Variational Problem and Stationary Functions

We are asked to find the stationary functions (extremals) for the variational problem defined by the functional:

\(J(y(x)) = \displaystyle\int_0^1 [(y')^2 − y|y| y' + xy] dx\)

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

The integrand is \(F(x, y, y') = (y')^2 − y|y| y' + xy\).

Euler-Lagrange Equation for Stationary Functions

Stationary functions must satisfy the Euler-Lagrange equation:

\(\dfrac{\partial F}{\partial y} − \dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right) = 0\)

We need to calculate the partial derivatives of \(F\) with respect to \(y\) and \(y'\). The term \(|y|\) requires us to consider the cases where \(y > 0\) and \(y < 0\).

Case 1: \(y > 0\)

If \(y > 0\), then \(|y| = y\). The integrand becomes \(F = (y')^2 − y^2 y' + xy\).

• \(\dfrac{\partial F}{\partial y} = −2yy' + x\)
• \(\dfrac{\partial F}{\partial y'} = 2y' − y^2\)
• \(\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right) = \dfrac{d}{dx}(2y' − y^2) = 2y'' − 2yy'\)

Substituting into the Euler-Lagrange equation:

\(−2yy' + x − (2y'' − 2yy') = 0\)

\(−2yy' + x − 2y'' + 2yy' = 0\)

\(x − 2y'' = 0 \implies 2y'' = x\)

Case 2: \(y < 0\)

If \(y < 0\), then \(|y| = −y\). The integrand becomes \(F = (y')^2 − y(−y) y' + xy = (y')^2 + y^2 y' + xy\).

• \(\dfrac{\partial F}{\partial y} = 2yy' + x\)
• \(\dfrac{\partial F}{\partial y'} = 2y' + y^2\)
• \(\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right) = \dfrac{d}{dx}(2y' + y^2) = 2y'' + 2yy'\)

Substituting into the Euler-Lagrange equation:

\(2yy' + x − (2y'' + 2yy') = 0\)

\(2yy' + x − 2y'' − 2yy' = 0\)

\(x − 2y'' = 0 \implies 2y'' = x\)

Case 3: \(y = 0\)

If \(y \equiv 0\), then \(y' = 0\) and \(y'' = 0\). The Euler-Lagrange equation \(2y'' = x\) becomes \(0 = x\), which is not true for all \(x \in [0, 1]\). This suggests that \(y \equiv 0\) is not a solution satisfying the Euler-Lagrange equation in the classical sense on the entire interval.

To formally check if \(y \equiv 0\) is a stationary function, we can examine the first variation of the functional around \(y \equiv 0\). Let \(y(x) = 0 + \epsilon\eta(x)\), where \(\eta(0) = \eta(1) = 0\). The Gateaux variation at \(y=0\) in the direction \(\eta\) is given by:

\(\delta J(0, \eta) = \left.\dfrac{d}{d\epsilon} J(\epsilon\eta)\right|_{\epsilon=0}\)

\(J(\epsilon\eta) = \int_0^1 [(\epsilon\eta')^2 − (\epsilon\eta)|\epsilon\eta|(\epsilon\eta') + x(\epsilon\eta)] dx = \int_0^1 [\epsilon^2(\eta')^2 − \epsilon^3 \eta|\eta|\eta' + x\epsilon\eta] dx\)

\(\dfrac{J(\epsilon\eta)}{\epsilon} = \int_0^1 [\epsilon(\eta')^2 − \epsilon^2 \eta|\eta|\eta' + x\eta] dx\)

\(\lim_{\epsilon\to 0} \dfrac{J(\epsilon\eta) - J(0)}{\epsilon} = \int_0^1 x\eta dx\) (since \(J(0) = \int_0^1 0 dx = 0\)).

For \(y \equiv 0\) to be a stationary function, the first variation must be zero for all admissible \(\eta\). \(\int_0^1 x\eta dx = 0\) for all \(\eta\) with \(\eta(0)=\eta(1)=0\). This implies \(x = 0\) on \((0, 1)\), which is false. Therefore, \(y \equiv 0\) is not a stationary function.

