Problem 3.1.
For a homogeneous isotropic elastic body occupying a region\(\mathrm{{B}}\subset {E}^3\) subject to zero body forces, the displacement equation of equilibrium takes the form [see Eq. (3.6) with \(\mathbf{b }=\mathbf{0 }\)]
$$\begin{aligned} \nabla ^2{\mathbf{u }}+\frac{1}{1-2\nu }\nabla (\mathrm{{div}}\,\mathbf{u })=\mathbf{0 }\quad \mathrm{{on}}\quad \mathrm{{B}} \end{aligned}$$
(3.81)
where \(\nu \) is Poisson’s ratio. Show that if \(\mathbf{u }=\mathbf{u }(\mathbf{x })\) is a solution to Eq. (3.81) then u also satisfies the equation
$$\begin{aligned} \nabla ^2\left[ {\mathbf{u }+\frac{\mathbf{x }}{2(1-2\nu )}(\mathrm{{div}}\,\mathbf{u })} \right] =\mathbf{0 }\quad \mathrm{{on}}\quad \mathrm{{B}} \end{aligned}$$
(3.82)
Solution.
Equations (3.81) and (3.82) in components take the forms
$$\begin{aligned} {u_{i,kk}}+\frac{1}{1-2{\nu }}{\mathrm{{u}}_{k,ki}}=0\quad {\mathrm{{on}}\ B} \end{aligned}$$
(3.83)
and
$$\begin{aligned} \left[ {u_{i}}+\frac{{x_{i}}}{2(1-2{\nu })}{u_{k,k}}\right] ,_{jj}=0\quad {\mathrm{{on}}\ B} \end{aligned}$$
(3.84)
respectively.
It follows from (3.83) that
$$\begin{aligned} {{u}_{i, ikk}}+\frac{1}{1-2{\nu }}{{u}_{k, kii}}=0\quad {\mathrm{{on}}\ B} \end{aligned}$$
(3.85)
or
$$\begin{aligned} \frac{2(1-{\nu })}{1-2{\nu }}\ {{u}_{i,ikk}}=0\quad {\mathrm{{on}}\ B} \end{aligned}$$
(3.86)
Since \(-1<{\nu }<1/2<1\) [see Eq. (2.50)]
$$\begin{aligned} \frac{2-2{\nu }}{1-2{\nu }}>0 \end{aligned}$$
(3.87)
and (3.86) implies that
$$\begin{aligned} {{u}_{i,ikk}}=0\quad {\mathrm{{on}}\ B} \end{aligned}$$
(3.88)
In addition
$$\begin{aligned} ({x_{i}\ u_{k,k}}),_{jj} =(\delta _{ij}\ {u_{k,k}}+{x_{i}\ u_{k,kj}}),_{j} =2\delta _{ij}\ {u_{k,kj}}+{x_{i}\ u_{k,kjj}} \end{aligned}$$
(3.89)
Hence, it follows from Eqs. (3.88) and (3.89) that
$$\begin{aligned} ({x_{i}\ u_{k,k}}),_{jj}=2{u_{k},_{ki}} \end{aligned}$$
(3.90)
Substituting (3.90) into (3.84) we obtain (3.83), and this completes solution of Problem 3.1.
Problem 3.2.
An alternative form of Eq. (3.81) in Problem 3.1 reads [see Eq. (3.7) in which \(\lambda =2\mu \nu \,/(1-2\nu )\;\;\mathrm{{and}}\;\;\mathbf{b }=\mathbf{0 }\)]
$$\begin{aligned} \nabla (\mathrm{{div}}\,\mathbf{u })-\frac{1-2\nu }{2-2\nu }\mathrm{{curl}}\,\mathrm{{curl}}\,\mathbf{u }=\mathbf{0 }\quad \mathrm{{on}}\quad \mathrm{{B}} \end{aligned}$$
(3.91)
Show that if \(\mathbf{u }=\mathbf{u }(\mathbf{x })\) is a solution to Eq. (3.91) then
$$\begin{aligned} \displaystyle&\int \limits _\mathrm{{B}}{\left[ {\mathrm{{div}}\,\mathbf{u })^2+\dfrac{1-2\nu }{2-2\nu }(\mathrm{{curl}}\,\mathbf{u })^2} \right] } \,\mathrm{{dv}} \nonumber \\ \quad&=\displaystyle \int \limits _{\partial \mathrm{{B}}}{\mathbf{u }\cdot \left[ {(\mathrm{{div}}\,\mathbf{u })\,\mathbf{n }+\dfrac{1-2\nu }{2-2\nu }(\mathrm{{curl}}\,\mathbf{u })\,\times \mathbf{n }} \right] } \,\mathrm{{da}} \end{aligned}$$
(3.92)
where n is the unit outward normal vector field on \(\partial \mathrm{{B}}\).
Hint. Multiply Eq. (3.91) by u in the dot product sense, integrate the result over B, and use the divergence theorem.
Note. Since \(-1<\nu <1/2\) [see Eq. (2.50)] then Eq. (3.92) implies that a displacement boundary value problem of homogeneous isotropic elastostatics may have at most one solution.
Solution.
In components Eq. (3.91) takes the form
$$\begin{aligned} {u_{k, ki}}-\frac{1-2{\nu }}{2-2{\nu }}\varepsilon _{iab}\ \varepsilon _{bcd}\ {u_{d,ca}}=0 \end{aligned}$$
(3.93)
Since
$$\begin{aligned} {u_{i}\ u_{k,ki}}&={(u_{i}\ u_{k,k}),_{i}}-({u_{i,i}})^{2} \end{aligned}$$
(3.94)
$$\begin{aligned} {u_{i}}\ \varepsilon _{iab}\ \varepsilon _{bcd}\ {u_{d,ca}}&=\varepsilon _{iab}\ \varepsilon _{bcd}[{(u_{d,c}\ u_{i}),_{a}}-{u_{d,c}\ u_{i,a}}] \end{aligned}$$
(3.95)
and
$$\begin{aligned} \varepsilon _{iab}\ \varepsilon _{bcd}\ {u_{d,c}\ u_{i,a}}=-(\varepsilon _{bai}\ {u_{i,a}})(\varepsilon _{bcd}\ {u_{d,c}}) \end{aligned}$$
(3.96)
therefore, multiplying (3.93) by \({u_{i}}\) we obtain
$$\begin{aligned} ({u_{i,i}})^{2}+\frac{1-2{\nu }}{2-2{\nu }}(\varepsilon _{bai}\ {u_{i,a}})(\varepsilon _{bcd}\ {u_{d,c}}) =\left( {u_{k,k}}\ {u_{a}}+\frac{1-2{\nu }}{2-2{\nu }}\varepsilon _{iab}\ \varepsilon _{bcd}\ {u_{d,c}\ u_{i}}\right) \!,_{a} \nonumber \\ \end{aligned}$$
(3.97)
Finally, integrating (3.97) over \(B\) and using the divergence theorem we obtain
$$\begin{aligned}&\int \limits _{B}\left[ ({u_{i,i}})^{2}+\frac{1-2{\nu }}{2-2{\nu }}(\varepsilon _{bai}\ {u_{i,a}})(\varepsilon _{bcd}\ {u_{d,c}})\right] {{dv}} \nonumber \\&\quad =\int \limits _{\partial {B}}{u_{a}}\left[ {u_{k,k}}\ {n_{a}}+\frac{1-2{\nu }}{2-2{\nu }}\varepsilon _{abi}(\varepsilon _{bcd}\ {u_{d,c}}){n_{i}}\right] {{da}} \end{aligned}$$
(3.98)
Equation (3.98) is equivalent to (3.92), and this completes solution of Problem 3.2.
Problem 3.3.
Show that for a homogeneous isotropic infinite elastic body subject to a temperature change \({T}={T}(\mathbf x )\) its volume change is represented by the formula
$$\begin{aligned} \mathrm{{tr}}\,\mathbf{E }(\mathbf{x })=\frac{1+\nu }{1-\nu }\alpha \,{T}(\mathbf{x })\quad \mathrm{{for}}\quad \mathbf{x }\in {E}^3 \end{aligned}$$
(3.99)
where \(\nu \) and \(\alpha \) denote Poisson’s ratio and coefficient of thermal expansion, respectively.
Hint. Apply the reciprocal relation (3.28) to the external force-temperature systems \([\mathbf{b },\mathbf{s },{T}]=[\mathbf{0 },\mathbf{0 },{T}]\) and \([{\widetilde{\mathbf{b }}},\mathbf{\widetilde{s} },{\widetilde{T}}]=[\mathbf{0 },\mathbf{0 }, \delta (\mathbf{x }-\xi )]\) on \({E}^3\). Also note that for an isotropic body \(\mathbf{M }=-(3\lambda +2\mu )\alpha \,\mathbf{1 }\) and \(\mathrm{{tr}}\,{\widetilde{\mathbf{E }}}=[(3\lambda +2\mu )\,/(\lambda +2\mu )]\alpha \,{\widetilde{T}}\).
Solution.
Let \({s}=[{\mathbf{u }},\mathbf{E },\mathbf{S }]\) and \(\widetilde{{s}}=[\widetilde{\mathbf{u }},{\widetilde{\mathbf{E }}}, \widetilde{\mathbf{S }}]\) be the thermoelastic states produced on \({E}^{3}\) by the external thermomechanical loads \([\mathbf{{b}},\mathbf{{s}},{T}]=[\mathbf{{0}},\mathbf{0 },{T}(\mathbf{{x}})]\) and \([\widetilde{\mathbf{{b}}},\widetilde{\mathbf{{s}}},\widetilde{T}]=[\mathbf{{0}},\mathbf{{0}},\delta (\mathbf{{x}}-{\upxi })]\), respectively. Applying Eq. (3.29) to the states s and \({\widetilde{S}}\) we obtain
$$\begin{aligned} \int \limits _{{E}^{3}}{T}\mathbf{M }\cdot {\widetilde{\mathbf{E }}}\ {{dv}}=\int \limits _{{E}^{3}}\widetilde{T}\mathbf{M }\cdot \mathbf{E }\ {{dv}} \end{aligned}$$
(3.100)
For a homogeneous isotropic thermoelastic solid
$$\begin{aligned} \mathbf{M }=-(3\lambda +2\mu )\alpha \mathbf{1 } \end{aligned}$$
(3.101)
Therefore,
$$\begin{aligned} \mathbf{M }\cdot {{\widetilde{\mathbf{E }}}}=-(3\lambda +2\mu )\alpha (\mathrm{{tr}}\,{\widetilde{\mathbf{E }}}) \end{aligned}$$
(3.102)
and
$$\begin{aligned} \mathbf{M }\cdot \mathbf{E }=-(3\lambda +2\mu )\alpha (\mathrm{{tr}}\,\mathbf{E }) \end{aligned}$$
(3.103)
where \(\lambda \) and \(\mu \) are Lamé constants. Substituting (3.102) and (3.103) into (3.100) we get
$$\begin{aligned} \int \limits _{{E}^{3}}{T}(\mathrm{{tr}}\,{\widetilde{\mathbf{E }}}){{dv}}=\int \limits _{{E}^{3}}\widetilde{T}(\mathrm{{tr}}\,\mathbf{E }){{dv}} \end{aligned}$$
(3.104)
Since \(\widetilde{s}\) is the thermoelastic state produced by the temperature \(\widetilde{T}\) on \({E}^{3}\), \(\widetilde{\mathbf{u }}\) takes the form
$$\begin{aligned} \widetilde{\mathbf{u }}=\varvec{\nabla }\widetilde{\phi } \end{aligned}$$
(3.105)
where the thermoelastic potential \(\widetilde{\phi }\) satisfies Poisson’s equation
$$\begin{aligned} \nabla ^{2}\widetilde{\phi }=\frac{1+{\nu }}{1-{\nu }}\alpha \widetilde{T} \end{aligned}$$
(3.106)
Hence
$$\begin{aligned} \mathrm{{tr}}\,{\widetilde{\mathbf{E }}}=\nabla ^{2}\widetilde{\phi }=\frac{1+{\nu }}{1-{\nu }}\alpha \widetilde{T} \end{aligned}$$
(3.107)
Therefore, substituting (3.107) into (3.104) we obtain
$$\begin{aligned} \frac{1+{\nu }}{1-{\nu }}\alpha \int \limits _{{E}^{3}}{T}({\upxi })\delta (\mathbf{x }-{\upxi }){{dv}}({\upxi }) = \int \limits _{{E}^{3}}\delta (\mathbf{x }-{\upxi })[\mathrm{{tr}}\,\mathbf{E }({\upxi })]{{{dv}}}({\upxi }) \end{aligned}$$
(3.108)
or using the filtrating property of the delta function we arrive at Eq. (3.99). This completes solution of Problem 3.3.
Problem 3.4.
