Q101: If $\theta = t^n e^{-\frac{r^2}{4t}}$, then the value of $n$... - Mathematics - ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper | Agrigyan

Given $\theta = t^n e^{-r^2/(4t)}$. Let $u = -r^2/(4t)$, so $\theta = t^n e^u$.
Calculate spatial derivatives:
$\frac{\partial \theta}{\partial r} = t^n e^u \left(-\frac{2r}{4t}\right) = -\frac{r}{2} t^{n-1} e^u$.
Then $r^2 \frac{\partial \theta}{\partial r} = -\frac{1}{2} t^{n-1} r^3 e^u$.
Differentiating with respect to $r$:
$\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = -\frac{1}{2} t^{n-1} \left[3r^2 e^u + r^3 e^u \left(-\frac{r}{2t}\right)\right] = -\frac{1}{2} t^{n-1} e^u \left[3r^2 - \frac{r^4}{2t}\right]$.
Dividing by $r^2$:
$\text{LHS} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = e^u \left[-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
Now calculate time derivative:
$\text{RHS} = \frac{\partial \theta}{\partial t} = n t^{n-1} e^u + t^n e^u \left(\frac{r^2}{4t^2}\right) = e^u \left[n t^{n-1} + \frac{r^2}{4} t^{n-2}\right]$.
Equating LHS and RHS:
$-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2} = n t^{n-1} + \frac{r^2}{4} t^{n-2}$.
Comparing coefficients of $t^{n-1}$ yields:
$n = -\frac{3}{2}$.

Detailed analysis of options:

• 0 is incorrect.
• is incorrect.
• $-1/2$ corresponds to 1D diffusion, not 3D spherical coordinates.
• $-3/2$ is the correct value corresponding to 3-dimensional spherical diffusion.