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Theorem 6.3. (Rychlewski’s theorem) A tensor-valued function \( g\left( {\mathbf{A}}_{i}\right) \) is anisotropic with the symmetry group \( {\operatorname{Sorth}}^{n} = \mathfrak{g} \) defined by (6.30) if and only if there exists an isotropic tensor-valued function \( \widehat{g}\left( {{\mathbf{A}}_{i},{\mathbf{L}}_... | Proof. Let us define a new tensor-valued function by\n\n\[ \widehat{g}\left( {{\mathbf{A}}_{i},{\mathbf{X}}_{j}}\right) = {\mathbf{Q}}^{\prime \mathrm{T}}g\left( {{\mathbf{Q}}^{\prime }{\mathbf{A}}_{i}{\mathbf{Q}}^{\prime \mathrm{T}}}\right) {\mathbf{Q}}^{\prime }, \]\n\nwhere the tensor \( {\mathbf{Q}}^{\prime } \in {... | No |
Derivative of the power function \( {\mathbf{A}}^{k}\left( {k = 1,2,\ldots }\right) \) . The directional derivative (6.111) of the power function yields | \[ {\left. \frac{\mathrm{d}}{\mathrm{d}t}{\left( \mathbf{A} + t\mathbf{X}\right) }^{k}\right| }_{t = 0} = {\left. \frac{\mathrm{d}}{\mathrm{d}t}\left( {\mathbf{A}}^{k} + t\mathop{\sum }\limits_{{i = 0}}^{{k - 1}}{\mathbf{A}}^{i}\mathbf{X}{\mathbf{A}}^{k - 1 - i} + {t}^{2}\ldots \right) \right| }_{t = 0} \] \[ = \mathop... | Yes |
Example 2. Derivative of the transposed tensor \( {\mathbf{A}}^{\mathrm{T}} \) . | \[ {\left. \frac{\mathrm{d}}{\mathrm{d}t}{\left( \mathbf{A} + t\mathbf{X}\right) }^{\mathrm{T}}\right| }_{t = 0} = {\left. \frac{\mathrm{d}}{\mathrm{d}t}\left( {\mathbf{A}}^{\mathrm{T}} + t{\mathbf{X}}^{\mathrm{T}}\right) \right| }_{t = 0} = {\mathbf{X}}^{\mathrm{T}}. \] | Yes |
Example 3. Derivative of the inverse tensor \( {\mathbf{A}}^{-1} \), where \( \mathbf{A} \in {\operatorname{Inv}}^{n} \) . Consider the directional derivative of the identity \( {\mathbf{A}}^{-1}\mathbf{A} = \mathbf{I} \) . It delivers: | \[ {\left. \frac{\mathrm{d}}{\mathrm{d}t}{\left( \mathbf{A} + t\mathbf{X}\right) }^{-1}\left( \mathbf{A} + t\mathbf{X}\right) \right| }_{t = 0} = \mathbf{0}. \] Applying the product rule of differentiation (2.9) and using (6.116) we further write \[ {\left. \frac{\mathrm{d}}{\mathrm{d}t}{\left( \mathbf{A} + t\mathbf{X}... | Yes |
The right and left Cauchy-Green tensors are given in terms of the deformation gradient \( \mathbf{F} \) respectively by\n\n\[ \mathbf{C} = {\mathbf{F}}^{\mathrm{T}}\mathbf{F},\;\mathbf{b} = {\mathbf{{FF}}}^{\mathrm{T}}. \] | Of special interest in continuum mechanics is the derivative of these tensors with respect to \( \mathbf{F} \). With the aid of the product rule (6.123) and using (5.42), (5.77),(5.82),(5.83),(6.118) and (6.121) we obtain\n\n\[ \mathbf{C}{,}_{\mathbf{F}} = {\mathbf{F}}^{\mathrm{T}}{,}_{\mathbf{F}}\mathbf{F} + {\mathbf{... | Yes |
With the aid of the above differentiation rules we can easily express the derivatives of the spherical and deviatoric parts (1.153) of a second-order tensor by | \[ \operatorname{sph}\mathbf{A},\mathbf{A} = \left\lbrack {\frac{1}{n}\operatorname{tr}\left( \mathbf{A}\right) \mathbf{I}}\right\rbrack ,\mathbf{A} = \frac{1}{n}\mathbf{I} \odot \mathbf{I} = {\mathbf{P}}_{\text{sph }}, \] (6.129) \[ \operatorname{dev}\mathbf{A},\mathbf{A} = \left\lbrack {\mathbf{A} - \frac{1}{n}\opera... | No |
We shall find expressions for the motion given by the angle \( \theta \left( t\right) \) and the constraining forces \( N \) and \( B \) . The initial conditions are: \( \theta \left( 0\right) = {\theta }_{0},\dot{\theta }\left( 0\right) = 0 \) . The acceleration a consists of a tangential acceleration and a normal acc... | \[ \mathbf{a} = {a}_{t}\mathbf{t} + {a}_{n}\mathbf{n},\;{a}_{t} = \ddot{s} = \sqrt{{R}^{2} + {b}^{2}}\ddot{\theta } \] \[ {a}_{n} = \frac{{\dot{s}}^{2}}{\rho } = \frac{1}{\rho }{\left( \frac{ds}{d\theta }\dot{\theta }\right) }^{2} = \frac{R}{{R}^{2} + {b}^{2}}{\left( \sqrt{{R}^{2} + {b}^{2}}\dot{\theta }\right) }^{2} =... | Yes |
For the ensemble presented in Example 3.1 we shall find the force \( {\mathbf{f}}_{a} \) and the torque \( {\mathbf{m}}_{t} \), see Fig. 3.9, which must be supplied to the plate in order to obtain the prescribed motion. When the ensemble is at rest the plate is only subjected to its weight \( {mg} \) . | The acceleration of the mass center \( C \) of the plate is found in Example 3.1: \( {\mathbf{a}}_{C} = - {\omega }_{3}^{2}c{\mathbf{e}}_{1} \) . From symmetry it follows that the coordinate axes are principal axes of inertia and from the table of moments of inertia we get:\n\n\[ {I}_{11} = {I}_{1} = {I}_{22} = {I}_{2}... | Yes |
Theorem 3.1 If \( \mathbf{A} \) is an orthogonal tensor that represents a rotation \( \phi \) with an axis of rotation parallel to a unit vector \( \mathbf{a} \), then the \( \mathbf{Q} \) -rotation of \( \mathbf{A} \) is a new orthogonal tensor \( \mathbf{B} = {\mathbf{{QAQ}}}^{T} \) with the same rotation \( \phi \) ... | The proof of the theorem is given as Problem 3.5. | No |
Theorem 3.3 (a) An isotropic, symmetric second tensor-valued function B[A] of a symmetric second order tensor \( \mathbf{A} \) is coaxial to the argument tensor \( \mathbf{A} \) . | Proof of Part (a) of Theorem 3.3 We need to show that any principal direction of \( \mathbf{A} \), here denoted \( {\mathbf{a}}_{3} \), also is a principal direction of \( \mathbf{B} \) . We choose a particular \( \mathbf{Q} \) - rotation that has an axis of rotation parallel to \( {\mathbf{a}}_{3} \) and the angle of ... | Yes |
Theorem 3.4 Let B[A] be a second order symmetric tensor-valued function of a second symmetric tensor \( \mathbf{A} \) . The function \( \mathbf{B}\left\lbrack \mathbf{A}\right\rbrack \) is then isotropic if and only if the function has the representation:\n\n\[ \mathbf{B}\left\lbrack \mathbf{A}\right\rbrack = {\gamma }... | Proof that: \( \left( {3.9.4}\right) \Rightarrow \left( {3.9.3}\right) \) . The functions \( {\gamma }_{0},{\gamma }_{1} \), and \( {\gamma }_{2} \) are isotropic scalar-valued functions of \( \mathbf{A} \) . Thus according to the definition (3.9.3):\n\n\[ {\gamma }_{i}\left\lbrack {\mathbf{{QAQ}}}^{T}\right\rbrack = {... | No |
Potential vortex A solid circular cylinder with radius \( a \) that rotates about its axis with constant angular velocity \( \omega \) in a linearly viscous fluid, creates an irrotational vortex, i.e. a vortex without vorticity, Fig. 4.12. | The fluid particles move in concentric circles and the velocity field is:\n\n\[ \n{v}_{\theta } = \frac{\alpha }{R},\;{v}_{R} = {v}_{z} = 0,\;\alpha = \omega {a}^{2} \n\]\n\n\[ \n\Rightarrow {v}_{1} = - {v}_{\theta }\sin \theta = - \frac{\alpha {x}_{2}}{{x}_{1}^{2} + {x}_{2}^{2}},\;{v}_{2} = {v}_{\theta }\cos \theta = ... | Yes |
