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User:Egm6936.f09/Vorticity equation

Vorticity equation for an incompressible flow is obtained when the curl of the Navier-Stokes equation, Eq.(15), is taken. The vorticity equation is given as

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$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} = \nabla \mathbf u \cdot \boldsymbol \omega + \nu \nabla^2 \boldsymbol \omega $$     (1a)
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In component form,

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$$  \displaystyle
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\frac{D \omega_i}{Dt} = \frac{\partial u_i}{\partial x_j} \omega_j + \nu \nabla^2 \omega_i $$     (1b)
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where the component form of the gradient of the velocity is given in Eq.(21), which might be new for especially fluid mechanics people. To get used to two different conventions, the reader is referred to the article Gradient of vector: Two tensor conventions. In Eq.(1b), $$ \displaystyle \nu $$ is the kinematic viscosity. Vorticity equation is used to have a more convenient explanation of the physical phenomena. The left hand side represents the change of the vorticity and the second term on the right hand side is the diffusion of the vorticity. The first term on the right hand side represents the change of vorticity due to the vortex-line stretching (see Meaning of vortex-line stretching). Vortex lines are the axis of rotation and they are tangent to the local vorticity vector. Vorticity vector reflects the rotation of the fluid particles or the circulation and is defined as

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$$  \displaystyle \boldsymbol \omega := \nabla \times \mathbf u $$ (1c)
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The vorticity video (Experiments were carried out by A. H. Shapiro) from National Committee for Fluid Mechanics Films (CFM), 1969, which can be found in the open-access collection of education videos and film notes by the NCFM is very instructive to understand the meaning of the terms vorticity, rotation, and circulation. See also Schwarzchild 2010 on vortex lines in turbulent superfluid.

Examples 2 and 3 in Tritton (1988, p.82) are highly instructive and explain the distinction of the rotation, which is changing orientation in space and closed path motion (this type of flow is called irrotational). Example 2 illustrates solid body rotation, an example of closed path motion. Here, the streamlines are curved. However, the flow is irrotational and thus does not have vorticity. Example 3 is laminar flow in a channel with straight vertical walls. The only component of the velocity is in the axial direction, $$\displaystyle v_x $$, which vanishes on the wall at $$\displaystyle  y=0 $$ due to no-slip condition and this axial velocity depends solely on the lateral position, i.e. $$\displaystyle  v_x = v_x(y) $$. Each fluid element on one lateral position stays on the same lateral position along the axis of the flow. Therefore, the streamlines are straight. However, as the magnitude of the velocity is larger at some lateral position compared to another position, e.g. $$\displaystyle v_x(y=y_1)>v_x(y=y_2) $$ where $$\displaystyle  y_1>y_2 $$, the fluid particle rotates while flowing down the channel and the flow is rotational. Therefore, the flow does not have to be curved to have vorticity (see Fig. 6.8, p.83, Tritton, 1988 ). Vorticity is used to better understand the characteristics of some flows such as turbulence. Vorticity or an irregular distribution of the vortices is one of the main characteristics of turbulence as explained in the Film Notes for Turbulence by R.W. Stewart 1969.

Now, the aim is to derive the vorticity equation and then explain the meaning of the vortex-line stretching.

Derivation of the vorticity equation
This derivation does not use any vector identities. It is built on the first principles. In other words, it is implicitly deriving the vector identities used in the derivation Using vector identities. Therefore, it is longer than that section.

We will first derive the Navier-Stokes (NS) equation from the balance of the linear momentum equation, which is

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$$  \displaystyle \rho \frac {D \mathbf u}  {Dt} =  {\rm div} \boldsymbol \tau +  \mathbf b   \Longrightarrow \rho \frac {D u_j} {Dt} \mathbf e_j =  \frac {\partial \tau_{ij}} {\partial x_i} \mathbf e_j +  b_j \mathbf e_j $$     (1)
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Here, $$\displaystyle \rho $$ is the density, $$\displaystyle \mathbf u $$ the velocity field, $$\displaystyle \boldsymbol \tau $$ the stress tensor, $$\displaystyle \mathbf b $$ body force, For constant property Newtonian fluid

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$$  \displaystyle \boldsymbol \tau =  - p   \mathbf I   + 2 \mu \boldsymbol \gamma (\mathbf u) $$ ||  (2)
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Here, $$\displaystyle \mathbf I $$ is the identity tensor, $$\displaystyle \mu $$ the dynamic viscosity, and $$\displaystyle \boldsymbol \gamma $$ the strain rate defined as

$$\displaystyle \boldsymbol \gamma := \frac{1}{2} \left[ \frac{\partial u_i} {\partial x_j} + \frac{\partial u_j} {\partial x_i} \right ] $$

Note about the minus sign in front of the pressure term. $$\displaystyle p>0 $$, since stress is compressive and thus negative. The stress tensor in component form can be written as

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$$  \displaystyle \tau_{ij} =     -p \delta_{ij} +      \mu \left[ \frac{\partial u_i} {\partial x_j} + \frac{\partial u_j} {\partial x_i} \right ] $$     (3)
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The stress tensor can be decomposed into the spherical (isotropic) part, denoted as $$\displaystyle \mathbf s $$, and the deviatoric part, $$\displaystyle \mathbf d $$ as

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$$  \displaystyle \boldsymbol \tau =  \mathbf s +    \mathbf d $$ (4)
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and

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$$  \displaystyle \mathbf s = \bar \tau  \mathbf I $$ (5)
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where $$\displaystyle \bar \tau $$ is the mean stress defined in terms of the trace of the stress tensor, $$\displaystyle Tr (\boldsymbol \tau) $$, as

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$$ \displaystyle \bar \tau = \frac{1}{3}  \tau_{ii} = \frac{1}{3} Tr (\boldsymbol \tau) $$ (6)
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$$ \displaystyle Tr (\boldsymbol \tau) = \tau_{ii} = -p \underbrace{\delta_{ii}}_{=3} + 2 \mu \underbrace{\frac{\partial u_i} {\partial x_i}}_{= {\rm div} \mathbf u} $$ (7)
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For an incompressible flow: $$\displaystyle {\rm div} \mathbf u = \mathbf 0 $$. Then,

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$$\displaystyle Tr (\boldsymbol \tau) = -3p $$ (8)
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$$\displaystyle p = -\frac{1}{3} Tr (\boldsymbol \tau) $$ (9)
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Thus $$\displaystyle p $$ is the mean stress $$\displaystyle \bar \tau $$ with a minus sign. Hence,

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$$\displaystyle \mathbf s = -p \mathbf I $$ (10)
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$$\displaystyle \mathbf d = \boldsymbol \tau - \mathbf s = 2 \mu   \boldsymbol \gamma (\mathbf u) $$ (11)
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Body force $$\displaystyle \mathbf b $$ : Assume $$\displaystyle \mathbf b $$ is the gradient of a potential $$\displaystyle \Psi $$ (like gravitational potential), then $$\displaystyle \mathbf b = - \rho \, {\rm grad} \Psi $$. Using Eq.(2), Eq.(1) becomes

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$$  \displaystyle \rho \frac {D \mathbf u}  {Dt} =  {\rm div} (-p \mathbf I)  + 2 \mu \, {\rm div} \boldsymbol \gamma (\mathbf u) - \rho \, {\rm grad} \Psi $$ (12)
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But,

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$$  \displaystyle {\rm div} (p \mathbf I) = \left ( \frac{\partial } {\partial x_i} p \, \delta_{ij} \right ) \mathbf e_j = \frac{\partial p} {\partial x_j} \mathbf e_j = {\rm grad}\, p  $$ (13)
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Now, let’s focus on the $$\displaystyle {\rm div} \boldsymbol \gamma (\mathbf u) $$ term.

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$$  \displaystyle {\rm div} \boldsymbol \gamma (\mathbf u) = \frac{\partial } {\partial x_i} \gamma_{ij} \, \mathbf e_j = \frac{1}{2} \frac{\partial } {\partial x_i} \left ( \frac{\partial u_i} {\partial x_j} + \frac{\partial u_j} {\partial x_i} \right) \mathbf e_j = \frac{1}{2} \left [ \frac{\partial } {\partial x_j} \underbrace{ \left(\frac{\partial u_i} {\partial x_i} \right) }_{ {\rm div} \mathbf u = \mathbf 0} + \frac{\partial^2 u_j} {\partial x_i \partial x_i}
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\right] \mathbf e_j = \frac{1}{2} \nabla^2 \mathbf u

$$ (14)
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Therefore, for an incompressible flow and assuming constant properties

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$$  \displaystyle \rho \frac {D \mathbf u}  {Dt} =  -{\rm grad} \,  P     + \mu \nabla^2 \mathbf u $$ (15)
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where the modified pressure is defined as $$\displaystyle P = p + \rho \Psi $$. Eq.(15) is known as Navier-Stokes (NS) equation. We will take the curl of the NS equation to arrive to the vorticity transport equation, given as Eq.(2.60) (Pope, 2000, p.126)

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$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} = \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u + \nu \nabla^2 \boldsymbol \omega $$ (16)
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Here, the gradient of the velocity vector is represented with a right arrow on the nabla operator, which is expressed in component form as

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$$  \displaystyle \overset{\rightarrow}{\nabla} \mathbf u = \frac{\partial u_i}{\partial x_j} {\mathbf e_j} \otimes {\mathbf e_i} $$ (16b)
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This definition is the transpose of our definition given in Eq.(21). To have a clear understanding of the distinctions between the two component forms of the gradient of a vector, the reader is referred to the article Gradient of vector: Two tensor conventions. In Eq.(16), $$\displaystyle \boldsymbol \omega $$ is the vorticity, defined as $$\displaystyle \nabla \times \mathbf u = e_{ijk} \partial_i u_j \mathbf e_k $$. Note that our definition of the gradient is the transpose of Pope’s, the vorticity equation in terms of our gradient is given in Eq.(1).

