}_i)\delta_{jk}\cdot(V_{kl}\mathbf{e}_l) = U_{ij}V_{kl}\delta_{jk}\delta_{il} = U_{ij}V_{ji}.\]
A function of the position vector \(\mathbf{x}\) is called a field. Following discussion is in Cartesian Frame.
Gradient of a Scalar: \[\nabla\phi = \mathbf{e}_j\frac{\partial\phi}{x_j} = \mathbf{e}_1\frac{\partial\phi}{x_1} + \mathbf{e}_2\frac{\partial\phi}{x_2} + \mathbf{e}_3\frac{\partial\phi}{x_3}.\]
Gradient of a Vector: \[\nabla\mathbf{u} = \left(\mathbf{e}_j\frac{\partial}{x_j}\right)(u_j\mathbf{e}_j) = \mathbf{e}_i\mathbf{e}_j\frac{\partial \mathbf{u}_j}{\partial x_i}.\] In short \((\nabla\mathbf{u})_{ij} = \partial \mathbf{u}_j/ \partial x_i\), in some books defination is different (the transpose of a gradient).
Transpose of a Gradient: \[\nabla\mathbf{u}^T = \mathbf{e}_i\mathbf{e}_j\frac{\partial \mathbf{u}_i}{\partial x_j}.\]
Divergence of a Vector: \[\nabla\cdot\mathbf{u} = \left(\mathbf{e}_j\frac{\partial}{x_j}\right)\cdot(u_j\mathbf{e}_j) = \mathbf{e}_i\cdot\mathbf{e}_j\frac{\partial \mathbf{u}_j}{\partial x_i} = \delta_{ij}\frac{\partial \mathbf{u}_j}{\partial x_i}.\] That is, \[\nabla\cdot\mathbf{u} = \frac{\partial u_i}{x_i} = \frac{\partial u_1}{x_1} + \frac{\partial u_2}{x_2} + \frac{\partial u_3}{x_3}.\]
Divergence of a Tensor: \[\nabla\cdot\mathbf{S} = \left(\mathbf{e}_j\frac{\partial}{x_j}\right) \cdot(S_{ij}\mathbf{e}_i\mathbf{e}_j) = \mathbf{e}_j\frac{\partial S_{ij}}{\partial x_i}.\]
\[\mathbf{u}^\dagger \cdot [\nabla \cdot (\dot{\mathbf{C}}:\nabla \mathbf{u})] - \mathbf{u} \cdot [\nabla \cdot (\dot{\mathbf{C}}:\nabla \mathbf{u}^\dagger)] = \nabla\cdot(\mathbf{u}^\dagger\cdot\dot{\mathbf{C}} : \nabla \mathbf{u}) - \nabla\cdot(\mathbf{u}\cdot\dot{\mathbf{C}} : \nabla \mathbf{u}^\dagger).\]
For \(\mathbf{u}^\dagger \cdot [\nabla \cdot (\dot{\mathbf{C}}:\nabla \mathbf{u})]\): \[\begin{align*} \dot{\mathbf{C}}:\nabla \mathbf{u} &= (\dot C_{ijkl}\mathbf{e}_i\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l): \left(\left(\mathbf{e}_m\frac{\partial}{\partial x_m}\right)(u_n \mathbf{e}_n)\right) \\ &= (\dot C_{ijkl}\mathbf{e}_i\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l): \left(\mathbf{e}_m\mathbf{e}_n\frac{\partial u_n}{\partial x_m} \right)\\ &= \mathbf{e}_i\mathbf{e}_j \dot C_{ijkl} \frac{\partial u_n}{\partial x_m}\delta_{lm}\delta_{kn} = \mathbf{e}_i\mathbf{e}_j \dot C_{ijkl} \frac{\partial u_k}{\partial x_l}. \end{align*}\] Then, \[\begin{align*} \nabla \cdot (\dot{\mathbf{C}}:\nabla \mathbf{u}) &= \left(\mathbf{e}_m\frac{\partial}{\partial x_m}\right) \cdot \left(\mathbf{e}_i\mathbf{e}_j \dot C_{ijkl} \frac{\partial u_k}{\partial x_l}\right)\\ &=\mathbf{e}_j\frac{\partial}{\partial x_i} \left(\dot C_{ijkl}\frac{\partial u_k}{\partial x_l}\right). \end{align*}\] So, \[\begin{align*} \mathbf{u}^\dagger \cdot [\nabla \cdot (\dot{\mathbf{C}}:\nabla \mathbf{u})] &= (\mathbf{e}_m