Question

In case of radial symmetry, \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}\) is_________

(a) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r}  \frac{∂\overline{u}}{∂r}\)

(b) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r}  \frac{∂\overline{u}}{∂r}\)

(c) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=-\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r}  \frac{∂\overline{u}}{∂r}\)

(d) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=-\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r}  \frac{∂\overline{u}}{∂r}\)

The question was posed to me during an online exam.

My query is from Consolidation Equation in Polar Coordinates topic in division One Dimensional Consolidation of Geotechnical Engineering I

Answer 1

The correct choice is (a) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r}  \frac{∂\overline{u}}{∂r}\)

Easiest explanation: The term  \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}\)  in terms of r and θ is given by,

\(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r}  \frac{∂\overline{u}}{∂r}+\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}.\)

In case of radial symmetry, \overline{u} is independent of θ and hence we get,

\(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r}  \frac{∂\overline{u}}{∂r}\)