The curl of the NS equation, i.e., Eq.(15), becomes

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$$  \displaystyle \nabla \times \frac {D \mathbf u}  {Dt} =  \nabla \times \left( -\frac {1}{\rho} {\rm grad} \, P \right) +  \nabla \times \left(  \mu \nabla^2 \mathbf u \right) $$ (18)
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Let’s first focus on the left hand side, using the definition of the material time derivative

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$$  \displaystyle \nabla \times \frac {D \mathbf u}  {Dt} =  \nabla \times \left( \nabla \mathbf u \cdot \mathbf  u \right) +  \nabla \times \frac{\partial \mathbf u} {\partial t}
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$$ (19)
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Here, the last terms is

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$$  \displaystyle \nabla \times \frac{\partial \mathbf u} {\partial t} = \frac{\partial \boldsymbol \omega} {\partial t}
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$$ (20)
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Note that

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$$  \displaystyle \nabla \mathbf u = \frac{\partial u_i} {\partial x_j} \mathbf e_i \otimes \mathbf e_j $$ (21)
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and

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$$  \displaystyle \nabla \mathbf u \cdot \mathbf u = \frac{\partial u_i} {\partial x_j} u_j \mathbf e_i $$ (22)
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So, the second term coming from the material time derivative is

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$$  \displaystyle \nabla \times \left( \nabla \mathbf u \cdot \mathbf  u \right) = e_{ijk} \partial_i \left[ (\partial_p u_j) u_p \right] \mathbf e_k = e_{ijk} \left[ (\partial_i \partial_p u_j) u_p + (\partial_p u_j)( \partial_i u_p) \right] \mathbf e_k $$ (23)
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Let

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$$  \displaystyle \boldsymbol \alpha = e_{ijk} (\partial_i \partial_p u_j) u_p  \mathbf e_k = u_p \partial_p \underbrace{ ( e_{ijk} \partial_i u_j) \mathbf e_k}_{\nabla \times \mathbf u = \boldsymbol \omega} $$ (24)
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which gives

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$$  \displaystyle \boldsymbol \alpha = \nabla \boldsymbol \omega \cdot \mathbf u
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$$ (25)
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Let

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$$  \displaystyle \boldsymbol \beta = e_{ijk} (\partial_p u_j)( \partial_i u_p) \mathbf e_k $$ (26)
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To facilitate the writing, let’s define Jacobian as

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$$  \displaystyle \mathbf J = {\rm grad} \ \mathbf u = \frac{\partial u_j} {\partial x_i} \mathbf e_j\otimes \mathbf e_i $$ (27)
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Or, in component form

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$$  \displaystyle J_i^j = \frac{\partial u_j} {\partial x_i} $$ (28)
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Here, the superscript represents the row, and the subscript the column index. Also, let

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$$  \displaystyle \mathbf A = A_i^j \mathbf e_j\otimes \mathbf e_i = \mathbf J^2 $$ (29)
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Then, $$\displaystyle \boldsymbol \beta $$ can be rewritten as

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$$  \displaystyle \boldsymbol \beta = e_{ijk} A_i^j \mathbf e_k = \mathbf e_1 (A_2^3 - A_3^2) + \mathbf e_2 (A_3^1 - A_1^3) + \mathbf e_3 (A_1^2 - A_2^1) $$ (30)
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We can write the first component of $$\displaystyle \boldsymbol \beta $$ in terms of the Jacobian as

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$$  \displaystyle A_2^3 = \frac{\partial u_3} {\partial x_p} \frac{\partial u_p} {\partial x_2} = J_p^3 J_2^p = J_1^3 J_2^1 + J_2^3 J_2^2 + J_3^3 J_2^3 $$ (31)
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$$  \displaystyle A_3^2 = \frac{\partial u_3} {\partial x_p} \frac{\partial u_p} {\partial x_2} = J_p^2 J_3^p = J_1^2 J_3^1 + J_2^2 J_3^2 + J_3^2 J_3^3 $$ (32)
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Then,

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$$  \displaystyle A_2^3 - A_3^2 = \frac{\partial u_3} {\partial x_p} \frac{\partial u_p} {\partial x_2} = J_1^3 J_2^1 - J_3^1 J_1^2 + (J_2^2 + J_3^3)( J_2^3 - J_3^2) $$ (33)
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Similar expressions can be written for the other components of $$\displaystyle \boldsymbol \beta $$. To obtain the vorticity equation, we want to relate $$\displaystyle \boldsymbol \beta $$ to $$\displaystyle \boldsymbol \gamma $$, which is given as $$\displaystyle \nabla \mathbf u \cdot \boldsymbol \omega$$. Let’s first write the vorticity in ters of the Jacobian, then we will obtain $$\displaystyle \boldsymbol \gamma $$.

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$$  \displaystyle \boldsymbol \omega = \nabla \times \mathbf u = \underbrace{ (\partial_ 2 u_3 - \partial_3 u_2)}_{ J_2^3 - J_3^2} \mathbf e_1 + \underbrace{ (\partial_3 u_1 - \partial_1 u_3)}_{ J_3^1 J_1^3}\mathbf e_2 + \underbrace{ (\partial_1 u_2 - \partial_2 u_1) }_{ J_1^2 J_2^1} \mathbf e_3 $$ (34)
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$$  \displaystyle \boldsymbol \gamma = \nabla \mathbf u \cdot \boldsymbol \omega = \left( \frac{\partial u_i} {\partial x_j} \mathbf e_i\otimes \mathbf e_j \right) \cdot (\omega_k \mathbf e_k) = \frac{\partial u_i} {\partial x_j} \omega_j \mathbf e_i = \underbrace{ \frac{\partial u_i} {\partial x_j}}_{J_j^i} \left(e_{klj} \partial_k u_l \right) \mathbf e_i = a_i \mathbf e_i $$ (35)
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The components of $$\displaystyle \boldsymbol \gamma $$ can be calculated using matrix algebra as

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$$  \displaystyle \begin{vmatrix} J_1^1 & J_2^1 & J_3^1 \\ J_1^2 & J_2^2 & J_3^2 \\ J_1^3 & J_2^3 & J_3^3 \\ \end{vmatrix} \begin{vmatrix} J_2^3 - J_3^2 \\ J_3^1 - J_1^3 \\ J_1^2 - J_2^1 \\ \end{vmatrix}
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$$ (36)
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Let’s compute the first component of $$\displaystyle \boldsymbol \gamma $$, i.e., $$\displaystyle a_1 $$

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$$  \displaystyle a_1 = J_1^1 (J_2^3 - J_3^2) + J_2^1 (J_3^1 - J_1^3) + J_3^1 (J_1^2 - J_2^1) $$ (37)
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But, $$\displaystyle J_1^1 = -(J_2^2+J_3^3) $$ as $$\displaystyle J_i^i = 0 = {\rm div} \mathbf u $$. Therefore,

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$$  \displaystyle a_1 = -(J_2^2+J_3^3) (J_2^3 - J_3^2) + J_3^1 J_1^2 - J_2^1 J_1^3 $$ (38)
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Now, we can relate the components of $$\displaystyle \boldsymbol \beta $$ to those of $$\displaystyle \boldsymbol \gamma $$ as

 :{| style="width:100%" border="0" $$  \displaystyle A_2^3 - A_3^2 = - a_1 $$ (39)
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Similarly

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$$  \displaystyle A_3^1 - A_1^3 = - a_2 $$ (40)
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and

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$$  \displaystyle A_1^2 - A_2^1 = - a_3 $$ (41)
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Hence,

 :{| style="width:100%" border="0" $$  \displaystyle \boldsymbol \beta = - \boldsymbol \gamma = -\nabla \mathbf u \cdot \boldsymbol \omega $$ (42)
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whose component form is given in Eq.(35). Now, we will work with the right hand side of the NS. The first terms is the pressure one

 :{| style="width:100%" border="0" $$  \displaystyle \nabla \times (\nabla P) = \mathbf 0 $$ (43)
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and as the density is constant

 :{| style="width:100%" border="0" $$  \displaystyle \nabla \times (\frac {1}{\rho} \nabla P) = \mathbf 0 $$ (44)
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Next,

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$$  \displaystyle \nabla \times (\nabla^2 \mathbf u) = {\rm curl} ({\rm div} ({\rm grad^T}\mathbf u)) = e_{ijk} \partial_i (\partial_p \partial_p u_j) \mathbf e_k = \partial_p \partial_p \underbrace{(e_{ijk} \partial_i u_j \mathbf e_k )}_{ \nabla \times \mathbf u = \boldsymbol \omega} = \nabla^2 \boldsymbol \omega $$ (45)
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As the viscosity and the density are constant

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$$  \displaystyle \nabla \times (\frac{\mu}{\rho} \nabla^2 \mathbf u) = \nu \nabla^2 \boldsymbol \omega $$ (46)
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Combining all terms, we obtain the evolution equation for the vorticity for incompressible flows as

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$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} = \nabla \mathbf u \cdot \boldsymbol \omega + \nu \nabla^2 \boldsymbol \omega $$ (47)
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The vorticity equation in component form is

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$$  \displaystyle \frac{D \omega_i}{Dt} = \frac{\partial u_i}{\partial x_j} \omega_j + \nu \nabla^2 \omega_i $$ (48)
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Here, the first term is the time rate of change of vorticity and it represents the vorticity production when it is positive. $$\displaystyle \nabla \mathbf u \cdot \boldsymbol \omega $$ is the vorticity production due to vortex-line stretching and $$\displaystyle  \nu \nabla^2 \boldsymbol \omega $$ is the vorticity diffusion.

An alternative method
The Jacobian $$\displaystyle \mathbf J $$ can be decomposed as

 :{| style="width:100%" border="0" $$  \displaystyle J_j^i = \underbrace{\frac{1}{2}(J_j^i + J_i^j)}_{S_j^i} - \underbrace{\frac{1}{2}(J_i^j - J_j^i)}_{\Omega_j^i} $$ (49)
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So,

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$$  \displaystyle \mathbf J = \mathbf S + \boldsymbol \Omega
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$$ (50)
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where $$\displaystyle \mathbf S $$ is a small strain tensor and $$\displaystyle  \boldsymbol  \Omega$$ is a small rotation tensor. The vorticity can be expressed as

 :{| style="width:100%" border="0" $$  \displaystyle \boldsymbol \omega = \nabla \times \mathbf u = e_{ijk} \underbrace{\partial_i u_j}_{J_i^j} \mathbf e_k = \underbrace{(J_2^3 - J_3^2)}_{2 \Omega_2^3} \mathbf e_1 + \underbrace{(J_3^1 - J_1^3)}_{2 \Omega_3^1} \mathbf e_2 + \underbrace{(J_1^2 - J_2^1)}_{2 \Omega_1^2} \mathbf e_3 = 2 \Omega_i^j \mathbf e_k $$ (51)
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Let’s define an extraction operator, $$\displaystyle \zeta$$. When it operates on the Jacobian it extracts the pseudo vector $$\displaystyle \boldsymbol \omega $$ as

 :{| style="width:100%" border="0" $$  \displaystyle \zeta \mathbf J = e_{ijk} J_i^j \mathbf e_k = \boldsymbol \omega $$ (52)
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 :{| style="width:100%" border="0" $$  \displaystyle \zeta \mathbf J = \zeta \mathbf S + \zeta \boldsymbol \Omega = \zeta \boldsymbol \Omega = \boldsymbol \omega $$ (53)
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Note that as $$\displaystyle \mathbf  S$$  is symmetric $$\displaystyle  \zeta \mathbf S = \mathbf  0$$. Our goal is to find an expression for $$\displaystyle \boldsymbol \beta  $$, which is expressed as