u_m^\dagger)\cdot \left(\mathbf{e}_j\frac{\partial}{\partial x_i} \left(\dot C_{ijkl}\frac{\partial u_k}{\partial x_l}\right)\right)\\ &= u_j^\dagger \frac{\partial}{\partial x_i} \left(\dot C_{ijkl}\frac{\partial u_k}{\partial x_l}\right)\\ &= u_j^\dagger \frac{\partial \dot C_{ijkl}}{\partial x_i} \frac{\partial u_k}{\partial x_l} + u_j^\dagger \dot C_{ijkl} \frac{\partial^2 u_k}{\partial x_i \partial x_l}. \end{align*}\]
By the same way, \[\mathbf{u} \cdot [\nabla \cdot (\dot{\mathbf{C}}:\nabla \mathbf{u}^\dagger)] = u_j \frac{\partial \dot C_{ijkl}}{\partial x_i} \frac{\partial u_k^\dagger}{\partial x_l} + u_j \dot C_{ijkl} \frac{\partial^2 u_k^\dagger}{\partial x_i \partial x_l}.\]
For the \(\nabla\cdot(\mathbf{u}^\dagger\cdot\dot{\mathbf{C}} : \nabla \mathbf{u})\), \[\mathbf{u}^\dagger\cdot\dot{\mathbf{C}} = (\mathbf{e}_m u_m^\dagger) \cdot (\dot C_{ijkl} \mathbf{e}_i\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l) = \mathbf{e}_j\mathbf{e}_k\mathbf{e}_l u_i^\dagger \dot C_{ijkl}\]
Then, \[\begin{align*} \mathbf{u}^\dagger\cdot\dot{\mathbf{C}} : \nabla \mathbf{u} &= (\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l u_i^\dagger \dot C_{ijkl}) : \left(\left(\mathbf{e}_m\frac{\partial}{\partial x_m}\right)(u_n\mathbf{e}_n)\right)\\ &= (\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l u_i^\dagger \dot C_{ijkl}) : \left(\mathbf{e}_m\mathbf{e}_n\frac{\partial u_n}{\partial x_m}\right)\\ &= (\mathbf{e}_j\mathbf{e}_k u_i^\dagger \dot C_{ijkl})\delta_{lm}\cdot \left(\mathbf{e}_n\frac{\partial u_n}{\partial x_m}\right)\\ &= \mathbf{e}_j u_i^\dagger \dot C_{ijkl} \frac{\partial u_n}{\partial x_m} \delta_{lm}\delta_{kn}\\ &= \mathbf{e}_j u_i^\dagger \dot C_{ijkl} \frac{\partial u_k}{\partial x_l}. \end{align*}\] Then, \[\begin{align*} \nabla\cdot(\mathbf{u}^\dagger\cdot\dot{\mathbf{C}} : \nabla \mathbf{u}) &= \left(\mathbf{e}_m\frac{\partial}{\partial x_m}\right) \cdot \left(\mathbf{e}_j u_i^\dagger \dot C_{ijkl} \frac{\partial u_k}{\partial x_l}\right)\\ &= \frac{\partial}{\partial x_j}\left(u_i^\dagger \dot C_{ijkl} \frac{\partial u_k}{\partial x_l}\right)\\ &= \frac{\partial u_i^\dagger}{\partial x_j}\dot C_{ijkl}\frac{\partial u_k}{\partial x_l} + u_i^\dagger\frac{\partial \dot C_{ijkl}}{\partial x_j}\frac{\partial u_k}{\partial x_l} + u_i^\dagger \dot C_{ijkl}\frac{\partial^2 u_k}{\partial x_l\partial x_j}. \end{align*}\]
By the same way, \[\nabla\cdot(\mathbf{u}\cdot\dot{\mathbf{C}} : \nabla \mathbf{u}^\dagger) = \frac{\partial u_i}{\partial x_j}\dot C_{ijkl}\frac{\partial u_k^\dagger}{\partial x_l} + u_i\frac{\partial \dot C_{ijkl}}{\partial x_j}\frac{\partial u_k^\dagger}{\partial x_l} + u_i \dot C_{ijkl}\frac{\partial^2 u_k^\dagger}{\partial x_l\partial x_j}\]
So,
