**Notations**


## **Appendix**

• At points *M*ð Þ*<sup>j</sup>*

**Figure 7.**

*M*ð Þ*<sup>j</sup>* <sup>2</sup> *ξ*1, *η*

stress, respectively.

**5. Conclusion**

**178**

coordinates.

max v*<sup>t</sup>* j j<sup>&</sup>lt; max v*<sup>n</sup>* j j*:*

boundary conditions (10<sup>0</sup>

ð Þ*j* 1 

<sup>2</sup> *ξ*1, *η*

ð Þ*j* 1 

the boundary conditions (10) are satisfied.

, max *σ<sup>t</sup>*

*ξξ* 

*Solid State Physics - Metastable, Spintronics Materials and Mechanics of Deformable…*

• When *ξ*<sup>1</sup> ! ∞, then displacements and stresses tend to zero, that is,

) are satisfied.

• When *<sup>η</sup>*<sup>1</sup> ! 0 (in this case there is a crack), then (a) at points *<sup>M</sup>*ð Þ*<sup>j</sup>*

The main results of this chapter can be formulated as follows:

representation by harmonic functions.

parabolic coordinate system.

• When *η*<sup>1</sup> ! ∞, then displacements and stresses tend to zero, that is, the

<sup>&</sup>gt; max *<sup>σ</sup><sup>n</sup>*

*Infinite region bounded by parabola marked with points, when obtaining the above-presented numerical results.*

tangential stresses and normal displacements tend to ∞, but other components equal to zero. It can be seen from the boundary conditions (8a) (b) at points

Here superscript *t* and *n* denote the tangential and normal displacement or the

• The equilibrium equations and Hooke's law are written in terms of parabolic

• The solution of the equilibrium equations is obtained by the method of separation of variables. The solution is constructed using its general

• In parabolic coordinates, analytical solutions of 2D static boundary value problems for the elasticity are constructed for homogeneous isotropic finite and infinite bodies occupying domains bounded by coordinate lines of

that all components of the displacements and stresses tend to ∞.

*ξξ* 

, max *ut* j j<sup>&</sup>gt; max *un* j j,

<sup>1</sup> 0, *η* ð Þ*j* 1 

### **A. Some basic formulas in parabolic coordinates**

In orthogonal parabolic coordinate system *ξ*, *η*(�∞ < *ξ*< ∞, 0≤*η*< ∞, see **Figure A1**) [23, 24]; we have

$$h\_{\xi} = h\_{\eta} = h = c\sqrt{\xi^2 + \eta^2}, \quad \varkappa = c\left(\xi^2 - \eta^2\right)/2, \quad \jmath = c\xi\eta,$$

where *hξ*, *h<sup>η</sup>* are Lame's coefficients of the system of parabolic coordinates, *c* is a scale coefficient, *x*, *y* are the Cartesian coordinates.

The coordinate axes are parabolas

$$y^2 = -2c\xi\_0^2(\varkappa - c\xi\_0^2/2), \quad \xi\_0 = \text{const}, \quad y^2 = -2c\eta\_0^2(\varkappa + c\eta\_0^2/2), \quad \eta\_0 = \text{const}.$$

Laplace's equation Δ*f* ¼ 0, where *f* ¼ *f*ð Þ *ξ*, *η* , in the parabolic coordinates has the form

$$\left(f\_{,\xi\xi} + f\_{,\eta\eta}\right) / c^2 \left(\xi^2 + \eta^2\right) = \mathbf{0}.$$

We have to find solution of the equation in following form

$$f = X(\xi) \cdot E(\eta),$$

and then by separation of variables, we will receive

$$\frac{1}{c^2(\xi^2+\eta^2)} \left[ \frac{\mathcal{X}''}{\mathcal{X}} + \frac{E'}{E} \right] = \mathbf{0}.$$

From here

$$X'' + mX = 0, \qquad E'' - mE = 0,$$

General solution of the system (B2) can be written in the form *u* ¼ *ψ*1, v ¼ *ψ*2,

*ψ*1,*<sup>ξ</sup>* þ *ψ*2,*<sup>η</sup>* ¼ 0, *ψ*2,*<sup>ξ</sup>* � *ψ*1,*<sup>η</sup>* ¼ 0*:*

