Differentiation With Respect To Time
Differentiation with respect to time can be written in several forms.
\[
\qquad \; \; \text{velocity} \qquad = \qquad {d{\bf x} \over dt} \qquad = \qquad \left( {dx_1 \over dt} , {dx_2 \over dt} , {dx_3 \over dt} \right) \qquad \ = \qquad \dot{\bf x} \qquad = \qquad \dot{x}_i \qquad = \qquad x_{i,t}
\]
\[
\text{acceleration} \qquad = \qquad {d{\bf v} \over dt} \qquad = \qquad \left( {dv_1 \over dt} , {dv_2 \over dt} , {dv_3 \over dt} \right) \qquad \ = \qquad \dot{\bf v} \qquad = \qquad \dot{v}_i \qquad = \qquad v_{i,t}
\]
\[
\text{acceleration} \qquad = \qquad {d^2{\bf x} \over dt^2} \qquad = \qquad \left( {d^2x_1 \over dt^2} , {d^2x_2 \over dt^2} , {d^2x_3 \over dt^2} \right) \qquad \ = \qquad \ddot{\bf x} \qquad = \qquad \ddot{x}_i \qquad = \qquad x_{i,tt}
\]
One can use the derivative with respect to \(\;t\), or the dot, which is probably the most popular, or the comma notation, which is a popular subset of
tensor notation.
Note that the notation \(x_{i,tt}\) somewhat violates the tensor notation rule of double-indices automatically summing from 1 to 3. This is because time does not have
3 dimensions as space does, so it is understood that no summation is performed.
Differentiation of a Vector
Suppose \( {\bf v} = (5t^2, \sin t, e^{3t}) \). Then \( \dot{\bf v} \) equals
\[
\dot{\bf v} = (10t, \cos t, 3e^{3t})
\]
Helix Example
The position of an ant crawling around and up a pipe is given by
\( {\bf x} = (2 \cos t, 2 \sin t, 5t) \). The velocity, \( {\bf v} \), equals
\[
\dot{\bf x} = {\bf v} = (-2 \sin t, 2 \cos t, 5)
\]
and the acceleration is
\[
\ddot{\bf x} = \dot{\bf v} = {\bf a} = (-2 \cos t, -2 \sin t, 0)
\]
which always points toward the center of the pipe.
Differentiation With Respect To A Coordinate
Suppose you want to differentiate a function, \(f(x,y,z)\), with respect to \(y\).
This is written as
\[
{\partial f \over \partial y} \quad \qquad \text{or} \qquad \quad
{\partial f \over \partial x_2} \quad \qquad \text{or} \qquad \quad
f_{,2}
\]
where the comma is common
tensor notation for a derivative.
In the more general case, differentiation with respect to \(x_j\) is (yes, this is a gradient)
\[
{\partial f \over \partial x_j} \qquad \qquad \text{or} \qquad \qquad f,_j
\]
Differentiation of a
vector, \({\bf v}\), is
\[
{\partial {\bf v} \over \partial x_j} \qquad \qquad \text{or} \qquad \qquad
\left( {\partial \, v_x \over \partial x_j} , {\partial \, v_y \over \partial x_j} , {\partial \, v_z \over \partial x_j} \right)
\qquad \qquad \text{or} \qquad \qquad v_{i,j}
\]
Differentiation of a tensor, \(\boldsymbol{\sigma}\), is
\[
{\partial \boldsymbol{\sigma} \over \partial x_k} \qquad \qquad \text{or} \qquad \qquad \sigma_{ij,k}
\]
As with
vectors, every component of a tensor is differentiated.
Differentiation of a Vector
Suppose \( {\bf v} = (3x^2 - 2y, z^2+x, y^3 - z) \). Then \( \partial {\bf v} \over \partial y \) equals
\[
{\partial {\bf v} \over \partial y} = (-2, 0, 3y^2)
\]
Gradient
The gradient of a function, \( f({\bf x}) \), is written, \( \nabla f({\bf x}) \), and is a vector.
