Question

If \( u = f(r) \), \( r^2 = x^2 + y^2 + z^2 \), then \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = ? \]

• A. \( f''(r) \cdot \frac{3}{r} \)
• B. \( f''(r) + \frac{2}{r} f'(r) \)
• C. \( f''(r) \)
• D. \( f''(r) + \frac{3}{r} f'(r) \)

Hint:

Use spherical symmetry: \( \nabla^2 u = \dfrac{d^2 u}{dr^2} + \dfrac{2}{r} \dfrac{du}{dr} \) for \( u = f(r) \).

Answer 1

The Correct Option is B

To find \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \), we start by considering \( u = f(r) \) where \( r = \sqrt{x^2 + y^2 + z^2} \). Thus, \( r^2 = x^2 + y^2 + z^2 \).

First, find the partial derivatives of \( u \):

\[\frac{\partial u}{\partial x} = \frac{df}{dr}\cdot\frac{\partial r}{\partial x}\]

Given \( r = (x^2 + y^2 + z^2)^{1/2} \), the partial derivatives of \( r \) are:

\[\frac{\partial r}{\partial x} = \frac{x}{r},\quad \frac{\partial r}{\partial y} = \frac{y}{r},\quad \frac{\partial r}{\partial z} = \frac{z}{r}\]

Hence,

\[\frac{\partial u}{\partial x} = f'(r)\cdot\frac{x}{r},\quad \frac{\partial u}{\partial y} = f'(r)\cdot\frac{y}{r},\quad \frac{\partial u}{\partial z} = f'(r)\cdot\frac{z}{r}\]

Now, find the second derivatives:

\[\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(f'(r)\cdot\frac{x}{r})\]

Applying the product rule:

\[\frac{\partial^2 u}{\partial x^2} = f''(r)\cdot\left(\frac{x}{r}\right)^2 + f'(r)\cdot\left(\frac{1}{r} - \frac{x^2}{r^3}\right)\]

Similarly:

\[\frac{\partial^2 u}{\partial y^2} = f''(r)\cdot\left(\frac{y}{r}\right)^2 + f'(r)\cdot\left(\frac{1}{r} - \frac{y^2}{r^3}\right)\]

\[\frac{\partial^2 u}{\partial z^2} = f''(r)\cdot\left(\frac{z}{r}\right)^2 + f'(r)\cdot\left(\frac{1}{r} - \frac{z^2}{r^3}\right)\]

Adding these gives:

\[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = f''(r)\cdot\left(\frac{x^2 + y^2 + z^2}{r^2}\right) + f'(r)\cdot\left(\frac{3}{r} - \frac{x^2 + y^2 + z^2}{r^3}\right)\]

Since \( x^2 + y^2 + z^2 = r^2 \), the equation simplifies to:

\[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = f''(r) + \frac{2}{r}f'(r)\]

Hence, the correct answer is \( f''(r) + \frac{2}{r} f'(r) \).