Example 6.3. Vector Operators in Cylindrical Coordinates \( \left( {R,\theta, z}\right) \) . | In cylindrical coordinates: \( h = R,{h}_{1} = {h}_{3} = 1,{h}_{2} = R \) . Formula (6.5.109) yields:\n\n\[ \n{\nabla }^{2} = \frac{{\partial }^{2}}{\partial {R}^{2}} + \frac{1}{R}\frac{\partial }{\partial R} + \frac{1}{R}\frac{{\partial }^{2}}{\partial {\theta }^{2}} + \frac{{\partial }^{2}}{\partial {z}^{2}} \n\] \n\... | Yes |
In spherical coordinates \( \left( {r,\theta ,\phi }\right) \), what is the formula for the Laplacian operator \( {\nabla }^{2} \)? | \[ {\nabla }^{2} = \frac{{\partial }^{2}}{\partial {r}^{2}} + \frac{2}{r}\frac{\partial }{\partial r} + \frac{1}{{r}^{2}}\frac{{\partial }^{2}}{\partial {\theta }^{2}} + \frac{1}{{r}^{2}}\cot \theta \frac{\partial }{\partial \theta } + \frac{1}{{r}^{2}{\sin }^{2}\theta }\frac{{\partial }^{2}}{\partial {\phi }^{2}} \] | Yes |
Example 7.1 Physical Components of E in Cylindrical Coordinates. | \n\\(\\left( {R,\\theta, z}\\right) : \\mathbf{u} = \\left\\lbrack {{u}_{R},{u}_{\\theta },{u}_{z}}\\right\\rbrack ,{h}_{1} = 1,{h}_{2} = R,{h}_{3} = 1, h = R \\) . The formulas (7.3.28- 7.3.29) yield:\n\n\\[ \n\\left( {E\\left( {ij}\\right) }\\right) = \\left( \\begin{array}{lll} {\\varepsilon }_{R} & {\\gamma }_{R\\t... | Yes |
Example 7.2 Physical Components of E in Spherical Coordinates. | \[ \left( {E\left( {ij}\right) }\right) = \left( \begin{matrix} {\varepsilon }_{r} & {\gamma }_{r\theta }/2 & {\gamma }_{r\phi }/2 \\ {\gamma }_{\theta r}/2 & {\varepsilon }_{\theta } & {\gamma }_{\theta \phi }/2 \\ {\gamma }_{\phi r}/2 & {\gamma }_{\phi \theta }/2 & {\varepsilon }_{\phi } \end{matrix}\right) \] \[ {\v... | Yes |
The physical components \( T\left( {ki}\right) \) of the stress tensor \( \mathbf{T} \) in cylindrical coordinates are expressed in Fig. 7.6 and with alternative symbols in the formulas (7.7.15). | \[ \left( {T\left( {ij}\right) }\right) \equiv \left( \begin{matrix} {\sigma }_{R} & {\tau }_{R\theta } & {\tau }_{Rz} \\ {\tau }_{\theta R} & {\sigma }_{\theta } & {\tau }_{\theta z} \\ {\tau }_{zR} & {\tau }_{z\theta } & {\sigma }_{z} \end{matrix}\right) \equiv \left( \begin{matrix} {T}_{RR} & {T}_{R\theta } & {T}_{R... | Yes |
\[ \frac{\partial {\sigma }_{R}}{\partial R} + \frac{{\sigma }_{R} - {\sigma }_{\theta }}{R} + \frac{1}{R}\frac{\partial {\tau }_{R\theta }}{\partial \theta } + \frac{\partial {\tau }_{Rz}}{\partial z} + \rho {b}_{R} = \rho {a}_{R} \] | \[ \frac{d{\sigma }_{R}}{dR} + \frac{{\sigma }_{R} - {\sigma }_{\theta }}{R} + \rho {b}_{R} = \rho {a}_{R} \Leftrightarrow \frac{d}{dR}\left( {R{\sigma }_{R}}\right) - {\sigma }_{\theta } + {\rho R}{b}_{R} = {\rho R}{a}_{R} \] | No |
Example 7.8 Cauchy’s Equations in Spherical Coordinates. | \[ \frac{\partial {\sigma }_{r}}{\partial r} + \frac{2{\sigma }_{r} - {\sigma }_{\theta } - {\sigma }_{\phi }}{r} + \frac{1}{r\sin \theta }\left\lbrack {\frac{\partial }{\partial \theta }\left( {\sin \theta {\tau }_{r\theta }}\right) + \frac{\partial {\tau }_{r\phi }}{\partial \phi }}\right\rbrack + \rho {b}_{r} = \rho... | Yes |
Example 7.10 Navier's Equations in Spherical Coordinates | Applying the formula (6.5.117) for \( {\nabla }^{2}\mathbf{u} \) and formula (6.5.115) for \( \nabla \left( {\nabla \cdot \mathbf{u}}\right) \) we obtain:\n\n\[ \n{\nabla }^{2}{u}_{r} - \frac{2}{{r}^{2}}{u}_{r} - \frac{2}{{r}^{2}}\frac{\partial {u}_{\theta }}{\partial \theta } - \frac{2\cot \theta }{{r}^{2}}{u}_{\theta... | Yes |
Expressions for \( \nabla \left( {\nabla \cdot \mathbf{v}}\right) \) and \( {\nabla }^{2}\mathbf{v} \) in Cylindrical Coordinates \( \left( {R,\theta, z}\right) \) | \[ \mathbf{b} = \operatorname{grad}\operatorname{div}\mathbf{v} = \nabla \left( {\nabla \cdot \mathbf{v}}\right) = {b}_{R}{\mathbf{e}}_{R} + {b}_{\theta }{\mathbf{e}}_{\theta } + {b}_{z}{\mathbf{e}}_{z}\; \Rightarrow \]\n\n\[ {b}_{R} = \frac{{\partial }^{2}{v}_{R}}{\partial {R}^{2}} + \frac{1}{R}\frac{\partial {v}_{R}}... | Yes |
Theorem 9.1. Integration Independent of the Integration Path Let a(r) be a vector field and \( {\mathbf{r}}_{1} \) and \( {\mathbf{r}}_{2} \) two places in space. Then:\n\n(1) If the vector field may be expressed as the gradient of a scalar field \( \alpha \left( \mathbf{r}\right) \) :\n\n\[ \mathbf{a} = \operatorname{... | Proof of (1) The assumption (9.1.3) \( \Rightarrow \) the result (9.1.4. Formula (9.1.3) \( \Rightarrow \mathbf{a} \cdot d\mathbf{r} = \alpha ,{}_{i}d{x}_{i} = {d\alpha } \) . The integral from \( {\mathbf{r}}_{1} \) to \( {\mathbf{r}}_{2} \) of the vector field \( \mathbf{a} \) (r) becomes:\n\n\[ {\int }_{{\mathbf{r}}... | Yes |
Theorem 9.2 Let \( f\left( x\right) \) and \( g\left( x\right) \) be continuous functions of the variable \( x \) and \( I \) the integral:\n\n\[ I = {\int }_{{x}_{1}}^{{x}_{2}}f\left( x\right) g\left( x\right) {dx} \]\n\nIf the integral \( I \) vanishes for any function \( g\left( x\right) \) satisfying the conditions... | Proof Assume that: \( f\left( \bar{x}\right) > 0 \) for some value of \( \bar{x} \) in the interval: \( {x}_{1} \leq \bar{x} \leq {x}_{2} \) . Continuity of the function \( f\left( x\right) \) implies then that there exists a value \( \varepsilon > 0 \) such that:\n\n\[ f\left( x\right) > 0\;\text{ for }\bar{x} - \vare... | Yes |
Theorem 9.3. Gauss' Integral Theorem in a Plane Let \( A \) be a surface in the \( {x}_{1}{x}_{2} \) -plane and bordered by the curve \( C \), Fig. 9.1. The unit normal \( \mathbf{n} \) to \( C \) lies in the \( {x}_{1}{x}_{2} \) -plane and is pointing out from the curve \( C \) . Then for any field function \( f\left(... | Proof The proof will be given for the index value \( \alpha = 1 \) . First we consider the situation in Fig. 9.1, in which a straight line parallel to the \( {x}_{1} \) -axis only intersects the curve \( C \) in two points \( {\widetilde{x}}_{1}\left( {x}_{2}\right) \) and \( {\bar{x}}_{1}\left( {x}_{2}\right) \) . The... | Yes |
Theorem 9.4. The Divergence Theorem in a Plane Let \( A \) be a surface in the \( {x}_{1}{x}_{2} \) -plane and bordered by the curve \( C \), Figs. 9.1 and 9.2. The unit normal \( \mathbf{n} \) to \( C \) lies in the \( {x}_{1}{x}_{2} \) -plane and is pointing out from the curve \( C \) . Then for any vector field \( \... | Proof: Theorem 9.2 is applied to each components \( {a}_{\alpha }\left( {{x}_{1},{x}_{2}}\right) \) of the vector \( \mathbf{a} \) and the results are added to give formula (9.2.9). | No |
Theorem 9.5 Stokes’ Theorem in a Plane Let \( A \) be a surface in the \( {x}_{1}{x}_{2} \) -plane and bordered by the curve \( C \), Figs. 9.1 and 9.2. The unit normal vector to the plane area \( A \) is the base vector \( {\mathbf{e}}_{3} \) of the Cartesian coordinate system \( {Ox} \) . The unit tangent vector to t... | Proof Applying the formulas (9.2.3), with \( {\mathbf{e}}_{i} \) as the base vectors of the system \( {Ox} \) , we obtain for the tangent vector \( \mathbf{t} \) :\n\n\[ \n\mathbf{t} = \frac{d\mathbf{r}}{ds} = \frac{\partial \mathbf{r}}{\partial {x}_{\alpha }}\frac{d{x}_{\alpha }}{ds} = {\mathbf{e}}_{\alpha }\frac{d{x}... | Yes |