 :{| style="width:100%" border="0" $$  \displaystyle \boldsymbol \beta = e_{ijk} \underbrace{(\partial_p u_j)}_{J_p^j} \underbrace{(\partial_i u_p)}_{J_i^p} \mathbf e_k = \zeta \mathbf J^2 $$ (54) and
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 :{| style="width:100%" border="0" $$  \displaystyle \mathbf J^2 = (\mathbf S + \boldsymbol \Omega) (\mathbf S + \boldsymbol \Omega) = \mathbf S^2 + \boldsymbol \Omega^2 + \boldsymbol \Omega \mathbf S + \mathbf S \boldsymbol \Omega $$ (55)
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Here,

 :{| style="width:100%" border="0" $$  \displaystyle (\mathbf S^2)^T = \mathbf S^T \mathbf S^T = \mathbf S^2 $$ (56)
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 *

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 :{| style="width:100%" border="0" $$  \displaystyle (\boldsymbol \Omega^2)^T = \boldsymbol \Omega^T \boldsymbol \Omega^T = (-\boldsymbol \Omega) (-\boldsymbol \Omega) = \boldsymbol \Omega^2 $$ (57)
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As both $$\displaystyle \mathbf S^2 $$ and $$\displaystyle \boldsymbol \Omega^2 $$ are symmetric

 :{| style="width:100%" border="0" $$  \displaystyle \zeta \mathbf S^2 = \zeta \boldsymbol \Omega^2 = \mathbf 0 $$ (58)
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 *

 * }

 :{| style="width:100%" border="0" $$  \displaystyle (\boldsymbol \Omega \mathbf S)^T = \mathbf S^T \boldsymbol \Omega^T = \mathbf S (-\boldsymbol \Omega) = - \mathbf S \boldsymbol \Omega \Longrightarrow \mathbf S \boldsymbol \Omega = - (\boldsymbol \Omega \mathbf S)^T $$ (59)
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Eq.(59) implies

 :{| style="width:100%" border="0" $$  \displaystyle \boldsymbol \Omega \mathbf S + \mathbf S \boldsymbol \Omega = \boldsymbol \Omega \mathbf S - (\boldsymbol \Omega \mathbf S)^T = \mathbf S \boldsymbol \Omega - (\mathbf S \boldsymbol \Omega)^T $$ (60)
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Now, we can use the extraction operator

 :{| style="width:100%" border="0" $$  \displaystyle \zeta (\mathbf S \boldsymbol \Omega) = e_{ijk}S_p^j \underbrace{\Omega_i^p}_{\frac{1}{2}e_{ipq}\omega_q} \mathbf e_k = \frac{1}{2}\underbrace{e_{ijk}e_{ipq}}_{\delta_{jp}\delta_{kq}-\delta_{jq}\delta_{kp}} S_p^j \omega_q \mathbf e_k = \frac{1}{2}\left[\underbrace{S_j^j}_{=0}\omega_k - \underbrace{S_k^j}_{=S_j^k}\omega_j \right] \mathbf e_k = -\frac{1}{2} \mathbf S \cdot \boldsymbol \omega $$ (61)
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 :{| style="width:100%" border="0" $$  \displaystyle \zeta (\mathbf S \boldsymbol \Omega)^T = \underbrace{ e_{ijk}}_{-e_{jik}} S_p^i \Omega_j^p \mathbf e_k = -\frac{1}{2}\underbrace{e_{jik}e_{jpq}}_{\delta_{ip}\delta_{kq}-\delta_{iq}\delta_{kp}} S_p^i \omega_q \mathbf e_k = -\frac{1}{2}\left[\underbrace{S_i^i}_{=0}\omega_k - \underbrace{S_k^i}_{=S_i^k}\omega_i \right] \mathbf e_k = \frac{1}{2} \mathbf S \cdot \boldsymbol \omega $$ (62)
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Collecting the terms,

 :{| style="width:100%" border="0" $$  \displaystyle \beta = \zeta \mathbf J^2 = \zeta \left[ (\mathbf S \boldsymbol \Omega) - (\mathbf S \boldsymbol \Omega)^T \right] = -\mathbf S \cdot \Omega $$ (63)
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But, $$\displaystyle \mathbf S \cdot \boldsymbol \Omega = \underbrace{(\mathbf S + \boldsymbol \Omega)}_{\mathbf J}\cdot \boldsymbol \omega $$ as $$\displaystyle  \boldsymbol \Omega \boldsymbol \omega = \boldsymbol \omega \times \boldsymbol \omega = \mathbf 0$$. Therefore,

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$$  \displaystyle \beta = -\mathbf J \cdot \omega = -\nabla \mathbf u \cdot \boldsymbol \omega $$ (64)
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This completes the proof.

Using vector identities
Here, we use the vector calculus identities to derive the vorticity equation. It should be noted that the gradient used in the vector calculus identities is the form used in Eq.(16b), which is the transpose of the gradient, given in Eq.(21), used in this article. The reader is referred to the article Gradient of vector: Two tensor conventions for a clear understanding of the two conventions.

The curl of the Navier-Stokes equation, i.e. Eq.(15), gives

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$$  \displaystyle \nabla \times  \frac {D \mathbf u}  {Dt} = \nabla \times \left( -\frac{1}{\rho} \nabla P \right) + \nabla \times \left(\nu \nabla^2 \mathbf u \right) $$     (65)
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Let’s first focus on the left hand side

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$$  \displaystyle \nabla \times  \frac {D \mathbf u}  {Dt} = \nabla \times  \left(\frac{\partial \mathbf u}{\partial  t}\right) + \nabla \times \left(\mathbf u \cdot \overset{\rightarrow}{\nabla} \mathbf u \right) $$     (66)
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But, the partial derivative in Eq.(66) can be written as

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$$  \displaystyle \nabla \times  \left(\frac{\partial \mathbf u}{\partial  t}\right) = \frac{\partial}{\partial t} \underbrace{\left (\nabla \times \mathbf u \right)}_{ \boldsymbol \omega} $$     (67)
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Also, using one of the vector dot product identities

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$$  \displaystyle \mathbf{u} \cdot (\overset{\rightarrow}{\nabla} \mathbf{u}) = \frac{1}{2}\nabla \left(\mathbf u \cdot \mathbf u \right) - \mathbf u \times \underbrace{\left (\nabla \times \mathbf u \right)}_{ \boldsymbol \omega} $$     (68)
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Now, using one of the vector cross product identities

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$$  \displaystyle \nabla \times \left(\mathbf u \times \boldsymbol \omega \right) = \mathbf u \underbrace{\left( \nabla \cdot \boldsymbol \omega \right)}_{0} - \boldsymbol \omega \underbrace{\left( \nabla \cdot \mathbf u \right)}_{0} + \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u - \mathbf u \cdot \overset{\rightarrow}{\nabla}\boldsymbol \omega $$     (69)
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The first term on the right of Eq.(69) is zero because it is the divergence of a curl, and the second term is zero because of the incompressibility condition. As the curl of the gradient of a scalar field vanishes

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$$  \displaystyle \nabla \times \left( \frac{1}{2}\nabla \left(\mathbf u \cdot \mathbf u \right) \right) = \mathbf 0 $$     (70)
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Grouping all terms on the left hand side

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$$  \displaystyle \nabla \times  \frac{D \mathbf u}{Dt} = \frac{\partial \boldsymbol \omega}{\partial t} - \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u + \mathbf u \cdot \overset{\rightarrow}{\nabla}\boldsymbol \omega = \frac{D \boldsymbol \omega}{Dt} - \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u $$ (71)
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The first term on the right hand side of Eq.(65) vanishes as in Eq.(70), i.e.,

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$$  \displaystyle \nabla \times \left( -\frac{1}{\rho} \nabla P \right) = \mathbf 0 $$     (72)
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Finally, the last term in Eq.(65) is

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$$  \displaystyle \nabla \times \left(\nu \nabla^2 \mathbf u \right) = \nu \nabla^2 \left(\nabla \times \mathbf u \right) = \nu \nabla^2 \boldsymbol \omega $$     (73)
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Combining the left and the right hand sides, we obtain

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$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} - \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u = \nu \nabla^2 \boldsymbol \omega $$  (74)
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Meaning of vortex-line stretching
Vortex-line is a line along the vorticity vector $$\displaystyle \mathbf \omega $$ at a point $$\displaystyle x $$. In a uniform flow field, $$\displaystyle \nabla \mathbf u = \mathbf 0$$, therefore two points on an infinitesimally small vorticity line have the same velocity and the line stays undistorted. However, in a non-uniform field the vortex line is stretched as different velocities are assumed at different points of the line. Therefore, $$\displaystyle \nabla \mathbf u \cdot \boldsymbol \omega $$ represents the vorticity production by vortex-line stretching.

Evolution equation for an infinitesimally small line element
The mapping at time $$\displaystyle t $$ between the spatial position $$\displaystyle \mathbf x $$ and the material position $$\displaystyle \mathbf X$$ is given with

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$$  \displaystyle \mathbf x = \phi (\mathbf X,t) = \phi_t (\mathbf X) $$ (75)
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Then a small displacement with series approximation can be written as,

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$$  \displaystyle \mathbf x + \mathbf h = \phi_t(\mathbf X+ \mathbf H) = \phi_t(\mathbf X) + \frac{\partial \phi(\mathbf X)}{\partial X_k} H_k + h.o.t = \underbrace{ \phi_t(\mathbf X)}_{ \mathbf x} + \underbrace{\left [\rm grad_X \phi_t(\mathbf X) \right] \cdot  \mathbf  H(\mathbf X)}_{\mathbf  h} $$ (76)
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The material time derivative is defined at constant material point as

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$$  \displaystyle
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\frac{D \mathbf h(\mathbf x, t)}{Dt} = \left. \frac{d \mathbf h(\mathbf x, t)}{dt} \right|_{X \ fixed} = \left[ \rm grad_X \underbrace{ \frac{\partial \phi (\mathbf X, t) }{\partial t } }_{\mathbf V(\mathbf X, t)= \mathbf v(\mathbf x, t) } \right] $$     (77)
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Note that the material description of the velocity field, $$ \displaystyle \mathbf V (\mathbf X, t) $$ is equal to the spatial description of the velocity field, $$ \displaystyle \mathbf v (\mathbf x, t) $$, i.e., $$ \displaystyle \mathbf V (\mathbf X, t) =\mathbf v (\mathbf x, t) $$ and using the chain rule

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$$  \displaystyle \frac{\partial \mathbf v}{\partial x_i} = \frac{\partial \mathbf V}{\partial X_k} \frac{\partial X_k}{\partial x_i} $$     (78)
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To write the gradient of the velocity field with material description in terms of that for spatial description, first the definition for the gradient is introduced as

