To solve this first-order partial differential equation $2u_t + 3u_x = 0$ with the auxiliary condition $u = \sin(\sin(x))$ when $t = 0$, we can use the method of characteristics.

Step 1: Write Down the Characteristic Equations

Given the PDE: $2u_t + 3u_x = 0,$ we can express this in terms of characteristics: $\frac{dt}{2} = \frac{dx}{3} = \frac{du}{0}.$

Step 2: Solve the Characteristic Equations

From $\frac{du}{0} = 0$, we find that $u$ is constant along the characteristics. Therefore, $u$ does not change along each characteristic line.

From $\frac{dt}{2} = \frac{dx}{3}$, we solve for the characteristic lines: $\frac{dx}{dt} = \frac{3}{2}.$ This gives: $x = \frac{3}{2}t + C,$ where $C$ is a constant that labels each characteristic.

Step 3: Apply the Initial Condition

At $t = 0$, we have the initial condition $u(x, 0) = \sin(\sin(x))$. This implies that along each characteristic line labeled by $C$, the solution $u$ remains constant and is given by the initial condition at $t = 0$: $u(x, t) = \sin(\sin(C)) = \sin(\sin\left(x - \frac{3}{2}t\right)).$

Final Solution

Thus, the solution to the PDE is: $u(x, t) = \sin\left(\sin\left(x - \frac{3}{2}t\right)\right).$

Would you like further explanation or details on any of these steps?

Related Questions:

1. How does the method of characteristics apply to different types of first-order PDEs?
2. What are the limitations of using characteristics in solving PDEs?
3. Can the solution to this PDE be interpreted in terms of a wave or transport phenomenon?
4. What other initial or boundary conditions could be applied to this type of PDE?
5. How does the solution change if the coefficients of $u_t$ and $u_x$ vary with time or space?

Tip:

For first-order linear PDEs, the method of characteristics is often a powerful tool, as it transforms the PDE into a set of ODEs along specific curves in the $(x, t)$ plane.