Assume \({T}_0\) to be a constant temperature, and let \({a}_{i} \;(\mathrm{{i}}=1,2,3)\) be positive constants of the length dimension. Show that for a homogeneous isotropic infinite elastic body subject to the temperature change
$$\begin{aligned} {T}(\mathbf x )=&\;{T}_0 [{H}({x}_1 +{a}_1 )-{H}({x}_1 -{a}_1 )]\times \,[{H}({x}_2 +{a}_2 )-{H}({x}_2 -{a}_2 )] \nonumber \\&\;\times [{H}({x}_3 +{a}_3 )-{H}({x}_3 -{a}_3 )] \end{aligned}$$
(3.109)
where \({H}={H}({x})\) denotes the Heaviside function defined by: \({H}({x})=1\) for \({x}>0\) and \({H}({x})=0\) for \({x}<0\); the stress components \({S}_{{ij}}\) are represented by the formulas
$$\begin{aligned} {S}_{{ij}} (\mathbf{x })=&\;\mathrm{{A}}_0 \!\!\displaystyle \int \limits _{-\mathrm{{a}}_1}^{\mathrm{{a}}_1}{\!}{{d}\xi _1} \displaystyle \!\!\int \limits _{-{a}_2 }^{{a}_2}{\!}{{d}\xi _2} \displaystyle \!\!\int \limits _{-{a}_3}^{{a}_3}{{d}\xi _3} \dfrac{\partial ^2}{\partial \xi _i \partial \xi _j}\left[ {({x}_1\,{-}\,\xi _1 )^2\,{+}\,({x}_2\,{-}\,\xi _2 )^2+({x}_3 -\xi _3 )^2} \right] ^{-1/2} \nonumber \\&\;+ 4\pi \delta _{ij}\mathrm{{A}}_0 \displaystyle \int \limits _{-\mathrm{{a}}_1}^{\mathrm{{a}}_1}{{d}\xi _1} \displaystyle \int \limits _{-{a}_2}^{{a}_2}{{d}\xi _2} \displaystyle \int \limits _{-{a}_3}^{{a}_3}{{d}\xi _3\,} \delta ({x}_1 -\xi _1 )\delta ({x}_2 -\xi _2 )\delta ({x}_3 -\xi _3 ) \end{aligned}$$
(3.110)
where
$$\begin{aligned} \mathrm{{A}}_0 =-\frac{\mu }{2\pi }\frac{1+\nu }{1-\nu }\alpha \,{T}_0 \end{aligned}$$
(3.111)
Also, show that the integrals on the RHS of Eq. (3.110) can be calculated in terms of elementary functions, and for the exterior of the parallelepiped
$$\begin{aligned} \left| {{x}_1} \right| \le {a}_1, \quad \left| {{x}_2} \right| \le {a}_2 \quad \left| {{x}_3} \right| \le {a}_3 \end{aligned}$$
(3.112)
we obtain
$$\begin{aligned} {S}_{12} =&\;{A}_0 \ln \left[ {\frac{({x}_3 +{a}_3 +{r}_{+1.+2.-3} )}{({x}_3 -{a}_3 +{r}_{+1.+2.+3} )}\frac{({x}_3 -{a}_3 +{r}_{+1.-2.+3} )}{({x}_3 +{a}_3 +{r}_{+1.-2.-3} )}} \right. \nonumber \\&\qquad \qquad \times \left. {\frac{({x}_3 -{a}_3 +{r}_{-1.+2.+3} )}{({x}_3 +{a}_3 +{r}_{-1.+2.-3})}\frac{({x}_3 +{a}_3 +{r}_{-1.-2.-3} )}{({x}_3 -{a}_3 +{r}_{-1.-2.+3} )}} \right] \end{aligned}$$
(3.113)
$$\begin{aligned} {S}_{23} =&\;{A}_0 \ln \left[ {\frac{({x}_1 +{a}_1 +{r}_{-1.+2.+3} )}{({x}_1 -{a}_1 +{r}_{+1.+2.+3} )}\frac{({x}_1 -{a}_1 +{r}_{+1.+2.-3} )}{({x}_1 +{a}_1 +{r}_{-1.+2.-3} )}} \right. \nonumber \\&\qquad \qquad \times \left. {\frac{({x}_1 -{a}_1 +{r}_{+1.-2.+3} )}{({x}_1 +{a}_1 +{r}_{-1.-2.+3})}\frac{({x}_1 +{a}_1 +{r}_{-1.-2.-3} )}{({x}_1 -{a}_1 +{r}_{+1.-2.-3} )}} \right] \end{aligned}$$
(3.114)
$$\begin{aligned} {S}_{31} =&\;{A}_0 \ln \left[ {\frac{({x}_2 +{a}_2 +{r}_{+1.-2.+3} )}{({x}_2 -{a}_2 +{r}_{+1.+2.+3} )}\frac{({x}_2 -{a}_2 +{r}_{-1.+2.+3} )}{({x}_2 +{a}_2 +{r}_{-1.-2.+3} )}} \right. \nonumber \\&\qquad \qquad \times \left. {\frac{({x}_2 -{a}_2 +{r}_{+1.+2.-3} )}{({x}_2 +{a}_2 +{r}_{+1.-2.-3} )}\frac{({x}_2 +{a}_2 +{r}_{-1.-2.-3} )}{({x}_2 -{a}_2 +{r}_{-1.+2.-3} )}} \right] \end{aligned}$$
(3.115)
and
$$\begin{aligned} {S}_{11} =&\;{A}_0 \left[ {\tan ^{-1}\left( {\frac{{x}_2 +{a}_2}{{x}_1 -{a}_1}\frac{{x}_3 +{a}_3}{{r}_{+1.-2.-3}}} \right) } -\tan ^{-1}\left( {\frac{{x}_2 +{a}_2}{{x}_1 -{a}_1}\frac{{x}_3 -{a}_3}{{r}_{+1.-2.+3}}} \right) \right. \nonumber \\&\qquad \qquad -\left. \tan ^{-1}\left( {\frac{{x}_2 -{a}_2}{{x}_1 -{a}_1 }\frac{{x}_3 +{a}_3}{{r}_{+1.+2.-3}}} \right) +\tan ^{-1}\left( {\frac{{x}_2 -{a}_2 }{{x}_1 -{a}_1}\frac{{x}_3 -{a}_3}{{r}_{+1.+2.+3}}} \right) \right. \nonumber \\&\qquad \qquad -\left. \tan ^{-1}\left( {\frac{{x}_2 +{a}_2}{{x}_1 +{a}_1 }\frac{{x}_3 +{a}_3}{{r}_{-1.-2.-3}}} \right) +\tan ^{-1}\left( {\frac{{x}_2 +{a}_2 }{{x}_1 +{a}_1}\frac{{x}_3 -{a}_3}{{r}_{-1.-2.+3}}} \right) \right. \nonumber \\&\qquad \qquad +\left. \tan ^{-1}\left( {\frac{{x}_2 -{a}_2}{{x}_1 +{a}_1 }\frac{{x}_3 +{a}_3}{{r}_{-1.+2.-3}}} \right) -\tan ^{-1}\left( {\frac{{x}_2 -{a}_2 }{{x}_1 +{a}_1}\frac{{x}_3 -{a}_3}{{r}_{-1.+2.+3}}} \right) \right] \end{aligned}$$
(3.116)
$$\begin{aligned} {S}_{22} =&\;{A}_0 \left[ {\tan ^{-1}\left( {\frac{{x}_3 +{a}_3}{{x}_2 -{a}_2}\frac{{x}_1 +{a}_1}{{r}_{-1.+2.-3}}} \right) } -\tan ^{-1}\left( {\frac{{x}_3 +{a}_3}{{x}_2 -{a}_2}\frac{{x}_1 -{a}_1}{{r}_{+1.+2.-3}}} \right) \right. \nonumber \\&\qquad \qquad -\left. \tan ^{-1}\left( {\frac{{x}_3 -{a}_3}{{x}_2 -{a}_2}\frac{{x}_1 +{a}_1}{{r}_{-1.+2.+3}}} \right) +\tan ^{-1}\left( {\frac{{x}_3 -{a}_3}{{x}_2 -{a}_2}\frac{{x}_1 -{a}_1 }{{r}_{+1.+2.+3}}} \right) \right. \nonumber \\&\qquad \qquad -\left. \tan ^{-1}\left( {\frac{{x}_3 +{a}_3}{{x}_2 +{a}_2}\frac{{x}_1 +{a}_1}{{r}_{-1.-2.-3}}} \right) +\tan ^{-1}\left( {\frac{{x}_3 +{a}_3}{{x}_2 +{a}_2}\frac{{x}_1 -{a}_1 }{{r}_{+1.-2.-3}}} \right) \right. \nonumber \\&\qquad \qquad +\left. \tan ^{-1}\left( {\frac{{x}_3 -{a}_3}{{x}_2 +{a}_2}\frac{{x}_1 +{a}_1 }{{r}_{-1.-2.+3}}} \right) -\tan ^{-1}\left( {\frac{{x}_3 -{a}_3}{{x}_2 +{a}_2 }\frac{{x}_1 -{a}_1}{{r}_{+1.-2.+3}}} \right) \right] \end{aligned}$$
(3.117)
$$\begin{aligned} {S}_{33} =&\;{A}_0 \left[ {\tan ^{-1}\left( {\frac{{x}_1 +{a}_1}{{x}_3 -{a}_3}\frac{{x}_2 +{a}_2}{{r}_{-1.-2.+3}}} \right) } -\tan ^{-1}\left( {\frac{{x}_1 +{a}_1}{{x}_3 -{a}_3}\frac{{x}_2 -{a}_2}{{r}_{-1.+2.+3}}} \right) \right. \nonumber \\&\qquad \qquad -\left. \tan ^{-1}\left( {\frac{{x}_1 -{a}_1}{{x}_3 -{a}_3}\frac{{x}_2 +{a}_2}{{r}_{+1.-2.-3}}} \right) +\tan ^{-1}\left( {\frac{{x}_1 -{a}_1}{{x}_3 -{a}_3}\frac{{x}_2 -{a}_2 }{{r}_{+1.+2.+3}}} \right) \right. \nonumber \\&\qquad \qquad -\left. \tan ^{-1}\left( {\frac{{x}_1 +{a}_1}{{x}_3 +{a}_3}\frac{{x}_2 +{a}_2}{{r}_{-1.-2.-3}}} \right) +\tan ^{-1}\left( {\frac{{x}_1 +{a}_1}{{x}_3 +{a}_3}\frac{{x}_2 -{a}_2 }{{r}_{-1.+2.-3}}} \right) \right. \nonumber \\&\qquad \qquad +\left. \tan ^{-1}\left( {\frac{{x}_1 -{a}_1}{{x}_3 +{a}_3}\frac{{x}_2 +{a}_2 }{{r}_{+1.-2.-3}}} \right) -\tan ^{-1}\left( {\frac{{x}_1 -{a}_1}{{x}_3 +{a}_3 }\frac{{x}_2 -{a}_2}{{r}_{+1.+2.-3}}} \right) \right] \end{aligned}$$
(3.118)
where
$$\begin{aligned} {r}_{\pm 1.\pm 2.\pm 3} =[({x}_1 \mp {a}_1 )^2+({x}_2 \mp {a}_2 )^2+({x}_3 \mp {a}_3 )^2]^{1/2} \end{aligned}$$
(3.119)
Note that Eq. (3.114) follows from Eq. (3.113) by the transformation of indices
$$\begin{aligned} 1\rightarrow 2,\quad 2\rightarrow 3,\quad 3\rightarrow 1 \end{aligned}$$
and Eq. (3.115) follows from Eq. (3.114) by the transformation of indices
$$\begin{aligned} 2\rightarrow 3,\quad 3\rightarrow 1,\quad 1\rightarrow 2 \end{aligned}$$
Also, Eq. (3.117) follows from Eq. (3.116) by the transformation of indices
$$\begin{aligned} 1\rightarrow 2,\quad 2\rightarrow 3,\quad 3\rightarrow 1 \end{aligned}$$
and Eq. (3.118) follows from Eq. (3.117) by the transformation of indices
$$\begin{aligned} 2\rightarrow 3,\quad 3\rightarrow 1,\quad 1\rightarrow 2. \end{aligned}$$
Hint. To find \({S}_{12} \) use the formula
$$\begin{aligned} \int {\frac{{du}}{\sqrt{{u}^2+{a}^2}}} =\ln \,\left( {{u}+\sqrt{{u}^2+{a}^2}} \right) \end{aligned}$$
(3.120)
and to calculate \({S}_{11} \) take advantage of the formulas
$$\begin{aligned} \int {\frac{{du}}{(\sqrt{{u}^2+{a}^2} )^3}} =\frac{1}{{a}^2}\frac{{u}}{\sqrt{{u}^2+{a}^2}} \end{aligned}$$
(3.121)
and
$$\begin{aligned} \int {\frac{{du}}{({u}^2+{b}^2)\sqrt{{u}^2+{a}^2}}} =\frac{1}{{b}\sqrt{{a}^2-{b}^2}}\,\tan ^{-1}\left( {\frac{{u}\sqrt{{a}^2-{b}^2}}{{b}\sqrt{{u}^2+{a}^2}}} \right) \end{aligned}$$
(3.122)
where a and b are constants subject to the conditions
$$\begin{aligned} {a}\ne 0,\quad {b}\ne 0,\quad \left| {a} \right| >\left| {b} \right| \end{aligned}$$
(3.123)
Solution.
To show (3.110) we note that
$$\begin{aligned} {S}_{{ij}} (\mathbf{x })=\int \limits _{-{a}_1}^{{a}_1}{{d}\xi _1} \int \limits _{-{a}_2}^{{a}_2}{{d}\xi _2} \int \limits _{-{a}_3}^{{a}_3}{{d}\xi _3} \,{S}_{{ij}}^*(\mathbf{x }\xi ) \end{aligned}$$
(3.124)
where
$$\begin{aligned} {S}_{{ij}}^*(\mathbf{x },\xi )=2\mu \left( {\phi _{,ij}^*-\delta _{ij} \phi _{,kk}^*} \right) \end{aligned}$$
(3.125)
and
$$\begin{aligned} \phi _{,kk}^*=\frac{1+\nu }{1-\nu }\alpha \,\delta ({x}_1 -\xi _1 )\delta ({x}_2 -\xi _2 )\delta ({x}_3 -\xi _3 ) \end{aligned}$$
(3.126)
Since
$$\begin{aligned} \phi ^*(\mathbf x ,\xi )=-\frac{1}{4\pi }\frac{1+\nu }{1-\nu }\alpha \,\frac{1}{\left| \mathbf{x -\xi } \right| } \end{aligned}$$
therefore
$$\begin{aligned} {S}_{{ij}}^{*} (\mathbf{x }\,\xi )={A}_0 \left\{ {\frac{\partial ^2}{\partial \xi _i \partial \xi _j}\frac{1}{\left| {\mathbf{x }-\xi } \right| }+4\pi \delta _{ij} \delta ({x}_1 -\xi _1 )\delta ({x}_2 -\xi _2 )\delta ({x}_3 -\xi _3 )} \right\} \end{aligned}$$
where \({A}_0\) is given by (3.111). Hence, substituting \({S}_{{ij}}^{*}\) into Eq. (3.124) we obtain (3.110).
