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Let

$$\vec{f}(x_1,x_2) = g_1(x_1,x_2) \hat{i} + g_2(x_1,x_2) \hat{j} + g_3(x_1,x_2) \hat{k}$$

then using the definition of divergence we get,

$$\mathrm{div} f = \sum_{i = 1}^{2} \frac{\partial{g_i}}{\partial x_i}$$

what happens to $ \frac{\partial g_3}{\partial x_{what}}$, since I don't have third co-ord to take derivative with?

How is divergence defined for such cases if at all?

Pratik Deoghare
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2 Answers2

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In your example, the divergence does not exist.

Divergence is a characteristic defined for functions of the form $\mathbf{F}:\mathbb{R}^3\rightarrow\mathbb{R^3}$. See the WA and Wiki articles.

The function in your example is of the form $\mathbf{G}:\mathbb{R}^2\rightarrow\mathbb{R}^3$. You cannot take $\text{div}\,\mathbf{G}$.

Pockets
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The divergence is given by $$\left\langle \nabla , f \right\rangle = \sum_{i=1}^3 \frac{\partial g_i}{\partial x_i}$$ $$=\frac{\partial g_1}{\partial x_1}+\frac{\partial g_2}{\partial x_2}+0$$

As your last term is independent of $x_3$.

William T.
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  • But I don't have any $x_3$!? – Pratik Deoghare May 26 '14 at 01:04
  • You can extend $f$ to be a function of three variables but it is no longer the same $f$ as stated in the problem. The divergence simply does not exist since it doesn't make sense (as OP is indicating with his/her difficulties). – Cameron Williams May 26 '14 at 01:05
  • You can argue that $f$ can be extended by defining some $f'(x_1,x_2,x_3)=f(x_1,x_2)$, but this answer in no way even indicates that. – Pockets May 26 '14 at 22:42