Question

Consider the one–dimensional wave (advection) equation \[ \frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=0, -\infty<x<\infty,\ t\ge 0. \] For the initial condition \(u(x,0)=e^{-x^{2}}\), the solution at \(t=1\) is:

• A. \(u(x,1)=e^{-(x-1)^2}\)
• B. \(u(x,1)=e^{-1}\)
• C. \(u(x,1)=e^{-x^{2}}\)
• D. \(u(x,1)=e^{-(x+1)^2}\)

Hint:

For linear advection \(u_t+c\,u_x=0\), simply shift the initial profile: \(u(x,t)=u_0(x-ct)\).

Answer 1

The Correct Option is A

Step 1: Characteristics.
For \(u_t+u_x=0\), the characteristic ODEs are \[ \frac{dx}{dt}=1, \frac{du}{dt}=0. \] Hence \(x-t=\text{const}\) and \(u\) is constant along each line \(x-t=c\).
\(\Rightarrow\) The general solution is \(u(x,t)=f(x-t)\) for some profile \(f\).

Step 2: Fit the initial data.
At \(t=0\), \(u(x,0)=f(x)=e^{-x^2}\).
Therefore \(u(x,t)=e^{-(x-t)^2}\).

Step 3: Evaluate at \(t=1\).
\[ u(x,1)=e^{-(x-1)^2}. \] (Check) \(u_t=2(x-t)e^{-(x-t)^2}(-1)= -2(x-t)e^{-(x-t)^2}\), \(u_x=2(x-t)e^{-(x-t)^2}\). Hence \(u_t+u_x=0\) as required.

Final Answer:
\[ \boxed{u(x,1)=e^{-(x-1)^2}} \]