Example 9.2 Area of the Surface of a Sphere | Figure 9.7 shows an eighth of a sphere of radius \( r \) . Spherical coordinates are chosen as curvilinear surface coordinates: \( {u}^{1} = \theta ,\;{u}^{2} = \phi \) . The mapping (9.2.12) is expressed by:\n\n\[ \n{x}_{\alpha } = {x}_{\alpha }\left( {{u}^{1},{u}^{2}}\right) \Rightarrow {x}_{1} = r\sin \theta \cos \p... | Yes |
Theorem 9.6 Gauss' Integral Theorem in Space Let \( V \) represent a volume in three-dimensional space bounded by the surface \( A \) with an outward unit normal vector \( \mathbf{n} \) . Then for any field \( f\left( \mathbf{r}\right) \) : | Proof for \( i = 3 \) : Consider first the case, Fig. 9.9, where a straight line parallel to the \( {x}_{3} \) -axis intersects the surface \( A \) in only two points: \( {\widetilde{x}}_{3}\left( {{x}_{1},{x}_{2}}\right) \) and \( {\bar{x}}_{3}\left( {{x}_{1},{x}_{2}}\right) \) .\n\nThe sets of points \( {\widetilde{x... | Yes |
Theorem 9.7 The Divergence Theorem in Space Let \( V \) represent a volume in three-dimensional space bounded by the surface \( A \) with an outward unit normal vector \( \mathbf{n} \) . Then for any vector field \( \mathbf{a}\left( \mathbf{r}\right) \) :\n\n\[ {\int }_{V}\operatorname{div}\mathbf{a}{dV} = {\int }_{A}\... | Proof In Cartesian coordinates the theorem follows directly from Theorem 9.6 by replacing the scalar field \( f\left( \mathbf{r}\right) \) in the formula (9.3.5) by the vector components \( {a}_{i} \) and summing the results over the index \( \left( i\right) \) . Because the expressions div \( \mathbf{a} = {a}_{i},{}_{... | Yes |
Theorem 9.8 Stokes’ Theorem for a Curved Surface Let \( \mathbf{b}\left( \mathbf{r}\right) \) be a vector field and \( A \) a surface in space with an outward unit normal vector \( \mathbf{n} \) . The surface is bordered by the curve \( C \) . Then:\n\n\[ \mathop{\int }\limits_{A}\left( {\operatorname{rot}\mathbf{b}}\r... | Proof With respect to a Cartesian coordinate system \( {Ox} \), with base vectors \( {\mathbf{e}}_{i} \), and a surface coordinate system \( u \), with base vectors \( {\mathbf{a}}_{\alpha } \), we obtain for the vectors rot \( \mathbf{b} \) and \( \mathbf{n}{dA} \) :\n\n\[ \operatorname{rot}\mathbf{b} = {e}_{ijk}{b}_{... | Yes |
Theorem 9.9 The Mean Value Theorem Let \( f\left( \mathbf{r}\right) \) and \( g\left( \mathbf{r}\right) \) be two continuous field functions and V a volume in space. Then at least one point \( \overline{\mathbf{r}} \) in V exists such that:\n\n\[{\int }_{V}f\left( \mathbf{r}\right) g\left( \mathbf{r}\right) {dV} = f\le... | Proof Let \( {f}_{\min } \) be the maximum value and \( {f}_{\max } \) the maximum value of the function \( f\left( \mathbf{r}\right) \) in the volume \( V \) . Then because \( f\left( \mathbf{r}\right) \) is a continuous function in \( V \), at least one point \( \overline{\mathbf{r}} \) exists in \( V \) such that: \... | Yes |
Theorem 9.10 Let \( f\left( \mathbf{r}\right) \) be a continuous field function in a volume \( {V}_{1} \) in the space \( {E}_{3} \) . Then the following holds true:\n\n\[ \text{If:}{\int }_{V}f\left( \mathbf{r}\right) {dV} = 0\text{for any choice of volume}V\text{in}{V}_{1}\text{, then:}f\left( \mathbf{r}\right) = 0\t... | Proof If it is assumed that \( f\left( \overline{\mathbf{r}}\right) > 0 \) in a point \( \overline{\mathbf{r}} \) in \( {V}_{1} \), then according to the condition of continuity for the function \( f\left( \mathbf{r}\right) \), the function must be positive, i.e. \( f\left( \mathbf{r}\right) > 0 \), in a small volume \... | Yes |
A straight cone has the height \( h \), a circular base with radius \( r \), a symmetry axis along the \( {x}_{3} \) -axis of a Cartesian coordinate system \( {Ox} \), and its apex at the origin \( O \). We shall find the volume of the cone using cylindrical coordinates \( \left( {R,\theta, z}\right) \equiv \left( {{y}... | The mapping is:\n\n\[ \n{x}_{1} = R\cos \theta \equiv {y}^{1}\cos {y}^{2},\;{x}_{2} = R\sin \theta \equiv {y}^{1}\sin {y}^{2},\;{x}_{2} = z \equiv {y}^{3} \]\n\nThe Jacobian \( J \) when transforming from the Cartesian coordinates \( {x}_{i} \) to the cylindrical coordinates becomes:\n\n\[ \nJ = \det \left( \frac{\part... | Yes |
Theorem 9.12 From Volume Integration to Surface Integration A body of a continuum has at the time \( t \) the volume \( V \) and the surface \( A \) with the outward unit normal vector \( \mathbf{n} \) . Let \( f\left( {\mathbf{r}, t}\right) \) be an intensive quantity defined per unit volume at the place \( \mathbf{r}... | Proof Let the volume \( V \) be subdivided into the volumes \( {V}_{1} \) and \( {V}_{2} \) by the interface surface \( {A}^{\prime } \), Fig. 9.12. Equation (9.3.29) is now applied to the three volumes \( V,{V}_{1} \), and \( {V}_{2} \) and the integral contributions from the volumes \( {V}_{1} \) and \( {V}_{2} \) ar... | Yes |
Theorem 9.13 Necessary and Sufficient Condition for a Vector Field to be Irrotational A vector field \( \mathbf{v}\left( {\mathbf{r}, t}\right) \) is irrotational if rot \( \mathbf{v}\left( {\mathbf{r}, t}\right) \equiv \nabla \times \mathbf{v}\left( {\mathbf{r}, t}\right) = \mathbf{0} \) . A necessary and sufficient c... | \[ \operatorname{rot}\mathbf{v}\left( {\mathbf{r}, t}\right) \equiv \nabla \times \mathbf{v}\left( {\mathbf{r}, t}\right) = \mathbf{0} \Leftrightarrow \mathbf{v}\left( {\mathbf{r}, t}\right) = \operatorname{grad}\phi \left( {\mathbf{r}, t}\right) \equiv \nabla \phi \left( {\mathbf{r}, t}\right) \] Proof of the implicat... | Yes |
Theorem 9.14 Material Derivative of an Extensive Quantity Let \( f\left( {\mathbf{r}, t}\right) \) be a place function representing an intensive physical quantity defined per unit mass and with \( F\left( t\right) \) as the corresponding extensive quantity for a material body of volume \( V\left( t\right) \), such that... | Proof of formula (9.3.40): Sect. 2.1.3 presents a physical argument as a proof for the theorem. Here we shall use Theorem 9.11 Change of variables in the volume integral, to present a somewhat more stringent mathematical proof of Theorem 9.14.\n\nThe motion of the body may be interpreted as a one-to-one mapping between... | Yes |
Theorem 9.15 Reynolds' Transport Theorem Let, at the present time t, B(t) be an extensive quantity for a body of a continuous medium with the volume \( V\\left( t\\right) \) and surface \( A\\left( t\\right) \), and let \( \\beta \\left( t\\right) \) be the intensive quantity related to \( B\\left( t\\right) \) and exp... | Proof The formula (9.3.19) in Theorem 9.11 is first used to transform the integral in Eq. (9.3.44):\n\n\[ B\\left( t\\right) = {\\int }_{V\\left( t\\right) }{\\beta dV} = {\\int }_{{V}_{o}}{\\beta Jd}{V}_{o} \]\n\n(9.3.46)\n\n\( {V}_{o} \) is the volume of the body in the reference configuration \( {K}_{\\mathrm{o}} \)... | Yes |