To show (3.113)–(3.118) we note that for the exterior of the parallelepiped
$$\begin{aligned} |{x}_{1}|\le {a}_{1},\quad |{x}_{2}|\le {a}_{2},\quad |{x}_{3}|\le {a}_{3} \end{aligned}$$
(3.127)
Equation (3.110) reduces to
$$\begin{aligned} {S_{ij}}(\mathbf{x })=&\;{A}_{0}\int \limits _{-{a}_{1}}^{{a}_{1}}{d}\xi _{1}\ \int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\ \int \limits _{-{a}_{3}}^{{a}_{3}}\ {d}\xi _{3} \nonumber \\&\; \times \,\frac{\partial ^{2}}{\partial \xi _{i}\ \partial \xi _{j}}[({x}_{1}-\xi _{1})^{2}+({x}_{2}-\xi _{2})^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2}\nonumber \\&\qquad {i,j}=1,\ 2,\ 3. \end{aligned}$$
(3.128)
Letting \({i}=1,\ {j}=2\) in (3.128) we obtain
$$\begin{aligned} {S}_{12}(\mathbf{x })=&\;{A}_{0}\int \limits _{-{a}_{3}}^{{a}_{3}}{d}\xi _{3}\ \int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\ \frac{\partial }{\partial \xi _{2}} \nonumber \\&\;\times \left\{ [({x}_{1}-{a}_{1})^{2}+({x}_{2}-\xi _{2})^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2}\right. \nonumber \\&\quad \qquad -\left. [({x}_{1}+{a}_{1})^{2}+({x}_{2}-\xi _{2})^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2}\right\} \end{aligned}$$
(3.129)
or
$$\begin{aligned} {S}_{12}(\mathbf{x })=&\;{A}_{0}\int \limits _{-{a}_{3}}^{{a}_{3}}\ {d}\xi _{3} \times \left\{ [({x}_{1}-{a}_{1})^{2}+({x}_{2}-{a}_{2})^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2}\right. \nonumber \\&\;\qquad \qquad \qquad \qquad -[({x}_{1}-{a}_{1})^{2}+({x}_{2}+{a}_{2})^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2} \nonumber \\&\;\qquad \qquad \qquad \qquad -[({x}_{1}+{a}_{1})^{2}+({x}_{2}-{a}_{2})^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2}\nonumber \\&\;\qquad \qquad \qquad \qquad +\left. [({x}_{1}+{a}_{1})^{2}+({x}_{2}+{a}_{2})^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2}\right\} \end{aligned}$$
(3.130)
Since for every \({b}>0\)
$$\begin{aligned} \int \limits _{-{a}_{3}}^{{a}_{3}}\ [{b}^{2}+({x}_{3}-\xi _{3})^{2}]^{-1/2}\ {d}\xi _{3}=\int \limits _{{x}_{3}-{a}_{3}}^{{x}_{3}+{a}_{3}}\ ({b}^{2}+{u}^{2})^{-1/2}\ {{du}} \end{aligned}$$
(3.131)
and by virtue of (3.121)
$$\begin{aligned} \int ({b}^{2}+{u}^{2})^{-1/2}\ {{du}}=\ln \left( {u}+\sqrt{{u}^{2}+{b}^{2}}\right) \end{aligned}$$
(3.132)
if follows from (3.130) that
$$\begin{aligned} {S}_{12}=&\;{A}_{0}\ln \left\{ \frac{({x}_{3}+{a}_{3}+r_{+1.\,+2.\,-3})}{({x}_{3}-{a}_{3}+r_{+1.\,+2.\,+3})}\ \frac{({x}_{3}-{a}_{3}+r_{+1.\,-2.\,+3})}{({x}_{3}+{a}_{3}+r_{+1.\,-2.\,-3})}\right. \nonumber \\&\qquad \qquad \times \left. \frac{({x}_{3}-{a}_{3}+r_{-1.\,+2.\,+3})}{({x}_{3}+{a}_{3}+r_{-1.\,+2.\,-3})}\ \frac{({x}_{3}+{a}_{3}+r_{-1.\,-2.\,-3})}{({x}_{3}-{a}_{3}+r_{-1.\,-2.\,+3})}\right\} \end{aligned}$$
(3.133)
where \(r_{\pm 1.\,\pm 2.\,\pm 3}\) is defined by (3.119). This completes proof of (3.113). The components \({S}_{23}\) and \({S}_{31}\), respectively, are obtained from (3.113) by the transformation of the indices
$$\begin{aligned} 1\rightarrow 2,\quad 2\rightarrow 3,\quad 3\rightarrow 1 \end{aligned}$$
(3.134)
and
$$\begin{aligned} 2\rightarrow 3,\quad 3\rightarrow 1,\quad 1\rightarrow 2 \end{aligned}$$
(3.135)
By letting \({i}=1,\ {j}=1\) in (3.128) we obtain
$$\begin{aligned} {S}_{11}(\mathbf{x })=&\;{A}_{0}\int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\ \int \limits _{-{a}_{3}}^{{a}_{3}}{d}\xi _{3}\ \int \limits _{-{a}_{1}}^{{a}_{1}}{d}\xi _{1}\ \frac{\partial ^{2}}{\partial \xi _{1}^{2}} \nonumber \\&\; \times [({x}_{2}-\xi _{2})^{2}+({x}_{3}-\xi _{3})^{2}+({x}_{1}-\xi _{1})^{2}]^{{-1}/{2}} \end{aligned}$$
(3.136)
Since
$$\begin{aligned} \frac{\partial }{\partial \xi _{1}}\ [({x}_{1}-\xi _{1})^{2}+\alpha ^{2}]^{{-1}/{2}}=({x}_{1}-\xi _{1})[({x}_{1}-\xi _{1})^{2}+\alpha ^{2}]^{{-3}/{2}} \end{aligned}$$
(3.137)
where
$$\begin{aligned} \alpha ^{2}=({x}_{2}-\xi _{2})^{2}+({x}_{3}-\xi _{3})^{2} \end{aligned}$$
(3.138)
therefore Eq. (3.136) takes the form
$$\begin{aligned} {S}_{11}(\mathbf{x })={A}_{0}\int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\ \int \limits _{-{a}_{3}}^{{a}_{3}}{d}\xi _{3} \times \Bigg \{\frac{({x}_{1}-{a}_{1})}{[({x}_{1}-{a}_{1})^{2}+\alpha ^{2}]^{3/2}}-\frac{({x}_{1}+{a}_{1})}{[({x}_{1}+{a}_{1})^{2}+\alpha ^{2}]^{3/2}}\Bigg \} \end{aligned}$$
(3.139)
Now, because of (3.121),
$$\begin{aligned}&\int \limits _{-{a}_{3}}^{{a}_{3}}{d}\xi _{3}\frac{1}{[({x}_{1}-{a}_{1})^{2}+\alpha ^{2}]^{3/2}}=\int \limits _{{x}_{3}-{a}_{3}}^{x_{3}+a_{3}}\frac{{du}}{[({x}_{1}-{a}_{1})^{2}+({x}_{2}-\xi _{2})^{2}+{u}^{2}]^{3/2}}\nonumber \\&\quad =\frac{1}{({x}_{1}-{a}_{1})^{2}+({x}_{2}-\xi _{2})^{2}}\Bigg \{\frac{({x}_{3}+{a}_{3})}{[({x}_{1}-{a}_{1})^{2}+({x}_{2}-\xi _{2})^{2}+({x}_{3}+{a}_{3})^{2}]^{1/2}} \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \quad -\frac{({x}_{3}-{a}_{3})}{[({x}_{1}-{a}_{1})^{2}+({x}_{2}-\xi _{2})^{2}+({x}_{3}-{a}_{3})^{2}]^{1/2}}\Bigg \} \end{aligned}$$
(3.140)
Also, using (3.122), we obtain
$$\begin{aligned}&\int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\frac{1}{({x}_{2}-\xi _{2})^{2}+({x}_{1}-{a}_{1})^{2}}\ \frac{1}{[({x}_{1}-{a}_{1})^{2}+({x}_{3}+{a}_{3})^{2}+({x}_{2}-\xi _2)^{2}]^{1/2}} \nonumber \\&\quad =\int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\frac{1}{({x}_{2}-\xi _{2})^{2}+{b}^{2}}\ \frac{1}{[({x}_{2}-\xi _{2})^{2}+{a}^{2}]^{1/2}} =\int \limits _{{x}_{2}-{a}_{2}}^{{x}_{2}+{a}_{2}}\ \frac{1}{{u}^{2}+{b}^{2}}\ \frac{1}{\sqrt{{u}^{2}+{a}^{2}}}{{du}} \nonumber \\&\quad =\frac{1}{{b}\sqrt{{a}^{2}-{b}^{2}}}\ \left\{ \tan ^{-1}\ \frac{{u}\sqrt{{a}^{2}-{b}^{2}}}{{b}\sqrt{{u}^{2}+{a}^{2}}}\right\} _{{u}={x}_{2}-{a}_{2}}^{{u}={x}_{2}+{a}_{2}} \end{aligned}$$
(3.141)
where
$$\begin{aligned} {b}^{2}=({x}_{1}-{a}_{1})^{2},\quad {a}^{2}=({x}_{1}-{a}_{1})^{2}+({x}_{3}+{a}_{3})^{2} \end{aligned}$$
(3.142)
By letting \({x}_{1}>{a}_{1}\) and \({x}_{3}>-{a}_{3}\) we receive
$$\begin{aligned} {b}={x}_{1}-{a}_{1},\quad \sqrt{{a}^{2}-{b}^{2}}={x}_{3}+{a}_{3} \end{aligned}$$
(3.143)
and reduce Eq. (3.141) to
$$\begin{aligned}&\int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\frac{1}{({x}_{2}-\xi _{2})^{2}+{b}^{2}}\ \frac{1}{[({x}_{2}-\xi _{2})^{2}+{a}^{2}]^{1/2}}\nonumber \\&\quad =\frac{1}{({x}_{1}-{a}_{1})({x}_{3}+{a}_{3})}\left\{ \tan ^{-1}\frac{{x}_{2}+{a}_{2}}{{x}_{1}-{a}_{1}}\ \frac{{x}_{3}+{a}_{3}}{r_{+1.\,-2.\,-3}} -\tan ^{-1} \frac{{x}_{2}-{a}_{2}}{{x}_{1}-{a}_{1}}\ \frac{{x}_{3}+{a}_{3}}{r_{+1.\,+2.\,-3}}\right\} \end{aligned}$$
(3.144)
It follows from Eq. (3.139) that
$$\begin{aligned} {S}_{11}(\mathbf{x }) =&\;{A}_{0}\int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\ ({x}_{1}-{a}_{1})\ \frac{1}{({x}_{1}-{a}_{1})^{2}+({x}_{2}-\xi _{2})^{2}} \nonumber \\&\quad \times \left\{ \frac{({x}_{3}+{a}_{3})}{[({x}_{2}-\xi _{2})^{2}+({x}_{1}-{a}_{1})^{2}+({x}_{3}+{a}_{3})^{2}]^{1/2}}\right. \nonumber \\&\quad \qquad \left. -\frac{({x}_{3}-{a}_{3})}{[({x}_{2}-\xi _{2})^{2}+({x}_{1}-{a}_{1})^{2}+({x}_{3}-{a}_{3})^{2}]^{1/2}}\right\} \nonumber \\&\quad -{A}_{0}\int \limits _{-{a}_{2}}^{{a}_{2}}{d}\xi _{2}\ ({x}_{1}+{a}_{1})\ \frac{1}{({x}_{1}+{a}_{1})^{2}+\left( {x}_{2}-\xi _{2}^{2}\right) ^{2}} \nonumber \\&\quad \times \left\{ \frac{({x}_{3}+{a}_{3})}{[({x}_{2}-\xi _{2})^{2}+({x}_{1}+{a}_{1})^{2}+({x}_{3}+{a}_{3})^{2}]^{1/2}}\right. \nonumber \\&\qquad \quad \left. -\frac{({x}_{3}-{a}_{3})}{[({x}_{2}-\xi _{2})^{2}+({x}_{1}+{a}_{1})^{2}+({x}_{3}-{a}_{3})^{2}]^{1/2}}\right\} \end{aligned}$$
(3.145)
Therefore, using Eq. (3.141) as well as equations obtained from Eq. (3.141) by suitable choice of \(a\) and \(b\), we obtain (3.116).
The components \({S}_{22}\) and \({S}_{33}\) are obtained from Eq. (3.116) by suitable transformation of indices.
Problem 3.5.
Let \({\mathbf{u }}={\mathbf{u }}({\mathbf{x }},\mathrm{{t}})\) be a solution of the vector equation
$$\begin{aligned} \nabla ^2{\mathbf{u }}-\frac{1}{{c}^2}\frac{\partial ^2{\mathbf{u }}}{\partial {t}^2}=-\frac{\mathbf{f }}{{c}^2}\quad \mathrm{{on}}\quad \mathrm{{B}}\times (0,\infty ) \end{aligned}$$
(3.146)
subject to the initial conditions
$$\begin{aligned} \mathbf{u }(\mathbf{x },0)=\mathbf{u }_0 (\mathbf{x }),\quad \dot{\mathbf{u }}(\mathbf{x },0)={\dot{\mathbf{u }}}_0 (\mathbf{x })\quad \mathrm{{for}}\quad \mathbf{x }\in {\bar{\mathrm{{B}}}} \end{aligned}$$
(3.147)
where \(\mathbf{f }=\mathbf{f }(\mathbf{x },{t})\) is a prescribed vector field on \({\bar{\mathrm{{B}}}}\times [0,\infty )\); and \(\mathbf{u }_0 (\mathbf{x })\) and \({\dot{\mathbf{u }}}_0 (\mathbf{x })\) are prescribed vector fields on \({\bar{\mathrm{{B}}}}\); and \({c}>0\).