*u* ¼ *φ*,*<sup>ξ</sup>* � *κφ*1*ξ* þ *ψ*1, v ¼ *φ*,*<sup>η</sup>* þ *κφ*1*η* þ *ψ*2, (B6)

� *ξηφ*1,*<sup>η</sup>* � ð Þ *κ* � 1 *φ*1*ξ* þ *ψ*1,

þ *ξηφ*1,*<sup>ξ</sup>* þ ð Þ *κ* � 1 *φ*1*η* þ *ψ*2*:*

<sup>0</sup>*<sup>K</sup>* <sup>¼</sup> *κμ φ*1,*ηξ* <sup>þ</sup> *<sup>φ</sup>*1,*ξη* � �,

<sup>2</sup> *<sup>φ</sup>*1,*<sup>η</sup>* <sup>þ</sup> ð Þ *<sup>κ</sup>* � <sup>1</sup> *<sup>φ</sup>*1*<sup>η</sup>* <sup>þ</sup> *<sup>ψ</sup>*2*:*

The full solution of equation system (B2) is written in the following form:

*<sup>φ</sup>* <sup>¼</sup> *<sup>ξ</sup>*<sup>2</sup> � *<sup>η</sup>*<sup>2</sup>

<sup>2</sup> *<sup>φ</sup>*1,

Without losing the generality, the expression in brackets can be taken as zero, because we already have in *u* and v of the solutions Laplacian (we mean *ψ*<sup>1</sup> and *ψ*2).

ð Þ<sup>c</sup> *<sup>u</sup>* ¼ �*ξηφ*1,*<sup>η</sup>* � ð Þ *<sup>κ</sup>* � <sup>1</sup> *<sup>φ</sup>*1*<sup>ξ</sup>* <sup>þ</sup> *<sup>ψ</sup>*1, dð Þ <sup>v</sup> <sup>¼</sup> *ξηφ*1,*<sup>ξ</sup>* <sup>þ</sup> ð Þ *<sup>κ</sup>* � <sup>1</sup> *<sup>φ</sup>*1*<sup>η</sup>* <sup>þ</sup> *<sup>ψ</sup>*2*:* (B7)

Now we have to write down three versions of *ψ*<sup>1</sup> and *ψ*<sup>2</sup> function representation.

*φ*1, *φ*~1, *φ*<sup>2</sup> are harmonic functions; in addition, *φ*1, *φ*~<sup>1</sup> are selected so that at *η* ¼ *α*,

�*ξηφ*1,*<sup>η</sup>* � ð Þ *κ* � 1 *φ*1*ξ* þ *φ*1,*<sup>η</sup>* þ *φ*~1,*<sup>η</sup>* ¼ 0, *ξηφ*1,*<sup>ξ</sup>* þ ð Þ *κ* � 1 *φ*1*ξ* þ *φ*1,*<sup>ξ</sup>* þ *φ*~1,*<sup>ξ</sup>* ¼ 0,

*ψ*<sup>1</sup> ¼ *φ*1,*<sup>η</sup>* þ *φ*~1,*<sup>η</sup>* þ *φ*2,*<sup>η</sup>*, *ψ*<sup>2</sup> ¼ *φ*1,*<sup>ξ</sup>* þ *φ*~1,*<sup>ξ</sup>* þ *φ*2,*<sup>ξ</sup>*, (B8)

<sup>þ</sup> *<sup>ξ</sup>*<sup>2</sup> � *<sup>η</sup>*<sup>2</sup>

<sup>þ</sup> *<sup>ξ</sup>*<sup>2</sup> � *<sup>η</sup>*<sup>2</sup>

<sup>2</sup> *<sup>φ</sup>*2,*<sup>ξ</sup>* <sup>þ</sup> *ξηφ*2,*<sup>η</sup>*,

(B9)

<sup>2</sup> *<sup>φ</sup>*2,*<sup>η</sup>* � *ξηφ*2,*<sup>ξ</sup>*,

where *φ* is the partial solution of the (B5).