\[\begin{align*} \mathrm{LEFT} &= u_j^\dagger \frac{\partial \dot C_{ijkl}}{\partial x_i} \frac{\partial u_k}{\partial x_l} + u_j^\dagger \dot C_{ijkl} \frac{\partial^2 u_k}{\partial x_i \partial x_l} - \left(u_j \frac{\partial \dot C_{ijkl}}{\partial x_i} \frac{\partial u_k^\dagger}{\partial x_l} + u_j \dot C_{ijkl} \frac{\partial^2 u_k^\dagger}{\partial x_i \partial x_l} \right). \end{align*}\]
And, \[\begin{align*} \mathrm{RIGHT} &= \frac{\partial u_i^\dagger}{\partial x_j}\dot C_{ijkl}\frac{\partial u_k}{\partial x_l} + u_i^\dagger\frac{\partial \dot C_{ijkl}}{\partial x_j}\frac{\partial u_k}{\partial x_l} + u_i^\dagger \dot C_{ijkl}\frac{\partial^2 u_k}{\partial x_l\partial x_j}\\ &- \left(\frac{\partial u_i}{\partial x_j}\dot C_{ijkl}\frac{\partial u_k^\dagger}{\partial x_l} + u_i\frac{\partial \dot C_{ijkl}}{\partial x_j}\frac{\partial u_k^\dagger}{\partial x_l} + u_i \dot C_{ijkl}\frac{\partial^2 u_k^\dagger}{\partial x_l\partial x_j} \right). \end{align*}\]
Since \(\mathbf{C}\) is symmetric, that is \(C_{ijkl} = C_{klij} = C_{jikl}\). Then, \[\mathrm{LEFT} = \mathrm{RIGHT}.\]
Define an object function (misfit function) by, \[\chi(\mathbf{m}) = \int_G\int_T \frac{1}{2} [\mathbf{u}^0(\mathbf{x},t) - \mathbf{u}(\mathbf{m};\mathbf{x},t)]^2 \delta(\mathbf{x} - \mathbf{x^r})\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x}.\]
Let \(\chi_1 = \frac{1}{2} [\mathbf{u}^0(\mathbf{x},t) - \mathbf{u}(\mathbf{m};\mathbf{x},t)]^2 \delta(\mathbf{x} - \mathbf{x^r})\), then, \[\chi(\mathbf{m}) = \int_G\int_T \chi_1(\mathbf{u}(\mathbf{m};\mathbf{x},t))\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} = <\chi_1(\mathbf{m})>,\] with \(<\cdot>\) shows the integral over \(G\times T\).
Our target is to compute \(\nabla_m\chi\delta\mathbf{m}\). \[\begin{align*} \nabla_m\chi\delta\mathbf{m} = \nabla_u\chi\delta\mathbf{u} = \nabla_u\chi(\nabla_m\mathbf{u}\delta\mathbf{m}). \end{align*}\]
And also, \[\begin{align*} \nabla_m\chi\delta\mathbf{m} &= (\delta\mathbf{m} \cdot \nabla_m)\chi \\ &= (\sum_i \delta m_i \partial_{m_i})\chi \\ &= (\sum_i \delta m_i \partial_{m_i}) \int_G\int_T \chi_1(\mathbf{u}(\mathbf{m};\mathbf{x},t))\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} \\ &= \int_G\int_T (\sum_i \delta m_i \partial_{m_i}) \chi_1(\mathbf{u}(\mathbf{m};\mathbf{x},t))\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} \\ &= \int_G\int_T \nabla_m\chi_1\delta\mathbf{m} \ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} \\ &= \int_G\int_T \nabla_u\chi_1\delta\mathbf{u} \ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} \\ &= \int_G\int_T (\delta\mathbf{u}\cdot \nabla_u)\chi_1 \ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} \\ &= \int_G\int_T \delta\mathbf{u} \cdot \nabla_u\chi_1 \ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} = <\delta\mathbf{u} \cdot \nabla_u\chi_1>. \end{align*}\]
We can not compute \(\nabla_m\chi\delta\mathbf{m}\) directly, so we need the adjoint method.