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$$  \displaystyle \rm grad_X \mathbf V = \frac{\partial V_i}{\partial X_j} \mathbf e_i \otimes \mathbf e_j $$     (79)
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Again, using the chain rule,

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$$  \displaystyle \frac{\partial V_i}{\partial X_j} = \underbrace{\frac{\partial V_i}{\partial x_p}}_{\frac{\partial v_i}{\partial x_j}} \underbrace{\frac{\partial x_p}{\partial X_j}}_{F_j^p} $$     (80)
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Now, the gradient of the material description of the velocity vector can be expressed in terms of the gradient of the spatial description of the velocity vector as

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$$  \displaystyle \rm grad_X \mathbf V = \rm grad_x \mathbf v \cdot \mathbf F $$ (81)
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where $$ \displaystyle \mathbf F $$ is defined in Eq.(80). Recall that the material or the spatial descriptions of the velocity fields are equal and can be expressed as

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$$  \displaystyle \mathbf V = V_i \mathbf E_i = \mathbf v= v_i \mathbf e_i $$     (82)
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The basis vectors are equal

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$$  \displaystyle \mathbf e_i = \mathbf E_i \, \ i=1,2,3 $$     (83)
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Hence, the components of the velocities are equal to each other as

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$$  \displaystyle v_i(\mathbf x,t) = V_i (\mathbf X,t) $$     (84)
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The term $$ \displaystyle \mathbf F $$ in Eq.(80) can be written in component form as

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$$  \displaystyle \mathbf F = F_j^i \mathbf e_i \otimes \mathbf E_j $$     (85)
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Combining Eq.(79) and Eq.(83), the gradient of the material description of the velocity field is

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$$  \displaystyle \nabla_X \mathbf V = \frac{\partial V_i}{\partial X_j} \mathbf E_i \otimes \mathbf E_j $$     (86)
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and the gradient of the spatial description of the velocity field is

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$$  \displaystyle \nabla_x \mathbf v = \frac{\partial v_i}{\partial x_j} \underbrace{ \mathbf e_i }_{\mathbf E_i} \otimes \; \mathbf e_j $$     (87)
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Then,

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$$  \displaystyle \nabla_X \mathbf V \cdot \mathbf H = \left(\nabla_x \mathbf v \cdot \mathbf F \right) \cdot \mathbf H = \nabla_x(\mathbf v) \cdot \underbrace{\left(\mathbf F \cdot \mathbf H \right) }_{\mathbf h} $$ (88)
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Hence, a small change in the spatial position is expressed as

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$$  \displaystyle \frac{D \mathbf h}{Dt} = \left( \nabla_x \mathbf v \right) \cdot \mathbf h $$ (89)
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The corresponding expression in Pope (2000, p.14, Eq.(2.16)) is given as

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$$  \displaystyle \frac{d \mathbf s}{dt} = \mathbf s \cdot \nabla \mathbf u $$ (90)
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and the component form is given as Pope (2000, p.13, Eq.(2.12))

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$$  \displaystyle \underbrace{\mathbf s \cdot \nabla \mathbf u}_{\nabla \mathbf u \cdot \mathbf s \ \rm our \; \rm notation} = s_i \frac{ \partial u_j}{ \partial x_i} \mathbf e_j $$     (91)
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= References =

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm6936.f09/Vorticity equation


text:
User:Manueldias

Cours 1

Introduction aux instruments de musique du 17ème siècle

Cours 2

Découverte des principaux compositeurs

Cours 3

Analyse de partitions du 17ème siècle

title: User:Manueldias


text:
User:Egm6321.f10.team2.zou/HW4

= Problem 1: Show Exactness of Second Order Linear Ordinary Differential Equations with Varying Coefficients (L2_ODE_VC) =

From the lecture slide Mtg 22-3

Given
For the following equation:
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$$ \displaystyle
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F\left(x,y,y',y\right)=\left(cos x \right)y+\left(x^{2}-sin x\right)y'+ 2xy=0

$$ (eq 1.1)
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(1) Show that Eq. 1.1 is exact

(2) Find the first integral,Φ, and

(3) Solve for y(x)

(1) Proof of Exactness
To be exact, the equation must satisfy two conditions.

1st Condition of Exactness
First, the equation must be presented in the following format
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$$ \displaystyle
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F\left(x,y,y',y\right)=g\left(x,y,y'\right)+f\left(x,y,y'\right)y

$$ (eq 1.2) By letting y' = p, the first condition of exactness is satisfied as shown in the following equation:
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$$ \displaystyle
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F\left(x,y,p,y\right)=\underbrace{\left(x^{2}-sin x \right)p+ 2xy}_{g\left(x,y,p\right)}+\underbrace{\left(cos x \right)}_{f\left(x,y,p\right)}y

$$

(eq 1.3)
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2nd Condition of Exactness
In order to meet the second condition of exactness, the following equations must be satisfied:

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$$ \displaystyle
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f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}

$$

(eq 1.4)
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$$ \displaystyle
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f_{xp}+pf_{yp}+2f_{y}=g_{pp}

$$

(eq 1.5) Each element of equations 1.4 and 1.5 are calculated to be the following:
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$$ \displaystyle f_{xx}=-cos\left(x\right)$$
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$$\displaystyle f_{xy}=f_{yy}=f_{y}=f_{xp}=f_{yp}=0$$

$$\displaystyle g_{xp}=2x-cos\left(x\right)$$

$$\displaystyle g_{y}=2x$$

$$\displaystyle g_{yp}=g_{pp}=0 $$

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Plugging each relevant element into equation 1.4 yields the following:
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$$ \displaystyle
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-cos\left(x\right)+2p\cancel{f_{xy}}^{0}+p^{2}\cancel{f_{yy}}^{0}=2x-cos\left(x\right)+p\cancel{g_{yp}}^{0}-2x $$

$$-cos\left(x\right)=-cos\left(x\right)$$ (eq 1.6)
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Plugging each relevant element into equation 1.5 yields the following:
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$$ \displaystyle
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\cancel{f_{xp}}^{0}+p\cancel{f_{yp}}^{0}+2\cancel{f_{y}}^{0}=\cancel{g_{pp}}^{0} $$

$$ \displaystyle 0=0$$ (eq 1.7) Equations 1.4 and 1.5 have been satisfied and therefore the equation is exact.
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(2) Solving for the First Integral,Φ
The first integral is defined by the following:

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$$ \displaystyle
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\phi=h\left(x,y\right)+\underbrace{\int \underbrace{f\left(x,y,p\right)}_{cos\left(x\right)}\,dp}_{pcos\left(x\right)}

$$

(eq 1.8)
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From equation 1.8 the following equations can be derived:
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$$ \displaystyle
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\phi_{x}=h_{x}+	\frac{\partial}{\partial x}\left[pcos\left(x\right)\right]=h_{x}-psin\left(x\right)

$$

(eq 1.9)
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$$ \displaystyle
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\phi_{y}=h_{y}+	\frac{\partial}{\partial y}\left[pcos\left(x\right)\right]=h_{y}

$$

(eq 1.10) Using the definition of g to solve for h, yields the following:
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$$ \displaystyle
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g\left(x,y,p\right)=\phi_{y}p+\phi_{x}=h_{y}p+\left[h_{x}-psin\left(x\right)\right]

$$

$$g\left(x,y,p\right)=\left[h_{y}-sin\left(x\right)\right]p+h_{x}$$

(eq 1.11) By equating equation 1.11 with the g(x,y,p) shown in equation 1.3, the partial derivatives of h can be solved as shown:
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$$ \displaystyle
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\left[h_{y}-sin\left(x\right)\right]p+h_{x}=\left[x^{2}-sin\left(x\right)\right]p+2xy $$ (eq 1.12)
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$$\displaystyle h_{y}=x^{2}$$ (eq 1.13)
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$$\displaystyle h_{x}=2xy$$
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(eq 1.14) Next, equations 1.13 and 1.14 can be integrated separately to solve for h(x,y) as follows:
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$$\displaystyle h=\int h_{x}\,dx +w\left(y\right)=x^{2}y+ w\left(y\right)$$
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(eq 1.15)
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$$\displaystyle h=\int h_{y}\,dy +v\left(x\right)=x^{2}y+ v\left(x\right)$$
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(eq 1.16)
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Equating equations 1.15 and 1.16 show that
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$$\displaystyle h=x^{2}y$$ (eq 1.17) Therefore, equation 1.8 can be rewritten as:
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$$ \displaystyle
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\phi=x^{2}y+{{y}'}cos\left(x\right)+k_{1}

$$

(eq 1.18)
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(3) Solving for y(x)
When combined with the definition of the first integral:
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$$ \displaystyle
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\phi=k_{2} $$

(eq 1.19)
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Equation 1.18 can be rewritten as:

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$$ \displaystyle
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\phi=x^{2}y+cos\left(x\right)y'=K $$

Where $$\displaystyle K=k_{2}-k_{1}$$ (eq 1.20) The equation is divided by cos(x) in order to find an integrating factor:
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$$ \displaystyle
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y'+\frac{x^{2}}{cos\left(x\right)}y=\frac{K}{cos\left(x\right)} $$

(eq 1.21)
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Multiply the entire equation by $$e^{\int \frac{x^{2}}{cos(x)}\,dx}$$ yields:
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$$ \displaystyle
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e^{\int \frac{x^{2}}{cos(x)}\,dx}y'+\frac{x^{2}}{cos\left(x\right)}e^{\int \frac{x^{2}}{cos(x)}\,dx}y=e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{K}{cos\left(x\right)} $$

(eq 1.22) The left hand side of the equation can be condensed to the following:
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$$ \displaystyle
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(e^{\int \frac{x^{2}}{cos(x)}\,dx}y)'=e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{K}{cos\left(x\right)} $$

(eq 1.23) Integrating both sides with respect to x gives:
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$$ \displaystyle
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e^{\int \frac{x^{2}}{cos(x)}\,dx}y=K\int e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{1}{cos\left(x\right)}\,dx $$

(eq 1.24) Dividing both sides by $$e^{\int \frac{x^{2}}{cos(x)}\,dx}$$ yields the solution for y:
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$$ \displaystyle
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y=Ke^{-\int \frac{x^{2}}{cos(x)}\,dx}\int e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{1}{cos\left(x\right)}\,dx $$

(eq 1.25)
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 * }

= Problem 2: Exactness of Bessel Equation =

From the lecture slide Mtg 24-1

Given
Bessel equation has the form as follows :

 * {| style="width:100%" border="0"

$$ \displaystyle {{x}^{2}}{y}''+x{y}'+({{x}^{2}}-{{\nu }^{2}})y=0$$ (Eq 2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle \nu $$ is some positive real number.

