So given picture I have to check whether divergence and curl is positive, negative or zero.
So I check vector field at some point ; for figure b I can see that vector field goes outward so divergence is positive but for figure a I am not able to predict. Also how to conclude about curl?
Best Answer
One approach to doing this is to model the vector fields by rotating them clockwise by 45-degrees. If you do this, then it’s not a bad approximation to say that
$$ F_1(x,y)=(y,0)\hspace{2pc} F_2(x,y)=(x, 0). $$
Even if it’s not exact, using the above vector fields can still give you the qualitative details about curl and divergence. Clearly the divergence of $F_1$ is zero while that of $F_2$ is positive. For curl, the only term that would appear would $\nabla\times F=\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\hat{z}$. From this you can clearly see that the curl of $F_1$ will be positive and that of $F_2$ will be zero.
Showing Curl/Divergence are Rotationally Invariant
We can demonstrate that the curl and divergence are rotationally-invariant properties using these particular vector fields. The vector fields are actually modeled after the following vector fields by rotating by 45 degrees ccw: $$ \widetilde{F}_1(x,y) = (y,0) \hspace{3pc} \widetilde{F}_2 = (x,0). $$
If $R$ is a rotation matrix (i.e. orthogonal with determinant 1) then the correct way to transform a vector field $\widetilde{F}$ into another is via the formula: $F(v) = R\widetilde{F}\left (R^Tv\right )$, which is a result obtained from this post on math.se. The matrix $R$ that we want to rotate in the plane by 45 degrees is given by: $$ R \;\; = \;\; \left [ \begin{array}{cc} 1/\sqrt{2} & - 1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \\ \end{array} \right ] $$
The vector field $F_1$ in your prompt can then be found as: \begin{eqnarray*} F_1(x,y) & = & \left [ \begin{array}{cc} 1/\sqrt{2} & - 1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \\ \end{array} \right ] \widetilde{F}_1 \left (\left [ \begin{array}{cc} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2} \\ \end{array} \right ]\left [ \begin{array}{c} x \\ y \\ \end{array} \right ] \right ) \\ & = & \left [ \begin{array}{cc} 1/\sqrt{2} & - 1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \\ \end{array} \right ] \widetilde{F}_1 \left (\left [ \begin{array}{c} \frac{1}{\sqrt{2}}(y+x) \\ \frac{1}{\sqrt{2}}(y-x) \\ \end{array} \right ] \right ) \\ & = & \left [ \begin{array}{cc} 1/\sqrt{2} & - 1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \\ \end{array} \right ]\left [ \begin{array}{c} \frac{1}{\sqrt{2}}(y-x) \\ 0 \\ \end{array} \right ] \\ & = & \frac{1}{2}\left [ \begin{array}{c} y-x \\ y-x \\ \end{array} \right ]. \end{eqnarray*}
Now, let's compare the curls and divergences of $F_1(x,y) = \frac{1}{2}(y-x,y-x)$ and $\widetilde{F}_1(x,y) = (y,0)$.
\begin{eqnarray*} \nabla \times F_1 & = & \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ \frac{1}{2}(y-x) & \frac{1}{2}(y-x) & 0 \\ \end{array} \right |\;\; =\;\; \left (-\frac{1}{2} - \frac{1}{2}\right )\hat{k} \;\; =\;\; -\hat{k} \\ \nabla \times \widetilde{F}_1 & = & \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ y & 0 & 0 \\ \end{array} \right | \;\; =\;\; - \hat{k} \\ \nabla\cdot F_1 & = & \frac{\partial}{\partial x}\frac{1}{2}(y-x) + \frac{\partial}{\partial y} \frac{1}{2}(y-x) \;\; =\;\; -\frac{1}{2} + \frac{1}{2} \;\; =\;\; 0 \\ \nabla \cdot \widetilde{F}_1 & = & \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(0) \;\; =\;\; 0. \end{eqnarray*}
Similar computations follow for $F_2$ and $\widetilde{F}_2$.