*2D Elastostatic Problems in Parabolic Coordinates DOI: http://dx.doi.org/10.5772/intechopen.91057*

and (B6) formula will receive the following form:

<sup>2</sup> *<sup>φ</sup>*1,*<sup>ξ</sup>* <sup>þ</sup> *ξηφ*1,*<sup>η</sup>* � �

<sup>2</sup> *<sup>φ</sup>*1,*<sup>η</sup>* � *ξηφ*1,*<sup>ξ</sup>* � �

<sup>0</sup>*<sup>D</sup>* <sup>¼</sup> *κμ φ*1,*ηη* � *<sup>φ</sup>*1,*ξξ* � �, bð Þ *<sup>h</sup>*<sup>2</sup>

Therefore, the solutions of system (2) are given in the following form:

where *α* ¼ *η*<sup>1</sup> or *α* ¼ *η*2, the following equations will be satisfied:

<sup>2</sup> *<sup>φ</sup>*1,*<sup>ξ</sup>* <sup>þ</sup> *ξηφ*1,*<sup>η</sup>* !

!

<sup>2</sup> *<sup>φ</sup>*1,*<sup>η</sup>*

<sup>2</sup> *<sup>φ</sup>*1,*<sup>ξ</sup>* � ð Þ *<sup>κ</sup>* � <sup>1</sup> *<sup>φ</sup>*1*<sup>ξ</sup>* <sup>þ</sup> *<sup>ψ</sup>*1, <sup>v</sup> <sup>¼</sup> *<sup>ξ</sup>*<sup>2</sup> � *<sup>η</sup>*<sup>2</sup>

If we take *κ* ¼ const, then

*<sup>u</sup>* <sup>¼</sup> *<sup>ξ</sup>*<sup>2</sup> � *<sup>η</sup>*<sup>2</sup>

<sup>v</sup> <sup>¼</sup> *<sup>ξ</sup>*<sup>2</sup> � *<sup>η</sup>*<sup>2</sup>

*<sup>u</sup>* <sup>¼</sup> *<sup>ξ</sup>*<sup>2</sup> � *<sup>η</sup>*<sup>2</sup>

From here

ð Þ<sup>a</sup> *<sup>h</sup>*<sup>2</sup>

In the first version

In the second version

**181**

*<sup>ψ</sup>*<sup>1</sup> ¼ �*<sup>α</sup> <sup>ξ</sup>*<sup>2</sup> � ð Þ *<sup>η</sup>* � *<sup>α</sup>* <sup>2</sup>

*<sup>ψ</sup>*<sup>2</sup> <sup>¼</sup> *α ξηφ*1,*<sup>ξ</sup>* � *<sup>ξ</sup>*<sup>2</sup> � ð Þ *<sup>η</sup>* � *<sup>α</sup>* <sup>2</sup>

where

#### **Figure A1.** *Parabolic coordinate system.*

where *m* is any constant, their solutions are [25]

$$\begin{array}{l} X = C\_1 \cos \left( m\xi \right) + C\_2 \sin \left( m\xi \right), \quad E = C\_3 e^{m\eta} + C\_4 e^{-m\eta}, \\ = C\_3^\* \cosh \left( m\eta \right) + C\_4^\* \sinh \left( m\eta \right). \end{array}$$

So

$$\begin{aligned} f(\xi, \eta) &= (\mathcal{C}\_3 e^{m\eta} + \mathcal{C}\_4 e^{-m\eta})(\mathcal{C}\_1 \cos \left(m\xi\right) + \mathcal{C}\_2 \sin \left(m\xi\right)) \\ &\text{or} \\ f(\xi, \eta) &= \left(\mathcal{C}\_3^\* \cosh \left(m\eta\right) + \mathcal{C}\_4^\* \sinh \left(m\eta\right)\right)(\mathcal{C}\_1 \cos \left(m\xi\right) + \mathcal{C}\_2 \sin \left(m\xi\right)), \end{aligned}$$