Problem 1.1\n\n(a) Validate the identity (1.1.19) through some test examples. | ## Solution\n\n(a) Validation of the identity (1.1.19): \( {e}_{ijk}{e}_{rsk} = {\delta }_{ir}{\delta }_{js} - {\delta }_{is}{\delta }_{jr} \)\n\n\[ \n\text{Test (1)}i = j = 1 \Rightarrow {e}_{11k}{e}_{rsk} = {\delta }_{1r}{\delta }_{1s} - {\delta }_{1s}{\delta }_{1r} = 0 \n\]\n\n\[ \n\text{Test (2)}\left. {i = 1, j = ... | Yes |
Problem 1.2 Determine the inverse matrix \( {U}^{-1} \) of the matrix: \( U = \)\n\n\[ \left( \begin{matrix} 1 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & \sqrt{2} \end{matrix}\right) \frac{1}{\sqrt{2}} \] | Solution Formula (1.1.29) implies:\n\n\[ {U}^{-1}\det U = \operatorname{Co}{U}^{T} \Rightarrow {U}^{-1} = \frac{1}{\det U}\operatorname{Co}{U}^{T} \]\n\n\[ \det U = \left\lbrack {\left( {1 \cdot 3 \cdot \sqrt{2} - 1 \cdot 0 \cdot 0}\right) + \left( {1 \cdot 0 \cdot 0 - 1 \cdot 1 \cdot \sqrt{2}}\right) + \left( {0 \cdot... | Yes |
Problem 1.3 Referred to the Cartesian coordinate system \( {Ox} \), the Cartesian coordinate system \( \bar{O}\bar{x} \) has the base vectors: \( {\overline{\mathbf{e}}}_{1} = \left\lbrack {1,1,1}\right\rbrack \left( {1/\sqrt{3}}\right) \) , \( {\overline{\mathbf{e}}}_{2} = \left\lbrack {1,0, - 1}\right\rbrack \left( {... | Solution The base vector \( \;{\overline{\mathbf{e}}}_{3} = {\overline{\mathbf{e}}}_{1} \times {\overline{\mathbf{e}}}_{2} = \det \left( \begin{matrix} {\mathbf{e}}_{1} & {\mathbf{e}}_{2} & {\mathbf{e}}_{3} \\ 1 & 1 & 1 \\ 1 & 0 & - 1 \end{matrix}\right) \left( {\frac{1}{\sqrt{2}}\frac{1}{\sqrt{3}}}\right) \Rightarrow ... | Yes |
Show that \( Q = \bar{Q}\widetilde{Q} \) | Proof that \( Q = \bar{Q}\widetilde{Q} : \;{\overline{\mathbf{e}}}_{i} = {\bar{Q}}_{ik}{\widetilde{\mathbf{e}}}_{k} \Rightarrow {Q}_{ij} = {\overline{\mathbf{e}}}_{i} \cdot {\mathbf{e}}_{j} = \left( {{\bar{Q}}_{ik}{\widetilde{\mathbf{e}}}_{k}}\right) \cdot {\mathbf{e}}_{j} = {\bar{Q}}_{ik}{\widetilde{Q}}_{kj} \Rightarr... | Yes |
Problem 1.5 The rotation rot \( \mathbf{a} \) of a vector field \( \mathbf{a} \) may be defined as the vector represented by the Cartesian components defined in formula (1.6.15). Show that rot a is a proper vector, i.e. the components obey the transformation rule (1.3.9). | Solution Let \( {Ox} \) and \( \bar{O}\bar{x} \) be two Cartesian coordinate systems with base vectors \( {\mathbf{e}}_{i} \) and \( {\overline{\mathbf{e}}}_{r} \) . The transformation matrix for the transformation from \( {Ox} \) to \( \bar{O}\bar{x} \) is \( {Q}_{ri} \) . With \( \mathbf{a} \) and \( \mathbf{b} = \) ... | Yes |
Problem 2.1 A velocity field for a continuous material in motion is given as:\n\n\[ \n{v}_{1} = \frac{\alpha {x}_{1}}{t - {t}_{0}},\;{v}_{2} = - \frac{\alpha {x}_{2}}{t - {t}_{0}},\;{v}_{3} = 0,\;\alpha \text{ and }{t}_{0}\text{ are constants }\n\]\n\n(a) Show that the flow is isochoric \( = \) volume preserving, i.e. ... | (a) The divergence of the velocity field \( \mathbf{v} \) :\n\n\[ \n\operatorname{div}\mathbf{v} = {v}_{i, i} = \frac{\partial {v}_{1}}{\partial {x}_{1}} + \frac{\partial {v}_{2}}{\partial {x}_{2}} + \frac{\partial {v}_{3}}{\partial {x}_{3}} = \frac{\alpha }{t - {t}_{0}} + \frac{-\alpha }{t - {t}_{0}} + 0 \Rightarrow \... | Yes |
Problem 2.2 The state of stress in a particle is represented by the following stress matrix with respect to the Cartesian coordinate system \( {Ox} \) :\n\n\[ T = \left( \begin{matrix} {90} & - {30} & 0 \\ - {30} & {120} & - {30} \\ 0 & - {30} & {90} \end{matrix}\right) \mathrm{{MPa}} \]\n\n(a) Determine the stress vec... | (a) By definition: Stress vector: a) \( \mathbf{t} = \mathbf{T} \cdot \mathbf{n} = {t}_{i}{\mathbf{e}}_{i} \Rightarrow {t}_{i} = {T}_{ik}{n}_{k} \Rightarrow \)\n\nnormal stress: \( \sigma = \mathbf{n} \cdot \mathbf{t} = {n}_{i}{T}_{ik}{n}_{k} \), shear stress: \( \tau = \sqrt{\mathbf{t} \cdot \mathbf{t} - {\sigma }^{2}... | Yes |
Problem 2.3 A thin-walled circular tube has middle radius \( r = {150}\mathrm{\;{mm}} \) and wall thickness \( h = 7\mathrm{\;{mm}} \) . The tube is closed in both ends, and is subjected to an internal pressure \( p = 8\mathrm{{MPa}} \), a torque \( {m}_{t} = {60}\mathrm{{kNm}} \), and an axial force \( N = {180}\mathr... | (a) Non-zero coordinate stresses and the stress matrix \( T \) :\n\n\[ \n{T}_{11} = \frac{N}{2\pi rh} + \frac{r}{2h}p = {113.0}\mathrm{{MPa}},\;{T}_{22} = \frac{r}{h}p = {171.4}\mathrm{{MPa}},\;{T}_{12} = \frac{{m}_{t}}{{2\pi }{r}^{2}h} = {60.6}\mathrm{{MPa}} \n\]\n\n\[ \nT = \left( \begin{matrix} {T}_{11} & {T}_{12} &... | Yes |
Problem 3.1 Show that a completely antisymmetric tensor of third order \( \mathbf{A} \) only has one distinct component different from zero, and that the tensor is represented by the product of a scalar \( \alpha \) and the permutation tensor \( \mathbf{P} \), i.e. \( \mathbf{A} = \alpha \mathbf{P} \) . | Solution Antisymmetry implies:\n\n\[ \n{A}_{ijk} = - {A}_{kji} = {A}_{kij} = - {A}_{ikj} \Rightarrow {A}_{123} = - {A}_{312} = {A}_{231} = - {A}_{321} = - {A}_{123} = - {A}_{213} = - {A}_{132} \n\]\n\nWith no summation with respect to \( k : {A}_{kjk} = - {A}_{jkk} = 0,\;{A}_{ikk} = - {A}_{ikk} = 0 \), \n\n\[ \n{A}_{kk... | Yes |
Problem 3.2 Show that an isotropic tensor of second order \( {\mathbf{I}}_{2} \) always is a product of a scalar and the unit tensor as given by formula (3.2.8): \( {\mathbf{I}}_{2} = \alpha \mathbf{1} \) . | Solution Let \( \mathbf{B} \) be an isotropic tensor of second order which implies that the tensor matrix is the same in any Cartesian coordinate system. Three Cartesian coordinate systems are now introduced: \( {Ox} \) with base vectors \( {\mathbf{e}}_{i}, O\bar{x} \) with base vectors \( {\overline{\mathbf{e}}}_{i} ... | Yes |
Problem 3.3 Show by induction that the result (3.3.32) follows from the formula (3.3.30). | Solution In a Cartesian coordinate system \( {Ox} \) with base vectors \( {\mathbf{e}}_{k} \equiv {\mathbf{a}}_{k} = {\delta }_{ki}{\mathbf{e}}_{i} \) the tensor \( \mathbf{S} \) is in accordance with the formula (3.3.30) represented by the components:\n\n\[ {S}_{ij} = \mathop{\sum }\limits_{k}{\sigma }_{k}{a}_{ki}{a}_... | Yes |
Problem 3.4 Show that the definition (3.3.33) of a positive definite and symmetric second order tensor \( \mathbf{S} \) implies that all the principal values and the principal invariants of the tensor are positive. | Solution Let \( \mathbf{S} \) be a positive and symmetric second order tensor. Then by definition:\n\n\[ \mathbf{c} \cdot \mathbf{S} \cdot \mathbf{c} > 0\;\text{ for all vectors }\mathbf{c} \neq 0\;\left( {3.3.33}\right) \]\n\nLet the principal values and the principal directions of \( \mathbf{S} \) be \( {\sigma }_{k}... | Yes |
