Question

If $u=f\left(r\right)$, where $r^2=x^2+y^2$ then $(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2})=$

• A. $f^{\prime }\left(r\right)$
• B. $f^{\prime \prime }\left(r\right)+f^{\prime }\left(r\right)$
• C. $f^{\prime \prime }\left(r\right)+\frac{1}{r}{f}^{\prime }\left(r\right)$
• D. $f^{\prime }\left(r\right)+r{f}^{\prime }\left(r\right)$

Answer 1

Answer: (C)

$f^{\prime \prime }\left(r\right)+\frac{1}{r}{f}^{\prime }\left(r\right)$

$u=f\left(r\right),r^2=x^2+y^2$

then $(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2})=?$

$u_x=f^{\prime }\left(r\right)\left(\frac{x}{r}\right)$

$u_{xx}=f^{\prime \prime }\left(r\right)\frac{x^2}{r^2}+f^{\prime }\left(r\right)\frac{r-x\cdot \frac{x}{r}}{r^2}$

$u_{xx}=f^{\prime \prime }\left(r\right)\cdot \frac{x^2}{r^2}+f^{\prime }\left(r\right)\frac{(r^2-x^2)}{r^3}....\left(1\right)$

Similarly,

$u_{yy}=f^{\prime \prime }\left(r\right)\frac{y^2}{r^2}+f^{\prime \prime }\left(r\right)\frac{r^2-y^2}{r^3}.....\left(2\right)$

Adding equation (1) and (2), we get

$u_{xx}+u_{yy}=\frac{f^{\prime \prime }\left(r\right)}{r^2}(x^2+y^2)+\frac{f^{\prime }\left(r\right)}{r^3}[r^2-x^2+r^2-y^2]$

$=f^{\prime \prime }\left(r\right)+\frac{f^{\prime }\left(r\right)}{r}$