Define the wave equation be \(\mathbf{L}(\mathbf{u}, \mathbf{m}) = \mathbf{f}\), we do the derivative with respect to \(\mathbf{m}\), \[\nabla_m\mathbf{L}\delta\mathbf{m} + \nabla_u\mathbf{L}\delta\mathbf{u} = \mathbf{0},\]
With \(\delta\mathbf{u} = \nabla_m\mathbf{u}\delta\mathbf{m}\). Suppose there exist a \(\mathbf{u}^\dagger\), then, \[<\mathbf{u}^\dagger \cdot (\nabla_m\mathbf{L}\delta\mathbf{m} + \nabla_u\mathbf{L}\delta\mathbf{u})> = 0.\]
That is, \[<\mathbf{u}^\dagger \cdot \nabla_m\mathbf{L}\delta\mathbf{m}> + <\mathbf{u}^\dagger \cdot \nabla_u\mathbf{L}\delta\mathbf{u}> = 0.\]
By introduce adjoint operator \(\nabla_u\mathbf{L}^\dagger\), we have, \[<\mathbf{u}^\dagger \cdot \nabla_m\mathbf{L}\delta\mathbf{m}> + <\delta\mathbf{u} \cdot \nabla_u\mathbf{L}^\dagger\mathbf{u}^\dagger> = 0.\]
And since, \[\begin{align*} \nabla_m\chi\delta\mathbf{m} &= <\delta\mathbf{u} \cdot \nabla_u\chi_1> \\ &= <\delta\mathbf{u} \cdot \nabla_u\chi_1> + <\mathbf{u}^\dagger \cdot \nabla_m\mathbf{L}\delta\mathbf{m}> + <\delta\mathbf{u} \cdot \nabla_u\mathbf{L}^\dagger\mathbf{u}^\dagger> \\ & = <\delta\mathbf{u} \cdot (\nabla_u\chi_1 + \nabla_u\mathbf{L}^\dagger\mathbf{u}^\dagger)> + <\mathbf{u}^\dagger \cdot \nabla_m\mathbf{L}\delta\mathbf{m}>. \end{align*}\]
Then we have the adjoint equation, \[\nabla_u\mathbf{L}^\dagger\mathbf{u}^\dagger = \nabla_u\chi_1.\]
By solving the adjoint equation, we can have the wave field \(\mathbf{u}^\dagger\), then, \[\nabla_m\chi\delta\mathbf{m} = <\mathbf{u}^\dagger \cdot \nabla_m\mathbf{L}\delta\mathbf{m}>.\]
The following problems are:
In this part, we prove the form of \(\mathbf{L}^\dagger\).
For the elastic wave equation: \[\begin{align*} \rho(\mathbf{x})\ddot{\mathbf{u}}(\mathbf{x},t) - \nabla\cdot\boldsymbol{\sigma}(\mathbf{x},t) &= \mathbf{f}(\mathbf{x},t), \quad \mathbf{x}\in G\subset\mathbb{R}^3, \quad t\in[t_0,\infty]\subset\mathbb{R},\\ \boldsymbol{\sigma}(\mathbf{x},t) &= \int_{t_0}^\infty \dot{\mathbf{C}}(\mathbf{x},t-\tau): \nabla\mathbf{u}(\mathbf{x},\tau)\ \mathrm{d}\tau. \end{align*}\]
Thus we have, \[\mathbf{L}(\mathbf{u}, \rho, \mathbf{C}) = \rho(\mathbf{x})\ddot{\mathbf{u}}(\mathbf{x},t) - \nabla\cdot \int_{t_0}^\infty \dot{\mathbf{C}}(\mathbf{x},t-\tau): \nabla\mathbf{u}(\mathbf{x},\tau)\ \mathrm{d}\tau,\] with the initial and boundary conditions, \[\mathbf{u}|_{t\leq t_0} = \dot{\mathbf{u}}|_{t\leq t_0} = \mathbf{0},\quad \mathbf{n}\cdot\boldsymbol{\sigma} = 0.\]
For the adjoint operator \(\mathbf{L}^\dagger\), we have defined, \[\begin{align*} <\mathbf{u}\cdot\mathbf{L}^\dagger\mathbf{u}^\dagger> &= <\mathbf{u}^\dagger\cdot \mathbf{L}\mathbf{u}> \\ &= \int_G\int_T \rho\mathbf{u}^\dagger\cdot\ddot{\mathbf{u}}\ \mathrm{d}t\ \mathbf{d}^3\mathbf{x} - \int_G\int_T \mathbf{u}^\dagger\cdot \left[\nabla\cdot \int_{t_0}^\infty \dot{\mathbf{C}}(t-\tau): \nabla\mathbf{u}(\tau)\ \mathrm{d}\tau\right]\ \mathrm{d}t\ \mathbf{d}^3\mathbf{x} \end{align*}\]