#### **B. Solution of system of partial differential equations**

We solve the system of partial differential equations (2). We have introduced *φ*<sup>1</sup> harmonic function, and if we take

$$\mathbf{B}(\mathbf{a}) \cdot D = \frac{\kappa \mu}{h\_0^2} \left(\wp\_{1,\eta}\eta - \wp\_{1,\xi}\xi\right), \qquad \text{(b)} \\ K = \frac{\kappa \mu}{h\_0^2} \left(\wp\_{1,\eta}\xi + \wp\_{1,\xi}\eta\right), \tag{B1}$$

then Eqs. (2a) and (2b) will be satisfied identically, while Eqs. (2c) and (2d) will receive the following form:

$$\mathbf{u} \cdot (\mathbf{a}) \cdot \overline{\boldsymbol{\mathfrak{u}}}\_{,\xi} + \overline{\boldsymbol{\mathfrak{v}}}\_{,\eta} = (\kappa - 2) \big( \boldsymbol{\varrho}\_{1,\eta} \eta - \boldsymbol{\varrho}\_{1,\xi} \overline{\boldsymbol{\xi}} \big), \quad \text{(b)} \\ \overline{\boldsymbol{\mathfrak{v}}}\_{,\xi} - \overline{\boldsymbol{\mathfrak{u}}}\_{,\eta} = \kappa \big( \boldsymbol{\varrho}\_{1,\eta} \xi + \boldsymbol{\varrho}\_{1,\xi} \eta \big), \quad \text{(B2)}$$

$$\begin{array}{ll} \text{(a)} \quad \overline{u}\_{,\xi} + \overline{v}\_{,\eta} = (\kappa - 2) \big( \rho\_{1,\eta} \eta - \rho\_{1,\xi} \overline{\xi} \big), \quad \text{(b)} \big( \overline{v} - \kappa \rho\_1 \eta \big)\_{,\xi} = \left( \overline{u} + \kappa \rho\_1 \xi \right)\_{,\eta}. \end{array} \text{(\bf B3} \big)$$

From equation (B3b) imply that exists such type harmonic function *φ*, for which fulfill the following

$$
\overline{\boldsymbol{\mu}} = \boldsymbol{\rho}\_{,\underline{\xi}} - \kappa \boldsymbol{\rho}\_1 \underline{\xi}, \qquad \qquad \overline{\mathbf{v}} = \boldsymbol{\rho}\_{,\underline{\eta}} + \kappa \boldsymbol{\rho}\_1 \boldsymbol{\eta}. \tag{\mathbb{B}4}
$$

Considering (B4), from Equation (B3a), the following will be obtained:

$$\begin{array}{l} h^2 \Delta \rho = \rho\_{\text{,}\tilde{\xi}} + \rho\_{\text{,}\eta \eta} = \kappa \rho\_1 + \kappa \rho\_{1,\xi} \xi - \kappa \rho\_1 - \kappa \rho\_{1,\eta} \eta + (\kappa - 2) \left( \rho\_{1,\eta} \eta - \rho\_{1,\xi} \xi \right) \\\ = 2 \left( \rho\_{1,\xi} \xi - \rho\_{1,\eta} \eta \right). \end{array} \tag{B5}$$

General solution of the system (B2) can be written in the form *u* ¼ *ψ*1, v ¼ *ψ*2, where

$$
\mu\_{1,\xi} + \mu\_{2,\eta} = \mathbf{0}, \qquad \mu\_{2,\xi} - \mu\_{1,\eta} = \mathbf{0}.
$$

The full solution of equation system (B2) is written in the following form:

$$
\overline{\mathfrak{u}} = \mathfrak{o}\_{,\mathfrak{F}} - \kappa \mathfrak{o}\_1 \mathfrak{F} + \mathfrak{w}\_1, \qquad \overline{\mathfrak{v}} = \mathfrak{o}\_{,\mathfrak{v}} + \kappa \mathfrak{o}\_1 \mathfrak{v} + \mathfrak{w}\_2,\tag{\text{B6}}
$$

where *φ* is the partial solution of the (B5). If we take *κ* ¼ const, then

$$
\rho = \frac{\xi^2 - \eta^2}{2} \rho\_1,
$$

and (B6) formula will receive the following form:

$$
\overline{u} = \frac{\xi^2 - \eta^2}{2} \rho\_{1,\xi} - (\kappa - 1)\rho\_1 \xi + \nu\_1, \qquad \overline{\mathbf{v}} = \frac{\xi^2 - \eta^2}{2} \rho\_{1,\eta} + (\kappa - 1)\rho\_1 \eta + \nu\_2.
$$