Solving the Euler-Lagrange Equation

Away from \(y=0\), the Euler-Lagrange equation is \(2y'' = x\). We solve this second-order linear ordinary differential equation:

\(y'' = \dfrac{x}{2}\)

Integrate once:

\(y' = \int \dfrac{x}{2} dx = \dfrac{x^2}{4} + C_1\)

Integrate again:

\(y = \int \left(\dfrac{x^2}{4} + C_1\right) dx = \dfrac{x^3}{12} + C_1x + C_2\)

Applying Boundary Conditions

We apply the boundary conditions \(y(0) = 0\) and \(y(1) = 0\):

• \(y(0) = \dfrac{0^3}{12} + C_1(0) + C_2 = C_2\). Since \(y(0) = 0\), we get \(C_2 = 0\).
• \(y(1) = \dfrac{1^3}{12} + C_1(1) + C_2 = \dfrac{1}{12} + C_1 + 0\). Since \(y(1) = 0\), we get \(\dfrac{1}{12} + C_1 = 0\), which means \(C_1 = −\dfrac{1}{12}\).

Substituting the values of \(C_1\) and \(C_2\), the unique solution to the ODE with boundary conditions is:

\(y(x) = \dfrac{x^3}{12} − \dfrac{x}{12} = \dfrac{x(x^2 − 1)}{12}\)

Consistency Check and Uniqueness

We found this solution by assuming \(y \neq 0\) and derived \(2y''=x\). Let's check the sign of this solution on \([0,1]\). For \(x \in (0,1)\), \(x > 0\), \(x^2-1 < 0\). Thus, \(y(x) = \frac{x(x^2-1)}{12} < 0\) for \(x \in (0,1)\). At \(x=0\) and \(x=1\), \(y(x)=0\).

Since \(y(x) \le 0\) for all \(x \in [0,1]\), this solution is consistent with the derivation of the Euler-Lagrange equation for the case \(y < 0\).

Could there be other solutions, particularly ones that cross the \(y=0\) axis? The Euler-Lagrange equation derivation relies on sufficient smoothness of the integrand \(F(x, y, y')\). The term \(y|y|\) makes \(\dfrac{\partial F}{\partial y}\) potentially discontinuous at \(y=0\). However, the term \(\dfrac{\partial F}{\partial y'} = 2y' - y|y|\) is continuous if \(y\) and \(y'\) are continuous. For a function to be a smooth extremal, the condition for a possible "corner" (where \(y'\) might be discontinuous) is the continuity of \(\dfrac{\partial F}{\partial y'}\). If a solution crosses \(y=0\) at \(x_0 \in (0,1)\), i.e., \(y(x_0) = 0\), the continuity of \(\dfrac{\partial F}{\partial y'}\) at \(x_0\) implies \(2y'(x_0^-) - y(x_0^-)|y(x_0^-)| = 2y'(x_0^+) - y(x_0^+)|y(x_0^+)|\). Since \(y(x_0^-) = y(x_0^+) = 0\), this simplifies to \(2y'(x_0^-) = 2y'(x_0^+)\), which means \(y'\) must be continuous at \(x_0\). Thus, any stationary function must be continuously differentiable (\(C^1\)). Since \(2y''=x\) is a second-order ODE, any \(C^1\) solution to this ODE must be \(C^2\). This implies stationary functions must be smooth (\(C^2\)).

As shown in the derivation, any smooth solution \(y(x)\) away from \(y=0\) must satisfy \(2y'' = x\). We showed that the unique smooth solution to \(2y''=x\) with the given boundary conditions is \(y(x) = \frac{x^3-x}{12}\). This solution is not identically zero, satisfies the boundary conditions, and satisfies the required Euler-Lagrange equation (specifically the \(y < 0\) case) over the interval \((0,1)\) because the solution itself is negative there. The analysis regarding crossing \(y=0\) shows that no other smooth solution satisfying the boundary conditions exists.

Therefore, the variational problem has a unique stationary function (extremal) which is not identically equal to 0, given by \(y(x) = \frac{x^3-x}{12}\).

Evaluating the Options

• Option 1: (P) has no stationary function (extremal). Incorrect, we found one candidate.
• Option 2: y ≡ 0 is the only stationary function (extremal) for (P). Incorrect, \(y \equiv 0\) is not a stationary function.
• Option 3: (P) has a unique stationary function (extremal) y not identically equal to 0. Correct, we found a unique non-zero candidate that satisfies the conditions.
• Option 4: (P) has infinitely many stationary functions (extremal). Incorrect, the derived ODE with boundary conditions yields a unique solution.