\[If\ \overrightarrow{F}={F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k}\]
is a vector function, defined and differentiable at each point (x, y, z)in a certain region of space [i.e., A defines a vector field], then the divergence of (abbreviated as ‘Div ‘) is defined as,
\[Div\ \overrightarrow{F} = \nabla\ . \overrightarrow{F}\]
\[ = (\overrightarrow{i}\frac{\partial}{\partial\ x} + \overrightarrow{j}\frac{\partial}{\partial\ y} + \overrightarrow{k}\frac{\partial}{\partial\ z})\ . \ ({F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k})\]
\[ = (\frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z})\]
Basic properties of Divergence:
If A, B are vector functions and ‘f’ is a scalar function, then
\[1) \nabla\ . ( A + B) = \nabla\ .A + \nabla\ .B\]
\[2) \nabla\ . ( fA) = (\nabla\ f) . A + (f.\nabla\ A)\]
\[3)\overrightarrow{F}\ is\ solenoidal\ if\ \nabla\ . \overrightarrow{F}=0\]
Example:
\[If\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 – zy^3)\overrightarrow{k}, then\ find\ div\ \overrightarrow{F}\]
Soln:
\[Let\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 – zy^3)\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(xyz)}+ \frac{\partial}{\partial\ y} {(3x^2y)}+ \frac{\partial}{\partial\ z} {(xy^2-zy^3)}\]
= yz + 3x2 ( 1 ) + ( 0 – y3 )
= yz + 3x2 – y3
Example:
\[If\ \overrightarrow{F}=x^2y\overrightarrow{i} + xy^2z\overrightarrow{j} + xyyz\overrightarrow{k}, then\ find\ div\ \overrightarrow{F}\ at\ the\ point\ (1,-1,2)\]
Soln:
\[Let\ \overrightarrow{F}=x^2y\overrightarrow{i} + xy^2z\overrightarrow{j} + xyyz\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(x^2y)}+ \frac{\partial}{\partial\ y} {(xy^2z)}+ \frac{\partial}{\partial\ z} {(xyyz)}\]
\[\nabla\ . \overrightarrow{F} = 2xy + 2xyz + xy\]
At ( 1, -1, 2)
\[\nabla\ . \overrightarrow{F} = 2(1)(-1) + 2(1)(-1)(2) + (1)(-1)\]
= -2 – 4 – 1
= -7
Example:
\[Show\ that\ \overrightarrow{F}=3y^4z^2\overrightarrow{i} + 4x^3z^2\overrightarrow{j} + 6x^2y^3\overrightarrow{k} is\ solenoidal\]
Soln:
\[Let\ \overrightarrow{F}=3y^4z^2\overrightarrow{i} + 4x^3z^2\overrightarrow{j} + 6x^2y^3\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(3y^4z^2)}+ \frac{\partial}{\partial\ y} {(4x^3z^2)}+ \frac{\partial}{\partial\ z} {(6x^2y^3)}\]
\[\nabla\ . \overrightarrow{F} = 0+ 0 + 0 = 0\]
\[\overrightarrow{F}\ is\ solenoidal\]
Example:
\[Show\ that\ \overrightarrow{F}=(x + 3y)\overrightarrow{i} + (x – 3z)\overrightarrow{j} + (x -z)\overrightarrow{k} is\ solenoidal\]
Soln:
\[Let\ \overrightarrow{F}=(x + 3y)\overrightarrow{i} + (x – 3z)\overrightarrow{j} + (x -z)\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(x + 3y)}+ \frac{\partial}{\partial\ y} {(x – 3z)}+ \frac{\partial}{\partial\ z} {(x -z)}\]
\[\nabla\ . \overrightarrow{F} = (1+ 0) + (0 -0) + (0 -1) = 0\]
\[\overrightarrow{F}\ is\ solenoidal\]
Example:
\[If\ \overrightarrow{F}=2xy\overrightarrow{i} + 3x^2y\overrightarrow{j} – 3pyz\overrightarrow{k} is\ solenoidal\ at\ (1,1,1)\, find\ ‘p’\]
Soln:
\[Let\ \overrightarrow{F}=2xy\overrightarrow{i} + 3x^2y\overrightarrow{j} – 3pyz\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(2xy)}+ \frac{\partial}{\partial\ y} {(3x^2y)}- \frac{\partial}{\partial\ z} {(3pyz)}\]
\[\nabla\ . \overrightarrow{F} = 2y + 3x^2- 3py\]
At ( 1, 1, 1)
\[\nabla\ . \overrightarrow{F} = 2(1) + 3(1)^2- 3p(1)\]
\[\nabla\ . \overrightarrow{F} = 2+ 3 – 3p\]
\[Given\ \overrightarrow{F}\ is\ solenoidal\]
\[i.e\ \nabla\ . \overrightarrow{F} =0\]
2 + 3 – 3p = 0
5 – 3p = 0
3p = 5
p = 5/3
Curl of a vector function:
\[If\ \overrightarrow{F}={F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k}\]
is a vector function,
\[then\ curl\ \overrightarrow{F}=\nabla\ × \overrightarrow{F}\]
\[curl\ \overrightarrow{F} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
\frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\
{F_1} & {F_2} & {F_3}\\
\end{vmatrix}\]
Irrotational vector :