It is formed by differentiating the function with respect to each coordinate.
\[
\nabla f({\bf x}) = \left( {\partial f \over \partial x_1} ,
{\partial f \over \partial x_2} , {\partial f \over \partial x_3} \right)
\]
Tensor Notation
The gradient can also be written as \( { \partial f \over \partial x_i} \), or simply as
\( f_{,i} \).
Gradient Example
Suppose \( f({\bf x}) = 3x^2 - 2 y z^2 \). Then the gradient is
\[
\nabla f({\bf x}) = ( 6x, -2 z^2, -4 y z )
\]
The gradient of a scalar function tells how much the function increases along
each global coordinate. In the above example, \(f\) increases at the rate of
\(6x\) along the \(x\) axis, \(-2 z^2\) along the \(y\) axis, and
\(-4 y z\) along the \(z\) axis.
Coincidentally, the gradient also gives the direction, or orientation, in space
that corresponds to the greatest rate of increase. The following example,
in 2D space, demonstrates this.
2nd Gradient Example
Take for example the paraboloid, \( f(x,y) = 2 x^2 + y^2 \). The gradient
at \( {\bf x} = (5,3) \) is
\[
\nabla f(x,y) \quad = \quad ( 4x, 2y ) \quad = \quad (20, 6)
\]
Therefore, at \( {\bf x} = (5,3) \), \(f\) is increasing at the rate of 20
along the \(x\) axis, and at the rate of 6 along the \(y\) axis.
\( 20{\bf i} + 6 {\bf j} \) also corresponds to the direction in the \(x,y\) plane
along which \(f\) will increase the most quickly.
Gradients of
vectors can also be computed. The result will be
a 2nd order tensor. For example, the gradient of a velocity field is written as
\( \nabla {\bf v} \). Writing this in
tensor notation \( v_{i,j} \) shows
more clearly that the result is a 2nd order tensor because of the presence
of the \(i\) and \(j\) subscripts. Gradients arise in mechanical deformation
and heat conduction applications. Mechanical strains are related to
gradients of displacements
and heat conduction is related to the gradient of the temperature distribution.
Divergence
The divergence of a
vector is a scalar result, and the divergence of a
2nd order tensor is a vector. The divergence of a vector is written as
\( \nabla \cdot {\bf v} \), or \( v_{i,i} \) in
tensor notation. It is
computed as
\[
\begin{eqnarray}
\nabla \cdot {\bf v} & = & (
{\partial \over \partial x} {\bf i} +
{\partial \over \partial y} {\bf j} +
{\partial \over \partial z} {\bf k} )
\cdot( v_x {\bf i} + v_y {\bf j} + v_z {\bf k} ) \\
\\
& = &
{\partial v_x \over \partial x} +
{\partial v_y \over \partial y} +
{\partial v_z \over \partial z}
\end{eqnarray}
\]
Tensor Notation
As stated above, the divergence is written in
tensor notation as \( v_{i,i}\).
It is very important that both subscripts are the same because this dictates
that they are automatically summed from 1 to 3. They can in fact be any
letter one desires, so long as they are both the same letter.
Divergence Example
If \( {\bf v} = (3x^2 - 2y, z^2+x, y^3 - z) \), then the dot product is
\[
\nabla \cdot {\bf v} \quad = \quad {\partial \over \partial x}(3x^2 - 2y) +
{\partial \over \partial y}(z^2 + x) + {\partial \over \partial z}(y^3 - z)
\quad = \quad 6x - 1
\]
The divergence of velocity vectors often arises in the discussion of incompressibility
and
conservation of mass.
Curl
The
curl of a
vector is the cross product of partial derivatives with the vector.
Curls arise when rotations are important, just as cross products of vectors tend to do.
Rotations of solids automatically imply large displacements, which in turn automatically
imply nonlinear analyses. And this is why one seldom comes across curls... because
most analyses are linear.