From here

where *m* is any constant, their solutions are [25]

*<sup>m</sup><sup>η</sup>* <sup>þ</sup> *<sup>C</sup>*4*<sup>e</sup>*

<sup>3</sup> cosh ð Þþ *<sup>m</sup><sup>η</sup> <sup>C</sup>*<sup>∗</sup>

**B. Solution of system of partial differential equations**

We solve the system of partial differential equations (2). We have introduced *φ*<sup>1</sup> harmonic function, and if we take

*<sup>φ</sup>*1,*ηη* � *<sup>φ</sup>*1,*ξξ* , bð Þ*<sup>K</sup>* <sup>¼</sup> *κμ*

<sup>3</sup> cosh ð Þþ *<sup>m</sup><sup>η</sup> <sup>C</sup>*<sup>∗</sup>

<sup>¼</sup> *<sup>C</sup>*<sup>∗</sup>

*f*ð Þ¼ *ξ*, *η C*3*e*

*<sup>f</sup>*ð Þ¼ *<sup>ξ</sup>*, *<sup>η</sup> <sup>C</sup>*<sup>∗</sup>

ð Þ<sup>a</sup> *<sup>D</sup>* <sup>¼</sup> *κμ*

receive the following form:

fulfill the following

*h*2

**180**

*h*2 0

or

So

**Figure A1.**

*Parabolic coordinate system.*

*X* ¼ *C*<sup>1</sup> cosð Þþ *mξ C*<sup>2</sup> sin ð Þ *mξ* , *E* ¼ *C*3*e*

*Solid State Physics - Metastable, Spintronics Materials and Mechanics of Deformable…*

�*m<sup>η</sup>* ð Þð Þ *<sup>C</sup>*<sup>1</sup> cosð Þþ *<sup>m</sup><sup>ξ</sup> <sup>C</sup>*<sup>2</sup> sin ð Þ *<sup>m</sup><sup>ξ</sup>*

<sup>4</sup> sinh ð Þ *mη :*

<sup>4</sup> sinh ð Þ *<sup>m</sup><sup>η</sup>* ð Þ *<sup>C</sup>*<sup>1</sup> cosð Þþ *<sup>m</sup><sup>ξ</sup> <sup>C</sup>*<sup>2</sup> sin ð Þ *<sup>m</sup><sup>ξ</sup>* ,

*h*2 0

*u* ¼ *φ*,*<sup>ξ</sup>* � *κφ*1*ξ*, v ¼ *φ*,*<sup>η</sup>* þ *κφ*1*η:* (B4)

then Eqs. (2a) and (2b) will be satisfied identically, while Eqs. (2c) and (2d) will

ð Þ<sup>a</sup> *<sup>u</sup>*,*<sup>ξ</sup>* <sup>þ</sup> v,*<sup>η</sup>* <sup>¼</sup> ð Þ *<sup>κ</sup>* � <sup>2</sup> *<sup>φ</sup>*1,*ηη* � *<sup>φ</sup>*1,*ξξ* , bð Þ v,*<sup>ξ</sup>* � *<sup>u</sup>*,*<sup>η</sup>* <sup>¼</sup> *κ φ*1,*ηξ* <sup>þ</sup> *<sup>φ</sup>*1,*ξη* , (B2) ð Þ<sup>a</sup> *<sup>u</sup>*,*<sup>ξ</sup>* <sup>þ</sup> v,*<sup>η</sup>* <sup>¼</sup> ð Þ *<sup>κ</sup>* � <sup>2</sup> *<sup>φ</sup>*1,*ηη* � *<sup>φ</sup>*1,*ξξ* , bð Þð Þ <sup>v</sup> � *κφ*1*<sup>η</sup>* ,*<sup>ξ</sup>* <sup>¼</sup> *<sup>u</sup>* <sup>þ</sup> *κφ*<sup>1</sup> ð Þ*<sup>ξ</sup>* ,*<sup>η</sup>:* (B3)