For the first part, we use integration by parts, \[\begin{align*} <\mathbf{u}^\dagger \cdot \rho\ddot{\mathbf{u}}> &= \int_G\int_T \rho\mathbf{u}^\dagger\cdot\ddot{\mathbf{u}}\ \mathrm{d}t\ \mathbf{d}^3\mathbf{x} \\ &= \int_G\int \rho\mathbf{u}^\dagger\cdot\ \mathrm{d}\dot{\mathbf{u}}\ \mathbf{d}^3\mathbf{x} \\ &= \int_G \rho \left[\mathbf{u}^\dagger\cdot\dot{\mathbf{u}}|_{t_0}^{t_1} - \int_T \dot{\mathbf{u}} \cdot \dot{\mathbf{u}}^\dagger \ \mathrm{d}t \right]\ \mathbf{d}^3\mathbf{x} \\ &= \int_G \rho \left[\mathbf{u}^\dagger\cdot\dot{\mathbf{u}}|_{t_0}^{t_1} - \int \dot{\mathbf{u}}^\dagger \cdot\ \mathrm{d}\mathbf{u} \right]\ \mathbf{d}^3\mathbf{x} \\ &= \int_G \rho \left[\mathbf{u}^\dagger\cdot\dot{\mathbf{u}}|_{t_0}^{t_1} - \mathbf{u}\cdot\dot{\mathbf{u}}^\dagger|_{t_0}^{t_1} + \int_T \ddot{\mathbf{u}}^\dagger \cdot \mathbf{u}\ \mathrm{d}t \right]\ \mathbf{d}^3\mathbf{x}. \end{align*}\] We introduce the terminal conditions for adjoint equation, \(\mathbf{u}^\dagger|_{t\geq t_0} = \dot{\mathbf{u}}^\dagger|_{t\geq t_0} = \mathbf{0}\). We have, \[<\mathbf{u}^\dagger \cdot \rho\ddot{\mathbf{u}}> = \int_G\int_T\rho\ddot{\mathbf{u}}^\dagger \cdot \mathbf{u}\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} = <\mathbf{u}\cdot\rho\ddot{\mathbf{u}}^\dagger>.\]
Let \(\gamma:= <\mathbf{u}^\dagger \cdot (\nabla\cdot\mathbf{\sigma})>\), \[\begin{align*}\gamma &= \int_T\int_G \mathbf{u}^\dagger(t) \cdot \left[ \nabla\cdot \int_{t_0}^{t}\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau)\ \mathrm{d}\tau\right]\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t \\ &= \int_T\int_G\int_{t_0}^{t} \mathbf{u}^\dagger(t) \cdot (\nabla\cdot(\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau)))\ \mathrm{d}\tau\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t \\ &= \int_T\int_G\int_{t_0}^{t} \nabla\cdot(\mathbf{u}^\dagger(t)\cdot\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau)) - \nabla\cdot(\mathbf{u}(\tau)\cdot\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t)) + \mathbf{u}(\tau)\cdot[\nabla\cdot(\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t))]\ \mathrm{d}\tau\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t \\ &= \int_T\int_G\int_{t_0}^{t} \nabla\cdot(\mathbf{u}^\dagger(t)\cdot\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau))\ \mathrm{d}\tau\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t \\ &- \int_T\int_G\int_{t_0}^{t} \nabla\cdot(\mathbf{u}(\tau)\cdot\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t))\ \mathrm{d}\tau\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t \\ &+ \int_T\int_G\int_{t_0}^{t} \mathbf{u}(\tau)\cdot[\nabla\cdot(\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t))]\ \mathrm{d}\tau\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t. \end{align*}\]
By Gauss’ Theorem, \[\begin{align*} \gamma &= \int_{\partial G} \left\{ \int_{t_0}^{t_1}\int_{t_0}^{t} \mathbf{u}^\dagger(t)\cdot\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau)\ \mathrm{d}\tau\ \mathrm{d}t\ \right\} \cdot \mathbf{n}\ \mathrm{d}^2\mathbf{x} \\ &- \int_{\partial G} \left\{ \int_{t_0}^{t_1}\int_{t_0}^{t} \mathbf{u}(\tau)\cdot\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t)\ \mathrm{d}\tau\ \mathrm{d}t\ \right\} \cdot \mathbf{n}\ \mathrm{d}^2\mathbf{x} \\ &+ \int_T\int_G\int_{t_0}^{t} \mathbf{u}(\tau)\cdot[\nabla\cdot(\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t))]\ \mathrm{d}\tau\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t. \end{align*}\]
