Which of the following is an extremal of the functional \(I\left( y \right) = \int_{ - 1}^1 {\left( {{y^{'2}} - 2xy} \right)} \,dx\) that satisfies the boundary conditions y(-1) = -1 and y(1) = 1?
Extremal of a Functional
To find the extremal of the given functional \(I\left( y \right) = \int_{ - 1}^1 {\left( {{y^{'2}} - 2xy} \right)} \,dx\), we use the Euler-Lagrange equation. The functional is in the form \(I(y) = \int_{x_1}^{x_2} F(x, y, y') \,dx\), where \(F(x, y, y') = y'^2 - 2xy\), the integration limits are \(x_1 = -1\) and \(x_2 = 1\), and the boundary conditions are \(y(-1) = -1\) and \(y(1) = 1\).
Euler-Lagrange Equation
The Euler-Lagrange equation is the necessary condition for a function \(y(x)\) to be an extremal of the functional \(I(y)\). The equation is given by:
\[ \frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0 \]Finding the Differential Equation
First, we find the partial derivatives of \(F(x, y, y') = y'^2 - 2xy\) with respect to \(y\) and \(y'\):
- Partial derivative with respect to \(y\): \[ \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(y'^2 - 2xy) = -2x \]
- Partial derivative with respect to \(y'\): \[ \frac{\partial F}{\partial y'} = \frac{\partial}{\partial y'}(y'^2 - 2xy) = 2y' \]
Now, substitute these into the Euler-Lagrange equation:
\[ -2x - \frac{d}{dx}(2y') = 0 \] \[ -2x - 2y'' = 0 \]Dividing by -2, we get the second-order differential equation for the extremal \(y(x)\):
\[ y'' = -x \]Solving the Differential Equation
We integrate the differential equation \(y'' = -x\) twice to find the general form of the extremal function \(y(x)\).
First integration to find \(y'\):
\[ y' = \int (-x) \,dx = -\frac{x^2}{2} + C_1 \]Second integration to find \(y(x)\):
\[ y(x) = \int \left(-\frac{x^2}{2} + C_1\right) \,dx = -\frac{x^3}{6} + C_1 x + C_2 \]Here, \(C_1\) and \(C_2\) are constants of integration.
Applying Boundary Conditions
We use the given boundary conditions \(y(-1) = -1\) and \(y(1) = 1\) to find the values of \(C_1\) and \(C_2\).
- Using the condition \(y(-1) = -1\): \[ -1 = -\frac{(-1)^3}{6} + C_1(-1) + C_2 \] \[ -1 = -\frac{-1}{6} - C_1 + C_2 \] \[ -1 = \frac{1}{6} - C_1 + C_2 \] \[ -1 - \frac{1}{6} = -C_1 + C_2 \] \[ -\frac{7}{6} = -C_1 + C_2 \quad \text{(Equation 1)} \]
- Using the condition \(y(1) = 1\): \[ 1 = -\frac{(1)^3}{6} + C_1(1) + C_2 \] \[ 1 = -\frac{1}{6} + C_1 + C_2 \] \[ 1 + \frac{1}{6} = C_1 + C_2 \] \[ \frac{7}{6} = C_1 + C_2 \quad \text{(Equation 2)} \]
Now we have a system of two linear equations for \(C_1\) and \(C_2\):
| Equation | Expression |
|---|---|
| 1 | \( -C_1 + C_2 = -\frac{7}{6} \) |
| 2 | \( C_1 + C_2 = \frac{7}{6} \) |
Adding Equation 1 and Equation 2:
\[ (-C_1 + C_2) + (C_1 + C_2) = -\frac{7}{6} + \frac{7}{6} \] \[ 2C_2 = 0 \] \[ C_2 = 0 \]Substitute \(C_2 = 0\) into Equation 2:
\[ C_1 + 0 = \frac{7}{6} \] \[ C_1 = \frac{7}{6} \]The Extremal Function
Substitute the values of \(C_1\) and \(C_2\) back into the general solution for \(y(x)\):
\[ y(x) = -\frac{x^3}{6} + \frac{7}{6}x + 0 \] \[ y(x) = -\frac{x^3}{6} + \frac{7x}{6} \]This function is the extremal of the given functional that satisfies the boundary conditions.
Important Questions from Euler-lagrange Equation
The minimum value of the functional
\(I(y)=\int^{\pi}_0\left(\frac{dy}{dx} \right)^2 dx,\)
subject to \(\int^{\pi}_0 y^2(x)dx = 1, y(0) = 0 = у(π)\) is equal to
Let y = y(x) ∈ C4 ([0, 1]) be an extremizing function for the functional \(I(y)=\int^{1}_0\left[\left(\frac{d^2y}{dx^2} \right)^2 -2y\right] dx,\) satisfying y(0) = 0 = y(1). Then an extremal y(x), satisfying the given conditions at 0 and 1 together with the natural boundary conditions, is given by
The extremal of the functional
\(J(y)=\int_0^1\left[2\left(y^{\prime}\right)^2+x y\right] d x,\) y(0) = 0, y(1) = 1, y ∈ C2[0, 1] is
The Euler equations satisfied by the extremals of the functional
\(J(y)=\int_0^5\left[y^2+x^3 y^{\prime}\right] d x\)
define a solution curve in the (x, y) - plane which is
Let B be the unit ball in \(\mathbb{R}\)3 centered at origin. The Euler-Lagrange equation corresponding to the functional \(I(u)=\frac{1}{2} \int_B|\nabla u|^2 d x-\frac{1}{5} \int_B|u|^5 d x\) is