From equation (B3b) imply that exists such type harmonic function *φ*, for which

<sup>¼</sup> <sup>2</sup> *<sup>φ</sup>*1,*ξξ* � *<sup>φ</sup>*1,*ηη :* (B5)

Considering (B4), from Equation (B3a), the following will be obtained:

<sup>Δ</sup>*<sup>φ</sup>* <sup>¼</sup> *<sup>φ</sup>*,*ξξ* <sup>þ</sup> *<sup>φ</sup>*,*ηη* <sup>¼</sup> *κφ*<sup>1</sup> <sup>þ</sup> *κφ*1,*ξξ* � *κφ*<sup>1</sup> � *κφ*1,*ηη* <sup>þ</sup> ð Þ *<sup>κ</sup>* � <sup>2</sup> *<sup>φ</sup>*1,*ηη* � *<sup>φ</sup>*1*ξξ*

*<sup>m</sup><sup>η</sup>* <sup>þ</sup> *<sup>C</sup>*4*<sup>e</sup>*

�*mη*

*<sup>φ</sup>*1,*ηξ* <sup>þ</sup> *<sup>φ</sup>*1,*ξη* , (B1)

$$
\overline{u} = \left(\frac{\xi^2 - \eta^2}{2}\rho\_{1,\xi} + \xi\eta\rho\_{1,\eta}\right) - \xi\eta\rho\_{1,\eta} - (\kappa - 1)\rho\_1\xi + \psi\_1,
$$

$$
\overline{\mathbf{v}} = \left(\frac{\xi^2 - \eta^2}{2}\rho\_{1,\eta} - \xi\eta\rho\_{1,\xi}\right) + \xi\eta\rho\_{1,\xi} + (\kappa - 1)\rho\_1\eta + \psi\_2.
$$

Without losing the generality, the expression in brackets can be taken as zero, because we already have in *u* and v of the solutions Laplacian (we mean *ψ*<sup>1</sup> and *ψ*2). Therefore, the solutions of system (2) are given in the following form:

$$\begin{aligned} \mathbf{(a)} \quad h\_0^2 D &= \kappa \mu \left( \rho\_{1,\eta} \eta - \rho\_{1,\xi} \xi \right), & \mathbf{(b)} \, h\_0^2 \mathbf{K} &= \kappa \mu \left( \rho\_{1,\eta} \xi + \rho\_{1,\xi} \eta \right), \\ \mathbf{(c)} \overline{\boldsymbol{\mathfrak{u}}} &= -\xi \eta \rho\_{1,\eta} - (\kappa - 1) \rho\_1 \xi + \nu\_1, & \mathbf{(d)} \overline{\boldsymbol{\mathfrak{v}}} &= \xi \eta \rho\_{1,\xi} + (\kappa - 1) \rho\_1 \eta + \nu\_2. \end{aligned} \tag{B7}$$

Now we have to write down three versions of *ψ*<sup>1</sup> and *ψ*<sup>2</sup> function representation. In the first version

$$\psi\_1 = \overline{\varphi}\_{1,\eta} + \tilde{\rho}\_{1,\eta} + \rho\_{2,\eta}, \qquad \qquad \psi\_2 = \overline{\varphi}\_{1,\xi} + \tilde{\rho}\_{1,\xi} + \rho\_{2,\xi}, \tag{\text{B8}}$$

*φ*1, *φ*~1, *φ*<sup>2</sup> are harmonic functions; in addition, *φ*1, *φ*~<sup>1</sup> are selected so that at *η* ¼ *α*, where *α* ¼ *η*<sup>1</sup> or *α* ¼ *η*2, the following equations will be satisfied:

$$-\xi\eta\rho\_{1,\eta} - (\kappa - 1)\rho\_1\xi + \overline{\rho}\_{1,\eta} + \tilde{\rho}\_{1,\eta} = 0,\qquad \xi\eta\rho\_{1,\xi} + (\kappa - 1)\rho\_1\xi + \overline{\rho}\_{1,\xi} + \tilde{\rho}\_{1,\xi} = 0,$$