Problem 3.5 The general form of the linear, symmetric isotropic tensor-valued function \( \mathbf{B}\left\lbrack \mathbf{A}\right\rbrack \) of a symmetric second order tensor \( \mathbf{A} \) is given by the formula:\n\n\[ \mathbf{B}\left\lbrack \mathbf{A}\right\rbrack = \left( {\gamma + \lambda \operatorname{tr}\mathb... | Solution The tensors \( \mathbf{A} \) and \( \mathbf{B} \) are decomposed into deviators and isotrops:\n\n\[ \mathbf{B} = {\mathbf{B}}^{\prime } + {\mathbf{B}}^{o},\;{\mathbf{B}}^{\prime } = \mathbf{B} - {\mathbf{B}}^{o},\;{\mathbf{B}}^{o} = \frac{1}{3}\left( {\operatorname{tr}\mathbf{B}}\right) \mathbf{1},\;\mathbf{A}... | Yes |
Show that in homogeneous deformation of a body of continuous material, formula (4.2.26), planes and straight lines in the reference configuration \( {K}_{0} \) deform into planes and straight lines in the present configuration \( K \) . | Homogeneous deformation is defined by:\n\n\[ \mathbf{r}\left( {{\mathbf{r}}_{0}, t}\right) = {\mathbf{u}}_{0}\left( t\right) + \mathbf{F}\left( t\right) \cdot {\mathbf{r}}_{0}\;\left( {4.2.26}\right) \]\n\nA material plane in \( {K}_{0} \) through any two particles \( {\mathbf{r}}_{0} \) and \( {\mathbf{r}}_{10} \) and... | Yes |
Problem 4.3 Derive the formula:\n\n\[ \dot{J} = J\operatorname{div}\mathbf{v}\;\left( {4.5.33}\right) . \] | Solution The formulas (1.1.24) and (1.1.27) imply:\n\n\[ \frac{\partial \left( {\det \mathbf{F}}\right) }{\partial {F}_{ik}} = \operatorname{Co}{F}_{ik},\;\left( {\operatorname{Co}{F}_{ik}}\right) {F}_{jk} = \left( {\det \mathbf{F}}\right) {\delta }_{ij} \]\n\n(1)\n\nBy the definition (4.5.32): \( J = \det \mathbf{F} \... | Yes |
Problem 5.1 Develop the (5.2.7) of Hooke's law from the form (5.2.5). | Solution The formula (5.2.5) implies:\n\n\[ \operatorname{tr}\mathbf{E} = \frac{1 + v}{\eta }\operatorname{tr}\mathbf{T} - \frac{v}{\eta }\left( {\operatorname{tr}\mathbf{T}}\right) \operatorname{tr}\mathbf{1} = \frac{1 - {2v}}{\eta }\operatorname{tr}\mathbf{T} \Rightarrow \operatorname{tr}\mathbf{T} = \frac{\eta }{1 -... | Yes |
Problem 5.2 Develop the decomposition (5.2.14)-(5.2.16) from Hooke's law (5.2.7). | Solution The formula (5.2.7) implies:\n\n\[ \operatorname{tr}\mathbf{T} = \frac{\eta }{1 + v}\left\lbrack {\operatorname{tr}\mathbf{E} + \frac{v}{1 - {2v}}\left( {\operatorname{tr}\mathbf{E}}\right) \operatorname{tr}\mathbf{1}}\right\rbrack = \frac{\eta }{1 + v}\left\lbrack {\frac{1 - {2v}}{1 - {2v}} + \frac{3v}{1 - {2... | Yes |
Problem 6.2 Show that the symbols \( {\varepsilon }_{ijk} \) and \( {\varepsilon }^{ijk} \) defined by the formulas (6.2.22) and (6.2.24) are components in a general coordinate system \( y \) of the permutation tensor \( \mathbf{P} \) defined by the formula: | Solution The formulas (3.1.25) and (6.2.25) imply that:\n\n\[ {P}_{ijk} = \mathbf{P}\left\lbrack {{\mathbf{g}}_{i},{\mathbf{g}}_{j},{\mathbf{g}}_{k}}\right\rbrack = {\varepsilon }_{ijk},\;{P}^{ijk} = \mathbf{P}\left\lbrack {{\mathbf{g}}^{i},{\mathbf{g}}^{j},{\mathbf{g}}^{k}}\right\rbrack = {\varepsilon }^{ijk} \]\n\nQE... | Yes |
Problem 6.3 Prove the identity: \( {\varepsilon }^{ijk}{\varepsilon }_{rsk} = {\delta }_{r}^{i}{\delta }_{s}^{j} - {\delta }_{s}^{i}{\delta }_{r}^{j} \) (6.4.23), using the identity: | Solution By the formulas (6.2.24) and (6.2.22):\n\n\[ {\varepsilon }^{ijk} = {J}_{x}^{y}{e}_{ijk},\;{\varepsilon }_{rsk} = {J}_{y}^{x}{e}_{rsk} \]\n\nThe formulas (6.2.4) imply that: \( {J}_{x}^{y}{J}_{y}^{x} = 1 \) . Applying the identity (1.1.19) we obtain:\n\n\[ {\varepsilon }^{ijk}{\varepsilon }_{rsk} = {J}_{x}^{y}... | Yes |
Problem 6.4 Prove the formulas (6.5.3), (6.5.4), and (6.5.5). | Solution Proof of the formulas (6.5.3). By the definition of the Christoffel symbols: \( {\mathbf{g}}_{i, j} = {\Gamma }_{ijk}{\mathbf{g}}^{k} = {\Gamma }_{ij}^{k}{\mathbf{g}}_{k} \) . From the formulas (6.5.1) it follows that:\n\n\[ {\mathbf{g}}_{i, j} \cdot {\mathbf{g}}^{l} = {\Gamma }_{ijk}{\mathbf{g}}^{k} \cdot {\m... | Yes |
Problem 6.5 Use the formulas \( {\left( {6.5}{.4}\right) }_{1},\left( {1.1.24}\right) \), and \( \left( {6.2.31}\right) \): To prove the formulas: \[ {\Gamma }_{ik}^{k} = \frac{1}{2g}g,{}_{i} = \frac{1}{\sqrt{g}}\left( \sqrt{g}\right) ,{}_{i}\;\left( {6.5.6}\right) \] | Solution We use the formulas (1.1.24) and (6.2.31) and obtain: \[ \frac{\partial g}{\partial {g}_{jk}} = \operatorname{Co}{\mathrm{g}}_{jk} = {\mathrm{{gg}}}^{jk} \] (1) From the formulas (1), \( {\left( {6.5}{.4}\right) }_{1} \), and (6.5.3) we get: \[ g{,}_{i} = \frac{\partial g}{\partial {g}_{jk}}{g}_{jk}{,}_{i} = \... | Yes |
Problem 6.6 Derive the transformation rule (6.5.10) for the Christoffel symbols of the second kind. | Solution By the definitions (6.5.1):\n\n\[ \frac{\partial {\overline{\mathbf{g}}}_{i}}{\partial {\bar{y}}^{j}} \equiv {\overline{\mathbf{g}}}_{i}{,}_{j} = {\bar{\Gamma }}_{ij}^{k}{\overline{\mathbf{g}}}_{k},\;\frac{\partial {\mathbf{g}}_{r}}{\partial {y}^{s}} \equiv {\mathbf{g}}_{r}{,}_{s} = {\Gamma }_{rs}^{t}{\mathbf{... | Yes |
Problem 6.7 Use the formula: \( {\left. {a}_{k}\right| }_{j} = {a}_{k}{}_{, j} - {a}_{l}{\Gamma }_{kj}^{l} \) from the formula (6.5.31) to prove the last two equalities in Eq. (6.5.50). | Solution Symmetry and antisymmetry imply:\n\n\[{\Gamma }_{kj}^{l} = {\Gamma }_{jk}^{l},\;{\varepsilon }^{ijk} = - {\varepsilon }^{ikj} \Rightarrow {\varepsilon }^{ijk}{\Gamma }_{kj}^{l} = - {\varepsilon }^{ikj}{\Gamma }_{kj}^{l} = - {\varepsilon }^{ijk}{\Gamma }_{jk}^{l} = - {\varepsilon }^{ijk}{\Gamma }_{kj}^{l} = 0\]... | Yes |
Problem 6.8 Derive the formulas (6.5.51) for the divergence of a vector field by using the formulas (6.5.49), \( {\left( {6.2}{.31}\right) }_{1} \), and (6.5.6). | Solution We use the formulas (6.5.49), \( {\left( {6.2}{.31}\right) }_{1} \), and (6.5.6) to write:\n\n\[ \operatorname{div}\mathbf{a} = {\left. {a}^{i}\right| }_{i} = {a}^{i}{,}_{i} + {a}^{k}{\Gamma }_{ki}^{i} = {a}^{i}{,}_{i} + {a}^{k}\frac{1}{\sqrt{g}}\left( \sqrt{g}\right) {,}_{k} = \frac{1}{\sqrt{g}}\left\lbrack {... | Yes |
Problem 6.9 Derive the formula (6.5.99) using the formula (6.5.51):\n\n\[ \n{\nabla }^{2}\alpha = \frac{1}{\sqrt{g}}\left( {\sqrt{g}{g}^{ij}\alpha ,{}_{j}}\right) ,{}_{i}\;\left( {6.5.99}\right) ,\;\operatorname{div}\mathbf{a} = \nabla \cdot \mathbf{a} = \frac{1}{\sqrt{g}}\left( {\sqrt{g}{a}^{i}}\right) ,{}_{i} \n\]\n\... | Solution We define the vector: \( \mathbf{a} = \operatorname{grad}\alpha \equiv \nabla \alpha = {g}^{ij}{\alpha }_{\cdot j}{\mathbf{g}}_{i} \) . The formula (6.5.51) gives the result:\n\n\[ \n{\nabla }^{2}\alpha = \nabla \cdot \nabla \alpha = \nabla \cdot \mathbf{a} = \frac{1}{\sqrt{g}}\left( {\sqrt{g}{g}^{ij}\alpha ,{... | Yes |
