Functional Minimum Value Calculation

This problem asks for the minimum value of a functional \(I(y)\) subject to a constraint, which is a classic example of an isoperimetric problem in the calculus of variations.

The functional to minimize is \(I(y)=\int^{\pi}_0\left(\frac{dy}{dx} \right)^2 dx\).

The constraint is \(\int^{\pi}_0 y^2(x)dx = 1\).

The boundary conditions are \(y(0) = 0\) and \(y(\pi) = 0\).

Solving the Isoperimetric Problem

To solve this, we use the Euler-Lagrange equation for isoperimetric problems. We introduce a Lagrange multiplier \(\lambda\) and form the augmented Lagrangian function \(L\):

\(L(x, y, y') = \left(\frac{dy}{dx}\right)^2 + \lambda y^2 = (y')^2 + \lambda y^2\)

The Euler-Lagrange equation is given by:

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

First, calculate the partial derivatives:

• \(\frac{\partial L}{\partial y} = \frac{\partial}{\partial y}(\!(y')^2 + \lambda y^2\!) = 2\lambda y\)
• \(\frac{\partial L}{\partial y'} = \frac{\partial}{\partial y'}(\!(y')^2 + \lambda y^2\!) = 2y'\)

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

\(\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) = \frac{d}{dx}(2y') = 2y''\)

Substitute these into the Euler-Lagrange equation:

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

This simplifies to a second-order linear homogeneous differential equation:

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

Applying Boundary Conditions and Constraint

We need to solve \(y'' - \lambda y = 0\) subject to \(y(0)=0\) and \(y(\pi)=0\), and the constraint \(\int^{\pi}_0 y^2(x)dx = 1\).

The characteristic equation is \(r^2 - \lambda = 0\).

If \(\lambda > 0\), let \(\lambda = k^2\) (\(k > 0\)). The general solution is \(y(x) = A e^{kx} + B e^{-kx}\). Applying boundary conditions \(y(0)=0\) and \(y(\pi)=0\) leads to \(A=B=0\), which gives the trivial solution \(y(x)=0\). This does not satisfy the constraint.

If \(\lambda = 0\), the equation is \(y'' = 0\). The general solution is \(y(x) = Ax + B\). Applying boundary conditions \(y(0)=0\) and \(y(\pi)=0\) leads to \(A=B=0\), which gives the trivial solution \(y(x)=0\). This does not satisfy the constraint.

If \(\lambda < 0\), let \(\lambda = -k^2\) (\(k > 0\)). The characteristic equation is \(r^2 + k^2 = 0\), with roots \(r = \pm ik\). The general solution is \(y(x) = A \cos(kx) + B \sin(kx)\).

Apply boundary condition \(y(0)=0\):

\(y(0) = A \cos(0) + B \sin(0) = A = 0\)

So, the solution becomes \(y(x) = B \sin(kx)\).

Apply boundary condition \(y(\pi)=0\):

\(y(\pi) = B \sin(k\pi) = 0\)

For a non-trivial solution (where \(B \neq 0\)), we must have \(\sin(k\pi) = 0\). This occurs when \(k\pi = n\pi\) for some integer \(n\). Since \(k > 0\), we have \(k = n\), where \(n\) is a positive integer (\(n=1, 2, 3, \dots\)). If \(n=0\), \(k=0\), leading back to \(\lambda=0\).

So, the possible solutions are of the form \(y(x) = B \sin(nx)\) for \(n=1, 2, 3, \dots\).

Satisfying the Constraint

Now, we use the constraint \(\int^{\pi}_0 y^2(x)dx = 1\) to find the value of \(B\):

\(\int^{\pi}_0 (B \sin(nx))^2 dx = 1\)

\(B^2 \int^{\pi}_0 \sin^2(nx) dx = 1\)

Using the identity \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\):

\(B^2 \int^{\pi}_0 \frac{1 - \cos(2nx)}{2} dx = 1\)

\(B^2 \left[ \frac{x}{2} - \frac{\sin(2nx)}{4n} \right]_0^{\pi} = 1\)

\(B^2 \left[ \left(\frac{\pi}{2} - \frac{\sin(2n\pi)}{4n}\right) - \left(\frac{0}{2} - \frac{\sin(0)}{4n}\right) \right] = 1\)

\(B^2 \left[ \frac{\pi}{2} - 0 - 0 + 0 \right] = 1\)

\(B^2 \frac{\pi}{2} = 1\)

\(B^2 = \frac{2}{\pi}\)

We can choose \(B = \sqrt{\frac{2}{\pi}}\). The functions satisfying the boundary conditions and the constraint are \(y_n(x) = \sqrt{\frac{2}{\pi}} \sin(nx)\) for \(n=1, 2, 3, \dots\).

Evaluating the Functional

Now, we evaluate the functional \(I(y)\) for these functions \(y_n(x)\):

\(y_n(x) = \sqrt{\frac{2}{\pi}} \sin(nx)\)

\(\frac{dy_n}{dx} = y_n'(x) = \sqrt{\frac{2}{\pi}} (n \cos(nx))\)

\(\left(\frac{dy_n}{dx}\right)^2 = \left(\sqrt{\frac{2}{\pi}} n \cos(nx)\right)^2 = \frac{2}{\pi} n^2 \cos^2(nx)\)

\(I(y_n) = \int^{\pi}_0 \frac{2}{\pi} n^2 \cos^2(nx) dx\)

\(I(y_n) = \frac{2n^2}{\pi} \int^{\pi}_0 \cos^2(nx) dx\)

Using the identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\):

\(I(y_n) = \frac{2n^2}{\pi} \int^{\pi}_0 \frac{1 + \cos(2nx)}{2} dx\)

\(I(y_n) = \frac{n^2}{\pi} \int^{\pi}_0 (1 + \cos(2nx)) dx\)

\(I(y_n) = \frac{n^2}{\pi} \left[ x + \frac{\sin(2nx)}{2n} \right]_0^{\pi}\)

\(I(y_n) = \frac{n^2}{\pi} \left[ \left(\pi + \frac{\sin(2n\pi)}{2n}\right) - \left(0 + \frac{\sin(0)}{2n}\right) \right]\)

\(I(y_n) = \frac{n^2}{\pi} \left[ \pi + 0 - 0 - 0 \right]\)

\(I(y_n) = \frac{n^2}{\pi} \pi = n^2\)

The possible values of the functional are \(n^2\) for \(n=1, 2, 3, \dots\). These values are \(1^2=1, 2^2=4, 3^2=9, \dots\).

Minimum Value

The minimum value among the possible values \(1, 4, 9, \dots\) is the smallest one, which corresponds to \(n=1\).

The minimum value is \(1^2 = 1\).

The final answer is \(\boxed{1}\).