GATE CE 2016 Set 1 — Question 15

The given partial differential equation is the one-dimensional heat equation:
$\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}$We can solve this using the method of separation of variables. Let $u(x,t) = X(x)T(t)$.
Substituting this into the PDE:
$X(x)T'(t) = \alpha X''(x)T(t)$Dividing by $\alpha X(x)T(t)$:
$\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)} = -k^2$ (where $-k^2$ is the separation constant)This gives two ordinary differential equations:So, $u(x,t) = C_T e^{-\alpha k^2 t} (C_1 \cos(kx) + C_2 \sin(kx))$. This form is for oscillatory solutions in space and decaying in time, typically for boundary conditions like $u(0,t)=u(L,t)=0$.However, the options provided suggest a different form for $X(x)$ and $T(t)$. Let's re-evaluate the separation constant. If we choose the separation constant as $k^2$ (positive) instead of $-k^2$:
$\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)} = k^2$This gives:Combining these, the general solution would be $u(x,t) = C_T e^{\alpha k^2 t} (C_1 e^{kx} + C_2 e^{-kx})$.Let's compare this with the given options. Let $k_{new} = \sqrt{\alpha} k$. Then the time part is $e^{k_{new}^2 t}$.
Or, if we use $k$ as the constant in the options, then the time part is $e^{kt}$ and the spatial part involves $\sqrt{k/\alpha}$.Let's assume the separation constant is $k$ (as used in the options for the time component).
So, $\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)} = k$.Let's assume the separation constant is such that $T(t)$ is $e^{kt}$. This means $\frac{T'(t)}{T(t)} = k$. So, $\frac{X''(x)}{X(x)} = \frac{k}{\alpha}$.Then, $X''(x) = \frac{k}{\alpha} X(x)$. The characteristic equation is $r^2 - \frac{k}{\alpha} = 0$, so $r = \pm \sqrt{\frac{k}{\alpha}}$.
Thus, $X(x) = C_1 e^{\sqrt{k/\alpha} x} + C_2 e^{-\sqrt{k/\alpha} x}$.Combining $T(t) = C_T e^{kt}$ and $X(x) = C_1 e^{\sqrt{k/\alpha} x} + C_2 e^{-\sqrt{k/\alpha} x}$, the solution is:
$u(x,t) = C_T e^{kt} (C_1 e^{\sqrt{k/\alpha} x} + C_2 e^{-\sqrt{k/\alpha} x})$.This matches option (B) if we absorb $C_T$ into the constants $C_1$ and $C_2$, or consider $C_T$ as the $C$ outside the bracket.Option (A) and (D) have $\cos(kt)$ or $\sin(kt)$ for the time component, which would arise if the separation constant for $T(t)$ was negative, leading to a decaying exponential for $X(x)$ (not hyperbolic sines/cosines or real exponentials).
Option (C) has $e^{kt}$ for time, but $\cos$ and $\sin$ for space, which would arise if the separation constant for $X(x)$ was negative, i.e., $X''(x) = -\frac{k}{\alpha} X(x)$. This would lead to $X(x) = C_1 \cos(\sqrt{k/\alpha} x) + C_2 \sin(\sqrt{k/\alpha} x)$. This is also a valid form depending on boundary conditions, but option (B) is also a valid general form.Given the options, option (B) is a standard form for solutions to the heat equation, particularly for problems with infinite domains or specific boundary conditions where exponential growth/decay in space is possible.