Find

 * i)Verify the exactness of Bessel Equation Eq 2.1 using method of Mtg15-3

 * ii)Veryfy the exactness of Bessel Equation Eq 2.1 using method of Mtg22-4

 * iii)If Bessel equation Eq 2.1 can't be made exact, try integrating factor method with the factor of $$ \displaystyle {{x}^{m}}{{y}^{n}}$$ where $$ \displaystyle m$$ and $$ \displaystyle n$$ are all constants.

i) verifying exactness by method given by Mtg15-3
By Eq 2.1 we define,

 * $$\begin{align}

& f\equiv {{x}^{2}} \\ & g\equiv x{y}'+({{x}^{2}}-{{\nu }^{2}})y \\ \end{align}$$

thus Eq 2.1 can be written as,

 * {| style="width:100%" border="0"

$$ \displaystyle f{y}''+g=0$$ (Eq 2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

which satisfy the first condition of exactness in method of Mtg15-3.

From the above equation we derive,

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & {{f}_{xx}}=2 \\ & {{f}_{xy}}=0 \\ & {{f}_{yy}}=0 \\ & {{f}_{xp}}=0 \\ & {{f}_{yp}}=0 \\ & {{f}_{y}}=0 \\ & {{g}_{xp}}=1 \\ & {{g}_{yp}}=0 \\ & {{g}_{pp}}=0 \\ & {{g}_{y}}={{x}^{2}}-{{\nu }^{2}} \\ \end{align}$$ (Eqs 2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute Eqs 2.3 into (1)15-3 and (2)15-3 for the second condition of exactness, we derive,

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{matrix} {{x}^{2}}-{{\nu }^{2}}+1=0 \\ 0=0 \\ \end{matrix}$$ (Eq 2.4)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

which can't be satisfied because $$ \displaystyle x$$ is a variable while $$ \displaystyle \nu $$ is a constant. Thus The original Bessel Equation is not exact.

ii) verifying exactness by method given by Mtg24-4
Define according to the Bessel Equation,

 * {| style="width:100%" border="0"

$$ \displaystyle F\equiv {{x}^{2}}{y}''+x{y}'+({{x}^{2}}-{{\nu }^{2}})y$$ so as for the form the Eq 2.1 satisfy the first condition in method of Mtg22-2.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We derive from the previous definition that,

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & \frac{\partial F}{\partial y}={{x}^{2}}-{{\nu }^{2}} \\ & \frac{\partial F}{\partial {y}'}=x \\ & \frac{d}{dx}\left( \frac{\partial F}{\partial {y}'} \right)=1 \\ & \frac{\partial F}{\partial {y}''}={{x}^{2}} \\ & \frac{d{{x}^{2}}}\left( \frac{\partial F}{\partial {y}''} \right)=2 \\ \end{align}$$ (Eqs 2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We learn from Mtg22-4 that the second condition of this method with respect to 2nd order ODE is,

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial F}{\partial y}+\frac{d}{dx}\left( \frac{\partial F}{\partial {y}'} \right)+\frac{d{{x}^{2}}}\left( \frac{\partial F}{\partial {y}''} \right)=0$$ (Eq 2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting Eqs 2.5 into Eq 2.6 we have,

 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle {{x}^{2}}-{{\nu }^{2}}+1=0$$ (Eq 2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This equation won't be satisfied for the same reason as Eq 2.4. Thus second method also shows that Bessel Equation is not exact.

iii) the trial of IFM method
Define integrating factor $$ \displaystyle h$$ as ,

 * $$\displaystyle h\equiv {{x}^{m}}{{y}^{n}}$$

where $$ \displaystyle m$$ and $$ \displaystyle n$$ are some constants.

Then we multiple $$ \displaystyle h$$ to both sides of Eq 2.1 changing it as,

 * {| style="width:100%" border="0"

$$ \displaystyle {{F}_{1}}={{x}^{m+2}}{{y}^{n}}{y}''+{{x}^{m+1}}{{y}^{n}}{y}'+({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{n+1}}=0$$ (Eq 2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This of course satisfy the first condition of exactness of method of Mtg22-4.

Then we have,

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & \frac{\partial {{F}_{1}}}{\partial y}=n{{x}^{m+2}}{{y}^{n-1}}{y}''+n{{x}^{m+1}}{{y}^{n-1}}{y}'+(n+1){{x}^{m+2}}{{y}^{n}}-(n+1){{\nu }^{2}}{{x}^{m}}{{y}^{n}} \\ & \frac{\partial {{F}_{1}}}{\partial {y}'}={{x}^{m+1}}{{y}^{n}} \\ & \frac{d}{dx}\left( \frac{\partial {{F}_{1}}}{\partial {y}'} \right)=(m+1){{x}^{m}}{{y}^{n}} \\ & \frac{\partial {{F}_{1}}}{\partial {y}''}={{x}^{m+2}}{{y}^{n}} \\ & \frac{d{{x}^{2}}}\left( \frac{\partial {{F}_{1}}}{\partial {y}''} \right)=(m+2)(m+1){{x}^{m}}{{y}^{n}} \\ \end{align}$$ (Eqs 2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then substitute Eq 2.9 into Eq 2.6, we get,

 * {| style="width:100%" border="0"

$$ \displaystyle n{{x}^{m+2}}{{y}^{n-1}}{y}''+n{{x}^{m+1}}{{y}^{n-1}}{y}'+{{x}^{m}}{{y}^{n}}\left[ (n+1)({{x}^{2}}-{{\nu }^{2}})+{{(m+1)}^{2}} \right]=0$$ (Eq 2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since this equation is supposed to be free from $$ \displaystyle {{y}}$$, thus we must eliminate terms with $$ \displaystyle {{y}}$$, so we derive,

 * {| style="width:100%" border="0"

$$ \displaystyle n=0$$ (Eq 2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and then,

 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle ({{x}^{2}}-{{\nu }^{2}})+{{(m+1)}^{2}}=0$$ (Eq 2.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

which can't be satisfied because $$ \displaystyle x$$ is a variable but $$ \displaystyle m$$ and $$ \displaystyle \nu$$ are all constants. This means the recipe of Mtg19 doesn't make Bessel Equation exact.

iiii) another perspective of explanation
To illustrate our ideas in another way, we can firstly assume that Eq 2.1 is exact and see whether we can find any $$ \displaystyle \Phi $$ such that,

 * $$\displaystyle \frac{d\Phi }{dx}=F$$

Since we know that ,

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & f={{x}^{2}}={{\Phi }_{p}} \\ & g=x{y}'+({{x}^{2}}-{{\nu }^{2}})y={{\Phi }_{x}}+{{\Phi }_{y}}{y}' \\ \end{align}$$ (Eq 2.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$ \displaystyle p$$ denotes $$ \displaystyle {{y}'}$$.

we can write $$ \displaystyle \Phi $$ as,

 * {| style="width:100%" border="0"

$$ \displaystyle \Phi =H(x,y)+{{x}^{2}}p$$ (Eq 2.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and from that we get,

 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} & {{\Phi }_{x}}={{H}_{x}}+2xp \\ & {{\Phi }_{y}}={{H}_{y}} \\ \end{align}$$ (Eq 2.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting Eq 2.15 back into Eq 2.13 we get,

 * $$\displaystyle {{H}_{x}}+{{H}_{y}}p+xp=({{x}^{2}}-{{\nu }^{2}})y$$

or

 * {| style="width:100%" border="0"

$$ \displaystyle {{H}_{x}}=-{{H}_{y}}p-xp+({{x}^{2}}-{{\nu }^{2}})y$$ (Eq 2.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

while we always have,

 * {| style="width:100%" border="0"

$$ \displaystyle dH={{H}_{x}}dx+{{H}_{y}}dy$$ (Eq 2.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If we substitute Eq 2.16 into Eq 2.17, we derive,

 * {| style="width:100%" border="0"

$$ \displaystyle dH=({{x}^{2}}-{{\nu }^{2}})ydx-xdy$$ (Eq 2.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can see that the term $$ \displaystyle {{H}_{y}}$$ is cancelled and there is no $$ \displaystyle {{H}_{x}}$$ left either which means the function $$\displaystyle H$$ always have the form of Eq 2.18 regardless of what you assume $$ \displaystyle {{H}_{x}}$$ or $$ \displaystyle {{H}_{y}}$$ to be. This also answers for the "2 choices" tricks of Mtg18 and the last problem of HW3("Finishing the story").

Besides that, we know that the prerequisite of the existense of $$ \displaystyle H$$ is that the RHS of Eq 2.18 is exact, meaning,

 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle \begin{align} & \frac{\partial ({{x}^{2}}-{{\nu }^{2}})y}{\partial y}=\frac{\partial (-x)}{\partial x} \\ & \Rightarrow {{x}^{2}}-{{\nu }^{2}}=-1 \\ \end{align}$$ (Eq 2.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Unfortunately Eq 2.19 is not satisfied thus this $$\displaystyle H$$ never exist and so is $$ \displaystyle \Phi $$. Therefore equations from Eq 2.13 to Eq 2.19(except Eq 2.17) they all become meaningless and we conclude that Bessel Equation is non-exact. In fact those equations are all based on the assumption that "Eq 2.1 is exact". If we know Eq 2.1 is non-exact in advance, then the effort to find $$ \displaystyle H$$ is predestined to fail.

One may argue that the integrating factor may turn Eq 2.1 Bessel Equation into exact one.

From Eq 2.8 we difine,

 * $$\displaystyle \begin{align}

& {{f}_{1}}\equiv {{x}^{m+2}}{{y}^{n}} \\ & {{g}_{1}}\equiv {{x}^{m+1}}{{y}^{n}}{y}'+{{x}^{m+2}}{{y}^{n+1}}-{{\nu }^{2}}{{x}^{m+1}}{{y}^{n+1}} \\ \end{align}$$

in which,

 * $$\begin{align}

& {{f}_{1}}={{\Phi }_{P}} \\ & {{g}_{1}}={{\Phi }_{x}}+{{\Phi }_{y}}{y}' \\ \end{align}$$

So we can write $$ \displaystyle \Phi $$ in such form:

 * {| style="width:100%" border="0"

$$ \displaystyle {{\Phi }_{}}={{H}_{1}}+{{x}^{m+2}}{{y}^{n}}p$$ (Eq 2.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

or

 * {| style="width:100%" border="0"

$$ \displaystyle d\Phi =d{{H}_{1}}+d({{x}^{m+2}}{{y}^{n}}p)$$ (Eq 2.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Note that,

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{d\Phi }{dx}={{F}_{1}}={{x}^{m+2}}{{y}^{n}}{y}''+{{x}^{m+1}}{{y}^{n}}{y}'+({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{n+1}}$$ (Eq 2.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then we derive $$ \displaystyle d{{H}_{1}}$$ without calculating its partial derivatives.