Also, let \({\widetilde{\mathbf{u }}}={\widetilde{\mathbf{u }}}({\mathbf{x }},{t})\) be a solution of the vector equation
$$\begin{aligned} \nabla ^2{\widetilde{\mathbf{u }}}-\frac{1}{{c}^2}\frac{\partial ^2{\widetilde{\mathbf{u }}}}{\partial {t}^2}=-\frac{{\widetilde{\mathbf{f }}}}{{c}^2}\quad \mathrm{{on}}\quad \mathrm{{B}}\times (0,\infty ) \end{aligned}$$
(3.148)
subject to the initial conditions
$$\begin{aligned} {\widetilde{\mathbf{u }}}({\mathbf{x }},0)={\widetilde{\mathbf{u }}}_0 ({\mathbf{x }}),\quad \dot{{\widetilde{\mathbf{u }}}}({\mathbf{x }},0)=\dot{{\widetilde{\mathbf{u }}}}_0 ({\mathbf{x }})\quad \mathrm{{for}}\quad \mathbf{x }\in {\bar{\mathrm{{B}}}} \end{aligned}$$
(3.149)
where \({\widetilde{\mathbf{f }}}={\widetilde{\mathbf{f }}}(\mathbf{x },{t})\ne \mathbf{f }(\mathbf{x }, \mathrm{{t}})\), \({\widetilde{\mathbf{u }}}_0 (\mathbf{x })\ne \mathbf{u }_0 (\mathbf{x })\), and \({\dot{\widetilde{\mathbf{u }}}}_0 (\mathbf{x })\ne {\dot{\mathbf{u }}}_0 (\mathbf{x })\) are prescribed functions on \({\bar{\mathrm{{B}}}}\times [0,\infty )\), \({\overline{\mathrm{{B}}}}\), and \({\bar{\mathrm{{B}}}}\), respectively. Show that the following reciprocal relation holds true
$$\begin{aligned} \dfrac{1}{{c}^2}&\int \limits _\mathrm{{B}}{(\mathbf{u }*{\widetilde{\mathbf{f }}}+\mathbf{u }\cdot {\dot{\widetilde{\mathbf{u }}}}_0 +{\dot{\mathbf{u }}}\cdot {\widetilde{\mathbf{u }}}_0 )\,{dv}} +\displaystyle \int \limits _{\partial \mathrm{{B}}}{\mathbf{u }*\dfrac{\partial {\widetilde{\mathbf{u }}}}{\partial \mathrm{{n}}}{da}} \nonumber \\ \quad&=\dfrac{1}{{c}^2}\int \limits _\mathrm{{B}}{({\widetilde{\mathbf{u }}}*\mathbf{f }+{\widetilde{\mathbf{u }}}\cdot {\dot{\mathbf{u }}}_0 +{\dot{\widetilde{\mathbf{u }}}}\cdot \mathbf{u }_0 )\,{dv}} +\displaystyle \int \limits _{\partial \mathrm{{B}}}{{\widetilde{\mathbf{u }}}*\dfrac{\partial \mathbf{u }}{\partial \mathrm{{n}}}{da}} \end{aligned}$$
(3.150)
where \(*\) represents the inner convolutional product, that is, for any two vector fields \(\mathbf{a }=\mathbf{a }(\mathbf{x },{t})\) and \(\mathbf{b }=\mathbf{b }(\mathbf{x },{t})\) on \({\bar{\mathrm{{B}}}}\times [0,\infty )\)
$$\begin{aligned} \mathbf{a }*\mathbf{b }=\int \limits _0^{t} {\mathbf{a }(\mathbf{x },{t}-\tau )\,\cdot \,} \mathbf{b }(\mathbf{x },\tau )\,{d}\tau \end{aligned}$$
(3.151)
Solution.
Let \(\overline{f}(\mathbf{x },{p})\) denote the Laplace transform of a function \({f}={f}(\mathbf{x },{t})\) defined by
$$\begin{aligned} {Lf}\equiv \bar{f}(\mathbf{x },{p})=\int \limits _{0}^{\infty }{e}^{-{pt}}{f}(\mathbf{x },{t}){dt} \end{aligned}$$
(3.152)
Then
$$\begin{aligned}&\overline{\dot{f}(\mathbf{x },{t})}={p}\ \bar{f}(\mathbf{x },{p})-{f}(\mathbf{x },0)\end{aligned}$$
(3.153)
$$\begin{aligned}&\overline{\ddot{f}(\mathbf{x },{t})}={p}^{2}\ \bar{f}(\mathbf{x },{p})-\dot{f}(\mathbf{x },0)-{{pf}}(\mathbf{x }, 0) \end{aligned}$$
(3.154)
Now, Eqs. (3.146) and (3.147) in components take the forms
$$\begin{aligned} {u_{i,kk}}-\frac{1}{{c}^{2}}\ddot{u}_{i}=-\frac{f_{i}}{{c}^{2}} \end{aligned}$$
(3.155)
and
$$\begin{aligned} {u_{i}}(\mathbf{x },0)={u_{0i}}(\mathbf{x }),\ \dot{u}_{i}(\mathbf{x },0)=\dot{u}_{0{i}}(\mathbf{x }) \end{aligned}$$
(3.156)
Taking the Laplace transform of Eq. (3.155) and using (3.156) we obtain
$$\begin{aligned} \bar{u}_{i,kk}-\frac{1}{{c}^{2}}({p}^{2}\bar{u}_{i}-\dot{u}_{0{i}}-{p\ u}_{0{i}})=-\frac{\bar{f}_{i}}{{c}^{2}} \end{aligned}$$
(3.157)
Similarly, Eqs. (3.148) and (3.149) imply that
$$\begin{aligned} \bar{\widetilde{u}}_{i, kk}-\frac{1}{{c}^{2}}({p}^{2}\bar{\widetilde{u}}_{i}-\dot{\widetilde{u}}_{0{i}}-{p}\ \widetilde{u}_{0{i}})=\frac{\bar{\widetilde{f}}_{i}}{{c}^{2}} \end{aligned}$$
(3.158)
Multiplying (3.157) by \(\overline{\widetilde{u}_{i}}\) and (3.158) by \(\overline{u_{i}}\), respectively, we obtain
$$\begin{aligned}&\bar{\widetilde{u}}_{i}\ \bar{u}_{i,kk}-\frac{1}{{c}^{2}}({p}^{2}\bar{\widetilde{u}}_{i}\bar{u}_{i}-\bar{\widetilde{u}}_{i}\dot{u}_{0{i}}-{p}\ \bar{\widetilde{u}}_{i}{u}_{0{i}})=-\frac{\bar{f}_{i}\ \bar{\widetilde{u}}_{i}}{{c}^{2}} \end{aligned}$$
(3.159)
$$\begin{aligned}&\bar{u}_{i}\ \bar{\widetilde{u}}_{i,kk}-\frac{1}{{c}^{2}}({p}^{2}\bar{u}_{i}\bar{\widetilde{u}}_{i}-\bar{u}_{i}\dot{\widetilde{u}}_{0{i}}-{p}\ \bar{u}_{i}\widetilde{u}_{0{i}})=-\frac{\bar{\widetilde{f}}_{i}\ \bar{u}_{i}}{{c}^{2}} \end{aligned}$$
(3.160)
Since
$$\begin{aligned} \bar{\widetilde{u}}_{i}\bar{u}_{i,kk}=(\bar{\widetilde{u}}_{i}\bar{u}_{i, k}),_{k}-\bar{\widetilde{u}}_{i,k}\bar{u}_{i,k} \end{aligned}$$
(3.161)
and
$$\begin{aligned} \bar{u}_{i}\bar{\widetilde{u}}_{i,kk}=(\bar{u}_{i}\bar{\widetilde{u}}_{i, k}),_{k}-\bar{u}_{i,k}\bar{\widetilde{u}}_{i,k} \end{aligned}$$
(3.162)
therefore, subtracting (3.160) from (3.159), and using the divergence theorem we obtain
$$\begin{aligned}&\int \limits _{\partial {B}}(\bar{\widetilde{u}}_{i}\bar{u}_{i,k}\ {n_{k}}-\bar{u}_{i}\bar{\widetilde{u}}_{i,k}\ {n_{k}}){{da}} +\frac{1}{{c}^{2}}\int \limits _{B}(\dot{u}_{0{i}}\bar{\widetilde{u}}_{i}+{p\ u_{0i}}\bar{\widetilde{u}}_{i}-\dot{\widetilde{u}}_{0{i}}\bar{u}_{i}-{p}\widetilde{u}_{0{i}}\bar{u}_{i}){dv} \nonumber \\&\quad =-\frac{1}{{c}^{2}}\int \limits _{B}(\bar{f}_{i}\bar{\widetilde{u}}_{i}-\bar{\widetilde{f}}_{i}\bar{u}_{i}){{dv}} \end{aligned}$$
(3.163)
or
$$\begin{aligned}&\int \limits _{\partial {B}}\bar{u}_{i}\bar{\widetilde{u}}_{i,k}{n_{k}}\ {{da}} +\frac{1}{{c}^{2}}\int \limits _{B}[\bar{u}_{i}\bar{\widetilde{f}}_{i}+\bar{u}_{i}\dot{\widetilde{u}}_{0{i}}+({p}\bar{u}_{i}-{u}_{0{i}})\widetilde{u}_{0{i}}+{u}_{0{i}}\widetilde{u}_{0{i}}]{{dv}} \nonumber \\&\quad =\int \limits _{\partial {B}}\bar{\widetilde{u}}_{i}\bar{u}_{i,k}\ {n_{k}\ {{da}}} +\frac{1}{{c}^{2}}\int \limits _{B}[\bar{\widetilde{u}}_{i}\bar{f}_{i}+\bar{\widetilde{u}}_{i}\dot{u}_{0{i}}+({p}\bar{\widetilde{u}}_{i}-\widetilde{u}_{0{i}}){u}_{0{i}}+ \widetilde{u}_{0{i}}{u}_{0{i}}]{{dv}} \end{aligned}$$
(3.164)
Using the formula
$$\begin{aligned} {L}^{-1}(\bar{f}\ \bar{g})={f}*{g} \end{aligned}$$
(3.165)
and applying the operator \({L}^{-1}\) to Eq. (3.164) we arrive at Eq. (3.150). This completes solution of Problem 3.5.
Problem 3.6.
Let \({\widetilde{\mathbf{U }}}={\widetilde{\mathbf{U }}}(\mathbf{x },\mathrm{{t}})\) be a symmetric second-order tensor field that satisfies the wave equation
$$\begin{aligned} \left( {\nabla ^2-\frac{1}{{c}^2}\frac{\partial ^2}{\partial {t}^2}} \right) \,{\widetilde{\mathbf{U }}}=-\frac{{\widetilde{\mathbf{F }}}}{{c}^2}\quad \mathrm{{on}}\quad \mathrm{{B}}\times (0,\infty ) \end{aligned}$$
(3.166)
subject to the initial conditions
$$\begin{aligned} {\widetilde{\mathbf{U }}}(\mathbf{x },0)={\widetilde{\mathbf{U }}}_0 (\mathbf{x }),\quad \dot{{\widetilde{\mathbf{U }}}}(\mathbf{x },0)={\dot{\widetilde{\mathbf{U }}}}_0 (\mathbf{x })\quad \mathrm{{for}}\quad \mathbf{x }\in {\bar{\mathrm{{B}}}} \end{aligned}$$
(3.167)
where \({\widetilde{\mathbf{F }}}={\widetilde{\mathbf{F }}}(\mathbf{x },{t})\), \({\widetilde{\mathbf{U }}}_0 (\mathbf{x })\), and \({\dot{\widetilde{\mathbf{U }}}}_0 (\mathbf{x })\) are prescribed functions on \({\bar{\mathrm{{B}}}}\times [0,\infty )\), \({\bar{\mathrm{{B}}}}\), and \({\bar{\mathrm{{B}}}}\), respectively. Let \(\mathbf{u }=\mathbf{u }(\mathbf{x },{t})\) be a solution to Eq. (3.166) and (3.167) of Problem 3.5. Show that
$$\begin{aligned}&\dfrac{1}{{c}^2}\int \limits _\mathrm{{B}}{({\widetilde{\mathbf{F }}}*\mathbf{u }+{\dot{\widetilde{\mathbf{U }}}}_0\mathbf{u }+{\widetilde{\mathbf{U }}}_0{\dot{\mathbf{u }}})\,{d{v}}} +\displaystyle \int \limits _{\partial \mathrm{{B}}}{\dfrac{\partial {\widetilde{\mathbf{U }}}}{\partial \mathrm{{n}}}*\mathbf{u }\,{da}} \nonumber \\&\quad =\dfrac{1}{{c}^2}\int \limits _\mathrm{{B}}{({\widetilde{\mathbf{U }}}*\mathbf{f }+{\dot{\widetilde{\mathbf{U }}}}\mathbf{u }_0 +{\widetilde{\mathbf{U }}}{\dot{\mathbf{u }}}_0 )\,{d{v}}} +\displaystyle \int \limits _{\partial \mathrm{{B}}}{{\widetilde{\mathbf{U }}}*\dfrac{\partial \mathbf{u }}{\partial \mathrm{{n}}}{da}} \end{aligned}$$
(3.168)
where for any tensor field \(\mathbf{T }=\mathbf{T }(\mathbf{x },{t})\) on \({\bar{\mathrm{{B}}}}\times [0,\infty )\) and for any vector field \(\mathbf{v }=\mathbf{v }(\mathbf{x },{t})\) on \({\bar{\mathrm{{B}}}}\times [0,\infty )\)
$$\begin{aligned} \mathbf{T }*\mathbf{v }=\int \limits _0^{t} \mathbf{T }(\mathbf{x },{t}-\tau )\mathbf{v }(\mathbf{x },\tau )\,{d}\tau \end{aligned}$$
(3.169)
Solution.
Equation (3.150) of Problem 3.5 in components takes the form
$$\begin{aligned}&\frac{1}{{c}^{2}}\int \limits _{B}(\widetilde{f}_{i}*{u_{i}}+\dot{\widetilde{u}}_{0{i}}{u_{i}}+\widetilde{u}_{0{i}}\dot{u}_{i}){{{dv}}} +\int \limits _{\partial {B}}\frac{\partial \widetilde{u}_{i}}{\partial {n}}*{u_{i}{{da}}} \nonumber \\&\quad =\frac{1}{{c}^{2}}\int \limits _{B}({f_{i}}*\widetilde{u}_{i}+\dot{u}_{0{i}}\widetilde{u_{i}}+{u}_{0{i}}\dot{\widetilde{u}}_{i}){{dv}} +\int \limits _{\partial {B}}\frac{\partial {u_{i}}}{\partial {n}}*\widetilde{u}_{i}{{da}} \end{aligned}$$
(3.170)
It follows from the formulation of Problems 3.5 and 3.6 that Eq. (3.170) holds also true if for a fixed index j we let
$$\begin{aligned} {\widetilde{u}}_{i}={\widetilde{U}}_{ij},\quad {\widetilde{u}}_{0{i}} =\widetilde{U}_{0{ij}},\quad \dot{\widetilde{u}}_{0{i}}=\dot{\widetilde{U}}_{0{ij}},\quad \widetilde{f}_{i}=\widetilde{F}_{ij} \end{aligned}$$
(3.171)
Therefore, substituting (3.171) into (3.170) we obtain
$$\begin{aligned}&\frac{1}{{c}^{2}}\int \limits _{B}(\widetilde{F}_{ij}*{u_{i}}+\dot{\widetilde{U}}_{0{ij}}{u_{i}}+\widetilde{U}_{0{ij}}\dot{u}_{i}){{dv}} +\int \limits _{B}\frac{\partial \widetilde{U}_{ij}}{\partial {n}}*{u_{i}}{{da}}\nonumber \\&\quad =\frac{1}{{c}^{2}}\int \limits _{B}({f_{i}}*\widetilde{U}_{ij}+\dot{u}_{0{i}}\ \widetilde{U}_{ij}+{u}_{0{i}}\ \dot{\widetilde{U}}_{ij}){{dv}} +\int \limits _{B}\frac{\partial {u_{i}}}{\partial {n}}*\widetilde{U}_{ij}\ {{da}} \end{aligned}$$
(3.172)
Finally, the symmetry of tensors \(\widetilde{U}_{ij},\ \widetilde{F}_{ij},\ \widetilde{U}_{0ij}\), and \(\dot{\widetilde{U}}_{0ij}\), as well as the relation
$$\begin{aligned} {a}*{b}={b}*{a} \end{aligned}$$
(3.173)
valid for arbitrary functions \({a}={a}(\mathbf{x },{t})\) and \({b}={b}(\mathbf{x },{t})\), imply that Eq. (3.172) is equivalent to Eq. (3.168). This completes solution to Problem 3.6.
