## 3 Study on Scattered Surface Wave from Surface Crack Based on Reciprocity Theorem
### Reciprocity Theorem for Solving the Amplitude of Scattered Surface Waves
For J\({}_{1}\), the substitution of \(u_{x}^{n-}\),\(u_{z}^{n-}\),\(x_{xx}^{n-}\),\(\tau_{xz}^{n-}\) (represented in Equations (17) and (18)), \(u_{x}^{s-}\),\(u_{z}^{s-}\),\(\tau_{xx}^{s-}\),\(\tau_{xz}^{s-}\) (represented in Equations (19) and (20)) and \(u_{x}^{vi}\),\(u_{z}^{vi}\),\(\tau_{xx}^{vi}\),\(\tau_{xz}^{vi}\) (represented in Equations (22) and (23)) into Equation (24) yields:  
\[J_{1}^{n}=\int_{0}^{\infty}\left(u_{x}^{n-}\tau_{xx}^{vi}+u_{z}^{n-}\tau_{xz}^ {vi}-u_{x}^{vi-}u_{z}^{vi}\tau_{xz}^{n-}\right)|_{x=a}*(-1)dz=2 iBA_{\varsigma_{ x}}^{-}I\]  
\[J_{1}^{s}=\int_{0}^{\infty}\left(u_{x}^{s-}\tau_{xx}^{vi}+u_{z}^{s-}\tau_{xz}^ {vi}-u_{x}^{vi}\tau_{xx}^{s-}-u_{z}^{v}\tau_{xz}^{s-}\right)|_{x=a}*(-1)dz=-2 iBA_{\varsigma_{x}}^{-}I \tag{26}\]  
The superscript \(n\) and \(s\) of the integral \(J_{1}\) represent the scattered surface waves generated by normal stresses and shear stress, respectively, and \(I\) is written as follows:  
\[I=\int_{0}^{\infty}\left(U^{R}(z)T_{xx}(z)-W^{R}(z)T_{xz}(z)\right) \tag{27}\]  
It should be noted that \(I\) can be determined by Equations (6) and (10); thus, it can be written as:  
\[I=\mu J \tag{28}\]  
where  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]0  
and \(J\) is dimensionless. Next, the contributions from the integration along \(J_{5}\) and \(J_{6}\) are considered. For the normal stresses condition, we have:  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]1  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]2  
where  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]3  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]4  
For the normal stresses, the crack opens up symmetrically, so we have \(u_{x}^{n+}=-u_{x}^{n-}\), \(u_{z}^{n+}=u_{z}^{n-}\),\(\tau_{xx}^{n+}=\tau_{xx}^{n-}\), \(\tau_{xz}^{n+}=\tau_{xz}^{n-}=0\). As a consequence, only the first term in Equation (31) remains, and \(J_{5}^{n}+J_{6}^{n}\) reduces to:  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]5  
Similarly, the crack opens up anti-symmetrically under the shear stresses, so we have \(u_{x}^{s+}=-u_{x}^{s-}\),\(u_{z}^{s+}=-u_{z}^{s-}\),\(\tau_{xx}^{s+}=\tau_{xx}^{s-}=0\), \(\tau_{xz}^{s+}=\tau_{xz}^{s-}\). Thus, for shear stresses, Equation (32) changes to:  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]6  
where \(\Delta u_{x}^{n}\) and \(\Delta u_{z}^{s}\) are the CODs in the \(x\) and \(z\) directions, respectively. The superscript \(n\) and \(s\) mean that the CODs are generated by normal and shear traction on the surface crack, respectively. Since the total contour integral is zero as expressed in Equation (25), the amplitudes \(A_{\varsigma_{xn}}^{-}\) and \(A_{\varsigma_{x}}^{-}\) can be solved by Equations (25), (26), (32) and (33), which are finally simplified as follows:  
\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]7\[u_{x}^{vi}=iBU^{R}(z)e^{ikx},u_{z}^{vi}=BW^{R}(z)e^{ikx} \tag{22}\]8  
It is worth mentioning that the CODs \(\Delta u_{x}^{n}\) and \(\Delta u_{z}^{s}\) are, as yet, unknown. The crack-opening displacement will be obtained numerically in the next section.