For u t+u x=0, -<x<, t0, and u(x,0)=e-x2, the solution at t=1 is

GATE 2023 AE – Question 13

Engineering Mathematics · Differential Equations: Classification of PDEs, wave, Laplace and heat equations by separation of variables · 1 mark · Multiple choice

For $u_t+u_x=0$, $-\infty<x<\infty$, $t\ge0$, and $u(x,0)=e^{-x^2}$, the solution at t=1 is

$e^{-(x-1)^2}$
$e^{-1}$
$e^{-x^2}$
$e^{-(x+1)^2}$

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Correct answer: (A) $e^{-(x-1)^2}$

Explanation

The equation $u_t+u_x=0$ is the one-dimensional **linear advection (wave) equation** with speed 1.

**Characteristics.** Along the curves $\dfrac{dx}{dt}=1$ the solution is constant, because $\dfrac{du}{dt}=u_t+u_x\dfrac{dx}{dt}=0$. The characteristics are $x-t=\text{constant}$.

**Solution.** A value of $u$ at $(x,t)$ equals the initial value at the foot of the characteristic, $x_0=x-t$:
$$u(x,t)=u(x-t,0)=e^{-(x-t)^2}.$$

At $t=1$: $u(x,1)=e^{-(x-1)^2}$ (option A). The initial pulse has moved one unit to the right.