Problem 3.7.
Let \(\mathbf{G }=\mathbf{G }(\mathbf{x },\xi ;{t})\) be a symmetric second-order tensor field that satisfies the wave equation
$$\begin{aligned} \square _0^2\mathbf{G }=-\mathbf{1 }\delta (\mathbf{x }-\xi )\delta ({t})\quad \mathrm{{for}} \quad \mathbf{x }\in {E}^3,\xi \in {E}^3,\, \, {t}>0 \end{aligned}$$
(3.174)
subject to the homogeneous initial conditions
$$\begin{aligned} \mathbf{G }(\mathbf{x },\xi ;0)=\mathbf{0 }, \quad {\dot{\mathbf{G }}}(\mathbf{x },\xi ;0)=\mathbf{0 }\quad \mathrm{{for}} \quad \mathbf{x }\in {E}^3,\xi \in {E}^3 \end{aligned}$$
(3.175)
where
$$\begin{aligned} \square _0^2 =\,\frac{\partial ^2}{\partial {x}_{k} \partial {x}_{k} }-\frac{1}{{c}^2}\frac{\partial ^2}{\partial {t}^2}\quad ({k}=1,2,3) \end{aligned}$$
(3.176)
Show that a solution u to Eqs. (3.146) and (3.147) of Problem 3.5 admits the integral representation
$$\begin{aligned} \mathbf{u }(\mathbf{x },{t})=&\;\dfrac{1}{{c}^2}\displaystyle \int \limits _\mathrm{{B}}{(\mathbf{G }*\mathbf{f }+{\dot{\mathbf{G }}}\mathbf{u }_0 +\mathbf{G }{\dot{\mathbf{u }}}_0 )\,d{v}(\xi )} \nonumber \\&\; +\displaystyle \int \limits _{\partial \mathrm{{B}}}{\left( \mathbf{G }*{\dfrac{\partial \mathbf{u }}{\partial \mathrm{{n}}}-\dfrac{\partial \mathbf{G }}{\partial \mathrm{{n}}}*\mathbf{u }} \right) } \,{da}(\xi ) \end{aligned}$$
(3.177)
Hint. Apply the reciprocal relation (3.168) of Problem 3.6 in which \({\widetilde{\mathbf{F }}}/{c}^2=\mathbf{1 }\delta (\mathbf{x }-\xi )\delta ({t})\) and \({\widetilde{\mathbf{U }}}=\mathbf{G }(\mathbf{x },\xi ;{t})\).
Solution.
To solve this problem we let in Eq. (3.168) of Problem 3.6 the following
$$\begin{aligned} \widetilde{\mathbf{U }}&={\mathbf{G }}({\mathbf{x }},\ {\upxi };\ {t}),\quad \widetilde{\mathbf{U }}_{0}={\mathbf{G }}({\mathbf{x }},\ {\upxi };\ 0)={\mathbf{0 }} \nonumber \\ \dot{\widetilde{\mathbf{U }}}_{0}&=\dot{\mathbf{G }}({\mathbf{x }},\ {\upxi };\ 0)={\mathbf{0 }},\quad {\widetilde{\mathbf{{F} }}}/{c}^{2}={\mathbf{1 }}\delta ({\mathbf{x }}-{\upxi })\ \delta ({t}) \end{aligned}$$
(3.178)
and
$$\begin{aligned} {{dv}}={{dv}}({\upxi }),\quad {{da}}={{da}}({\upxi }) \end{aligned}$$
(3.179)
and obtain
$$\begin{aligned}&\int \limits _{B}[\mathbf{1 }\ \delta (\mathbf{x }-{\upxi })\delta ({t})]*\mathbf{u }({\upxi };\ {t}){{dv}}({\upxi }) +\int \limits _{\partial {B}}\left( \frac{\partial \mathbf{G }}{\partial {n}}\right) *\mathbf{u }\ {{da}}({\upxi }) \nonumber \\&\quad =\frac{1}{{c}^{2}}\int \limits _{B}(\mathbf{G }*\mathbf{f }+\dot{\mathbf{G }}\,\mathbf{u }_{0}+\mathbf{G }\ \dot{\mathbf{u }}_{0}){{dv}}({\upxi }) +\int \limits _{\partial {B}}\mathbf{G }*\frac{\partial \mathbf{u }}{\partial {n}}{{da}}({\upxi }) \end{aligned}$$
(3.180)
Finally, using the filtrating property of the delta function we find that (3.180) is equivalent to (3.177). This completes solution to Problem 3.7.
Problem 3.8.
Show that a unique solution to Eqs. (3.174) and (3.175) of Problem 3.7 takes the form
$$\begin{aligned} \mathbf{G }(\mathbf{x },\xi ;{t})=\frac{1}{4\pi \left| \mathbf{x -\xi } \right| }\delta \left( {{t}-\frac{\left| \mathbf{x -\xi } \right| }{{c}}} \right) \,\mathbf 1 \end{aligned}$$
(3.181)
and, hence, reduce Eq. (3.177) from Problem 3.7 to the Poisson-Kirchhoff integral representation
$$\begin{aligned} \mathbf{u }(\mathbf{x },{t}) =\,&\dfrac{1}{4\pi \,{c}^2}\displaystyle \int \limits _\mathrm{{B}}{\dfrac{\mathbf{f }(\xi ,{t}-\left| \mathbf{x -\xi } \right| \,/{c})}{\left| \mathbf{x -\xi } \right| }} \,d{v}(\xi )+\dfrac{\partial }{\partial {t}}[{t\,M}_{\mathbf{x },{ct}} (\mathbf{u }_0 )] +{t\,M}_{\mathbf{x },{ct}} ({\dot{\mathbf{u }}}_0 ) \nonumber \\&+\dfrac{1}{4\pi }\displaystyle \int \limits _{\partial \mathrm{{B}}} \left\{ \dfrac{1}{\left| \mathbf{x -\xi } \right| }\dfrac{\partial \mathbf{u }}{\partial \mathrm{{n}}} (\xi ,{t}-\left| \mathbf{x -\xi } \right| \,/{c})-\mathbf{u }(\xi ,{t}-\left| {\mathbf{x }-\xi } \right| \,/{c})\dfrac{\partial }{\partial \mathrm{{n}}}\dfrac{1}{\left| {\mathbf{x }-\xi } \right| } \right. \nonumber \\&\left. +\dfrac{1}{{c}\,\left| {\mathbf{x }-\xi } \right| }\left[ {\dfrac{\partial }{\partial \mathrm{{n}}}\left| {\mathbf{x }-\xi } \right| } \right] \,\left[ {\dfrac{\partial \mathbf{u }}{\partial {t}}(\xi ,{t}-\left| \mathbf{x -\xi } \right| \,/{c})} \right] \,\right\} \,{da}(\xi )\nonumber \\ \end{aligned}$$
(3.182)
where for any vector field \(\mathbf{v }=\mathbf{v }(\mathbf{x })\) on \(\mathrm{{B}}\subset {E}^3\) the symbol \(\mathrm{{M}}_{\mathbf{x },{ct}} (\mathbf{v })\) represents the mean value of v over the spherical surface with a center at x and of radius ct, that is,
$$\begin{aligned} \mathrm{{M}}_{\mathbf{x },{ct}} (\mathbf{v })=&\;\dfrac{1}{4\pi }\displaystyle \int \limits _0^{2\pi }{\mathrm{{d}}\varphi } \displaystyle \int \limits _0^\pi {{d}\theta \,\sin \,\theta } \nonumber \\&\;\times \mathbf{v }(\mathrm{{x}}_1 +{ct}\,\sin \theta \,\cos \varphi ,\mathrm{{x}}_2 +{ct}\,\sin \theta \,\sin \varphi ,{x}_3 +{ct}\,\cos \theta ) \end{aligned}$$
(3.183)
and we adopt the convention that all relevant quantities vanish for negative time arguments.
Note. If \(\mathrm{{B}}={E}^3\) and \(\mathbf{f }=\mathbf{0 }\,\;\mathrm{{on}}\;\;{E}^3\times [0,\infty )\) then Eq. (3.182) reduces to the form
$$\begin{aligned} \mathbf{u }(\mathbf{x },{t})=\frac{\partial }{\partial {t}}[{t\,M}_{\mathbf{x },{ct}} (\mathbf{u }_0 )]+{t\,M}_{\mathbf{x },\mathrm{{ct}}} ({\dot{\mathbf{u }}}_0 ) \end{aligned}$$
(3.184)
Solution.
Equation (3.174) of Problem 3.7 takes the form
$$\begin{aligned} \left( \nabla ^{2}-\frac{1}{{c}^{2}}\,\frac{\partial ^{2}}{\partial {t}^{2}}\right) \ \mathbf{G }=-\mathbf{1 }\ \delta (\mathbf{x }-{\upxi })\ \delta ({t}) \end{aligned}$$
(3.185)
Applying the Laplace transform to this equation and using the homogeneous initial conditions (3.175) of Problem 3.7 we obtain
$$\begin{aligned} \left[ \nabla ^{2}-\left( \frac{{p}}{{c}}\right) ^{2}\right] \ \bar{\mathbf{G }}=-\mathbf{1 }\ \delta (\mathbf{x }-{\upxi }) \end{aligned}$$
(3.186)
where
$$\begin{aligned} \bar{\mathbf{G }}=\bar{\mathbf{G }} (\mathbf{x }, {\upxi };\ {p})=\int \limits _{0}^{\infty }{{e}^{-pt}\mathbf{G }(\mathbf{x }, {\upxi },{t}){dt}} \end{aligned}$$
(3.187)
The only solution to Eq. (3.186) in \({E}^{3}\) that vanishes as \(|\mathbf{x }|\rightarrow \infty ,\ |{\upxi }|<\infty \), takes the form \(({p}>0)\)
$$\begin{aligned} \bar{\mathbf{G }}=\frac{1}{4\pi }\ \frac{1}{|\mathbf{x }-{\upxi }|}\ {e}^{-\frac{{p}}{{c}}|\mathbf{x }-{\upxi }|}\mathbf{1 } \end{aligned}$$
(3.188)
Hence, applying the operator \({L}^{-1}\) to (3.188) we obtain
$$\begin{aligned} \mathbf{G }(\mathbf{x },\ {\upxi };\ {t})=\frac{1}{4\pi \,|\mathbf{x }-{\upxi }|}\ \delta \bigg ({t}-\frac{1}{{c}}|\mathbf{x }-{\upxi }|\bigg )\mathbf{1 } \end{aligned}$$
(3.189)
This completes proof of (3.181). To show that (3.182) holds true, we split (3.177) of Problem 3.7 into the sum
$$\begin{aligned} \mathbf{u }(\mathbf{x },{t})=\mathbf{u }^{(1)}(\mathbf{x },{t})+\mathbf{u }^{(2)}(\mathbf{x },{t})+\mathbf{u }^{(3)}(\mathbf{x },{t}) \end{aligned}$$
(3.190)
where
$$\begin{aligned}&\mathbf{u }^{(1)}(\mathbf{x },{t})=\frac{1}{{c}^{2}}\ \int \limits _{B}\mathbf{G }*\mathbf{f }\ {{dv}}({\upxi }) \end{aligned}$$
(3.191)
$$\begin{aligned}&\mathbf{u }^{(2)}(\mathbf{x },{t})=\int \limits _{\partial {B}}\ \left( \mathbf{G }*\frac{\partial \mathbf{u }}{\partial {n}}-\frac{\partial \mathbf{G }}{\partial {n}}*\mathbf{u }\right) {{da}}({\upxi }) \end{aligned}$$
(3.192)
$$\begin{aligned}&\mathbf{u }^{(3)}(\mathbf{x },{t})=\frac{1}{{c}^{2}}\ \int \limits _{B}\ {(\dot{\mathbf{G }}\ \mathbf{u }_{0}+\mathbf{G }\ \dot{\mathbf{u }}_{0})\ {{dv}}}({\upxi }) \end{aligned}$$
(3.193)
If G from Eq. (3.189) is substituted into Eq. (3.191) we obtain
$$\begin{aligned} \mathbf{u }^{(1)}(\mathbf{x },{t})=\frac{1}{4\pi \,{c}^{2}}\ \int \limits _{B}{{{dv}}({\upxi })}\ \int \limits _{0}^{{t}}{{d\tau }} \frac{1}{|\mathbf{x }-{\upxi }|}\mathbf{f }({\upxi },{t}-\tau )\ \delta \left( \tau -\frac{1}{{c}}|\mathbf{x }-{\upxi }|\right) \end{aligned}$$
(3.194)
Using the filtrating property of the delta function
$$\begin{aligned} \int \limits _{0}^{{t}}{\delta (\tau -{t}_{0})}\ {g}({t}-\tau ){d}\tau ={g}({t}-{t}_{0}){H}({t}-{t}_0) \end{aligned}$$
(3.195)
where \({g}={g(t)}\) is an arbitrary function and \({H}={H(t)}\) is the Heaviside function
$$\begin{aligned} {H(t)}=\left\{ \begin{matrix} 1\quad {t}>0\\ 0\quad {t}<0\\ \end{matrix}\right\} \end{aligned}$$
(3.196)
we reduce Eq. (3.194) to the form
$$\begin{aligned} \mathbf{u }^{(1)}(\mathbf{x },{t})=\frac{1}{4\pi \,{c}^{2}}\ \int \limits _{B}{\frac{\mathbf{f }\left( {\upxi },{t}-\frac{{R}}{{c}}\right) }{{R}}}\ {H}\left( {t}-\frac{{R}}{{c}}\right) {{dv}}({\upxi }) \end{aligned}$$
(3.197)
where
$$\begin{aligned} {R}=|\mathbf{x }-{\upxi }| \end{aligned}$$
(3.198)
The function \(\mathbf{u }^{(1)}(\mathbf{x },{t})\) given by (3.197) is identical to the first integral on the RHS of Eq. (3.182) when the convention that \(\mathbf{f }(\mathbf{x },{t})\) vanishes for \({t}\le 0\) is adopted.