 * {| style="width:100%" border="0"

$$ \displaystyle d{{H}_{1}}=({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{n+1}}dx+\left[ -(m+1){{x}^{m+1}}{{y}^{n}}-np{{x}^{m+2}}{{y}^{n-1}} \right]dy$$ (Eq 2.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since $$ \displaystyle {{H}_{1}}$$ is supposed to be independent of $$ \displaystyle p$$ but integrating according to $$ \displaystyle dy$$ won't make $$ \displaystyle p$$ disappear. This means one may expect the term including $$ \displaystyle p$$ to be eliminated, thus $$ \displaystyle n=0$$.

If $$ \displaystyle {{H}_{1}}$$ really exist, the RHS of Eq 2.23 must be exact, meaning,

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial \left[ ({{x}^{2}}-{{\nu }^{2}}){{x}^{m}}{{y}^{0+1}} \right]}{\partial y}=\frac{\partial \left[ -(m+1){{x}^{m+1}}{{y}^{0}}-0*p{{x}^{m+2}}{{y}^{0-1}} \right]}{\partial x}$$ (Eq 2.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then we have,
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle {{(m+1)}^{2}}+({{x}^{2}}-{{\nu }^{2}})=0$$ (Eq 2.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

which can't be satisfied. We still can't find legitimate $$ \displaystyle m$$ and $$ \displaystyle n$$ to make $$ \displaystyle {{H}_{1}}$$ exist and the same with $$ \displaystyle \Phi$$ which means integrating factor can't save Bessel Equation from being non-exact.

=Problem 3: FindX(x) in terms of sin,cos,sinh,cosh=

From the lecture slide Mtg 25-2

Given

Equation of motion for Euler-Bernoulli Beam given as :

 * {| style="width:100%" border="0"

$$  \displaystyle -EI\frac{\partial^4 u}{\partial x^4}+f=m\frac{\partial^2 u}{\partial t^2} $$     (Eq 3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where

f=disrubuted load

u=transverse displacement

For f=0 Which equals the free vibration condition :

 * {| style="width:100%" border="0"

$$  \displaystyle -EI\frac{\partial^4 u}{\partial x^4}=m\frac{\partial^2 u}{\partial t^2} $$     (Eq 3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Euler thought that this equation has a solution in the form of exponential.

Method of variable separation

Since $$\displaystyle u=u(x,t) $$ we have to use separation of variables method separation of variables to separate u into two functions both depends on its own variable.Then we can conclude that:

 * {| style="width:100%" border="0"

$$  \displaystyle u(x,t)=X(x).T(t) $$     (Eq 3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle -EIX^{(4)}T=mX\ddot{T} $$     (Eq 3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Assuming that

 * {| style="width:100%" border="0"

$$  \displaystyle T(t)=e^{iwt} $$     (Eq 3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \ddot{T}=-w^{2}T $$     (Eq 3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{EI}{mw^{2}}X^{(4)}=X $$     (Eq 3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle K^{4}=\frac{EI}{mw^{2}} $$     (Eq 3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle K^{4}X^{(4)}=X $$     (Eq 3.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Method Of Trial Solution:Assuming that the solution in the form of:

 * {| style="width:100%" border="0"

$$  \displaystyle X(x)=exp(rx) $$     (Eq 3.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

r=unknown coefficient

r's are can be found by solving ths characteristic equation. Since the characteristic equation has a power of 4 we have four roots.

 * {| style="width:100%" border="0"

$$  \displaystyle K^{4}r^{4}=1 $$     (Eq 3.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle r^{4}=\frac{1}{K^{4}} $$     (Eq 3.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle r_{1}=\frac{1}{K} $$     (Eq 3.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle r_{2}=-\frac{1}{K} $$     (Eq 3.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle r_{3}=\frac{i}{K} $$     (Eq 3.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle r_{4}=-\frac{i}{K} $$     (Eq 3.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Show X(x) in terms of sin,cos,sinh,cosh

Solution
If we set up solution

 * {| style="width:100%" border="0"

$$  \displaystyle X(x)=c_{1}exp(\frac{x}{k})+c_{2}exp(-\frac{x}{k})+c_{3}exp(i\frac{x}{k})+c_{4}exp(-i\frac{x}{k}) $$     (Eq 3.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle e^{i\phi }=cos\phi +isin\phi $$     (Eq 3.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle e^{-i\phi }=cos\phi-isin\phi $$     (Eq 3.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we can manupulate equation by part by part.

 * {| style="width:100%" border="0"

$$  \displaystyle c_{3}e^{i\frac{x}{k}}=c_{3}\left [ cos\frac{x}{k}+isin\frac{x}{k} \right ] $$     (Eq 3.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle c_{4}e^{-i\frac{x}{k}}=c_{4}\left [ cos\frac{x}{k}-isin\frac{x}{k} \right ] $$     (Eq 3.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle c_{3}e^{i\frac{x}{k}}+c_{4}e^{-i\frac{x}{k}}=(c_{3}+c_{4})cos\frac{x}{k}+i(c_{3}-c_{4})sin\frac{x}{k} $$     (Eq 3.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle c_{1}exp(\frac{x}{k})=\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{1}exp(\frac{x}{k}) $$     (Eq 3.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle c_{2}exp(-\frac{x}{k})=\frac{1}{2}c_{2}exp(-\frac{x}{k})+\frac{1}{2}c_{2}exp(-\frac{x}{k}) $$     (Eq 3.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle c_{1}exp(\frac{x}{k})+c_{2}exp(-\frac{x}{k})=\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{2}exp(-\frac{x}{k})+\frac{1}{2}c_{2}exp(-\frac{x}{k}) $$     (Eq 3.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Let's add and subtract $$\displaystyle\frac{1}{2}c_{1}exp(-\frac{x}{k})$$ and $$ \displaystyle \frac{1}{2}c_{2}exp(\frac{x}{k}) $$

Hyperbolic forms of sin and cos are given as:

 * {| style="width:100%" border="0"

$$  \displaystyle sinhx=\frac{1}{2}(e^{x}-e^{-x}) $$     (Eq 3.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle coshx=\frac{1}{2}(e^{x}+e^{-x}) $$     (Eq 3.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle c_{1}cosh\frac{x}{k}=\frac{1}{2}c_{1}exp(\frac{x}{k})+\frac{1}{2}c_{1}exp(-\frac{x}{k}) $$     (Eq 3.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle c_{1}sinh\frac{x}{k}=\frac{1}{2}c_{1}exp(\frac{x}{k})-\frac{1}{2}c_{1}exp(-\frac{x}{k}) $$     (Eq 3.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then we can conclude same things for c2 and if we put all terms in the Eq.(3.17)

 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$  \displaystyle X(x)=(c_{1}+c_{2})cosh\frac{x}{k}+(c_{1}-c_{2})sinh\frac{x}{k}+(c_{3}+c_{4})cos\frac{x}{k}+i(c_{3}-c_{4})sin\frac{x}{k}
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 3.30)
 *

 * }

 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$  \displaystyle X(x)=A_{1}cosh\frac{x}{k}+A_{2}sinh\frac{x}{k}+A_{3}cos\frac{x}{k}+A_{4}sin\frac{x}{k}
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 3.31)
 *

 * }

Where

 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$  \displaystyle A_{1}=(c_{1}+c_{2}) ,
 * style="width:95%" |
 * style="width:95%" |

A_{2}=(c_{1}-c_{2}) ,

A_{3}=(c_{3}+c_{4}) ,

A_{4}=i(c_{3}-c_{4})

$$
 *

 * }

=Contributions Of Team Members=

 * Problem 1:    Solved by: Patterson, Rasul

Author: Patterson

Proofread by: Rasul, Dube

 * Problem 2:    Solved by: Zou, Oztekin

Author: Zou, Yaakov, Ismail

Proofread by: Yaakov, Ismail

 * Problem 3:    Solved by: Oztekin

Author: Oztekin, Dube

Proofread by: Dube

= References =

title: User:Egm6321.f10.team2.zou/HW4


text: File:Egm6321.f10.team2.Soundwaveplate.png


text:
User:Abdurramanemane

Title: Geogebra to Mozambican schools

1.	Summary GeoGebra is dynamic mathematics software for schools that joins geometry, algebra, and calculus. On the one hand, GeoGebra is an interactive geometry system. You can do constructions with points, vectors, segments, lines, and conic sections as well as functions while changing them dynamically afterwards. On the other hand, equations and coordinates can be entered directly. Thus, GeoGebra has the ability to deal with variables for numbers, vectors, and points. It finds derivatives and integrals of functions and offers commands like Root or Vertex. These two views are characteristic of GeoGebra: an expression in the algebra window corresponds to an object in the geometry window and vice verse. As a way to contribute to for dynamic teaching and learning of mathematics, introducing a program that encourages the student to like math and still make it easier to understand the various process ranging happening along the lessons, the GeoGebra being a dynamic mathematics software for schools, which joins geometry, algebra and calculus, with demonstrations of real cases shall encourage and promote the teaching and learning of mathematics in Mozambicans schools.

2.	Main objective •	Introduce Geogebra in Mozambicans schools

2.1 Specific objectives •	Study ways and methods for introducing a software (Geogebra) •	Examine to what extent can be an advantage this software. •	Studying way of making the student using this program in mathematics.

3.	Problematization Mozambican schools will be able to work with geogebra and does your usage won't disturb more students and depart the increasingly of mathematics.

3.1	Hypotheses As to the material, the Mozambican schools are in the process of the computerization of their respective computer rooms. With regard to stimulation of students, geogebra will provoke in students an eagerness to learn and demonstrate mathematical processes apply in everyday’s life.

title: User:Abdurramanemane


text:
User:Nhabique-1

Espero seja bem vindo neste lugar!

title: User:Nhabique-1


text:
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title: User talk:Alzac


text:
User talk:Nhabique-1

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title: User talk:Nhabique-1


text:
User:Egm6321.f10.team4/HW 4

= Problem 1 -  Computing $$\phi$$ and y from an exact L2_ODE =

From

Given

 * {| style="width:100%" border="0" align="left"

$$\displaystyle F={cos(x)y''+(x^2-sin(x))p+2xy}=0 $$ $$
 * $$\displaystyle (Eq. 1-1)
 * }
 * }

Find
a) Is F exact?

b) Find $$\phi$$

c) Solve for $$y$$

Solution
Part a

In order for F to be exact, the following two equations need to be satisfied.

