Variational Problem and Euler-Lagrange Equations

The problem asks for the stationary functions (extremals) of a given variational problem. The functional is:

\(J(y(x), z(x)) = \displaystyle\int_0^1 [(y')^2 + (z')^2 + y'z']dx\)

The boundary conditions are \(y(0) = 1\), \(y(1) = 0\), \(z(0) = -1\), and \(z(1) = 2\).

The integrand is \(F(y', z') = (y')^2 + (z')^2 + y'z'\). Since the integrand \(F\) depends only on \(y'\) and \(z'\), the Euler-Lagrange equations for stationary functions are given by:

• \(\frac{\partial F}{\partial y'} = c_1\) (constant)
• \(\frac{\partial F}{\partial z'} = c_2\) (constant)

Let's calculate the partial derivatives of \(F\):

\(\frac{\partial F}{\partial y'} = \frac{\partial}{\partial y'} [(y')^2 + (z')^2 + y'z'] = 2y' + z'\)

\(\frac{\partial F}{\partial z'} = \frac{\partial}{\partial z'} [(y')^2 + (z')^2 + y'z'] = 2z' + y'\)

So, the Euler-Lagrange equations become:

1. \(2y' + z' = c_1\)
2. \(y' + 2z' = c_2\)

Solving the System of Equations

We have a system of two linear first-order differential equations with constant coefficients. We can solve for \(y'\) and \(z'\).

Multiply the first equation by 2 and subtract the second equation:

\(2(2y' + z') - (y' + 2z') = 2c_1 - c_2\)

\(4y' + 2z' - y' - 2z' = 2c_1 - c_2\)

\(3y' = 2c_1 - c_2\)

\(y' = \frac{2c_1 - c_2}{3}\)

Let \(A = \frac{2c_1 - c_2}{3}\). So, \(y'(x) = A\), where \(A\) is a constant.

Multiply the second equation by 2 and subtract the first equation:

\(2(y' + 2z') - (2y' + z') = 2c_2 - c_1\)

\(2y' + 4z' - 2y' - z' = 2c_2 - c_1\)

\(3z' = 2c_2 - c_1\)

\(z' = \frac{2c_2 - c_1}{3}\)

Let \(B = \frac{2c_2 - c_1}{3}\). So, \(z'(x) = B\), where \(B\) is a constant.

Finding Stationary Functions y(x) and z(x)

Integrating \(y'(x) = A\) and \(z'(x) = B\) with respect to \(x\), we get:

\(y(x) = Ax + D_1\)

\(z(x) = Bx + D_2\)

where \(D_1\) and \(D_2\) are integration constants.

Applying Boundary Conditions

We use the given boundary conditions to find the values of the constants \(A\), \(D_1\), \(B\), and \(D_2\).

For \(y(x)\):

• \(y(0) = 1 \Rightarrow A(0) + D_1 = 1 \Rightarrow D_1 = 1\)
• \(y(1) = 0 \Rightarrow A(1) + D_1 = 0 \Rightarrow A + 1 = 0 \Rightarrow A = -1\)

So, \(y(x) = -x + 1 = 1 - x\).

For \(z(x)\):

• \(z(0) = -1 \Rightarrow B(0) + D_2 = -1 \Rightarrow D_2 = -1\)
• \(z(1) = 2 \Rightarrow B(1) + D_2 = 2 \Rightarrow B - 1 = 2 \Rightarrow B = 3\)

So, \(z(x) = 3x - 1\).

The stationary functions are \(y(x) = 1 - x\) and \(z(x) = 3x - 1\).

Checking the Statements

Now, let's check each statement using \(y(x) = 1 - x\) and \(z(x) = 3x - 1\).

Statement 1: \(z(x) + 3y(x) = 2\) for \(x \in [0, 1]\)

Substitute \(y(x)\) and \(z(x)\):

\((3x - 1) + 3(1 - x) = 3x - 1 + 3 - 3x = 2\)

The statement is correct.

Statement 2: \(3z(x) + y(x) = 2\) for \(x \in [0, 1]\)

Substitute \(y(x)\) and \(z(x)\):

\(3(3x - 1) + (1 - x) = 9x - 3 + 1 - x = 8x - 2\)

This is not equal to 2 for all \(x \in [0, 1]\). The statement is incorrect.

Statement 3: \(y(x) + z(x) = 2x\) for \(x \in [0, 1]\)

Substitute \(y(x)\) and \(z(x)\):

\((1 - x) + (3x - 1) = 1 - x + 3x - 1 = 2x\)

The statement is correct.

Statement 4: \(y(x) + z(x) = x\) for \(x \in [0, 1]\)

Substitute \(y(x)\) and \(z(x)\):

\((1 - x) + (3x - 1) = 2x\)

This is not equal to \(x\) for all \(x \in [0, 1]\). The statement is incorrect.

Based on the analysis, Statements 1 and 3 are correct.