An alternative form of (3.197) reads
$$\begin{aligned} \mathbf{u }^{(1)}(\mathbf{x },{t})=\frac{1}{4\pi \,{c}^{2}}\ \int \limits _{{B}\cap {S}(\mathbf{x },{ct})}\ \frac{1}{|\mathbf{x }-{\upxi }|}\mathbf{f }\bigg ({\upxi },{t}-\frac{1}{{c}}|\mathbf{x }-{\upxi }|\bigg )\ {{dv}}({\upxi }) \end{aligned}$$
(3.199)
where
$$\begin{aligned} {S}(\mathbf{x },{ct})=\{{\upxi }:|{\upxi }-\mathbf{x }|<{ct}\}. \end{aligned}$$
To show that \(\mathbf{u }^{(2)}(\mathbf{x },{t})\) is identical to the last integral on the RHS of (3.182), we apply the Laplace transform to Eq. (3.192), use Eq. (3.188), and obtain
$$\begin{aligned} \bar{\mathbf{u }}^{(2)}(\mathbf{x },{p})=\frac{1}{4\pi }\int \limits _{\partial {B}}\Bigg \{\frac{1}{{R}}\ {e}^{-\frac{{p}}{{c}}{R}}\frac{\partial \bar{\mathbf{u }}}{\partial {n}} -\left[ \frac{\partial }{\partial {n}}\left( \frac{1}{{R}}\right) \ {e}^{-\frac{{p}}{{c}}{R}}-\frac{1}{{cR}}\ \frac{\partial {R}}{\partial {n}}\ {e}^{-\frac{{p}}{{c}}{R}}{p}\right] \bar{\mathbf{u }}\Bigg \}{{da}}({\upxi }) \end{aligned}$$
(3.200)
Applying the inverse Laplace transform to Eq. (3.200) we obtain
$$\begin{aligned} \mathbf{u }^{(2)}(\mathbf{x },{t})=&\;\frac{1}{4\pi }\int \limits _{\partial {B}}\left\{ \frac{1}{{R}}\right. \ \frac{\partial \mathbf{u }}{\partial {n}}\ \left( {\upxi },{t}-\frac{{R}}{{c}}\right) -\frac{\partial }{\partial {n}}\left( \frac{1}{{R}}\right) \ \mathbf{u }\ \left( {\upxi },{t}-\frac{{R}}{{c}}\right) \nonumber \\&\qquad \qquad \qquad +\frac{1}{{cR}}\ \frac{\partial {R}}{\partial {n}}\ \left. \dot{\mathbf{u }}\left( {\upxi },{t}-\frac{{R}}{{c}}\right) \right\} {H}\left( {t}-\frac{{R}}{{c}}\right) {{da}}({\upxi }) \end{aligned}$$
(3.201)
The function \(\mathbf{u }^{(2)}\) given by (3.201) is equivalent to the last integral on the RHS of (3.182), if the convention that \(\mathbf{u }(\mathbf{x },{t})=\mathbf{0 }\) for \({t}\le 0\) is adopted. An equivalent form of (3.201) reads
$$\begin{aligned} \mathbf{u }^{(2)}(\mathbf{x },{t})=&\;\frac{1}{4\pi }\ \int \limits _{\partial {B}\cap {S}(\mathbf{x },{ct})}\left\{ \frac{1}{{R}}\ \frac{\partial \mathbf{u }}{\partial {n}}\ \left( {\upxi },{t}-\frac{{R}}{{c}}\right) \right. -\frac{\partial }{\partial {n}}\left( \frac{1}{{R}}\right) \mathbf{u }\left( {\upxi },{t}-\frac{{R}}{{c}}\right) \nonumber \\&\qquad \qquad \;\qquad \qquad \qquad +\frac{1}{{cR}}\left. \frac{\partial {R}}{\partial {n}}\dot{\mathbf{u }}\left( {\upxi },{t}-\frac{{R}}{{c}}\right) \right\} {{da}}({\upxi }) \end{aligned}$$
(3.202)
To show that \(\mathbf{u }^{(3)}(\mathbf{x },{t})\) given by (3.193) is equal to a sum of the second and third terms on the RHS of (3.182), consider the integral
$$\begin{aligned} \mathbf{h }(\mathbf{x },{t})=\frac{1}{{c}^{2}}\ \int \limits _{B}\ {\mathbf{G \, \dot{\mathbf{u }}_{0}}\ {{dv}}({\upxi })} \end{aligned}$$
(3.203)
Since
$$\begin{aligned} \frac{1}{{R}}\ \delta \left( {t}-\frac{{R}}{{c}}\right) =\frac{{tc}^{2}}{{R}^{2}}\ \delta ({R}-{ct}) \end{aligned}$$
(3.204)
therefore, an alternative form of G given by (3.189) reads
$$\begin{aligned} \mathbf{G }(\mathbf{x }, {\upxi },{t})=\frac{{tc}^{2}}{4\pi \!{R}^{2}}\ \delta ({R}-{ct})\mathbf{1 } \end{aligned}$$
(3.205)
and the function \(\mathbf{h }=\mathbf{h }(\mathbf{x },{t})\) takes the form
$$\begin{aligned} \mathbf{h }(\mathbf{x },{t})=\frac{{t}}{4\pi }\ \int \limits _{B}\ \frac{\dot{\mathbf{u }}_{0}({\upxi })}{|\mathbf{x }-{\upxi }|^{2}}\ \delta (|\mathbf{x }-{\upxi }|-{ct}){{dv}}({\upxi }) \end{aligned}$$
(3.206)
Next, we let
$$\begin{aligned} \mathbf{R }={\upxi }-\mathbf{x } \end{aligned}$$
(3.207)
and introduce the spherical coordinates \(({R}, \varphi , \theta )\) with a center at x
$$\begin{aligned}&\qquad {R}_{1}={R}\cos \,\varphi \,\sin \theta \nonumber \\&\qquad {R}_{2}={R}\sin \,\varphi \,\sin \theta \nonumber \\&\qquad {R}_{3}={R}\cos \,\theta \end{aligned}$$
(3.208)
$$\begin{aligned}&0\le {R}<\infty ,\quad 0\le \varphi \le 2\pi ,\quad 0\le \theta <\pi \end{aligned}$$
(3.209)
Then the integral (3.206) takes the form
$$\begin{aligned} \mathbf{h }(\mathbf{x },{t})=\frac{t}{4\pi }\int \limits _{{B}^{*}}\frac{\dot{\mathbf{u }}_{0}(\mathbf{R }+\mathbf{x })}{{R}^{2}}\delta ({R}-{ct}){{dv}}(\mathbf{R }) \end{aligned}$$
(3.210)
where
$$\begin{aligned} {dv}(\mathbf{R })={R}^2\sin \theta \ {d}\varphi \ {d} \theta \ {dR} \end{aligned}$$
(3.211)
and
$$\begin{aligned} {B}^{*}=\{({R},\varphi ,\theta ):{R_a}<{R}<{R_b};\ 0\le \varphi \le 2\pi , 0\le \theta \le \pi \} \end{aligned}$$
(3.212)
The domain \({B}^{*}\) is a mapping of \(B\) under the transformation defined by Eqs. (3.207) and (3.208), and \({R_{a}}\) and \({R_{b}}\) are uniquely defined nonnegative numbers. Hence, Eq. (3.210) can also be written as
$$\begin{aligned} \mathbf{h }(\mathbf{x },{t})=&\;\frac{t}{4\pi }\int \limits _{{R_a}}^{{R_b}}{d\!R}\ \delta ({R-ct})\int \limits _{0}^{2\pi }{d}\varphi \int \limits _{0}^{\pi }{d}\theta \sin \theta \nonumber \\&\;\times \dot{\mathbf{u }}_0({x}_{1}+{R}\cos \varphi \sin \theta ,\ {x}_{2}+{R}\sin \varphi \sin \theta ,\ {x}_3+{R}\cos \theta ) \end{aligned}$$
(3.213)
or
$$\begin{aligned} \mathbf{h }(\mathbf{x },{t})=&\;\frac{t}{4\pi }\int \limits _{0}^{2\pi }{d}\varphi \int \limits _{0}^{\pi }{d}\theta \sin \theta \nonumber \\&\;\times \dot{\mathbf{u }}_0({x}_1+{ct}\cos \varphi \sin \theta ,\ {x}_{2}+{ct}\sin \varphi \sin \theta ,\ {x}_{3}+{ct}\cos \theta ) \end{aligned}$$
(3.214)
or
$$\begin{aligned} \mathbf{h }(\mathbf{x },{t})={t}\ \mathbf{M }_{\mathbf{x },{ct}}(\dot{\mathbf{u }}_{0}) \end{aligned}$$
(3.215)
where \(\mathbf{M }_{\mathbf{x },{ct}}(\dot{\mathbf{u }}_{0})\) is defined by (3.183). Finally, if we note that
$$\begin{aligned} \mathbf{g }(\mathbf{x },{t})\mathop {=}\limits ^\mathrm{{df}}\frac{1}{c^2}\int \limits _{B}\dot{\mathbf{G }}\mathbf{u }_0\ {{dv}}({\upxi }) \end{aligned}$$
(3.216)
can be written as
$$\begin{aligned} \mathbf{g }(\mathbf{x },{t})=\frac{\partial }{\partial {t}}\left\{ \frac{1}{{c}^2}\int \limits _{B}\mathbf{Gu }_0\ {{dv}}({\upxi })\right\} \end{aligned}$$
(3.217)
then computing the integral on the RHS of (3.217) in a way similar to that of the integral \(\mathbf{h }=\mathbf{h }(\mathbf{x },{t})\), and taking into account Eq. (3.193) we obtain
$$\begin{aligned} \mathbf{u }^{(3)}(\mathbf{x },{t})=\frac{\partial }{\partial {t}}\left[ {t}\mathbf{M }_{\mathbf{x },{ct}}(\mathbf{u }_0)\right] +{t}\mathbf{M }_{\mathbf{x },{ct}}(\dot{\mathbf{u }}_0) \end{aligned}$$
(3.218)
This completes proof of (3.182).
Problem 3.9.
Let \(\mathbf{G }^*=\mathbf{G }^*(\mathbf{x },\xi ;{t})\) be a solution to the initial-boundary value problem:
$$\begin{aligned} \square _0^2\mathbf{G }^*=\mathbf{- }\mathbf{1 }\delta (\mathbf{x }-\xi )\delta (t)\quad \mathrm{{for}} \quad \mathbf{x },\,\;\xi \in \mathrm{{B}},\;\;{t}>0 \end{aligned}$$
(3.219)
$$\begin{aligned} \mathbf{G }^*(\mathbf{x },\xi ;0)=\mathbf{0 }, \quad {\dot{\mathbf{G }}}^*(\mathbf{x },\xi ;0)=\mathbf{0 }\quad \mathrm{{for}} \quad \mathbf{x },\xi \in \mathrm{{B}} \end{aligned}$$
(3.220)
and
$$\begin{aligned} \mathbf{G }^*(\mathbf{x },\xi ;{t})=\mathbf{0 }\quad \mathrm{{for}}\quad \mathbf{x }\in \partial \mathrm{{B}},\quad {t}>0,\quad \xi \in {\bar{\mathrm{{B}}}} \end{aligned}$$
(3.221)
and let \(\mathbf{u }=\mathbf{u }(\mathbf{x },{t})\) be a solution to the initial-boundary value problem
$$\begin{aligned} \square _0^2\mathbf{u }=-\frac{\mathbf{f }}{{c}^2}\quad \mathrm{{on}}\quad \mathrm{{B}}\times (0,\infty ) \end{aligned}$$
(3.222)
$$\begin{aligned} \mathbf{u }(\mathbf x ,0)=\mathbf{u }_0 (\mathbf x ),\quad {\dot{\mathbf{u }}}(\mathbf{x },0)={\dot{\mathbf{u }}}_0 (\mathbf{x })\quad \mathrm{{for}}\quad \mathbf{x }\in {\bar{\mathrm{{B}}}} \end{aligned}$$
(3.223)
$$\begin{aligned} \mathbf{u }(\mathbf{x },{t})=\mathbf{g }(\mathbf{x },{t})\quad \mathrm{{on}}\quad \partial \mathrm{{B}}\times [0,\infty ) \end{aligned}$$
(3.224)
where the functions \(\mathbf{f },\mathbf{u }_0,{\dot{\mathbf{u }}}_0 \), and g are prescribed. Use the representation formula (3.177) of Problem 3.7 to show that
$$\begin{aligned} \mathbf{u }(\mathbf{x },{t})=&\;\dfrac{1}{{c}^2}\displaystyle \int \limits _\mathrm{{B}}{(\mathbf{G }^**\mathbf{f }+{\dot{\mathbf{G }}}^*\mathbf{u }_0 +\mathbf{G }^*{\dot{\mathbf{u }}}_0 )\,d{v}(\xi )} \nonumber \\&\; -\displaystyle \int \limits _{\partial \mathrm{{B}}}{\left( {\dfrac{\partial \mathbf{G }^*}{\partial \mathrm{{n}}}*\mathbf{g }} \right) } \,{da}(\xi ) \end{aligned}$$
(3.225)
Solution.
The representation formula (3.177) of Problem 3.7 reads
$$\begin{aligned} \mathbf{u }(\mathbf{x },{t})=&\;\dfrac{1}{{c}^2}\displaystyle \int \limits _\mathrm{{B}}{(\mathbf{G }*\mathbf{f }+\dot{\mathbf{G }}\mathbf{u }_0 +\mathbf{G }{\dot{\mathbf{u }}}_0 )\,d{v}(\xi )} \nonumber \\&\;+\displaystyle \int \limits _{\partial \mathrm{{B}}}{\left( \mathbf{G }*{\dfrac{\partial \mathbf{u }}{\partial \mathrm{{n}}}-\dfrac{\partial \mathbf{G }}{\partial \mathrm{{n}}}*\mathbf{u }} \right) } \,{da}(\xi ) \end{aligned}$$
(3.226)
where G satisfies Eqs. (3.174) and (3.175) of Problem 3.7, and u satisfies Eqs. (3.146) and (3.147) of Problem 3.5.
