The given problem is a first-order partial differential equation:

$2u_t + 3u_x = 0$

with the auxiliary condition:

$u(x, 0) = \sin(\sin(x)).$

This is a linear first-order PDE, and we can solve it using the method of characteristics. The general form of a linear first-order PDE is:

$a u_t + b u_x = 0,$

where $a = 2$ and $b = 3$ in this case.

Step 1: Solve the characteristic equations

The characteristic equations are given by:

$\frac{dt}{a} = \frac{dx}{b} = \frac{du}{0}.$

Substituting $a = 2$ and $b = 3$, we get:

$\frac{dt}{2} = \frac{dx}{3} = \frac{du}{0}.$

From $\frac{du}{0}$, we see that $u$ remains constant along the characteristic curves.

Next, solve for the characteristic curves:

$\frac{dt}{2} = \frac{dx}{3} \implies 3 dt = 2 dx \implies x - \frac{3}{2}t = \text{constant}.$

Thus, the characteristic curves are given by:

$x - \frac{3}{2} t = \xi,$

where $\xi$ is a constant along each characteristic.

Step 2: General solution

Since $u$ is constant along each characteristic, the solution must be a function of $x - \frac{3}{2}t$, i.e.,

$u(x,t) = f\left(x - \frac{3}{2}t\right),$

where $f$ is determined by the initial condition.

Step 3: Apply the initial condition

The initial condition is $u(x,0) = \sin(\sin(x))$. Using this in the general solution:

$u(x,0) = f(x) = \sin(\sin(x)).$

Thus, the function $f(x)$ is:

$f(x) = \sin(\sin(x)).$

Therefore, the solution to the PDE is:

$u(x,t) = \sin\left(\sin\left(x - \frac{3}{2}t\right)\right).$

Final Answer:

The solution for the PDE $2u_t + 3u_x = 0$ with the auxiliary condition $u(x,0) = \sin(\sin(x))$ is:

$u(x,t) = \sin\left(\sin\left(x - \frac{3}{2}t\right)\right).$

Would you like further explanation or have any questions about the method used?

Here are 5 related questions to extend this:

1. How do characteristic curves help in solving first-order PDEs?
2. What is the physical interpretation of characteristic lines?
3. How would the solution change if the coefficients of $u_t$ and $u_x$ were different?
4. Can this method be applied to nonlinear first-order PDEs?
5. What happens if a boundary condition is added to this problem?

Tip: For solving first-order linear PDEs, the method of characteristics is a powerful tool that reduces the partial differential equation to ordinary differential equations along characteristic curves.