Divide Monomials #2 Worksheet for 6th - 7th Grade | Lesson Planet - Free Printable

Problem Description:

The task involves analyzing a system of equations and determining whether the given conditions are consistent. Specifically, we need to verify if the following equations hold true under the given constraints:

1. \( \frac{\partial u}{\partial t} = -u_x \)
2. \( \frac{\partial v}{\partial t} = -v_x \)
3. \( u(x, 0) = v(x, 0) \)
4. \( u(0, t) = v(0, t) \)

Additionally, we are asked to determine if the condition \( u(x, t) = v(x, t) \) for all \( x \) and \( t \) is satisfied.

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Step-by-Step Solution:

#### Step 1: Analyze the PDEs
The given partial differential equations (PDEs) are:
1. \( \frac{\partial u}{\partial t} = -u_x \)
2. \( \frac{\partial v}{\partial t} = -v_x \)

These are first-order linear hyperbolic PDEs, which describe the transport of quantities \( u \) and \( v \) in the negative \( x \)-direction with constant speed. The general solution to such PDEs can be found using the method of characteristics.

#### Step 2: Solve the PDEs Using Characteristics
For the PDE \( \frac{\partial u}{\partial t} = -u_x \), the characteristic equations are:
- \( \frac{dt}{ds} = 1 \implies t = s + C_1 \)
- \( \frac{dx}{ds} = -1 \implies x = -s + C_2 \)
- \( \frac{du}{ds} = 0 \implies u = C_3 \)

From these, we can express \( u \) as a function of the initial condition. Along the characteristic curve, \( x + t = \text{constant} \). Thus, the solution can be written as:
\[ u(x, t) = u_0(x + t) \]
where \( u_0 \) is the initial condition \( u(x, 0) \).

Similarly, for the PDE \( \frac{\partial v}{\partial t} = -v_x \), the solution is:
\[ v(x, t) = v_0(x + t) \]
where \( v_0 \) is the initial condition \( v(x, 0) \).

#### Step 3: Apply Initial Conditions
The initial conditions are:
1. \( u(x, 0) = v(x, 0) \)

This implies:
\[ u_0(x) = v_0(x) \]

Thus, the solutions become:
\[ u(x, t) = u_0(x + t) \]
\[ v(x, t) = u_0(x + t) \]
(since \( u_0(x) = v_0(x) \)).

#### Step 4: Apply Boundary Conditions
The boundary condition is:
\[ u(0, t) = v(0, t) \]

Substituting the solutions:
\[ u(0, t) = u_0(0 + t) = u_0(t) \]
\[ v(0, t) = u_0(0 + t) = u_0(t) \]

This condition is automatically satisfied because \( u(x, t) \) and \( v(x, t) \) are identical functions.

#### Step 5: Verify \( u(x, t) = v(x, t) \)
From the above analysis, we have:
\[ u(x, t) = u_0(x + t) \]
\[ v(x, t) = u_0(x + t) \]

Thus, \( u(x, t) = v(x, t) \) for all \( x \) and \( t \).

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Final Answer:

The condition \( u(x, t) = v(x, t) \) is satisfied for all \( x \) and \( t \). Therefore, the final answer is:
\[
\boxed{u(x, t) = v(x, t)}
\]