By letting \(\mathbf{G }=\mathbf{G }^{*}\) in Eq. (3.226) and using the boundary conditions (3.221) and (3.224) we obtain (3.225).
This completes a solution to Problem 3.9.
Problem 3.10.
A tensor field S corresponds to the solution of a traction problem of classical elastodynamics if and only if
$$\begin{aligned} {\widehat{\nabla }}[\rho ^{-1}(\mathrm{{div}}\,\mathbf{S })]-{\varvec{\mathsf{K }}}[{\ddot{\mathbf{S }}}]=\mathbf - \mathbf{B }\quad \mathrm{{on}}\quad \mathrm{{B}}\times [0,\infty ) \end{aligned}$$
(3.227)
$$\begin{aligned} \mathbf{S }(\mathbf{x },0)=\mathbf{S }^{(0)}(\mathbf{x }),\quad {\dot{\mathbf{S }}}(\mathbf{x },0)=\dot{\mathbf{S }}^{(0)}(\mathbf{x })\quad \mathrm{{for}}\quad \mathbf{x }\in \mathrm{{B}} \end{aligned}$$
(3.228)
$$\begin{aligned} \mathbf{Sn }={\widehat{\mathbf{s }}}\quad \mathrm{{on}}\quad \partial \mathrm{{B}}\times [0,\infty ) \end{aligned}$$
(3.229)
[see Eqs. (3.71)–(3.73)] in which B is expressed in terms of a body force b, and \(\mathbf{S }^{(0)}\) and \({\dot{\mathbf{S }}}^{(0)}\) are defined in terms of two vector fields. A tensor field S corresponding to an external load \([\mathbf{B }, \mathbf{S }^{(0)}, {\dot{\mathbf{S }}}^{(0)}, {\widehat{\mathbf{s }}}]\) is said to be of a \(\sigma \)-type if S satisfies Eqs. (3.227) through (3.229) with an arbitrary symmetric second-order tensor field B and arbitrary symmetric initial tensor fields \(\mathbf{S }^{(0)}\) and \({\dot{\mathbf{S }}}^{(0)}\), not necessarily related to the data of classic elastodynamics. Show that if S and \({\widetilde{\mathbf{S }}}\) are two different tensorial fields of \(\sigma \)-type corresponding to the external loads \([\mathbf{B }, \mathbf{S }^{(0)}, {\dot{\mathbf{S }}}^{(0)}, {\widehat{\mathbf{s }}}]\) and \([{\widetilde{\mathbf{B }}},{\widetilde{\mathbf{S }}}^{(0)},{\dot{\widetilde{\mathbf{S }}}}^{(0)},{\widehat{\widetilde{\mathbf{s }}}}]\), respectively, then the following reciprocal relation holds true
$$\begin{aligned} \begin{array}{l} \displaystyle \int \limits _\mathrm{{B}}{\left\{ {\widetilde{\mathbf{B }}}*\mathbf{S }+{\widetilde{\mathbf{S }}}^{(0)}\cdot {\varvec{\mathsf{K }}}[{\dot{\mathbf{S }}}]+{\dot{\widetilde{\mathbf{S }}}}^{(0)}\cdot {\varvec{\mathsf{K }}}[\mathbf{S }]\right\} } \,d{v}\;\;+\;\;\displaystyle \int \limits _{\partial \mathrm{{B}}}{\rho ^{-1}(\mathrm{{div}}\,{\widetilde{\mathbf{S }}})*(\mathbf{Sn })\,{da}} \\ \quad =\displaystyle \int \limits _\mathrm{{B}}{\left\{ \mathbf{B }*{\widetilde{\mathbf{S }}}+\mathbf{S }^{(0)}\cdot {\varvec{\mathsf{K }}}[{\dot{\widetilde{\mathbf{S }}}}]+{\dot{\mathbf{S }}}^{(0)}\cdot {\varvec{\mathsf{K }}}[{\widetilde{\mathbf{S }}}]\right\} } \,d{v}\;\;+\;\;\displaystyle \int \limits _{\partial \mathrm{{B}}}{\rho ^{-1}(\mathrm{{div}}\,\mathbf{S })*({\widetilde{\mathbf{S }}}\mathbf{n })\,{da}} \end{array} \end{aligned}$$
(3.230)
Solution.
The tensor fields \({S_{ij}}\) and \({\widetilde{S}}_{ij}\), respectively, satisfy the equations
$$\begin{aligned}&({\rho }^{-1}{S}_{(ik,k),_{j}})-{K_{ijkl}}\ \ddot{S}_{kl}=-{B_{ij}}\quad \mathrm{{on}}\ {B}\times (0,\infty ) \end{aligned}$$
(3.231)
$$\begin{aligned}&{S_{ij}}(\mathbf{x },0)={S}_{ij}^{(0)}(\mathbf{x }),\quad {\dot{S}}_{ij}(\mathbf{x },0)={\dot{S}}_{ij}^{(0)}(\mathbf{x })\quad \mathrm{{on}}\ {B} \end{aligned}$$
(3.232)
$$\begin{aligned}&\qquad {S_{ij}n_j}=\widehat{s}_{i}\quad \mathrm{{on}}\ \partial {B}\times (0,\infty ) \end{aligned}$$
(3.233)
and
$$\begin{aligned}&({\rho }^{-1}{\widetilde{S}}_{(ik,k),_{j}})-{K_{ijkl}}\ \ddot{{\widetilde{S}}}_{kl}=-\widetilde{B}_{ij}\quad \mathrm{{on}}\ {B}\times (0,\infty ) \end{aligned}$$
(3.234)
$$\begin{aligned}&{\widetilde{S}}_{ij}(\mathbf{x },0)={\widetilde{S}}_{ij}^{(0)}(\mathbf{x }),\quad {{\dot{{\widetilde{S}}}}}_{ij}(\mathbf{x },0)={{\dot{{\widetilde{S}}}}}_{ij}^{(0)}(\mathbf{x })\quad \mathrm{{on}}\ {B} \end{aligned}$$
(3.235)
$$\begin{aligned}&\qquad {\widetilde{S}}_{ij}{n_{j}}={\widehat{{\widetilde{S}}}}_{i}\quad \mathrm{{on}}\ \partial {B} \times (0,\infty ) \end{aligned}$$
(3.236)
Applying the Laplace transform to Eqs. (3.231) and (3.234), and using the initial conditions (3.232) and (3.235), respectively, we obtain
$$\begin{aligned} ({\rho }^{-1}\bar{S}_{({ik,k}),_{j}})-{K_{ijkl}}\left( {p}^2\bar{S}_{kl}-{\dot{S}}_{kl}^{(0)}-{pS}_{kl}^{(0)}\right) =-\overline{B_{ij}} \end{aligned}$$
(3.237)
and
$$\begin{aligned} ({\rho }^{-1}\bar{\widetilde{S}}_{(ik,k),_{j}})-{K_{ijkl}}\left( {p}^2\bar{{\widetilde{S}}}_{kl}-\dot{\widetilde{{S}}}_{kl}^{(0)}-{p}{\widetilde{S}}_{kl}^{(0)}\right) =-\overline{\widetilde{B}_{ij}} \end{aligned}$$
(3.238)
Next, multiplying (3.237) by \(\bar{{\widetilde{S}}}_{ij}\) and (3.238) by \(-\bar{S}_{ij}\), and adding up the results we obtain
$$\begin{aligned}&\bar{{\widetilde{S}}}_{ij}({\rho }^{-1}\bar{S}_{ik,k}),_{j}-\bar{{\widetilde{S}}}_{ij}\ {K_{ijkl}}\left( {p}^{2}\bar{S}_{kl}-{\dot{S}}_{kl}^{(0)}-{pS}_{kl}^{(0)}\right) -\bar{{S}}_{ij}({\rho }^{-1}\bar{{\widetilde{S}}}_{ik,k}),_{j}\nonumber \\&\quad +\bar{{S}}_{ij}\ {K_{ijkl}}\left( {p}^{2}\bar{{\widetilde{S}}}_{kl}-\dot{\widetilde{{S}}}_{kl}^{(0)}-{p}{\widetilde{S}}_{kl}^{(0)}\right) +{\bar{B}}_{ij}\ \bar{{\widetilde{S}}}_{ij}-\bar{\widetilde{B}}_{ij}\ \bar{S}_{ij}=0 \end{aligned}$$
(3.239)
Since
$$\begin{aligned} \bar{{\widetilde{S}}}_{ij}({\rho }^{-1}\bar{S}_{ik,k}),_{j}-\bar{S}_{ij}({\rho }^{-1}\bar{{\widetilde{S}}}_{ik,k}),_{j} =(\bar{{\widetilde{S}}}_{ij}{\rho }^{-1}\bar{S}_{ik,k}-\bar{S}_{ij}{\rho }^{-1}\bar{{\widetilde{S}}}_{ik,k}),_{j} \end{aligned}$$
(3.240)
and
$$\begin{aligned} {K_{ijkl}}\ \bar{{\widetilde{S}}}_{ij}\ {\bar{S}_{kl}}={K_{ijkl}}\ \bar{S}_{ij}\ \bar{{\widetilde{S}}}_{kl} \end{aligned}$$
(3.241)
therefore, by integrating (3.239) over B and using the divergence theorem, we obtain
$$\begin{aligned}&\int \limits _{B}\left( {\bar{B}}_{ij}\ \bar{{\widetilde{S}}}_{ij}-\bar{\widetilde{B}}_{ij}\ \bar{S}_{ij}\right) {{dv}}({\upxi }) +\int \limits _{B}{K_{ijkl}}\left[ \bar{{\widetilde{S}}}_{ij}\ {\dot{S}}_{kl}^{(0)}+\left( {p}\bar{{\widetilde{S}}}_{ij}-\widetilde{S}_{ij}^{(0)}\right) {S}_{kl}^{(0)}+{\widetilde{S}}_{ij}^{0}\ {S}_{kl}^{(0)}\right. \nonumber \\&\qquad \quad \qquad \quad \qquad \quad \qquad \quad \qquad \quad \qquad \quad \qquad \quad -\left. \bar{S}_{ij}\ {\dot{{\widetilde{S}}}}_{kl}^{(0)}-\left( {p}\bar{S}_{ij}-{S}_{ij}^{(0)}\right) {\widetilde{S}}_{kl}^{(0)}-{S}_{ij}^{(0)}{\widetilde{S}}_{kl}^{(0)}\right] {{dv}}\nonumber \\&\quad +\int \limits _{\partial {B}}{\rho }^{-1}\left( \bar{{\widetilde{S}}}_{ij}\ \bar{S}_{ik,k}-\bar{S}_{ij}\ \bar{{\widetilde{S}}}_{ik,k}\right) {n_{j}}{{da}}({\upxi })=0 \end{aligned}$$
(3.242)
Finally, applying the inverse Laplace transform to (3.242), using the convolution theorem
$$\begin{aligned} \overline{{f}*{g}}=\bar{f}\bar{g} \end{aligned}$$
(3.243)
as well as the relation
$$\begin{aligned} {K_{ijkl}}\ {\widetilde{S}}_{ij}^{(0)}{S}_{kl}^{(0)}={K_{ijkl}}\ {S}_{ij}^{(0)}{\widetilde{S}}_{kl}^{(0)} \end{aligned}$$
(3.244)
we obtain
$$\begin{aligned}&\int \limits _{B}\!(\mathbf{B }*\widetilde{\mathbf{S }}\,{-}\,\widetilde{\mathbf{B }}*\mathbf{S }){{dv}}({\upxi }) \,{+}\,\!\int \limits _{B}\left\{ \widetilde{\mathbf{S }}\cdot {\varvec{\mathsf{K }}}[\dot{\mathbf{S }}^{(0)}]\,{+}\,\dot{\widetilde{\mathbf{S }}}\cdot {\varvec{\mathsf{K }}}[\mathbf{S }^{(0)}] \,{-}\,\mathbf{S }\cdot {\varvec{\mathsf{K }}}[\dot{\widetilde{\mathbf{S }}}^{(0)}]\,{-}\,\dot{\mathbf{S }}\cdot {\varvec{\mathsf{K }}}[\widetilde{\mathbf{S }}^{(0)}]\right\} {{dv}}({\upxi })\nonumber \\&\quad +\int \limits _{B}{\rho }^{-1}\left[ (\mathrm{{div}}\ \mathbf{S })*(\widetilde{\mathbf{S }}\mathbf{n })\,{-}\,(\mathrm{{div}}\ \widetilde{\mathbf{S }})*(\mathbf{Sn })\right] {{da}}({\upxi })=0 \end{aligned}$$
(3.245)
Since \({\varvec{\mathsf{K }}}\) is symmetric
$$\begin{aligned} \mathbf{A }\cdot {\varvec{\mathsf{K }}}[\mathbf{B }]=\mathbf{B }\cdot {\varvec{\mathsf{K }}}[\mathbf{A }]\quad \forall \ \mathbf{A } \ {\text{ and }}\ \mathbf{B } \end{aligned}$$
(3.246)
therefore, Eq. (3.245) is equivalent to Eq. (3.230), and this completes a solution to Problem 3.10.
Problem 3.11.