 * {| style="width:100%" border="0" align="left"

$$\displaystyle {g_{pp}}=f_{xp}+pf_{yp}+2f_{y} $$ $$
 * $$\displaystyle (Eq. 1-2)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle {f_{xx}}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}+g_{y} $$ $$
 * $$\displaystyle (Eq. 1-3)
 * }
 * }

Before computing the differentials let's first identify $$f$$ and $$g$$

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f(x,y,p)=cos(x) $$ $$
 * $$\displaystyle (Eq. 1-4)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle g(x,y,p)=(x^2-sin(x))p+2xy $$ $$
 * $$\displaystyle (Eq. 1-5)
 * }
 * }

The differentials are as follows

 * {| style="width:100%" border="0" align="left"

$$\displaystyle g_y=2x $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle g_{yp}=0 $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle g_x=2y+(2x-cox(x))p $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle g_{xp}=2x-cos(x) $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_x=-sin(x) $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_{xp}=0 $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_{xy}=0 $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_{xx}=-cos(x) $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle f_y=0 $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle g_{pp}=0 $$
 * }
 * }

If we take the above differentials and replace them in Eq (1-2)

 * {| style="width:100%" border="0" align="left"

$$\displaystyle 0=0 $$
 * }
 * }

thus the first condition of exactness is met. Next, let's take the corresponding differentials and replace them into Eq. (1-3)
 * {| style="width:100%" border="0" align="left"

$$\displaystyle -cos(x)=2x-cos(x)-2x $$
 * }
 * }

therefore

 * {| style="width:100%" border="0" align="left"

$$ 0=0 $$
 * }
 * }

Since both Eq (1-2) and Eq. (1-3) are satisfied, F= EXACT

Part b

We will first identify the different components of a L2_ODE in symbolic form

 * {| style="width:100%" border="0" align="left"

$$\displaystyle {\phi_{p}}=cos(x) $$ $$
 * $$\displaystyle (Eq. 1-6)
 * }
 * }

integrating Eq. (1-6)

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \phi=\int{\phi_{p}}dp=P(x)p+C(x,y) $$ $$
 * $$\displaystyle (Eq. 1-7)
 * }
 * }

Differentiating $$\phi$$ with respect to $$x$$
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \phi_{x}=P'(x)p+\frac{\delta(C)}{\delta(x)} $$ $$
 * $$\displaystyle (Eq. 1-8)
 * }
 * }

and with respect to $$y$$

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \phi_{y}=\frac{\delta(C)}{\delta(y)} $$ $$
 * $$\displaystyle (Eq. 1-9)
 * }
 * }

Now we need to connect the symbols with Eq. (1-1)

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P(x)=cos(x)=\phi_p $$ $$
 * $$\displaystyle (Eq. 1-10)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle Q(x)=P'(x)-\frac{\delta(C)}{\delta(y)}=x^2-sin(x) $$ $$
 * $$\displaystyle (Eq. 1-11)
 * }
 * }

and the last term

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{\delta(C)}{\delta(x)}=R(x)y=(2x)y $$ $$
 * $$\displaystyle (Eq. 1-12)
 * }
 * }

We will now proceed to calculate $$C$$

 * {| style="width:100%" border="0" align="left"

$$\displaystyle C(x,y)=y\int{2\xi}d\xi+K(y)=yx^2+k(y) $$ $$
 * $$\displaystyle (Eq. 1-13)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \phi=cos(x)p+x^2y+k(y) $$ $$
 * $$\displaystyle (Eq. 1-14)
 * }
 * }

But in ordet to get a complete solution we need to identify $$k(y)$$

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_x(x)+C_y(x,y)=Q(x) $$
 * }
 * }

Rearranging

 * {| style="width:100%" border="0" align="left"

$$\displaystyle C_y(x,y)=Q(x)-P_x(x) $$ which is only a function of $$x$$, that leads us to the conclusion that $$k=$$constant. Which will yield the final form of $$\phi$$
 * }
 * }

$$\displaystyle \phi(x,y,p)=cos(x)p+x^2y+k $$ $$
 * $$\displaystyle (Eq. 1-15)
 * }
 * }

Part c

Let's assume $$k=0$$ the

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \phi=cos(x)\frac{dy}{dx}+x^2y=0 $$ $$  Rearranging Eq. (1-16)
 * $$\displaystyle (Eq. 1-16)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle cos(x)y'=-x^2y $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{y'}{y}=-\frac{x^2}{cos(x)} $$ $$
 *
$$\displaystyle (Eq. 1-17)
 * }
 * }

Integrating Eq. (1-17)

 * {| style="width:100%" border="0" align="left"

$$\displaystyle ln(y)=\int{\frac{-x^2}{cos(x)}}dx+C_1 $$ $$
 *
$$\displaystyle (Eq. 1-18)
 * }
 * }

Solving for $$y$$

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(x)=C_2e^{-\int{\frac{-x^2}{cos(x)}}dx} $$ $$
 *
$$\displaystyle (Eq. 1-19)
 * }
 * }

Egm6321.f10.team4.petralanda.n 23:09, 16 October 2010 (UTC)egm6321.f10.team4.petralanda.n Egm6321.f10.team4.ejm 17:20, 20 October 2010 (UTC) - Editor

= Problem 2 : Bessel Equation(L2_ODE_VC) = From Meeting 13, p 24-1

Find
Part 1) Is the given Bessel Equation exact? where,
 * First condition of Exactness:

 * {| style="width:100%" border="0"

$$ \begin{align} F(x,y,p) = f(x,y,p)p' + g(x,y,p)=0 \end{align} $$     (Eq 2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * Second conditions of Exactness:

 * {| style="width:100%" border="0"

$$ \begin{align} f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \end{align} $$     (Eq 2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} f_{xp} + pf_{yp} + 2f_{y} = g_{pp} \end{align} $$     (Eq 2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2) If the Bessel Equation is not exact, find an integrating factor $$\displaystyle h(x,y)$$, if one exists, to make it exact.

Solution
PART 1: Exactness of Bessel Equation

 Applying the first exactness condition

The first condition of exactness for N2-ODE is showing whether the given equation has the following form,

 * {| style="width:100%" border="0"

$$ \begin{align} F(x,y,p) = f(x,y,p)p' + g(x,y,p)=0 \end{align} $$ (Eq 2-2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * where, $$\displaystyle p:=y' $$ and $$\displaystyle p':=y'' $$

above bessel equation can be transformed as follow,

 * {| style="width:100%" border="0"

$$ \begin{align} f(x,y,p)={{x}^{2}} \end{align} $$ (Eq 2-5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\begin{align} g(x,y,p)=x{p}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y \end{align}$$ (Eq 2-6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Hence, the first condition of exactness is satisfied.

 Applying the second exactness condition

Taking derivatives of f(x) yields,

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {f_x = 2x} & {f_{xx} = 2}\\ & {f_{xy} = 0}\\ & {f_{xp} = 0}\\ \\  {f_y  = 0}  & {f_{yy} = 0}\\ & {f_{yp} = 0}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Then, taking derivatives of g(x) yields,

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {g_x = p+2xy} & {g_{xp} = 1}\\ {{g}_{y}}={{x}^{2}}-{{\upsilon }^{2}} & {g_{yp} = 0 }\\ {g_p = x} & {g_{pp}=0}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Substituting the above derivatives into the second exactness conditions(the first and the second equation) yields,

 * {| style="width:100%" border="0"

$$ \begin{align} \left( 2 \right)+ \left(2p \times 0 \right) + \left(p^2 \times 0 \right) = \left(1\right) + \left(p \times 0 \right)- \left( x^2 - {{\upsilon }^{2}} \right) \end{align} $$     (Eq 2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} 2 \neq 1-{x}^{2}+{\upsilon }^{2} \end{align} $$     (Eq 2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * the first equation of the second exactness condition isn't satisfied.

 * {| style="width:100%" border="0"

$$ \begin{align} \left( 0 \right) + \left( p \times 0 \right) + \left( 2 \times 0 \right) = 0 \end{align} $$     (Eq 2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} 0=0 \end{align} $$     (Eq 2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * the second equation of the second exactness condition is satisfied.

Even if the second equation of the second exactness is satisfied, the first equation is not satisfied.

Thus, the Bessel equation(Eq.2.1) is not Exact(N2_ODE).

Alternatively, the exactness (or lack thereof) can be determined using the single relation

 * {| style="width:100%" border="0"

$$ \begin{align} f_0 +  \frac {df_1}{dx}  +  \frac {d^2 f_2}{dx^2}  =  0 \end{align} $$     (Eq 2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting the values $$\displaystyle p = y' $$  and  $$\displaystyle  q = y'' $$, our original equation becomes

 * {| style="width:100%" border="0"

$$ \begin{align} F(x,y,p,q) =  x^2 q + xp + (x^2 - \upsilon^2)y  =  0 \end{align} $$     (Eq 2.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where

 * $$\displaystyle

\begin{array}{*{20}l} f(x,y,p) = x^2, & & & g(x,y,p) = xp + (x^2 - \upsilon^2)y\\ \end{array} $$

Taking the appropriate derivatives

 * $$\displaystyle

\begin{array}{*{20}l} {f_y = 0} & {f_p = 0}\\ {g_y = (x^2 - \upsilon^2)} & {g_p = x}\\ \end{array} $$

and using these derivatives to determine expressions for $$\displaystyle f_0 $$, $$\displaystyle f_1 $$ , and $$\displaystyle f_2 $$

 * {| style="width:100%" border="0"

$$ \begin{align} f_0 =  \frac {dF}{dy^{(0)}}  =  f_y q + g_y  =  0 + (x^2 - \upsilon^2) \end{align} $$     (Eq 2.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} f_1 =  \frac {dF}{dy^{(1)}}  =  f_p q + g_p  =  0 + x \end{align} $$     (Eq 2.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} f_2 =  \frac {dF}{dy^{(2)}}  =  f  =  x^2 \end{align} $$     (Eq 2.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using our values for $$\displaystyle f_0 $$, $$\displaystyle f_1 $$ ,and $$\displaystyle f_2 $$ , in the exactness relation give

 * {| style="width:100%" border="0"

$$ \begin{align} x^2 - \upsilon^2 - 1 + 2 = 0 \end{align} $$     (Eq 2.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} x^2 - \upsilon^2 + 1 \neq 0 \end{align} $$     (Eq 2.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the Bessel Equation fails this exactness condition.