Problem 6.10 Derive the formula: \( \nabla \times \left( {\nabla \times \mathbf{a}}\right) = \frac{1}{\sqrt{g}}\frac{\partial }{\partial {y}^{k}}\left\lbrack {\sqrt{g}{g}^{kr}{g}^{is}\left( {\frac{\partial {a}_{r}}{\partial {y}^{s}} - \frac{\partial {a}_{s}}{\partial {y}^{r}}}\right) }\right\rbrack {\mathbf{g}}_{i} \) ... | Solution We define the vector \( \mathbf{b} = \nabla \times \mathbf{a} \) . The formula (6.5.50) gives:\n\n\[ \mathbf{b} = \nabla \times \mathbf{a} = {\varepsilon }^{ijk}{a}_{k}{\left| {}_{j}{\mathbf{g}}_{i} = {\varepsilon }_{ijk}{a}^{k}\right| }^{j}{\mathbf{g}}^{i} \Rightarrow {b}_{i} = {\varepsilon }_{ijk}{a}^{k}{\le... | Yes |
Problem 7.1 Derive the formulas (7.4.14) and (7.4.15). | Solution Derivation of the formula (7.4.14).\n\nFirst we develop the relations between the material line elements \( d{\mathbf{r}}_{0} \) of length \( d{s}_{0} \) and \( d\mathbf{r} \) of length \( {ds} \) :\n\n\[ \mathbf{r} = {\mathbf{r}}_{0} + \mathbf{u} \Rightarrow d\mathbf{r} = d{\mathbf{r}}_{0} + d\mathbf{u},\;{ds... | Yes |
Problem 7.2 Derive the formula (7.6.3).\n\n\[ \n{\dot{\mathbf{c}}}^{K} = - {\left. {v}^{K}\right| }_{L}{\mathbf{c}}^{L}\;\left( {7.6.3}\right) \n\] | Solution By definition of the base vectors \( {\mathbf{c}}^{K} \) and \( {\mathbf{c}}_{L} \), and by Eq. (7.6.2):\n\n\[ \n{\mathbf{c}}^{K} \cdot {\mathbf{c}}_{L} = {\delta }_{L}^{K}\;\text{ and }\;{\left. {\dot{\mathbf{c}}}_{L} = {v}^{N}\right| }_{L}{\mathbf{c}}_{N} \Rightarrow {\dot{\mathbf{c}}}^{K} \cdot {\mathbf{c}}... | Yes |
Problem 7.3 Show that Eq. (7.6.18) follows from Eq. (7.6.17). | Solution The following equations are used:\n\n\[ \left( {7.2.12}\right) \Rightarrow {\dot{B}}_{j}^{i} = {\left. \frac{\partial }{\partial t}{B}_{j}^{i} + {B}_{j}^{i}\right| }_{k}{v}^{k},\;{\left( {6.5.31}\right) }_{1} \Rightarrow {\left. {v}^{i}\right| }_{k} = {v}^{i},{}_{k} + {v}^{l}{\Gamma }_{lk}^{i} \]\n\n\[ \left( ... | Yes |
Problem 7.5 Derive the formula:\n\n\[ \n{\left. {\partial }_{c}{B}_{ij} = {\dot{B}}_{ij} + {\left. {B}_{kj}{v}^{k}\right| }_{i} + {B}_{ik}{v}^{k}\right| }_{j}\;{\left( {7.6.23}\right) }_{1} \n\] | Solution Let \( \mathbf{a} \) and \( \mathbf{b} \) be any two vector fields such that according to Eq. (7.6.8):\n\n\[ \n{\partial }_{c}{a}^{i} = {\dot{a}}^{i} - {\left. {a}^{k}{v}^{i}\right| }_{k},\;{\partial }_{c}{b}^{j} = {\dot{b}}^{j} - {\left. {b}^{k}{v}^{j}\right| }_{k} \n\]\n\n(1)\n\nLet a scalar field be defined... | Yes |
Problem 7.6 Show that the two sets of physical components of stress defined respectively by Eqs. (7.7.8) and (7.7.11) are related through Eq. (7.7.12). | Solution The formulas (7.7.8) and (7.7.11) give:\n\n\[ \n{\tau }_{ki} = {T}^{ki}\sqrt{\frac{{g}_{kk}}{{g}^{ii}}} = \mathop{\sum }\limits_{j}\left( {{T}_{j}^{k}{g}^{ij}}\right) \sqrt{\frac{{g}_{kk}}{{g}^{ii}}},\;{T}_{j}^{k} = T\left( {kj}\right) \sqrt{\frac{{g}_{jj}}{{g}_{kk}}} \n\]\n\nThen:\n\n\[ \n{\tau }_{ki} = {T}^{... | Yes |
Problem 7.7 Use the general Cauchy equations for orthogonal coordinates (7.7.18) to develop the Cauchy equations (7.7.19) in cylindrical coordinates. | Solution In cylindrical coordinates \( \left( {R,\theta, z}\right) : {h}_{1} = 1,{h}_{2} = R,{h}_{3} = 1, h = R \) . \n\n\[ \n\text{Physical coordinate stresses:}\left( {T\left( {ik}\right) }\right) = \left( \begin{matrix} {\sigma }_{R} & {\tau }_{R\theta } & {\tau }_{Rz} \\ {\tau }_{\theta R} & {\sigma }_{\theta } & {... | Yes |
Use the formula (6.4.23) and the definitions (1.1.16) and (8.1.11) for the permutation symbols\n\n\\[ \n{e}_{ijk},{\varepsilon }_{ijk},{\varepsilon }^{ijk},{e}_{\alpha \beta },{\varepsilon }_{\alpha \beta },\\;\\text{ and }\\;{\varepsilon }^{\alpha \beta }\n\\]\n\n(1)\n\nto prove the relationships:\n\n\\[ \n{\varepsilo... | (a) Proof of the formulas (8.1.12). From the definitions of the permutation symbols (1) it follows that:\n\n\\[ \n{\varepsilon }_{\alpha \beta } = {e}_{\alpha \beta }\\sqrt{\\alpha } = {e}_{\alpha ⓷}\\sqrt{\\alpha },\\;{\varepsilon }^{\alpha \beta } = {e}_{\alpha \beta }/\\sqrt{\\alpha } = {e}_{\alpha ⓷}/\\sqrt{\\alpha... | Yes |
Problem 8.2 The covariant components of the unit normal vector \( {\mathbf{a}}_{3} \) in a general curvilinear coordinate system \( y \) are denoted by \( {\alpha }_{i} \) such that \( {\mathbf{a}}_{3} = {\alpha }_{i}{\mathbf{g}}^{i} \) . Derive the formula (8.1.14): | Solution The following formulas are applied:\n\n\[ {\mathbf{a}}_{3}{\varepsilon }_{\alpha \beta } = {\mathbf{a}}_{\alpha } \times {\mathbf{a}}_{\beta } = {\varepsilon }_{\alpha \beta }{\mathbf{a}}_{3}\;{\left( {8.1}{.13}\right) }_{3},\;{\varepsilon }_{\alpha \beta }{\varepsilon }^{\gamma \beta } = {\delta }_{\alpha }^{... | Yes |
Problem 8.3 Derive the formulas (8.1.18)-(8.1.20). | Solution Derivation of the formulas (8.1.18).\n\nWe use the formula (8.1.15): \( {\mathbf{a}}_{\alpha }{}_{,\beta } = {\Gamma }_{\alpha \beta }^{\gamma }{\mathbf{a}}_{\gamma } + {B}_{\alpha \beta }{\mathbf{a}}_{3} \) to obtain:\n\n\[ {\mathbf{a}}_{3} \cdot {\mathbf{a}}_{\beta } = 0 \Rightarrow {\mathbf{a}}_{3},{}_{\alp... | No |
Problem 8.5 Let \( \mathbf{S} \) be a symmetric surface tensor field of second order. Show that the angle \( \phi \) between a principal direction \( \mathbf{b} \) corresponding to the principal value \( \sigma \) , and the base vector \( {\mathbf{a}}_{1} \) in a surface coordinate system \( u \) is given by the formul... | Solution From Fig. 8.6 we obtain:\n\n\[ \left. \begin{array}{l} \sin \phi = \mathbf{b} \cdot \frac{{\mathbf{a}}^{2}}{\sqrt{{a}^{22}}} = \left( {{b}_{\alpha }{\mathbf{a}}^{\alpha }}\right) \cdot \frac{{\mathbf{a}}^{2}}{\sqrt{{a}^{22}}} = \frac{{b}_{1}{a}^{12} + {b}_{2}{a}^{22}}{\sqrt{{a}^{22}}} \\ \cos \phi = \mathbf{b}... | Yes |
Problem 8.6 Show that for a geodesic coordinate system \( u \) for a point \( P \) on a surface \( {y}^{i}\left( u\right) \) the following hold true:\n\n\[ \n{a}_{{\alpha \beta },\gamma } = {a}^{\alpha \beta }{,}_{\gamma } = \alpha {,}_{\alpha } = {\varepsilon }_{\alpha \beta }{,}_{\gamma } = {\varepsilon }^{\alpha \be... | Solution Because the Christoffel symbols are zero in the pole \( P \) for a geodesic coordinate system \( u \), it follows from the formulas (8.1.9) and (8.1.20) that:\n\n\[ \n{a}_{{\alpha \beta },\gamma } = 0,\;\alpha ,\alpha = 0\text{ in }P \n\]\n\n(1)\n\nFrom the formulas (8.1.7) and (8.1.11) we write:\n\n\[ \n{a}_{... | Yes |
Problem 8.7 Use the results (8.5.29) to prove that the formulas (8.5.30). | Solution In the special geodesic coordinate system \( u \) for the pole \( P \) the formulas (8.5.29) represent the tensor equations:\n\n\[ {\left. {a}_{\alpha \beta }\right| }_{\gamma } = {\left. {a}^{\alpha \beta }\right| }_{\gamma } = {\left. {\varepsilon }_{\alpha \beta }\right| }_{\gamma } = {\left. {\varepsilon }... | No |