In the second version

$$\begin{aligned} \mathcal{Y}\_1 &= -a \left( \frac{\xi^2 - (\eta - a)^2}{2} \varrho\_{1,\xi} + \xi \eta \rho\_{1,\eta} \right) + \frac{\xi^2 - \eta^2}{2} \varrho\_{2,\xi} + \xi \eta \rho\_{2,\eta}, \\ \mathcal{Y}\_2 &= a \left( \xi \eta \rho\_{1,\xi} - \frac{\xi^2 - (\eta - a)^2}{2} \varrho\_{1,\eta} \right) + \frac{\xi^2 - \eta^2}{2} \varrho\_{2,\eta} - \xi \eta \rho\_{2,\xi}, \end{aligned} \tag{B9}$$

where *φ*<sup>2</sup> is the harmonic function. In the third version

$$\begin{split} \psi\_{1} &= -\alpha^{2} \left( \frac{\xi^{2} - \eta^{2}}{2} \rho\_{1,\xi} + \xi \eta \rho\_{1,\eta} \right) + \frac{\xi^{2} - \eta^{2}}{2} \rho\_{2,\xi} + \xi \eta \rho\_{2,\eta}, \\ \psi\_{2} &= \alpha^{2} \left( \xi \eta \rho\_{1,\xi} - \frac{\xi^{2} - \eta^{2}}{2} \rho\_{1,\eta} \right) + \frac{\xi^{2} - \eta^{2}}{2} \rho\_{2,\eta} - \xi \eta \rho\_{2,\xi}. \end{split} \tag{B10}$$

Inserting (B8) in (B7c and d), we will get

$$\begin{aligned} \mathbf{(a)}\overline{\boldsymbol{\mathfrak{u}}} &= -\xi\eta\rho\_{1,\eta} - (\kappa - \mathbf{1})\rho\_1\xi + \overline{\rho}\_{1,\eta} + \tilde{\rho}\_{1,\eta} + \rho\_{2,\eta}, \\ \mathbf{(b)}\overline{\boldsymbol{\mathfrak{v}}} &= \xi\eta\rho\xi + (\kappa - \mathbf{1})\rho\_1\xi + \overline{\rho}\_{1,\xi} + \tilde{\rho}\_{1,\xi} + \rho\_{2,\xi}. \end{aligned} \tag{B11}$$

*Finding such ξ*<sup>1</sup> ¼ *ξ*1*<sup>k</sup>, for which F*1*k=F*<sup>10</sup> <*ε.*

*2D Elastostatic Problems in Parabolic Coordinates DOI: http://dx.doi.org/10.5772/intechopen.91057*

*ε*>*F*1*<sup>k</sup>=F*<sup>10</sup> > *F*1*k*þ<sup>1</sup>*=F*<sup>10</sup> >*F*1*k*þ<sup>2</sup>*=F*<sup>10</sup> > … is valid, too.

**Author details**

Natela Zirakashvili

**183**

University, Tbilisi, Georgia

Distance *<sup>l</sup>* between surfaces *<sup>ξ</sup>* <sup>¼</sup> *<sup>ξ</sup>*<sup>1</sup> and *<sup>ξ</sup>* <sup>¼</sup> <sup>~</sup>*ξ*1, which gives the guarantee for condition *F*1*k=F*<sup>10</sup> <*ε* to be valid in the parabolic coordinate system, will be taken along the axis of the parabola , and the following expression will be obtained:

q

is purposeful to admit that *l=c* ¼ 4, 5, 6, … , which allows finding *ξ*<sup>1</sup> from the relevant equation. Let us note that when *l=c* ¼ 4, we will denote value *ξ*<sup>1</sup> by *ξ*11, when *l=c* ¼ 5; by *ξ*12, when *l=c* ¼ 6; by *ξ*13, etc. If after selecting *ξ*<sup>1</sup> ¼ *ξ*1*k*, inequality *F*1*k=F*<sup>10</sup> <*ε* is valid; in order to check the righteousness of the selection, it is necessary to once again make sure that, together with condition *F*1*k=F*<sup>10</sup> <*ε*, condition

I. Vekua Institute of Applied Mathematics of Iv. Javakhishvili Tbilisi State

© 2020 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium,

\*Address all correspondence to: natzira@yahoo.com

provided the original work is properly cited.