PART 2: Finding an Integrating Factor

Since our equation is in a power form, the integrating factor $$\displaystyle h(x,y) $$ should be in the form

 * {| style="width:100%" border="0"

$$ \begin{align} h(x,y) = x^m y^n \end{align} $$     (Eq 2.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the second equation in the second exactness condition, our integrating factor is multiplied by each of our functions $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$

 * {| style="width:100%" border="0"

$$ \begin{align} h(x,y) f(x,y,p) = (x^m y^n) * x^2 = x^{m + 2} y^n \end{align} $$     (Eq 2.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} h(x,y) g(x,y,p) = (x^m y^n) * ((x^2 - \upsilon^2)y + xp) = x^{m + 2} y^{n + 1} - x^m y^{n + 1} \upsilon^2 + x^{m + 1} y^n p\end{align} $$     (Eq 2.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * $$\displaystyle

\begin{array}{*{20}l} {f_x = (m + 2)x^{m + 1} y^n} & {f_{xp} = 0}\\ {f_y = n x^{m + 2} y^{n - 1}}  & {f_{yp} = 0}\\ {g_p = x^{m + 1} y^n} & {g_{pp} = 0}\\ \end{array} $$

Substituting these values into the second equation for the exactness condition yields

 * {| style="width:100%" border="0"

$$ \begin{align} 0 + p(0) + 2n x^{m + 1} y^n = 0 \end{align} $$     (Eq 2.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} 2n x^{m + 1} y^n = 0 \end{align} $$     (Eq 2.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

In order for this to be true,

 * $$\displaystyle n = 0 $$

Our integrating factor is now

 * $$\displaystyle x^m y^0 = x^m $$

Using this in combination with the first equation for the second exactness condition

 * {| style="width:100%" border="0"

$$ \begin{align} h(x,y) f(x,y,p) = x^m * x^2 = x^{m + 2} \end{align} $$     (Eq 2.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} h(x,y) g(x,y,p) = x^m * ((x^2 - \upsilon^2)y + xp = x^{m + 2} y - x^m y \upsilon^2 + x^{m + 1} p \end{align} $$     (Eq 2.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finding the derivatives of these two equations

 * $$\displaystyle

\begin{array}{*{20}l} {f_x = (m + 2)x^{m + 1} } & {f_{xx} = (m + 2)(m + 1) x^m }\\ & {f_{xy} = 0}\\ {f_y = 0}  & {f_{yy} = 0}\\ {g_x = (m + 2)x^{m + 1} y - m x^{m - 1} y \upsilon^2 + (m + 1)x^m p} & {g_{xp} = (m + 1)x^m}\\ {g_y = x^{m + 2} - \upsilon^2 x^m } & {g_{yp} = 0} \end{array} $$

Substituting these values into our equation yields

 * {| style="width:100%" border="0"

$$ \begin{align} (m + 2)(m + 1)x^m + 2p (0) + p^2 (0) = (m + 1)x^m + p (0) - x^{m + 2} + \upsilon^2 x^m\end{align} $$     (Eq 2.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} (m + 2)(m + 1) = (m + 1) - x^2 + \upsilon^2 \end{align} $$     (Eq 2.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} m^2 + 3m + 2 = m + 1 - x^2 + \upsilon^2 \end{align} $$     (Eq 2.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} m^2 + 2m + 1 = \upsilon^2 - x^2 \end{align} $$     (Eq 2.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} m + 1 = \pm \sqrt{\upsilon^2 - x^2} \end{align} $$     (Eq 2.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$ \begin{align} m = \pm \sqrt{\upsilon^2 - x^2} - 1 \end{align} $$     (Eq 2.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since $$\displaystyle m $$ must be a real number, and not a function, then there is no integrating factor which will make the Bessel Equation exact.

128.227.57.129 21:25, 19 October 2010 (UTC) EGM6321.f10.team4.osentowski - co-author &  Editor

Egm6321.f10.team4.Yoon 09:09, 20 October 2010 (UTC) - Co-Author & Editor

= Problem 3 -  Euler-Bernoulli Beam Equation =

From Meeting 25, p 25-2.

Given
The equation of motion of a Euler-Bernoulli Beam from Meeting 24, p 24-2:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle -EI\frac{\partial^4u }{\partial x^4}+f=m\frac{\partial^2u }{\partial t^2} $$ $$
 *
$$\displaystyle (Eq. 3-1)
 * }
 * }

Find

 * {| style="width:100%" border="0" align="left"

$$\displaystyle X(x) $$
 * }
 * }

In terms of

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \sin, \cos, \sinh, \cosh $$
 * }
 * }

Solution
In order to solve for $$\displaystyle X(x) $$ in terms of $$\displaystyle \sin, \cos, \sinh, \cosh $$, we must make a series of assumptions.

First, let's assume the beam is experiencing free vibration. This would mean the distributive load, $$\displaystyle f $$, in equation 3-1 would be zero.

Thus,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle -EI\frac{\partial^4u }{\partial x^4}=m\frac{\partial^2u }{\partial t^2} $$ $$
 *
$$\displaystyle (Eq. 3-2)
 * }
 * }

If we go on to assume the form of $$\displaystyle u $$ to be

 * {| style="width:100%" border="0" align="left"

$$\displaystyle u(x,t)=X(x)T(t) $$ $$
 *
$$\displaystyle (Eq. 3-3)
 * }
 * }

then substituting equation 3-3 into equation 3-2 and using the method of separation of variables yields:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle -EIX^{(4)}T=mX\ddot{T} $$ $$
 *
$$\displaystyle (Eq. 3-4)
 * }
 * }

Next let's assume $$\displaystyle T $$ to be in the form:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle T(t)=e^{iwt} $$ $$
 *
$$\displaystyle (Eq. 3-5)
 * }
 * }

Taking the the derivatives of equation 3-5 would result in

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \ddot{T}=(iw)^2e^{iwt}=-w^2T $$ $$
 *
$$\displaystyle (Eq. 3-6)
 * }
 * }

Substituting equation 3-6 into equation 3-4:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle -EIX^{(4)}T=-mw^2XT $$ $$
 *
$$\displaystyle (Eq. 3-7)
 * }
 * }

Equation 3-7 can be further reduced to:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{EI}{mw^2}X^{(4)}=X $$ $$
 *
$$\displaystyle (Eq. 3-8)
 * }
 * }

Next we can set the a variable $$\displaystyle K $$ to be to help simply equation 3-8

 * {| style="width:100%" border="0" align="left"

$$\displaystyle K^4:=\frac{EI}{mw^2} $$
 * }
 * }
 * }

equation 3-8 can be rewritten as

 * {| style="width:100%" border="0" align="left"

$$\displaystyle K^4X^{(4)}=X $$ $$
 *
$$\displaystyle (Eq. 3-9)
 * }
 * }

Note: equation 3-9 is a linear, 4th order ordinary differential equation with constant coefficients (L4_ODE_CC)

Using Euler's method of trial solution, one can assume

 * {| style="width:100%" border="0" align="left"

$$\displaystyle X(x)=e^{rx} $$ $$
 *
$$\displaystyle (Eq. 3-10)
 * }
 * }

Taking the the derivatives of equation 3-10 would result in

 * {| style="width:100%" border="0" align="left"

$$\displaystyle X^{(4)}=r^4e^{rx}=r^4X $$ $$
 *
$$\displaystyle (Eq. 3-11)
 * }
 * }

Substituting equation 3-11 into equation 3-9:

 * {| style="width:100%" border="0" align="left"

$$\displaystyle K^4r^4X=X $$ $$
 *
$$\displaystyle (Eq. 3-12)
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Equation 3-12 can be further reduced to:

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$$\displaystyle r^4=\frac{1}{K^4} $$ $$
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$$\displaystyle (Eq. 3-13)
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The 4 roots of $$\displaystyle r $$ are

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$$\displaystyle r=\frac{1}{K},\frac{-1}{K},\frac{i}{K},\frac{-i}{K} $$ where
 * }
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$$\displaystyle i=\sqrt{-1} $$
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$$\displaystyle X $$ can now be written as

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$$\displaystyle X(x)=C_1 e^{\frac{1}{K}x}+C_2 e^{\frac{-1}{K}x}+C_3 e^{\frac{i}{K}x}+C_4 e^{\frac{-i}{K}x} $$ $$ where $$\displaystyle C_1, C_2, C_3, $$ and $$\displaystyle C_4 $$ are constants.
 *
$$\displaystyle (Eq. 3-14)
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First, let's use Euler's formula to convert the exponential terms with imaginary powers to $$\displaystyle \sin $$ and $$\displaystyle \cos $$ terms:

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$$\displaystyle e^{ix} = \cos x + i\sin x $$ $$
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$$\displaystyle (Eq. 3-15)
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Equation 3-14 can now be written as:

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$$\displaystyle X(x)=C_1e^{\frac{1}{K}x}+C_2e^{\frac{-1}{K}x}+C_3\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right]+C_4\left[\cos {\left(-\frac{1}{K}x\right)} + i\sin {\left(-\frac{1}{K}x\right)}\right] $$ $$
 *
$$\displaystyle (Eq. 3-16)
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Using Trigonometric Symmetry Identities, Equation 3-16 can be rewritten as:

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$$\displaystyle X(x)=C_1e^{\frac{1}{K}x}+C_2e^{\frac{-1}{K}x}+C_3\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right]+C_4\left[\cos {\left(\frac{1}{K}x\right)} - i\sin {\left(\frac{1}{K}x\right)}\right] $$ $$
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$$\displaystyle (Eq. 3-17)
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The remaining exponential terms can be converted to $$\displaystyle \sinh $$ and $$\displaystyle \cosh $$ terms using properties of Hyperbolic functions:

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$$\displaystyle e^x = \cosh x + \sinh x $$ $$
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$$\displaystyle (Eq. 3-18)
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and

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$$\displaystyle e^{-x} = \cosh x - \sinh x $$ $$
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$$\displaystyle (Eq. 3-19)
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Thus Equation 3-17 can be written as:

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$$\displaystyle X(x)=C_1 \left[\cosh {\left(\frac{1}{K}x\right)} + \sinh {\left(\frac{1}{K}x\right)}\right]+C_2\left[\cosh {\left(\frac{1}{K}x\right)} - \sinh {\left(\frac{1}{K}x\right)}\right]+C_3\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right]+C_4\left[\cos {\left(\frac{1}{K}x\right)} + i\sin {\left(\frac{1}{K}x\right)}\right] $$ $$
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$$\displaystyle (Eq. 3-20)
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Finally, combing the like trigonometric terms, we can write:

$$\displaystyle X(x)=\left(C_1+C_2 \right) \cosh \left(\frac{1}{K}x\right) + \left(C_1-C_2 \right) \sinh \left(\frac{1}{K}x\right)+ \left(C_3+C_4 \right) \cos \left(\frac{1}{K}x\right)+\left(C_3-C_4 \right) i\sin \left(\frac{1}{K}x\right) $$ $$
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$$\displaystyle (Eq. 3-21)
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Egm6321.f10.team4.Auerbach 04:27, 20 October 2010 (UTC) - Primary Author

Egm6321.f10.team4.ejm 16:45, 20 October 2010 (UTC) - Editor

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