Curls are calculated as follows.
\[
\begin{eqnarray}
\nabla \times {\bf v} & = & (
{\partial \over \partial x} {\bf i} +
{\partial \over \partial y} {\bf j} +
{\partial \over \partial z} {\bf k} )
\times ( v_x {\bf i} + v_y {\bf j} + v_z {\bf k} ) \\
\\
& = &
\left|
\matrix {
{\bf i\;} & {\bf j\;} & {\bf k\;} \\
{\partial \over \partial x} &
{\partial \over \partial y} &
{\partial \over \partial z} \\
v_x & v_y & v_z
} \right| \\
\\
& = &
({\partial \, v_z \over \partial y} - {\partial \, v_y \over \partial z}) {\bf i}
+ ({\partial \, v_x \over \partial z} - {\partial \, v_z \over \partial x}) {\bf j}
+ ({\partial \, v_y \over \partial x} - {\partial \, v_x \over \partial y}) {\bf k}
\end{eqnarray}
\]
Tensor Notation of Curls
The curl of a
vector is written in
tensor notation as \( \epsilon_{ijk} v_{k,j} \).
It is critical to recognize that the vector is written as \( v_{k,j} \) here, not \( v_{j,k} \).
This is because the curl is \( \nabla \times {\bf v} \), not \( {\bf v} \times \nabla \).
An easy way to get the
tensor notation right is to think of
\( \nabla \times {\bf v} \) as \( \epsilon_{ijk} \nabla_j v_k \) and note the order of the
subscripts. Of course, this reduces to the correct result: \( \epsilon_{ijk} v_{k,j} \).
As with cross products, the fact that \(j\) and \(k\) both occur twice in \( \epsilon_{ijk} v_{k,j} \)
dictates that both are automatically summed from 1 to 3. The term expands to
\[
\matrix {
\epsilon_{ijk} v_{k,j} & = &
\epsilon_{i11} v_{1,1} & + & \epsilon_{i12} v_{2,1} & + & \epsilon_{i13} v_{3,1} & + & \\
& & \epsilon_{i21} v_{1,2} & + & \epsilon_{i22} v_{2,2} & + & \epsilon_{i23} v_{3,2} & + & \\
& & \epsilon_{i31} v_{1,3} & + & \epsilon_{i32} v_{2,3} & + & \epsilon_{i33} v_{3,3}
}
\]
Curls Using Tensor Notation
To obtain the y
^{th} component of a curl, set \(i\) equal to 2 in the above equation.
\[
\matrix {
\epsilon_{2jk} v_{k,j} & = &
\epsilon_{211} v_{1,1} & + & \epsilon_{212} v_{2,1} & + & \epsilon_{213} v_{3,1} & + & \\
& & \epsilon_{221} v_{1,2} & + & \epsilon_{222} v_{2,2} & + & \epsilon_{223} v_{3,2} & + & \\
& & \epsilon_{231} v_{1,3} & + & \epsilon_{232} v_{2,3} & + & \epsilon_{233} v_{3,3}
}
\]
All subscripts are now specified, and this permits evaluation of all alternating tensors.
All of them will equal zero except two, leaving
\[
\epsilon_{2jk} v_{k,j} \; = \; v_{1,3} - v_{3,1} \; = \;
\frac{ \partial \, v_x}{\partial z} - \frac{\partial \, v_z}{\partial x}
\]
which is again consistent with the determinant result (as it must be).
Results for the x
^{th} and z
^{th} components
are obtained by setting \(i\) equal to 1 and 3, respectively.
Curl Example - Rotating Disk
Consider a disk rotating about the \(z\) axis such that
\[
\begin{eqnarray}
x & = & X \cos (\omega \, t) - Y \sin (\omega \, t) \\
y & = & X \sin (\omega \, t) + Y \cos (\omega \, t) \\
z & = & Z
\end{eqnarray}
\]
where \({\bf X}\) is the vector of original coordinates of each point at \(t = 0\),
and \({\bf x}\) is the vector of that point's coordinates at any other time, \(t\).
Note that this is common in Continuum Mechanics to use \({\bf X}\) as the position
vector at \(t = 0\), the so-called
reference configuration, and \({\bf x}\)
for the position vector following any translations, rotations, and deformations,
the so-called
current configuration.
