We are asked to find the extremal function \(y(x)\) for a given functional and then determine which statements about this function are true. The functional is:

\(J(y(x)) = \displaystyle\int_0^1 (y(x)^2 - 4y(x)y'(x) + 4y'(x)^2)dx\)

with boundary conditions \(y(0) = 1\) and \(y'(0) = 1/2\).

This is a problem in the calculus of variations. The function \(y(x)\) that extremizes the functional must satisfy the Euler-Lagrange equation:

\(\frac{\partial f}{\partial y} - \frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right) = 0\)

where \(f(y, y', x) = y^2 - 4yy' + 4(y')^2\).

Finding the Euler-Lagrange Equation

First, we calculate the partial derivatives of \(f\) with respect to \(y\) and \(y'\):

• \(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(y^2 - 4yy' + 4(y')^2) = 2y - 4y'\)
• \(\frac{\partial f}{\partial y'} = \frac{\partial}{\partial y'}(y^2 - 4yy' + 4(y')^2) = -4y + 8y'\)

Next, we calculate the total derivative of \(\frac{\partial f}{\partial y'}\) with respect to \(x\):

\(\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right) = \frac{d}{dx}(-4y + 8y') = -4\frac{dy}{dx} + 8\frac{dy'}{dx} = -4y' + 8y''\)

Now, substitute these into the Euler-Lagrange equation:

\((2y - 4y') - (-4y' + 8y'') = 0\)

\(2y - 4y' + 4y' - 8y'' = 0\)

\(2y - 8y'' = 0\)

\(y - 4y'' = 0\)

\(y'' - \frac{1}{4}y = 0\)

Solving the Differential Equation

We have a second-order linear homogeneous differential equation: \(y'' - \frac{1}{4}y = 0\). The characteristic equation is \(r^2 - \frac{1}{4} = 0\). This gives \(r^2 = \frac{1}{4}\), so \(r = \pm \frac{1}{2}\).

The general solution is of the form:

\(y(x) = c_1 e^{x/2} + c_2 e^{-x/2}\)

where \(c_1\) and \(c_2\) are constants determined by the boundary conditions.

Applying Boundary Conditions

We are given \(y(0) = 1\) and \(y'(0) = 1/2\).

First, apply \(y(0) = 1\) to the general solution:

\(y(0) = c_1 e^{0/2} + c_2 e^{-0/2} = c_1 e^0 + c_2 e^0 = c_1 + c_2\)

\(c_1 + c_2 = 1\) (Equation 1)

Next, find the derivative of \(y(x)\):

\(y'(x) = \frac{d}{dx}(c_1 e^{x/2} + c_2 e^{-x/2}) = c_1 \cdot \frac{1}{2} e^{x/2} + c_2 \cdot (-\frac{1}{2}) e^{-x/2}\)

\(y'(x) = \frac{1}{2}c_1 e^{x/2} - \frac{1}{2}c_2 e^{-x/2}\)

Apply \(y'(0) = 1/2\) to the derivative:

\(y'(0) = \frac{1}{2}c_1 e^{0/2} - \frac{1}{2}c_2 e^{-0/2} = \frac{1}{2}c_1 e^0 - \frac{1}{2}c_2 e^0 = \frac{1}{2}c_1 - \frac{1}{2}c_2\)

\(\frac{1}{2}c_1 - \frac{1}{2}c_2 = \frac{1}{2}\)

Multiply by 2:

\(c_1 - c_2 = 1\) (Equation 2)

Now we solve the system of linear equations for \(c_1\) and \(c_2\):

\(c_1 + c_2 = 1\)

\(c_1 - c_2 = 1\)

Adding the two equations gives:

\((c_1 + c_2) + (c_1 - c_2) = 1 + 1\)

\(2c_1 = 2\)

\(c_1 = 1\)

Substitute \(c_1 = 1\) into Equation 1:

\(1 + c_2 = 1\)

\(c_2 = 0\)

So the extremal function is \(y(x) = 1 \cdot e^{x/2} + 0 \cdot e^{-x/2} = e^{x/2}\).

Analyzing the Statements

Now we check which of the given statements are true for \(y(x) = e^{x/2}\) on the interval \([0, 1]\).

Convex Function or Concave Function

To determine if \(y(x)\) is convex or concave, we need to find its second derivative.

\(y(x) = e^{x/2}\)

\(y'(x) = \frac{1}{2}e^{x/2}\)

\(y''(x) = \frac{d}{dx}(\frac{1}{2}e^{x/2}) = \frac{1}{2} \cdot \frac{1}{2} e^{x/2} = \frac{1}{4}e^{x/2}\)

For \(x \in [0, 1]\), \(e^{x/2}\) is always positive. Therefore, \(y''(x) = \frac{1}{4}e^{x/2}\) is always positive on \([0, 1]\).

Since \(y''(x) > 0\), the function \(y(x)\) is a convex function on \([0, 1]\).

• Statement 1: \(y\) is a convex function. This is true.
• Statement 2: \(y\) is a concave function. This is false.

Functional Properties of y(x)

Let's check the given functional equations for \(y(x) = e^{x/2}\).

• Statement 3: \(y(x_1 + x_2) = y(x_1) y(x_2)\) for all \(x_1, x_2\) in \([0, 1]\).

  Left side: \(y(x_1 + x_2) = e^{(x_1 + x_2)/2} = e^{x_1/2 + x_2/2}\)

  Right side: \(y(x_1) y(x_2) = e^{x_1/2} \cdot e^{x_2/2} = e^{x_1/2 + x_2/2}\)

  The left side equals the right side. This statement is true.

• Statement 4: \(y(x_1 x_2) = y(x_1) + y(x_2)\) for all \(x_1, x_2\) in \([0, 1]\).

  Left side: \(y(x_1 x_2) = e^{(x_1 x_2)/2}\)

  Right side: \(y(x_1) + y(x_2) = e^{x_1/2} + e^{x_2/2}\)

  These are not equal in general. For example, if \(x_1 = 1\) and \(x_2 = 1\), LHS is \(y(1) = e^{1/2}\) and RHS is \(y(1) + y(1) = e^{1/2} + e^{1/2} = 2e^{1/2}\). Since \(e^{1/2} \neq 2e^{1/2}\), the statement is false.

Based on our analysis, the true statements are Statement 1 and Statement 3.