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi *<sup>l</sup>=<sup>c</sup>* <sup>þ</sup> <sup>~</sup>*<sup>ξ</sup>* 2 1

By relying on the known solutions of the relevant plain problems of elasticity, it

*:*

*ξ*<sup>1</sup> ¼

Inserting (B9) in (B7c and d), we will have

$$\begin{aligned} \mathbf{b}(\mathbf{a})\overline{\boldsymbol{u}} &= -a \left( \frac{\overline{\boldsymbol{\xi}}^2 - (\boldsymbol{\eta} - a)^2}{2} \boldsymbol{\rho}\_{1,\overline{\boldsymbol{\xi}}} + \xi \boldsymbol{\eta} \boldsymbol{\rho}\_{1,\overline{\boldsymbol{\eta}}} \right) - \xi \boldsymbol{\eta} \boldsymbol{\rho}\_{1,\overline{\boldsymbol{\eta}}} - (\boldsymbol{\kappa} - 1) \boldsymbol{\rho}\_1 \overline{\boldsymbol{\xi}} + \frac{\overline{\boldsymbol{\xi}}^2 - \boldsymbol{\eta}^2}{2} \boldsymbol{\rho}\_{2,\overline{\boldsymbol{\xi}}} + \xi \boldsymbol{\eta} \boldsymbol{\rho}\_{2,\overline{\boldsymbol{\eta}}}, \\ \mathbf{b}(\mathbf{b})\overline{\boldsymbol{v}} &= a \left( \xi \boldsymbol{\eta} \boldsymbol{\rho}\_{1,\overline{\boldsymbol{\xi}}} - \frac{\overline{\boldsymbol{\xi}}^2 - (\boldsymbol{\eta} - a)^2}{2} \boldsymbol{\rho}\_{1,\overline{\boldsymbol{\eta}}} \right) + \xi \boldsymbol{\eta} \boldsymbol{\rho}\_{1,\overline{\boldsymbol{\xi}}} + (\boldsymbol{\kappa} - 1) \boldsymbol{\rho}\_1 \boldsymbol{\eta} + \frac{\overline{\boldsymbol{\xi}}^2 - \boldsymbol{\eta}^2}{2} \boldsymbol{\rho}\_{2,\overline{\boldsymbol{\eta}}} - \xi \boldsymbol{\eta} \boldsymbol{\rho}\_{2,\overline{\boldsymbol{\xi}}}. \end{aligned} \tag{B12}$$

Inserting (B10) in (B7c and d), we will get

$$\begin{aligned} \mathbf{b}(\mathbf{a})\overline{\boldsymbol{\mathfrak{u}}} &= -\alpha^2 \left( \frac{\xi^2 - \eta^2}{2} \boldsymbol{\varrho}\_{1,\xi} + \xi \boldsymbol{\eta} \boldsymbol{\varrho}\_{1,\eta} \right) - \xi \boldsymbol{\eta} \boldsymbol{\varrho}\_{1,\eta} - (\kappa - 1) \boldsymbol{\varrho}\_1 \xi + \frac{\xi^2 - \eta^2}{2} \boldsymbol{\varrho}\_{2,\xi} + \xi \boldsymbol{\eta} \boldsymbol{\varrho}\_{2,\eta}, \\ \mathbf{b}(\mathbf{b})\overline{\boldsymbol{\mathfrak{v}}} &= \alpha^2 \left( \xi \boldsymbol{\eta} \boldsymbol{\varrho}\_{1,\xi} - \frac{\xi^2 - \eta^2}{2} \boldsymbol{\varrho}\_{1,\eta} \right) + \xi \boldsymbol{\eta} \boldsymbol{\varrho}\_{1,\xi} + (\kappa - 1) \boldsymbol{\varrho}\_1 \boldsymbol{\eta} + \frac{\xi^2 - \eta^2}{2} \boldsymbol{\varrho}\_{2,\eta} - \xi \boldsymbol{\eta} \boldsymbol{\varrho}\_{2,\xi}. \end{aligned} \tag{B13}$$
