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1. |
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. | \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=-\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}\) |
Answer» B. \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}\) | |