\[A\ vector\ \overrightarrow{F}\ is\ said\ to\ be\ irrotational\ if\]
\[curl\ \overrightarrow{F} =0\]
\[i.e\ \nabla\ × \overrightarrow{F} =0\]
Example:
\[If\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} +(x y ^2- zy^3)\overrightarrow{k}\ then\ find\ curl\ \overrightarrow{F}\]
Soln:
\[Let\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (x y ^2- zy^3)\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
\frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\
{xyz} & {3x^2y} & {x y ^2- zy^3}\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {(x y ^2- zy^3)} – \frac{\partial}{\partial\ z} {(3x^2y)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(x y ^2- zy^3)} – \frac{\partial}{\partial\ z} {(xyz)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(3x^2y)} – \frac{\partial}{\partial\ y} {(xyz)})\]
\[ = \overrightarrow{i}( 2xy – 3zy^2 ) -\overrightarrow{j}((y^2 – 0) – xy)+\overrightarrow{k}(6xy – xz))\]
\[\nabla\ × \overrightarrow{F} = [2xy – 3zy^2] \overrightarrow{i} – [y^2 – xy] \overrightarrow{j} + [ 6xy – xz] \overrightarrow{k}\]
Example:
\[Show\ that\ \overrightarrow{F}= x\overrightarrow{i} +y^2\overrightarrow{j} +z^3\overrightarrow{k}\ is\ irrotational\]
Soln:
\[Let\ \overrightarrow{F}=x\overrightarrow{i} +y^2\overrightarrow{j} +z^3\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
\frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\
{x} & y^2 & z^3\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {(z^3)} – \frac{\partial}{\partial\ z} {(y^2)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(z^3)} – \frac{\partial}{\partial\ z} {(x)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(y^2)} – \frac{\partial}{\partial\ y} {(x)})\]
\[ = \overrightarrow{i}[ 0- 0 ] -\overrightarrow{j}[0 – 0]+\overrightarrow{k}[0- 0]\]
\[\nabla\ × \overrightarrow{F} = 0\]
\[ \overrightarrow{F}\ is\ irrotational\]
Example:
\[If\ \overrightarrow{F}=( x^2 – y^2 + 2xz)\overrightarrow{i} + (xz – xy + yz)\overrightarrow{j} + (z^2 + x^2)\overrightarrow{k}\ find\ \nabla\ × ( \nabla\ × \overrightarrow{F})\]
Soln:
\[Let\ \overrightarrow{F}=( x^2 – y^2 + 2xz)\overrightarrow{i} + (xz – xy + yz)\overrightarrow{j} + (z^2 + x^2)\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
\frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\
x^2 – y^2 + 2xz & xz – xy + yz & z^2 + x^2\\
\end{vmatrix}\]
\[ = \overrightarrow{i}[ \frac{\partial}{\partial\ y} {( z^2 + y^2 )} – \frac{\partial}{\partial\ z} {(xz – xy + yz)}]- \overrightarrow{j}[ \frac{\partial}{\partial\ x} {(z^2 + y^2)} – \frac{\partial}{\partial\ z} {(x^2 – y^2 + 2xz)}] + \overrightarrow{k}[ \frac{\partial}{\partial\ x} {(xz – xy + yz )} – \frac{\partial}{\partial\ y} {(x^2 – y^2 + 2xz)}]\]
\[ = \overrightarrow{i}[ (0 + 0) – (x – 0 +y) ] -\overrightarrow{j}[(0 + 0) – ( 0 – 0 + 2x)]+\overrightarrow{k}[ (z – y + 0) – ( 0 – 2y + 0)]\]
\[\nabla\ × \overrightarrow{F} = \overrightarrow{i} (-x – y ) – \overrightarrow{j}(-2x) + \overrightarrow{k}(z +2y)\]
\[\nabla\ × \overrightarrow{F} =- (x + y)\overrightarrow{i} + 2x\overrightarrow{j} + ( z + 2y)\overrightarrow{k}\]
\[\nabla\ × ( \nabla\ × \overrightarrow{F}) =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
\frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\
-(x + y) & 2x & z + 2y \\
\end{vmatrix}\]
\[ = \overrightarrow{i}[\frac{\partial}{\partial\ y} {(z + 2y)} – \frac{\partial}{\partial\ z} {(2x)}]- \overrightarrow{j}[\frac{\partial}{\partial\ x} {(z + 2y )} – \frac{\partial}{\partial\ z} {-(x + y)}] + \overrightarrow{k}[\frac{\partial}{\partial\ x} {(2x)} – \frac{\partial}{\partial\ y} {-(x + y)}]\]
\[ = \overrightarrow{i}[ (0 + 2) – (0)] -\overrightarrow{j}[( 0 + 0) + ( 0 + 0)]+\overrightarrow{k}[(2) + ( 0 +1 )]\]
\[ = \overrightarrow{i}[ 2] -\overrightarrow{j}[0]+\overrightarrow{k}[3]\]
\[\nabla\ × ( \nabla\ × \overrightarrow{F}) = 2\overrightarrow{i} + 3\overrightarrow{k}\]
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