Since \(\int_{t_0}^{t_1}\int_{t_0}^{t}\ \mathrm{d}\tau\ \mathrm{d}t = \int_{t_0}^{t_1}\int_{\tau}^{t_1}\ \mathrm{d}t\ \mathrm{d}\tau\), \[\begin{align*} \gamma &= \int_{\partial G} \left\{ \int_{t_0}^{t_1} \mathbf{u}^\dagger(t)\cdot \left[\int_{t_0}^{t} \dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau) \mathrm{d}\tau\right]\ \mathrm{d}t\ \right\} \cdot \mathbf{n}\ \mathrm{d}^2\mathbf{x} \\ &- \int_{\partial G} \left\{ \int_{t_0}^{t_1} \mathbf{u}(\tau)\cdot \left[\int_{\tau}^{t_1} \dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t) \mathrm{d}t \right]\ \mathrm{d}\tau\ \right\} \cdot \mathbf{n}\ \mathrm{d}^2\mathbf{x} \\ &+ \int_T\int_G\int_{t_0}^{t} \mathbf{u}(\tau)\cdot[\nabla\cdot(\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}^\dagger(t))]\ \mathrm{d}\tau\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t = A-B+C. \end{align*}\]
Thus,
\[\begin{align*} <\mathbf{u}\cdot\nabla_uL^\dagger \mathbf{u}^\dagger> &= <\mathbf{u}^\dagger\cdot \nabla_uL\mathbf{u}>\\ &= \int_T\int_G \mathbf{\rho}\mathbf{u}^\dagger(t)\cdot \ddot{\mathbf{u}}(t)\ \mathrm{d}^3 \mathbf{x} \mathrm{d}t - \int_T\int_G \mathbf{u}^\dagger(t) \cdot [ \nabla\cdot \int_{t_0}^{t}\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau)\ \mathrm{d}\tau]\ \mathrm{d}^3\mathbf{x}\ \mathrm{d}t \\ &= <\mathbf{u}^\dagger\cdot\rho\ddot{\mathbf{u}}> - <\mathbf{u}\cdot(\nabla\cdot\mathbf{\sigma}^\dagger)> \\ &= <\mathbf{u}\cdot\mathbf{\rho}\ddot{\mathbf{u}}^\dagger> - <\mathbf{u}\cdot(\nabla\cdot\mathbf{\sigma}^\dagger)> \\ &= <\mathbf{u}\cdot(\mathbf{\rho}\ddot{\mathbf{u}}^\dagger - (\nabla\cdot\mathbf{\sigma}^\dagger))>. \end{align*}\]
Then,
\[\nabla_uL^\dagger\mathbf{u}^\dagger = \mathbf{\rho}\ddot{\mathbf{u}}^\dagger - (\nabla\cdot\mathbf{\sigma}^\dagger) = -\nabla_u\chi_1^\dagger.\]
Suppose we are working on the \(L_2\) waveform misfit function, \[\begin{align*} \chi(\mathbf{m}) &= \int_G\int_T \frac{1}{2} [\mathbf{u}^0(\mathbf{x},t) - \mathbf{u}(\mathbf{m};\mathbf{x},t)]^2 \delta(\mathbf{x} - \mathbf{x^r})\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} \\ &= \int_G\int_T \chi_1(\mathbf{m};\mathbf{x},t)\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} \end{align*}\] with \(\chi_1(\mathbf{m};\mathbf{x},t) = \frac{1}{2} [\mathbf{u}^0(\mathbf{x},t) - \mathbf{u}(\mathbf{m};\mathbf{x},t)]^2 \delta(\mathbf{x} - \mathbf{x^r})\).
We let the adjoint source \(\mathbf{f}^\dagger(\mathbf{x},t)\) be, \[\mathbf{f}^\dagger(\mathbf{x},t) = -\nabla_u\chi_1 = [\mathbf{u}^0(\mathbf{x},t) - \mathbf{u}(\mathbf{m};\mathbf{x},t)]\delta(\mathbf{x} - \mathbf{x^r}).\]
Then the adjoint equation is, \[\begin{align*} \nabla_uL^\dagger\mathbf{u}^\dagger &= -\nabla_u\chi_1^\dagger, \\ \mathbf{\rho}\ddot{\mathbf{u}}^\dagger - (\nabla\cdot\mathbf{\sigma}^\dagger) &= [\mathbf{u}^0(\mathbf{x},t) - \mathbf{u}(\mathbf{m};\mathbf{x},t)]\delta(\mathbf{x} - \mathbf{x^r}). \end{align*}\] With the initial and boundary conditions, \[\mathbf{u}^\dagger |_{t\geq t_1} = \dot{\mathbf{u}}^\dagger |_{t\geq t_1} = \mathbf{0},\quad \mathbf{n}\cdot\mathbf{\sigma}^\dagger |_{\mathbf{x}\in\partial G} = 0.\] By solving adjoint equation with time reverse, we can have \(\mathbf{u}^\dagger\).