Let \({S}_{ij}^{({kl})} ={S}_{ij}^{({kl})} (\mathbf{x },\xi ;{t})\) be a solution of the following equation
$$\begin{aligned}&(\rho ^{-1}{S}_{({ik,k}),{j}}^{(kl)}) -{K}_{{ijpq}}{\ddot{S}}_{{pq}}^{({kl})} =0\nonumber \\&\mathrm{{for}}\quad \mathbf{x }\in {E}^3,\;\;\xi \in {E}^3;\quad {\text{ i },\text{ j },\text{ k },\text{ l }}=1,2,3 \end{aligned}$$
(3.247)
subject to the initial conditions
$$\begin{aligned}&{S}_{{ij}}^{({kl})} (\mathbf{x },\xi ;0)=0,\quad {\dot{S}}_{{ij}}^{({kl})} (\mathbf{x },\xi ;0)={C}_{{ijkl}} \delta (\mathbf{x }-\xi )\nonumber \\&\mathrm{{for}}\quad \mathbf{x }\in {E}^3,\;\;\xi \in {E}^3;\quad {\text{ i },\text{ j },\text{ k },\text{ l }}=1,2,3 \end{aligned}$$
(3.248)
where \({K}_{{ijkl}} \) denotes the components of the compliance tensor \({\varvec{\mathsf{K }}}\), and \({C}_{{ijkl}} \) stands for the components of elasticity tensor \({{\varvec{{\mathsf{C }}}}}\), that is,
$$\begin{aligned} {C}_{{ijkl}}{K}_{{klmn}} =\delta _{(im} \delta _{{nj})} \end{aligned}$$
(3.249)
Let \({S}_{{ij}} ={S}_{{ij}} (\mathbf{x },{t})\) be a solution of the equation
$$\begin{aligned} (\rho ^{-1}{S}_{({ik,k}),{j}}) -{K}_{{ijkl}}{\ddot{S}}_{{kl}} =0\quad \mathrm{{for}}\;\;\mathbf{x }\in \mathrm{{B}},\quad {t}>0 \end{aligned}$$
(3.250)
subject to the homogeneous initial conditions
$$\begin{aligned} {S}_{{ij}} (\mathbf{x },0)=0,\quad {\dot{S}}_{{ij}} (\mathbf{x },0)=0\quad \mathrm{{for}}\;\;\mathbf{x }\in \mathrm{{B}} \end{aligned}$$
(3.251)
and the boundary condition
$$\begin{aligned} {S}_{{ij}}{n}_{j} ={\widehat{s}}_{i} \quad \mathrm{{on}}\quad \partial \mathrm{{B}}\times [0,\infty ) \end{aligned}$$
(3.252)
Use the reciprocal relation (3.230) of Problem 3.10 to show that
$$\begin{aligned} {S}_{{kl}} (\mathbf{x },{t})=\int \limits _{\partial \mathrm{{B}}}{\rho ^{-1}\left( {S}_{{im},{m}} *{S}_{{ij}}^{({kl})}{n}_{j} -{\widehat{s}}_{i} *{S}_{{im,m}}^{({kl})} \right) } \;{da}(\xi ) \end{aligned}$$
(3.253)
Note. Equation (3.253) provides a solution to the traction initial-boundary value problem of classical elastodynamics if the field \({S}_{{im,m}} \;\;\mathrm{{on}}\;\;\partial \mathrm{{B}}\times [0,\infty )\) is found from an associated integral equation on \(\partial \mathrm{{B}}\times [0,\infty )\). The idea of solving a traction problem of elastodynamics in terms of displacements through an associated boundary integral equation is due to V.D. Kupradze.
Solution.
Note that for a fixed pair (\(k,l\)) \({\widetilde{S}}_{ij}={S}_{ij}^{({kl})}(\mathbf{x },\ {\upxi };\ {t})\) is a tensor field of\(\sigma \)-type corresponding to the data: \(\widetilde{B}_{ij}=0,\ \widehat{{\widetilde{s}}}_{i}\ne 0,\ {\widetilde{S}}_{ij}^{(0)}=0\), and \({\dot{{\widetilde{S}}}}_{ij}^{(0)}={C}_{ijkl}\delta (\mathbf{x }-{\upxi })\); and \({S_{ij}}={S_{ij}}(\mathbf{x },{t})\) is a tensor field of \(\sigma \)-type corresponding to the data: \({B}_{ij}=0,{\widehat{s}}_{i}\ne 0,\ {S}_{ij}^{(0)}=0\), and \({\dot{S}}_{ij}^{(0)}=0\). Therefore, using the reciprocal relation (3.230) of Problem 3.10. in which \({\widetilde{S}}_{ij}={S}_{ij}^{({kl})}(\mathbf{x },{\upxi };{t})\) and \({S_{ij}}={S_{ij}}(\mathbf{x },{t})\), we obtain
$$\begin{aligned} \int \limits _{B}{\dot{{\widetilde{S}}}}_{{ij}}^{(0)}K_{ijpq}\ {S}_{pq}{{dv}}({\upxi }) =\int \limits _{\partial {B}}{\rho }^{-1}({S}_{im,\,m}*{\widetilde{S}}_{ij}{n}_{j}-\widehat{s}_{i}*{\widetilde{S}}_{im,\,m})\ {{da}}({\upxi }) \end{aligned}$$
(3.254)
where
$$\begin{aligned} {\dot{{\widetilde{S}}}}_{ij}^{(0)}&={\dot{S}}_{ij}^{({kl})}(\mathbf{x },\ {\upxi }; 0)={ C}_{ijkl}\ \delta (\mathbf{x }-{\upxi }) \end{aligned}$$
(3.255)
$$\begin{aligned} {\widetilde{S}}_{ij}{n}_{j}&={S}_{ij}^{({kl})}(\mathbf{x },\ {\upxi };{t}){n}_{j}({\upxi }) \end{aligned}$$
(3.256)
$$\begin{aligned} {\widetilde{S}}_{im,\,m}&={S}_{im,\,m}^{({kl})} (\mathbf{x },\ {\upxi };\ {t}) \end{aligned}$$
(3.257)
Since
$$\begin{aligned} {\dot{{\widetilde{S}}}}_{ij}^{(0)}{K}_{ijpq}\ {S}_{pq}&={C}_{ijkl}\ {K}_{ijpq}\ S_{pq}\ \delta (\mathbf{x }-{\upxi }) ={C}_{klij}\ K_{ijpq}\ {S}_{pq}\ \delta (\mathbf{x }-{\upxi })\nonumber \\&=\delta _{({kp}}\ \delta _{ql)}\ {S}_{pq}\ \delta (\mathbf{x }-{\upxi })={S}_{kl} ({\upxi },{t})\delta (\mathbf{x }-{\upxi }) \end{aligned}$$
(3.258)
therefore, Eq. (3.254) takes the form
$$\begin{aligned}&\int \limits _{B}{S}_{kl}({\upxi },{t})\delta (\mathbf{x }-{\upxi }){{dv}}({\upxi })=\int \limits _{\partial {B}}{\rho }^{-1}\left( {S}_{im,\,m}*{S}_{ij}^{({kl})}{n}_{j}-\widehat{s}_{i}*{S}_{im,\,m}^{({kl})}\right) {{da}}({\upxi }) \nonumber \\ \end{aligned}$$
(3.259)
Finally, using the filtrating property of the delta function we obtain (3.253). This completes a solution to Problem 3.11.
Problem 3.12.
Consider the pure stress initial-boundary value problem of linear elastodynamics for a homogeneous isotropic incompressible elastic body B [see Eq. (3.55) with \(\mu >0\;\;\mathrm{{and}}\;\;\lambda \rightarrow \infty \)]. Find a tensor field \(\mathbf{S }=\mathbf{S }(\mathbf{x },{t})\;\;\mathrm{{on}}\mathrm{{B}}\times [0,\infty )\) that satisfies the equation
$$\begin{aligned} {\widehat{\nabla }}(\mathrm{{div}}\,\mathbf{S })-\frac{\rho }{2\mu }\left[ {\ddot{\mathbf{S }}}-\frac{1}{3}(\mathrm{{tr}}\;{\ddot{\mathbf{S }}})\,\mathbf{1 } \right] =-{\widehat{\nabla }}\mathbf{b }\quad \mathrm{{on}}\quad \mathrm{{B}}\times [0,\infty ) \end{aligned}$$
(3.260)
subject to the initial conditions
$$\begin{aligned} \mathbf{S }(\mathbf{x },0)=\mathbf{S }_0 (\mathbf{x }),\quad {\dot{\mathbf{S }}}(\mathbf{x },0)={\dot{\mathbf{S }}}_0 (\mathbf{x })\quad \mathrm{{for}}\quad \mathbf{x }\in \mathrm{{B}} \end{aligned}$$
(3.261)
and the traction boundary condition
$$\begin{aligned} \mathbf{Sn }={\widehat{\mathbf{s }}}\quad \mathrm{{on}}\quad \partial \mathrm{{B}}\times [0,\infty ) \end{aligned}$$
(3.262)
Here, \(\mathbf{b },{\widehat{\mathbf{s }}},\mathbf{S }_0, \;\mathrm{{and}}\;\;{\dot{\mathbf{S }}}_0,\) are prescribed functions (\(\mu >0,\;\rho >0\)). Show that the problem (3.260) through (3.262) may have at most one solution.
Solution.
We are to show that the field equation
$$\begin{aligned} {S}_{({ik,kj})}-\frac{{\rho }}{2\mu }\left( \ddot{S}_{ij}-\frac{1}{3}\ddot{S}_{kk}\ \delta _{ij}\right) =0\quad \text{ on }\ {B}\times [0,\infty ) \end{aligned}$$
(3.263)
subject to the homogeneous initial conditions
$$\begin{aligned} {S}_{ij}(\mathbf{x },\ 0)=0,\quad {\dot{S}}_{ij}(\mathbf{x },\ 0)=0\quad {\text{ on }}\ B \end{aligned}$$
(3.264)
and the homogeneous traction boundary condition
$$\begin{aligned} {S}_{ij}\ {n}_{j}=0\quad {\text{ on }}\ {\partial }{B}\times [0,\infty ) \end{aligned}$$
(3.265)
imply that
$$\begin{aligned} {S}_{ij}=0\quad {\text{ on }}\ {\bar{B}}\times [0, \infty ) \end{aligned}$$
(3.266)
To this end we multiply (3.263) by \({\dot{S}}_{ij}\) and obtain
$$\begin{aligned} {\mathrm{{S}}}_{({{ik,kj}})}\ {\dot{S}}_{ij}-\frac{{\rho }}{2\mu }\left( \ddot{S}_{ij}\ {\dot{S}}_{ij}-\frac{1}{3}\ddot{S}_{kk}\ {\dot{S}}_{ii}\right) =0 \end{aligned}$$
(3.267)
Since
$$\begin{aligned} {S}_{(ik,kj)}\ {\dot{S}}_{ij}={S}_{ik,kj}\ {\dot{S}}_{ij} =({S}_{ik,k}\ {\dot{S}}_{ij}),_{j}-{S}_{ik,k}\ {\dot{S}}_{ij,j} \end{aligned}$$
(3.268)
therefore (3.267) can be written in the form
$$\begin{aligned} ({S}_{ik,k}\ {\dot{S}}_{ij})_{,j}-{S}_{ik,k}\ {\dot{S}}_{ij,j} -\frac{{\rho }}{2\mu }\left[ \frac{1}{2}\frac{\partial }{\partial {t}}({\dot{S}}_{ij}\ \dot{S}_{ij})-\frac{1}{3}\frac{1}{2}\frac{\partial }{\partial {t}}({\dot{S}}_{aa})^{2}\right] =0 \end{aligned}$$
(3.269)
or
$$\begin{aligned} ({S}_{ik,k}\ {\dot{S}}_{ij})_{,j}-\frac{1}{2}\frac{\partial }{\partial {t}}({S}_{ik,k}\ {S}_{ij,j}) -\frac{{\rho }}{2\mu }\left[ \frac{1}{2}\frac{\partial }{\partial {t}}(\dot{S}_{ij}\ {\dot{S}}_{ij})-\frac{1}{3}\frac{1}{2}\frac{\partial }{\partial {t}}({\dot{S}}_{aa})^{2}\right] =0 \end{aligned}$$
(3.270)
Integrating Eq. (3.270) over the cartesian product \({B}\times [0, t]\), using the divergence theorem, the homogeneous initial conditions (3.264) as well as the boundary condition
$$\begin{aligned} {\dot{S}}_{ij}\ {n}_{j}=0\quad {\text{ on }}\ \partial {B}\times [0, \infty ) \end{aligned}$$
(3.271)
obtained by differentiation of (3.265) with respect to time, we obtain
$$\begin{aligned} \int _{B}\Bigg \{{S}_{ik,k}\ {S}_{ij,j}+\frac{{\rho }}{2\mu }\left[ {\dot{S}}_{ij}\ \dot{S}_{ij}-\frac{1}{3}({\dot{S}}_{aa})^{2}\right] \Bigg \}{{dv}}=0 \end{aligned}$$
(3.272)
Since
$$\begin{aligned} {\dot{S}}_{ij}=\dot{S}_{ij}^{({d})}+\frac{1}{3}{\dot{S}}_{aa}\ \delta _{ij} \end{aligned}$$
(3.273)
where
$$\begin{aligned} {\dot{S}}_{ij}^{({d})}={\dot{S}}_{ij}-\frac{1}{3}{\dot{S}}_{aa}\ \delta _{ij} \end{aligned}$$
(3.274)
and
$$\begin{aligned} {\dot{S}}_{ij}\ \dot{S}_{ij}={\dot{S}}_{ij}^{({d})}{\dot{S}}_{ij}^{({d})}+\frac{1}{3}({\dot{S}}_{aa})^{2} \end{aligned}$$
(3.275)
therefore, Eq. (3.272) can be written as
$$\begin{aligned} \int \limits _{B}\left( {S}_{ik,k}\ {S}_{ij,j}+\frac{\rho }{2\mu }{\dot{S}}_{ij}^{({d})}\,{\dot{S}}_{ij}^{({d})}\right) {{dv}}=0 \end{aligned}$$
(3.276)
Equation (3.276) implies that
$$\begin{aligned} {S}_{ik,k}=0,\quad {\dot{S}}_{ij}^{({d})}=0\quad \text{ on }\ {\bar{B}}\times [0, \infty ) \end{aligned}$$
(3.277)
Equation (3.277)\(_{2}\) together with the homogeneous initial conditions (3.264)\(_{1}\) imply that
$$\begin{aligned} {S}_{ij}=\frac{1}{3}{S}_{aa}\ \delta _{ij}\quad {\text{ on }}\ {\bar{B}}\times [0, \infty ) \end{aligned}$$
(3.278)
Equations (3.278) and (3.277)\(_{1}\) imply that
$$\begin{aligned} {S}_{aa,i}=0\quad {\text{ on }}\ {\bar{B}}\times [0, \infty ) \end{aligned}$$
(3.279)
which is equivalent to
$$\begin{aligned} {S}_{aa}(\mathbf{x },{t})={c}({t})\quad {\text{ on }}\ {\bar{B}}\times [0, \infty ) \end{aligned}$$
(3.280)
where \({c}={c(t)}\) is an arbitrary function of time.
Finally, Eqs. (3.265), (3.278), and (3.280) lead to
$$\begin{aligned} {c}({t}){n}_{i}(\mathbf{x })=0\quad {\text{ on }}\ \partial {B}\times [0, \infty ) \end{aligned}$$
(3.281)
Since \(|{n_{i}\ n_{i}}|=1\) on \(\partial {B}\), therefore Eq. (3.281) implies that
$$\begin{aligned} |{c}({t})|=0 \end{aligned}$$
(3.282)
Equation (3.282) together with Eq. (3.278) implies Eq. (3.266), and this completes a solution to Problem 3.12.
