Question

question 3
using the method of separation of variables, solve the following initial - value problem:
$\frac{partial u}{partial t}=3\frac{partial u}{partial x}$; $u(x,0)=8e^{-2x}$.

Explanation:

Step1: Separate variables

Assume $u(x,t)=X(x)T(t)$. Substitute into the PDE $\frac{\partial u}{\partial t}=3\frac{\partial u}{\partial x}$, we get $X(x)T'(t)=3X'(x)T(t)$. Then $\frac{T'(t)}{3T(t)}=\frac{X'(x)}{X(x)} = -\lambda$ (a separation - constant).
So we have two ODEs: $T'(t)+ 3\lambda T(t)=0$ and $X'(x)+\lambda X(x)=0$.

Step2: Solve ODE for $T(t)$

The general solution of $T'(t)+3\lambda T(t)=0$ is $T(t)=C_1e^{-3\lambda t}$.

Step3: Solve ODE for $X(x)$

The general solution of $X'(x)+\lambda X(x)=0$ is $X(x)=C_2e^{-\lambda x}$.
So $u(x,t)=Ce^{-\lambda x - 3\lambda t}$, where $C = C_1C_2$.

Step4: Use initial - condition

Given $u(x,0)=8e^{-2x}$, substitute $t = 0$ into $u(x,t)$: $u(x,0)=Ce^{-\lambda x}=8e^{-2x}$.
By comparing the coefficients, we have $\lambda = 2$ and $C = 8$.

Answer:

$u(x,t)=8e^{-2x-6t}$