Question

@ganeshie8

Answer 1

Find an expression involving the function \[\phi (x_1, x_2, x_3) \] that has a minimum average value of the square of its gradient within a certain volume V of space.

Answer 2

yay for triple integrals

Answer 3

a multivariate function is minimized when its 1st partial derivatives with respect to all of its arguments are simultaneously zero..

Answer 4

This requires Euler's equation as well, a bit tougher.

Answer 5

Ok, so you know the average value of the square gradient w/in a certian volume, V, is given by:
\[\sf I = \frac{1}{V} \int \int \int (\bigtriangledown \phi)^2 dx_1dx_2dx_3\]
so
\[v =\sf \frac{1}{V} \int \int \int [ (\frac{\partial \phi}{\partial x_1})^2+(\frac{\partial \phi}{\partial x_2)})^2 + (\frac{\partial \phi}{\partial x_3})^2 ]dx_1 dx_2 dx_3 \]
in order to make it a minumum,
\[f = \sf (\frac{\partial \phi}{\partial x_1})^2+(\frac{\partial \phi}{\partial x_2)})^2 + (\frac{\partial \phi}{\partial x_3})^2 \]
which must satisfy the euler equationn
so substitute that last equation in to it.
\[ \frac{\partial f}{\partial \phi} - \sum_{i=1}^{3} \frac{\partial }{\partial x_i} \left [ \frac{\partial f}{\partial [\frac{\partial \phi}{\partial x_i}]} \right ] = 0 \]

and now,
\[\sum_{i=1}^{3} \frac{\partial}{dx_1}~\frac{\partial \phi}{\partial x_i}=0\]
which is just laplace equation

Answer 6

and you therefore must have \(\phi\) satisfy laplaces equation in order that \(I\) have a minumum value.

Answer 7

Very nice!!

Answer 8

That's very interesting that it turned out to be Laplace equation mhm.