Question

Observe the following statements:

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

II: If $u=f\left(r\right),r^2=x^2+y^2+z^2$, then $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}=f\left(r\right)+\frac{1}{r}{f}^{\prime }\left(r\right)$.

Which of the above statements is correct?

• A. Only I
• B. Only II
• C. Both I and II
• D. Neither I nor II

Answer 1

Answer: (D)

Neither I nor II

The given information is: $u=f\left(r\right)$, $r^2=x^2+y^2$

So, $\Rightarrow r=\sqrt{x^2+y^2}$

Taking partial differentiation with respect to $x$, $y$ one at a time.

$\Rightarrow \frac{\partial u}{\partial x}=f^{\prime }\left(r\right).\frac{x}{\sqrt{x^2+y^2}}$

$\Rightarrow \frac{\partial u}{\partial y}=f^{\prime }\left(r\right).\frac{y}{\sqrt{x^2+y^2}}$

Again differentiating w.r.t to $x$, $y$ one at a time.

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}=f^{\prime }\left(r\right).\frac{\sqrt{x^2+y^2}-\frac{x^2}{\sqrt{x^2+y^2}}}{x^2+y^2}+f^{\prime \prime }\left(r\right).\frac{x}{\sqrt{x^2+y^2}}.\frac{x}{\sqrt{x^2+y^2}}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}=f^{\prime }\left(r\right).\frac{y^2}{(x^2+y^2{)}^{3/2}}+f^{\prime \prime }\left(r\right).\frac{x^2}{x^2+y^2}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial y^2}=f^{\prime }\left(r\right).\frac{\sqrt{x^2+y^2}-\frac{y^2}{\sqrt{x^2+y^2}}}{x^2+y^2}+f^{\prime \prime }\left(r\right).\frac{y}{\sqrt{x^2+y^2}}.\frac{y}{\sqrt{x^2+y^2}}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial y^2}=f^{\prime }\left(r\right).\frac{x^2}{(x^2+y^2{)}^{3/2}}+f^{\prime \prime }\left(r\right).\frac{y^2}{x^2+y^2}$

Adding all we get,

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial x^2}=f^{\prime }\left(r\right).\frac{x^2+y^2}{(x^2+y^2{)}^{3/2}}+f^{\prime \prime }\left(r\right).\frac{x^2+y^2}{x^2+y^2}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial x^2}=f^{\prime }\left(r\right).\frac{1}{\sqrt{x^2+y^2}}+f^{\prime \prime }\left(r\right)$

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial x^2}=\frac{f^{\prime }\left(r\right)}{r}+f^{\prime \prime }\left(r\right)$

Now looking into the reason part we have,

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

So, $r=\sqrt{x^2+y^2+z^2}$

Taking partial differentiation with respect to $x$, $y$ and $z$ one at a time.

$\Rightarrow \frac{\partial u}{\partial x}=f^{\prime }\left(r\right)\frac{x}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow \frac{\partial u}{\partial y}=f^{\prime }\left(r\right)\frac{y}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow \frac{\partial u}{\partial z}=f^{\prime }\left(r\right)\frac{z}{\sqrt{x^2+y^2+z^2}}$

Again differentiating w.r.t to $x$, $y$ and $z$ one at a time.

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}=f^{\prime }\left(r\right).\frac{\sqrt{x^2+y^2+z^2}-\frac{x^2}{\sqrt{x^2+y^2+z^2}}}{x^2+y^2+z^2}+f^{\prime \prime }\left(r\right).\frac{x}{\sqrt{x^2+y^2+z^2}}.\frac{x}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}=f^{\prime }\left(r\right).\frac{z^2+y^2}{(x^2+y^2+z^2{)}^{3/2}}+f^{\prime \prime }\left(r\right).\frac{x^2}{x^2+y^2+z^2}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial y^2}=f^{\prime }\left(r\right).\frac{\sqrt{x^2+y^2+z^2}-\frac{y^2}{\sqrt{x^2+y^2+z^2}}}{x^2+y^2+z^2}+f^{\prime \prime }\left(r\right).\frac{y}{\sqrt{x^2+y^2+z^2}}.\frac{y}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial y^2}=f^{\prime }\left(r\right).\frac{z^2+x^2}{(x^2+y^2+z^2{)}^{3/2}}+f^{\prime \prime }\left(r\right).\frac{y^2}{x^2+y^2+z^2}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial z^2}=f^{\prime }\left(r\right).\frac{\sqrt{x^2+y^2+z^2}-\frac{y^2}{\sqrt{x^2+y^2+z^2}}}{x^2+y^2+z^2}+f^{\prime \prime }\left(r\right).\frac{y}{\sqrt{x^2+y^2+z^2}}.\frac{y}{\sqrt{x^2+y^2+z^2}}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial z^2}=f^{\prime }\left(r\right).\frac{x^2+y^2}{(x^2+y^2+z^2{)}^{3/2}}+f^{\prime \prime }\left(r\right).\frac{z^2}{x^2+y^2+z^2}$

Adding all these we get,

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}=f^{\prime }\left(r\right).\frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2{)}^{3/2}}+f^{\prime \prime }\left(r\right).\frac{x^2+y^2+z^2}{x^2+y^2+z^2}$

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}=f^{\prime }\left(r\right).\frac{2}{\sqrt{x^2+y^2+z^2}}+f^{\prime \prime }\left(r\right)$

$\Rightarrow \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}=f^{\prime }\left(r\right).\frac{2}{r}+f^{\prime \prime }\left(r\right)$

So, neither I or II are correct.