Problem 8.9 Show that: \( {\left. {g}_{ij}\right| }_{\alpha } = {\left. {g}^{ij}\right| }_{\alpha } = 0,{\left. \;{\varepsilon }_{ijk}\right| }_{\alpha } = {\left. {\varepsilon }^{ijk}\right| }_{\alpha } = 0\;\left( {8.5.32}\right) \) | Solution In a Cartesian coordinate \( y \) -system and a geodesic \( u \) -system:\n\n\[ \n{g}_{ij} = {g}^{ij} = {\delta }_{ij},\;{\varepsilon }_{ijk} = {\varepsilon }^{ijk} = {e}_{ijk} \Rightarrow \n\]\n\n\[ \n{\left. {g}_{ij}\right| }_{\alpha } = {\left. {g}^{ij}\right| }_{\alpha } = \frac{\partial \left( {\delta }_{... | Yes |
Problem 8.10 Derive the formulas: \( \;{B}_{\alpha \beta } = {\left. \frac{\partial {y}^{i}}{\partial {u}^{\alpha }}\right| }_{\beta }{\underset{ \sim }{a}}_{i}\; \Leftrightarrow \) | \[ {\left. \frac{\partial {y}^{i}}{\partial {u}^{\alpha }}\right| }_{\beta } = {B}_{\alpha \beta }{\underset{ \sim }{a}}^{i}\;\left( {8.5.33}\right) \] Solution The following formulas will be used: \[ \begin{array}{l} \text{ (8.1.2) } \Rightarrow {\mathbf{a}}_{\alpha } = {\mathbf{g}}_{i}\frac{\partial {y}^{i}}{\partial... | Yes |
Problem 8.11 Let be \( \phi \left( y\right) \) scalar field on a surface \( {y}^{i}\left( u\right) \) . Show that:\n\n\[ \n{\left. \phi \right| }_{\alpha \beta } = {\left. \phi \right| }_{\beta \alpha }\;\left( {8.5.38}\right) \n\] | Solution By definition: \( {\left. \phi \right| }_{\alpha } = \phi ,{}_{\alpha } \) . Using the formula (8.5.3)2 and the symmetry of the Christoffel symbols we get:\n\n\[ \n{\left. \phi \right| }_{\alpha \beta } = {\left. {\left( {\left. \phi \right| }_{\alpha }\right) }_{,\beta } - {\left. \phi \right| }_{\gamma }{\Ga... | Yes |
Problem 8.13 Prove the formula for the Gauss curvature \( \gamma \) of a surface \( {y}^{i}\left( u\right) \) :\n\n\[ \gamma = \frac{1}{4}{\varepsilon }^{\delta \alpha }{\varepsilon }^{\beta \gamma }{R}_{\delta \alpha \beta \gamma }\;\left( {8.5.42}\right) \] | Solution We use the formulas:\n\n\[ \text{(8.1.11)} \Rightarrow {\varepsilon }_{\alpha \beta } = {e}_{\alpha \beta }\sqrt{\alpha },\;{\varepsilon }^{\alpha \beta } = {e}_{\alpha \beta }/\sqrt{\alpha },\;\left( {8.1.12}\right) \Rightarrow {\varepsilon }_{\alpha \beta }{\varepsilon }^{\gamma \beta } = {\delta }_{\alpha }... | Yes |
Problem 8.14 Derive Weingarten’s formula: \( {\left. {\underset{ \sim }{a}}^{i}\right| }_{\alpha } = - {B}_{\alpha }^{\beta }\frac{\partial {y}^{i}}{\partial {u}^{\beta }}\;({8.5.45} \) | Solution From the formulas:\n\n\[ \n{\mathbf{a}}_{3} = {}_{ \sim }^{a}{}^{i}{\mathbf{g}}_{i},\;\left( {8.1.18}\right) \Rightarrow {\mathbf{a}}_{3}{,}_{\alpha } = - {B}_{\alpha \beta }{\mathbf{a}}^{\beta } = - {B}_{\alpha }^{\beta }{\mathbf{a}}_{\beta },\;\left( {8.5.26}\right) \Rightarrow {\left. {\mathbf{g}}_{j}\right... | Yes |
Problem 8.15 Let \( \mathbf{b} \) be the space vector: \( \mathbf{b} = {b}_{\beta }{\mathbf{a}}^{\beta } + {b}_{3}{\mathbf{a}}_{3} \) . Derive the equation:\n\n\[ \mathbf{b},{}_{\alpha } = \left( {{\left. {b}_{\beta }\right| }_{\alpha } - {b}_{3}{B}_{\beta \alpha }}\right) {\mathbf{a}}^{\beta } + \left( {{b}_{3,\alpha ... | Solution We introduce a surface vector \( \mathbf{c} = {b}_{\beta }{\mathbf{a}}^{\beta }\mathbf{c} \) such that \( \mathbf{b} = \mathbf{c} + {b}_{3}{\mathbf{a}}_{3} \) . Then\n\nwe use the result (8.5.4) to obtain:\n\n\[ \mathbf{b},{}_{\alpha } = \mathbf{c},{}_{\alpha } + {b}_{3},{}_{\alpha }{\mathbf{a}}_{3} + {b}_{3}{... | Yes |
Show that the fundamental parameters of third order \( {C}_{\alpha \beta } \), of second order \( {B}_{\alpha \beta } \), and of first order \( {a}_{\alpha \beta } \) of a surface \( {y}^{i}\left( u\right) \) satisfy the equation:\n\n\[ \n{C}_{\alpha \beta } - {2\mu }{B}_{\alpha \beta } + \gamma {a}_{\alpha \beta } = 0... | Solution First we obtain from the formula (8.1.2):\n\n\[ \n{\mathbf{a}}_{\alpha } \cdot {\mathbf{a}}_{\beta } = \left( {\frac{\partial {y}^{i}}{\partial {u}^{\alpha }}{\mathbf{g}}_{i}}\right) \cdot \left( {\frac{\partial {y}^{j}}{\partial {u}^{\beta }}{\mathbf{g}}_{j}}\right) = \frac{\partial {y}^{i}}{\partial {u}^{\al... | Yes |
Property 1.1.2 Following Einstein summation convention, we get | \[ {\delta }_{i}^{i} = {\delta }_{1}^{1} + {\delta }_{2}^{2} + \cdots + {\delta }_{N}^{N} = 1 + 1 + \cdots + 1\left( {N\text{ times }}\right) = N. \] | Yes |
Property 1.1.4 Using the Einstein summation convention, we get\n\n\[ \n{\delta }_{j}^{i}{\delta }_{k}^{j} = {\delta }_{1}^{i}{\delta }_{k}^{1} + {\delta }_{2}^{i}{\delta }_{k}^{2} + \cdots + {\delta }_{i}^{i}{\delta }_{k}^{i} + \cdots + {\delta }_{N}^{i}{\delta }_{k}^{N} \n\] | \[ \n= 0{\delta }_{k}^{1} + 0{\delta }_{k}^{2} + \cdots + 1{\delta }_{k}^{i} + \cdots + 0{\delta }_{k}^{N} = {\delta }_{k}^{i}. \n\]\n\nUsing definition of \( {\delta }_{j}^{i} \) we get,\n\n\[ \n{\delta }_{j}^{i}{\delta }_{k}^{j} = \frac{\partial {x}^{i}}{\partial {x}^{j}}\frac{\partial {x}^{j}}{\partial {x}^{k}} = \f... | Yes |
Property 1.2.1 If a transformation (1.2) of co-ordinates possesses an inverse transformation (1.4) with respective Jacobians \( J \) and \( K \), respectively, then \( {JK} = 1 \) . | Proof: Since \( {\bar{x}}^{i}\mathrm{\;s} \) are independent and \( {x}^{i}\mathrm{\;s} \) are also independent functions of \( {\bar{x}}^{i}\mathrm{\;s} \), by the formula of partial differentiation and summation convention we can write\n\n\[ \n\frac{\partial {\bar{x}}^{i}}{\partial {\bar{x}}^{j}} = \frac{\partial {\b... | Yes |
Property 1.2.2 The Jacobian of the product transformation is equal to the product of the Jacobians of transformations entering in the product. | Proof: Let us consider any two admissible transformations\n\n\[ \n{T}_{1} : {\bar{x}}^{i} = {\bar{x}}^{i}\left( {{x}^{1},{x}^{2},\ldots ,{x}^{N}}\right) ;\;{T}_{2} : {\overline{\bar{x}}}^{i} = {\overline{\bar{x}}}^{i}\left( {{\bar{x}}^{1},{\bar{x}}^{2},\ldots ,{\bar{x}}^{N}}\right) , \n\]\n\nwhere \( i = 1,2,\ldots, N ... | Yes |
Property 1.2.3 The set of all admissible transformations of co-ordinates forms a group. | Proof: The set of all admissible transformations of co-ordinates forms a group if the following four axioms are satisfied:\n\n(i) The product of two admissible transformations is a transformation belonging to the set of admissible transformations. This property is known as the property of closure.\n\n(ii) The product t... | Yes |
Property 1.3.1 \( e \) -systems of second order: According to the definition of determinants, | \[ \left| {a}_{j}^{i}\right| = \left| \begin{array}{ll} {a}_{1}^{1} & {a}_{2}^{1} \\ {a}_{1}^{2} & {a}_{2}^{2} \end{array}\right| = {a}_{1}^{1}{a}_{2}^{2} - {a}_{2}^{1}{a}_{1}^{2} \] \[ = {e}_{12}{a}_{1}^{1}{a}_{2}^{2} + {e}_{21}{a}_{1}^{2}{a}_{2}^{1};\text{ as }{e}_{12} = 1,{e}_{21} = - 1 \] or \[ \left| {a}_{j}^{i}\r... | Yes |