The velocity vector is
\[
{\bf v} = {\partial {\bf x} \over \partial t} =
\left(
\begin{eqnarray}
& - \omega X \sin (\omega \, t) - \omega Y \cos (\omega \, t) \\
& + \omega X \cos (\omega \, t) - \omega Y \sin (\omega \, t) \\
& 0
\end{eqnarray}
\right)
\]
which simplifies to
\[
{\bf v} = ( - \omega \, y, \omega \, x, 0 )
\]
making the curl of the velocity vector relatively simple to compute.
\[
\nabla \times {\bf v} = ( 0, 0, 2 \omega )
\]
As stated above, the curl is related to
rotations.
It turns out that \( \nabla \times {\bf v} \) gives the axis of rotation, and
\( \frac{1}{2} | \nabla \times {\bf v} | \) is the rotational rate.
So \( \frac{1}{2} ( \nabla \times {\bf v} ) \) gives
\[
\frac{1}{2} ( \nabla \times {\bf v} ) = ( 0, 0, \omega )
\]
Laplacian
The Laplacian is the divergence of the gradient of a function. It often arises
in 2nd order partial differential equations and is usually written as
\(\nabla^2 \! f({\bf x})\). It can also be written in the less popular, but more
descriptive form of \(\nabla \cdot \nabla f({\bf x})\). Its definition is
\[
\nabla^2 \! f({\bf x}) \equiv {\partial^{\,2} \! f({\bf x}) \over \partial \, x^2} +
{\partial^{\,2} \! f({\bf x}) \over \partial \, y^2} +
{\partial^{\,2} \! f({\bf x}) \over \partial \, z^2}
\]
Tensor Notation
The Laplacian is written in
tensor notation simply as \(f,_{ii}\) where the two
\(i\) indices means that they are automatically summed from 1 to 3.
Laplacian Example
Determine the Laplacian of \( f({\bf x}) = 2 x^3 y - z \sin(y) \).
Start by calculating the gradient of \(f({\bf x})\).
\[
\nabla f({\bf x}) = \left( 6 x^2 y, 2 x^3 - z \cos(y), - \sin(y) \right)
\]
And the divergence of the gradient (which is the Laplacian after all) is
\[
\nabla^2 \! f({\bf x}) = \nabla \cdot \nabla f({\bf x}) = 12 x y + z \sin(y)
\]
Derivatives of Vector Products
Differentiation of
vector products (dot, cross, and diadic) follow the same rules as
differentiation of scalar products. For example, the derivative of a dot product is
\[
{d \over dt} ( {\bf a} \cdot {\bf b} ) = {d \, {\bf a} \over dt} \cdot {\bf b} + {\bf a} \cdot {d \, {\bf b} \over dt}
\]
while the derivative of a cross product is
\[
{d \over dt} ( {\bf a} \times {\bf b} ) = {d \, {\bf a} \over dt} \times {\bf b} + {\bf a} \times {d \, {\bf b} \over dt}
\]
and the derivative of a diadic product is
\[
{d \over dt} ( {\bf a} \otimes {\bf b} ) = {d \, {\bf a} \over dt} \otimes {\bf b} + {\bf a} \otimes {d \, {\bf b} \over dt}
\]
Dot Product Derivative Example
Suppose \( {\bf a} = (5t, \sin t, e^t) \) and \( {\bf b} = (t^2, \sin t, 6t) \), then
\[
{\bf a} \cdot {\bf b} = 5 t^3 + \sin^2 t + 6t e^t
\]
and the derivative is
\[
{d \over dt} ( {\bf a} \cdot {\bf b} ) = 15 t^2 + 2 \sin t \cos t + 6 (t + 1) e^t
\]
Applying the differentiation product rule gives the same result.
\[
\begin{eqnarray}
{d \, {\bf a} \over dt} \cdot {\bf b} + {\bf a} \cdot {d \, {\bf b} \over dt} & = &
(5, \cos t, e^t) \cdot (t^2, \sin t, 6t) + (5t, \sin t, e^t) \cdot (2t, \cos t, 6)\\
\\
& = & 15 t^2 + 2 \sin t \cos t + 6 (t + 1) e^t
\end{eqnarray}
\]
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