For the derivative of misfit function, \[\begin{align*} \nabla_m\chi\delta\mathbf{m} &= <\mathbf{u}^\dagger \cdot \nabla_mL\delta\mathbf{m}> \\ &= \int_G\int_T \mathbf{u}^\dagger(t) \cdot \left[\delta\boldsymbol{\rho}\ddot{\mathbf{u}}(t) - \nabla\cdot\int_{t_0}^t \delta\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau)\ \mathrm{d}\tau\right]\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} \\ &= \int_G\int_T \delta\boldsymbol{\rho}\mathbf{u}^\dagger(t)\cdot\ddot{\mathbf{u}}(t)\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} - \int_G\int_T \mathbf{u}^\dagger(t) \cdot \left[\nabla\cdot\int_{t_0}^t \delta\dot{\mathbf{C}}(t-\tau):\nabla\mathbf{u}(\tau)\ \mathrm{d}\tau\right]\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} \\ &= A - B. \end{align*}\] With the initial conditions of original equation and adjoint equation, we have, \[\begin{align*} A &= \int_G\int \delta\boldsymbol{\rho}\mathbf{u}^\dagger(t) \cdot \mathrm{d}\dot{\mathbf{u}}(t)\ \mathrm{d}^3\mathbf{x} \\ &= \int_G \delta\boldsymbol{\rho} \left[\mathbf{u}^\dagger(t)\cdot\dot{\mathbf{u}}(t)|_{t_0}^{t_1} - \int_T \dot{\mathbf{u}}^\dagger(t)\cdot\dot{\mathbf{u}}(t)\ \mathrm{d}t\right]\ \mathrm{d}^3\mathbf{x} \\ &= -\int_G\int_T \delta\boldsymbol{\rho}\dot{\mathbf{u}}^\dagger(t)\cdot\dot{\mathbf{u}}(t)\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x}. \end{align*}\] Since, \[\begin{align*} \mathbf{u}^\dagger \cdot (\nabla \cdot (\delta \dot{\mathbf{C}} : \nabla \mathbf{u})) &= \mathbf{u}^\dagger \cdot \left(\nabla \cdot \left((\delta C_{ijkl}\mathbf{e}_i\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l) : \left(\mathbf{e}_m\mathbf{e}_n\frac{\partial u_n}{\partial x_m}\right)\right)\right) \\ &= \mathbf{u}^\dagger \cdot \left(\nabla \cdot \left( \mathbf{e}_i\mathbf{e}_j \delta C_{ijkl} \frac{\partial u_k}{\partial x_l}\right)\right) \\ &= \mathbf{u}^\dagger \cdot \left(\left(\mathbf{e}_m \frac{\partial}{\partial x_m}\right) \cdot \left( \mathbf{e}_i\mathbf{e}_j \delta C_{ijkl} \frac{\partial u_k}{\partial x_l}\right)\right) \\ &= (\mathbf{e}_m u_m^\dagger) \cdot \left(\mathbf{e}_j \frac{\partial}{\partial x_i}\left(\delta C_{ijkl} \frac{\partial u_k}{\partial x_l} \right)\right) \\ &= u_j^\dagger \frac{\partial}{\partial x_i} \left(\delta C_{ijkl}\frac{\partial u_k}{\partial x_l}\right). \end{align*}\] And also, \[\begin{align*} \nabla\mathbf{u}^\dagger : \delta\dot{\mathbf{C}} : \nabla\mathbf{u} &= \left(\mathbf{e}_m\mathbf{e}_n\frac{\partial u_n^\dagger}{\partial x_m}\right) : (\mathbf{e}_i\mathbf{e}_j\mathbf{e}_k\mathbf{e}_l\delta C_{ijkl}) : \nabla\mathbf{u}\\ &= \left(\mathbf{e}_k\mathbf{e}_l\frac{\partial u_i^\dagger}{\partial x_j} \delta C_{ijkl}\right) : \left(\mathbf{e}_m\mathbf{e}_n\frac{\partial u_n}{\partial x_m}\right) \\ &= \frac{\partial u_i^\dagger}{\partial x_j} \delta C_{ijkl} \frac{\partial u_k}{\partial x_l}. \end{align*}\] Since in previous step, we have proved \(\mathbf{u}^\dagger \cdot (\nabla \cdot (\delta \dot{\mathbf{C}} : \nabla \mathbf{u})) + \nabla\mathbf{u}^\dagger : \delta\dot{\mathbf{C}} : \nabla\mathbf{u} = \nabla\cdot (\mathbf{u}^\dagger \cdot \dot{\mathbf{C}} : \nabla\mathbf{u})\). And the \(\iiint \nabla\cdot (\mathbf{u}^\dagger \cdot \dot{\mathbf{C}} : \nabla\mathbf{u}) \ \mathrm{d}\tau\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x} = 0\). Then, \[B = -\int_T\int_G\int_{t_0}^{t_1} \nabla\mathbf{u}^\dagger : \delta\dot{\mathbf{C}} : \nabla\mathbf{u} \ \mathrm{d}\tau\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x}.\] So, \[\nabla_m\chi\delta\mathbf{m} = A - B = -\int_G\int_T \delta\boldsymbol{\rho}\dot{\mathbf{u}}^\dagger(t)\cdot\dot{\mathbf{u}}(t)\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} + \int_T\int_G\int_{t_0}^{t_1} \nabla\mathbf{u}^\dagger : \delta\dot{\mathbf{C}} : \nabla\mathbf{u} \ \mathrm{d}\tau\ \mathrm{d}t \ \mathrm{d}^3\mathbf{x}\]
