Question

If \(\theta=t^n e^{-\frac{r^2}{4t}}\), then the value of \(n\) that will make \(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \theta}{\partial r}\right)=\frac{\partial \theta}{\partial t}\) is given by:

• A. 0
• B. 1
• C. $-1/2$
• D. $-3/2$

Hint:

This differential equation is related to the heat conduction equation in spherical coordinates. The parameter $n = -3/2$ is a standard result in physical sciences for diffusion in three dimensions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem requires the application of partial differentiation to a function of two variables, $r$ and $t$.
We need to calculate the partial derivatives on both sides of the given equation and equate them to find the value of the parameter $n$.
Key Formula or Approach:
The given equation is: \[ \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = \frac{\partial \theta}{\partial t} \] We will compute the Left-Hand Side (LHS) and Right-Hand Side (RHS) individually.

Step 2: Detailed Explanation:
Let $\theta = t^n e^{-\frac{r^2}{4t}}$.
First, let us find the derivative on the RHS: \[ \frac{\partial \theta}{\partial t} = \frac{\partial}{\partial t}\left( t^n e^{-\frac{r^2}{4t}} \right) \] Applying the product rule of differentiation: \[ \frac{\partial \theta}{\partial t} = n t^{n-1} e^{-\frac{r^2}{4t}} + t^n e^{-\frac{r^2}{4t}} \left( \frac{r^2}{4t^2} \right) \] \[ \frac{\partial \theta}{\partial t} = t^{n-1} e^{-\frac{r^2}{4t}} \left[ n + \frac{r^2}{4t} \right] \quad \text{--- (RHS)} \] Now, let us find the derivatives on the LHS.
First, find $\frac{\partial \theta}{\partial r}$: \[ \frac{\partial \theta}{\partial r} = t^n e^{-\frac{r^2}{4t}} \left( -\frac{2r}{4t} \right) = -\frac{r}{2} t^{n-1} e^{-\frac{r^2}{4t}} \] Multiply by $r^2$: \[ r^2 \frac{\partial \theta}{\partial r} = -\frac{r^3}{2} t^{n-1} e^{-\frac{r^2}{4t}} \] Now, differentiate this expression with respect to $r$: \[ \frac{\partial}{\partial r} \left( -\frac{r^3}{2} t^{n-1} e^{-\frac{r^2}{4t}} \right) = t^{n-1} \frac{\partial}{\partial r} \left( -\frac{r^3}{2} e^{-\frac{r^2}{4t}} \right) \] Applying the product rule: \[ = t^{n-1} \left[ -\frac{3r^2}{2} e^{-\frac{r^2}{4t}} - \frac{r^3}{2} e^{-\frac{r^2}{4t}} \left( -\frac{2r}{4t} \right) \right] \] \[ = t^{n-1} e^{-\frac{r^2}{4t}} \left[ -\frac{3r^2}{2} + \frac{r^4}{4t} \right] \] Divide the entire term by $r^2$ to find the LHS: \[ \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = t^{n-1} e^{-\frac{r^2}{4t}} \left[ -\frac{3}{2} + \frac{r^2}{4t} \right] \quad \text{--- (LHS)} \] Equating LHS and RHS: \[ t^{n-1} e^{-\frac{r^2}{4t}} \left[ -\frac{3}{2} + \frac{r^2}{4t} \right] = t^{n-1} e^{-\frac{r^2}{4t}} \left[ n + \frac{r^2}{4t} \right] \] Comparing the coefficients on both sides, we get: \[ n = -\frac{3}{2} \]

Step 3: Final Answer:
The value of $n$ is $-3/2$, which corresponds to Option (D).