Property 1.3.2 \( e \) -system of third order: If we take,\n\n\[ \n\left| {a}_{j}^{i}\right| = \left| \begin{array}{lll} {a}_{1}^{1} & {a}_{2}^{1} & {a}_{3}^{1} \\ {a}_{1}^{2} & {a}_{2}^{2} & {a}_{3}^{2} \\ {a}_{1}^{3} & {a}_{2}^{3} & {a}_{3}^{3} \end{array}\right| \n\]\n\nthen, by the similar arguments as in property ... | \[ \n\left. \begin{array}{ll} \left| {a}_{j}^{i}\right| & = {e}_{ijk}{a}_{1}^{i}{a}_{2}^{j}{a}_{3}^{k} \\ \left| {a}_{j}^{i}\right| & = {e}^{ijk}{a}_{i}^{1}{a}_{j}^{2}{a}_{k}^{3} \\ \left| {a}_{j}^{i}\right| {e}_{pqr} & = {e}_{ijk}{a}_{p}^{i}{a}_{q}^{j}{a}_{r}^{k} \\ \left| {a}_{j}^{i}\right| {e}^{pqr} & = {e}^{ijk}{a}... | Yes |
Property 1.3.3 The product of \( {e}^{ij} \) and \( {e}_{pq} \) is called the generalised Kronecker delta and is denoted by \( {\delta }_{pq}^{ij} \), i.e. | \[ {\delta }_{pq}^{ij} = {e}^{ij}{e}_{pq} \] | No |
Property 1.4.1 If \( {a}_{j}^{i}{b}_{p}^{j} = {c}_{p}^{i} \), then \( \left( {a}_{j}^{i}\right) \left( {b}_{p}^{j}\right) = \left( {c}_{p}^{i}\right) \) and \( \left| {a}_{j}^{i}\right| \left| {b}_{p}^{j}\right| = \left| {c}_{p}^{i}\right| \) . | Proof: Since we take a system of second order, so\n\n\[ \n{c}_{p}^{i} = {a}_{j}^{i}{b}_{p}^{j} = {a}_{1}^{i}{b}_{p}^{1} + {a}_{2}^{i}{b}_{p}^{2} \n\]\n\nor\n\[ \n\left( \begin{array}{ll} {c}_{1}^{1} & {c}_{2}^{1} \\ {c}_{1}^{2} & {c}_{2}^{2} \end{array}\right) = \left( \begin{array}{ll} {a}_{1}^{1}{b}_{1}^{1} + {a}_{2}... | Yes |
Property 1.4.2 If \( {a}_{ij}{b}^{ik} = {c}_{j}^{k} \), then \( {\left( {b}^{ik}\right) }^{T}\left( {a}_{ij}\right) = \left( {c}_{j}^{k}\right) \) and \( \left| {b}^{ik}\right| \left| {a}_{ij}\right| = \left| {c}_{j}^{k}\right| \), where \( {\left( {b}^{ik}\right) }^{T} \) is the transpose of \( \left( {b}^{ik}\right) ... | Proof: Since we take a system of second order, so\n\n\[ \n{c}_{j}^{k} = {a}_{ij}{b}^{ik} = {a}_{1j}{b}^{1k} + {a}_{2j}{b}^{2k} \n\]\n\nor\n\n\[ \n\left( \begin{array}{ll} {c}_{1}^{1} & {c}_{2}^{1} \\ {c}_{1}^{2} & {c}_{2}^{2} \end{array}\right) = \left( \begin{array}{ll} {a}_{11}{b}^{11} + {a}_{21}{b}^{21} & {a}_{12}{b... | Yes |
Property 1.4.3 Let the cofactor of the element \( {a}_{j}^{i} \) in the determinant \( \left| {a}_{j}^{i}\right| \) be denoted by the symbol \( {A}_{i}^{j} \) . Then by summation convention, | \[ {a}_{j}^{i}{A}_{k}^{j} = {a}_{1}^{i}{A}_{k}^{1} + {a}_{2}^{i}{A}_{k}^{2} + \cdots + {a}_{N}^{i}{A}_{k}^{N} = {\delta }_{k}^{i}\left| {a}_{j}^{i}\right| = {\delta }_{k}^{i}a, \] (1.24) and \[ {a}_{j}^{i}{A}_{j}^{k} = {a}_{j}^{1}{A}_{1}^{k} + {a}_{j}^{2}{A}_{2}^{k} + \cdots + {a}_{j}^{N}{A}_{N}^{k} = {\delta }_{j}^{k}... | Yes |
Property 1.4.4 Let us consider a system of \( n \) linear equations as\n\n\[ \n{a}_{j}^{i}{x}^{j} = {b}^{i};\;i, j = 1,2,\ldots, n \n\]\n\n(1.26)\n\nin \( n \) unknown \( {x}^{i} \), where \( \left| {a}_{j}^{i}\right| \neq 0 \) . | Multiplying both sides of equations in (1.26) by \( {A}_{i}^{k} \) , and sum with respect to \( i \) yields\n\n\[ \n{a}_{j}^{i}{A}_{i}^{k}{x}^{j} = {b}^{i}{A}_{i}^{k} \n\]\n\nor,\n\n\[ \na{\delta }_{j}^{k}{x}^{j} = {b}^{i}{A}_{i}^{k};\;\text{ using (1.25) } \n\]\n\nor,\n\n\[ \na{x}^{k} = {b}^{i}{A}_{i}^{k} \Rightarrow ... | Yes |
Property 1.4.5 Consider the determinant \( \left| {a}_{j}^{i}\right| = a \) . Let the elements \( {a}_{j}^{i}{A}_{k}^{j} \) be functions of the independent variables \( {x}_{1},{x}_{2},\ldots {x}_{N} \) then, | \[ \frac{\partial a}{\partial {x}_{1}} = \left| \begin{matrix} \frac{\partial {a}_{1}^{1}}{\partial {x}_{1}} & \frac{\partial {a}_{2}^{1}}{\partial {x}_{1}} & \ldots & \frac{\partial {a}_{N}^{1}}{\partial {x}_{1}} \\ {a}_{1}^{2} & {a}_{2}^{2} & \ldots & {a}_{N}^{2} \\ \vdots & \vdots & \ldots & \vdots \\ {a}_{1}^{N} & ... | Yes |
Property 1.4.6 Consider the transformations \( {z}^{i} = {z}^{i}\left( {y}^{k}\right) \) and \( {y}^{i} = {y}^{i}\left( {x}^{k}\right) \) (Figure 1.1). Let the \( N \) functions \( {z}^{i}\left( {{y}^{1},{y}^{2},\ldots ,{y}^{N}}\right) \) be independent on \( N \) variables \( {y}^{1},{y}^{2},\ldots ,{y}^{N} \) so that... | \[ \frac{\partial {z}^{i}}{\partial {x}^{k}} = \frac{\partial {z}^{i}}{\partial {y}^{1}}\frac{\partial {y}^{1}}{\partial {x}^{k}} + \frac{\partial {z}^{i}}{\partial {y}^{2}}\frac{\partial {y}^{2}}{\partial {x}^{k}} + \cdots + \frac{\partial {z}^{i}}{\partial {y}^{N}}\frac{\partial {y}^{N}}{\partial {x}^{k}} = \frac{\pa... | Yes |
Property 1.15.1 Symmetric property remains unchanged by tensor law of transformation, i.e. if a tensor is symmetric with respect to two contravariant or covariant indices in any co-ordinate system, then it remains so with respect to these two indices in any other co-ordinate system. | Proof: Let a tensor \( {A}_{ij} \) be symmetric in one co-ordinate system \( \left( {x}^{i}\right) \), i.e. \( {A}_{ij} = {A}_{ji} \) and \( {\bar{A}}_{ij} \) in another co-ordinate system \( \left( {\bar{x}}^{i}\right) \) . Now,\n\n\[ \n{\bar{A}}_{ij} = \frac{\partial {x}^{p}}{\partial {\bar{x}}^{i}}\frac{\partial {x}... | Yes |
Property 1.15.2 In an \( N \) dimensional space, a symmetric covariant tensor of second order has atmost \( \frac{N\left( {N + 1}\right) }{2} \) different components. | Proof: Let \( {A}_{ij} \) be a symmetric covariant tensor of second order, then it has \( {N}^{2} \) components in \( {V}_{N} \) . These components are\n\n\[ \n\begin{array}{llll} {A}_{11} & {A}_{12} & \cdots & {A}_{1N} \end{array} \n\]\n\n\[ \n\begin{array}{llll} {A}_{21} & {A}_{22} & \cdots & {A}_{2N} \end{array} \n\... | Yes |
Property 1.16.1 If a tensor is skew-symmetric with respect to a pair of contravariant or covariant indices in any co-ordinate system, then it remains so with respect to these two indices in any other co-ordinate system. | Proof: Let a tensor \( {A}_{ij} \) be skew-symmetric in one co-ordinate system \( \left( {x}^{i}\right) \), i.e. \( {A}_{ij} \) \( = - {A}_{ji} \) and \( {\bar{A}}_{ij} \) in another co-ordinate system \( \left( {\bar{x}}^{i}\right) \) . Now,\n\n\[ \n{\bar{A}}_{ij} = \frac{\partial {x}^{p}}{\partial {\bar{x}}^{i}}\frac... | Yes |
Property 1.16.2 In an \( N \) dimensional space, a skew-symmetric covariant tensor of second order has atmost \( \frac{N\\left( {N - 1}\\right) }{2} \) different components. | Proof: Let \( {A}_{ij} \) be a skew-symmetric covariant tensor of second order, then it has \( {N}^{2} \) components in \( {V}_{N} \). These components are\n\n\[ \n\\begin{array}{llll} 0 & {A}_{12} & \\cdots & {A}_{1N} \\end{array} \n\]\n\n\[ \n\\begin{array}{lll} {A}_{21} & 0 & \\cdots {A}_{2N} \\end{array} \n\]\n\n\[... | Yes |
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