For perfectly elastic and isotropic medium, we have the following simplification, \(\mathbf{C}(\mathbf{x},t) = \mathbf{C}(\mathbf{x})H(t)\) and \(\delta\mathbf{C}(\mathbf{x},t) = \delta\mathbf{C}(\mathbf{x})H(t)\). In this case, \[\sigma(\mathbf{x},t) = \int_{-\infty}^{\infty} \dot{\mathbf{C}}(\mathbf{x},t-t') : \nabla\mathbf{u}(\mathbf{x},t')\ \mathrm{d}t' = \mathbf{C} : \nabla\mathbf{u}(\mathbf{x},t).\] Then, \[\nabla_m\chi\delta\mathbf{m} = -\int_G\int_T \delta\boldsymbol{\rho}\dot{\mathbf{u}}^\dagger(t)\cdot\dot{\mathbf{u}}(t)\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} + \int_T\int_G \nabla\mathbf{u}^\dagger(t) : \delta\mathbf{C} : \nabla\mathbf{u}(t) \ \mathrm{d}t \ \mathrm{d}^3\mathbf{x}.\]
Since \(C_{ijkl} = \lambda\delta_{ij}\delta_{kl} + \mu\delta_{ik}\delta_{jl} + \mu\delta_{il}\delta_{jk}\), \(\lambda\) and \(\mu\) are Lamé parameters. We have proved, \[\begin{align*} \nabla\mathbf{u}^\dagger : \delta\mathbf{C} : \nabla\mathbf{u} &= \frac{\partial u_i^\dagger}{\partial x_j} \delta C_{ijkl} \frac{\partial u_k}{\partial x_l} \\ &= \frac{\partial u_i^\dagger}{\partial x_j} [\delta\lambda\delta_{ij}\delta_{kl} + \delta\mu\delta_{ik}\delta_{jl} + \delta\mu\delta_{il}\delta_{jk}] \frac{\partial u_k}{\partial x_l} \\ &= \delta\lambda\frac{\partial u_i^\dagger}{\partial x_j}\frac{\partial u_k}{\partial x_k} + \delta\mu\frac{\partial u_i^\dagger}{\partial x_j}\frac{\partial u_i}{\partial x_j} + \delta\mu\frac{\partial u_i^\dagger}{\partial x_j}\frac{\partial u_j}{\partial x_i} \\ &= \delta\lambda(\nabla\cdot\mathbf{u}^\dagger)(\nabla\cdot\mathbf{u}) + \delta\mu[(\nabla\mathbf{u}^\dagger):(\nabla\mathbf{u}) + (\nabla\mathbf{u}^\dagger) : (\nabla\mathbf{u})^T]. \end{align*}\] Then we have, \[\begin{align*} \nabla_m\chi\delta\mathbf{m} &= - \int_G\int_T\delta\boldsymbol{\rho}\dot{\mathbf{u}}^\dagger(t)\cdot\dot{\mathbf{u}}(t)\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} + \int_G\int_T \delta\boldsymbol{\lambda}(\nabla\cdot\mathbf{u}^\dagger)(\nabla\cdot\mathbf{u})\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} + \int_G\int_T \delta\boldsymbol{\mu}[(\nabla\mathbf{u}^\dagger):(\nabla\mathbf{u}) + (\nabla\mathbf{u}^\dagger) : (\nabla\mathbf{u})^T]\ \mathrm{d}t\ \mathrm{d}^3\mathbf{x} \\ &= \int_G K_\rho^0 \delta\boldsymbol{\rho}\ \mathrm{d}^3\mathbf{x} + \int_G K_\lambda^0 \delta\boldsymbol{\lambda}\ \mathrm{d}^3\mathbf{x} + \int_G K_\mu^0 \delta\boldsymbol{\mu}\ \mathrm{d}^3\mathbf{x} \\ &= \nabla_\rho\chi\delta\boldsymbol{\rho} + \nabla_\lambda\chi\delta\boldsymbol{\lambda} + \nabla_\mu\chi\delta\boldsymbol{\mu}. \end{align*}\] With sensitivity kernal: \[\begin{align*} K_\rho^0 &= -\int_T \dot{\mathbf{u}}^\dagger(t)\cdot\dot{\mathbf{u}}(t)\ \mathrm{d}t, \\ K_\lambda^0 &= \int_T (\nabla\cdot\mathbf{u}^\dagger)(\nabla\cdot\mathbf{u})\ \mathrm{d}t, \\ K_\mu^0 &= \int_T [(\nabla\mathbf{u}^\dagger):(\nabla\mathbf{u}) + (\nabla\mathbf{u}^\dagger) : (\nabla\mathbf{u})^T]\ \mathrm{d}t. \end{align*}\]