Q21Mathematics I

Question

If $u = f (r)$ , where $r^{2} = x^{2} + y^{2}$ , then prove that:

\frac{\partial ^{2} u}{\partial x ^{2}} + \frac{\partial ^{2} u}{\partial y ^{2}} = f^{''} (r) + \frac{1}{r} f^{'} (r)

Answer 1

The Laplacian operator, denoted as $\nabla^{2}$ (or sometimes $Δ$ ), is a second-order differential operator that is absolutely foundational in fields ranging from heat conduction and fluid mechanics to quantum mechanics (Schrödinger's equation). In standard two-dimensional Cartesian coordinates $(x, y)$ , it is strictly defined as the sum of the unmixed second partial derivatives:

\nabla^{2} u = \frac{\partial ^{2} u}{\partial x ^{2}} + \frac{\partial ^{2} u}{\partial y ^{2}} = u_{xx} + u_{y y}

The mathematical goal is to completely convert this Cartesian operator into polar coordinates $(r, θ)$ . To simplify the rigorous derivation, we are explicitly given that the function depends exclusively on the radial distance $r$ , meaning $u = f (r)$ . It possesses absolute angular symmetry, so it has no dependence whatsoever on the angle $θ$ .

Step 1: Establishing the Coordinate Transformation

The fundamental mathematical link between Cartesian and polar coordinates is governed by the Pythagorean distance formula:

r = x^{2} + y^{2} = (x^{2} + y^{2})^{1/2}

To employ the chain rule effectively, we must first calculate the first-order partial derivatives of $r$ with respect to both $x$ and $y$ :

\frac{\partial r}{\partial x} = \frac{1}{2} (x^{2} + y^{2})^{- 1/2} \cdot 2 x = \frac{x}{x ^{2} + y ^{2}} = \frac{x}{r}

\frac{\partial r}{\partial y} = \frac{1}{2} (x^{2} + y^{2})^{- 1/2} \cdot 2 y = \frac{y}{x ^{2} + y ^{2}} = \frac{y}{r}

Step 2: Calculating the First Partial Derivatives of $u$

Since $u = f (r)$ is essentially a function of a function when viewed from Cartesian space, we must apply the strict mathematical chain rule to find $u_{x}$ and $u_{y}$ :

u_{x} = \frac{\partial u}{\partial x} = \frac{df}{d r} \cdot \frac{\partial r}{\partial x} = f^{'} (r) \cdot \frac{x}{r}

u_{y} = \frac{\partial u}{\partial y} = \frac{df}{d r} \cdot \frac{\partial r}{\partial y} = f^{'} (r) \cdot \frac{y}{r}

Step 3: Calculating the Second Partial Derivatives of $u$

We must now differentiate $u_{x}$ again with respect to $x$ to find $u_{xx}$ . Because $u_{x} = f^{'} (r) \cdot \frac{x}{r}$ is a mathematical product of two separate terms that both depend on $x$ , we must meticulously apply both the Product Rule and the Quotient Rule simultaneously.

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

u_{xx} = (\frac{\partial}{\partial x} f^{'} (r)) \cdot \frac{x}{r} + f^{'} (r) \cdot (\frac{\partial}{\partial x} \frac{x}{r})

We evaluate the two separate derivative components: 1. Using the chain rule: $\frac{\partial}{\partial x} f^{'} (r) = f^{''} (r) \cdot \frac{\partial r}{\partial x} = f^{''} (r) \cdot \frac{x}{r}$ 2. Using the quotient rule: $\frac{\partial}{\partial x} (\frac{x}{r}) = \frac{1 \cdot r - x \cdot \frac{\partial r}{\partial x}}{r ^{2}} = \frac{r - x ( x / r )}{r ^{2}} = \frac{r ^{2} - x ^{2}}{r ^{3}}$

Substituting these rigorously back into the $u_{xx}$ equation yields:

u_{xx} = (f^{''} (r) \cdot \frac{x}{r}) \cdot \frac{x}{r} + f^{'} (r) \cdot (\frac{r ^{2} - x ^{2}}{r ^{3}}) = f^{''} (r) \frac{x ^{2}}{r ^{2}} + f^{'} (r) \frac{r ^{2} - x ^{2}}{r ^{3}}

By absolute mathematical symmetry, differentiating $u_{y}$ with respect to $y$ yields the exact parallel structure for $u_{y y}$ :

u_{y y} = f^{''} (r) \frac{y ^{2}}{r ^{2}} + f^{'} (r) \frac{r ^{2} - y ^{2}}{r ^{3}}

Step 4: Summing the Components to form the Laplacian

Finally, we mathematically add $u_{xx}$ and $u_{y y}$ together to construct the complete Laplacian $\nabla^{2} u$ :

\nabla^{2} u = (f^{''} (r) \frac{x ^{2}}{r ^{2}} + f^{'} (r) \frac{r ^{2} - x ^{2}}{r ^{3}}) + (f^{''} (r) \frac{y ^{2}}{r ^{2}} + f^{'} (r) \frac{r ^{2} - y ^{2}}{r ^{3}})

We group the $f^{''} (r)$ terms and the $f^{'} (r)$ terms together algebraically:

\nabla^{2} u = f^{''} (r) [\frac{x ^{2} + y ^{2}}{r ^{2}}] + f^{'} (r) [\frac{( r ^{2} - x ^{2} ) + ( r ^{2} - y ^{2} )}{r ^{3}}]

Applying the foundational relation $x^{2} + y^{2} = r^{2}$ , the first bracket simplifies to exactly $1$ . The numerator of the second bracket simplifies to $2 r^{2} - (x^{2} + y^{2}) = 2 r^{2} - r^{2} = r^{2}$ . The second bracket becomes $\frac{r ^{2}}{r ^{3}} = \frac{1}{r}$ .

Thus, the final, completely derived expression for the Laplacian of a radial function in polar coordinates is proven to be:

\nabla^{2} u = f^{''} (r) + \frac{1}{r} f^{'} (r)

Q21Mathematics I

Question

Answer

Step 1: Establishing the Coordinate Transformation

Step 2: Calculating the First Partial Derivatives of u

Step 3: Calculating the Second Partial Derivatives of u

Step 4: Summing the Components to form the Laplacian

Step 2: Calculating the First Partial Derivatives of $u$

Step 3: Calculating the Second Partial Derivatives of $u$