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
Extremal of the Functional J(y)
The problem asks for the extremal of the given functional \(J(y)\) subject to specific boundary conditions. An extremal function for a functional can be found by solving the Euler-Lagrange equation.
Functional and Lagrangian
The given functional is:
\(J(y)=\int_0^1\left[2\left(y^{\prime}\right)^2+x y\right] d x\)
The Lagrangian function \(L(x, y, y')\) is the integrand of the functional:
\(L(x, y, y') = 2(y')^2 + xy\)
The boundary conditions are \(y(0) = 0\) and \(y(1) = 1\).
Euler-Lagrange Equation
The Euler-Lagrange equation for a functional of the form \(\int_{x_1}^{x_2} L(x, y, y') dx\) is:
\(\frac{\partial L}{\partial y} - \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) = 0\)
First, we calculate the required partial derivatives of the Lagrangian \(L\):
- Partial derivative with respect to \(y\):
- \(\frac{\partial L}{\partial y} = \frac{\partial}{\partial y}(2(y')^2 + xy) = x\)
- Partial derivative with respect to \(y'\):
- \(\frac{\partial L}{\partial y'} = \frac{\partial}{\partial y'}(2(y')^2 + xy) = 4y'\)
Next, we calculate the total derivative with respect to \(x\) of \(\frac{\partial L}{\partial y'}\):
- \(\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) = \frac{d}{dx}(4y') = 4y''\)
Now, substitute these into the Euler-Lagrange equation:
\(x - 4y'' = 0\)
This gives us a second-order ordinary differential equation:
\(4y'' = x\)
\(y'' = \frac{x}{4}\)
Solving the Differential Equation
We integrate the differential equation twice to find \(y(x)\).
Integrating \(y''\) once gives \(y'\):
\(y' = \int y'' dx = \int \frac{x}{4} dx = \frac{x^2}{8} + C_1\)
Integrating \(y'\) once gives \(y\):
\(y = \int y' dx = \int \left(\frac{x^2}{8} + C_1\right) dx = \frac{x^3}{24} + C_1 x + C_2\)
Here, \(C_1\) and \(C_2\) are constants of integration.
Applying Boundary Conditions
We use the given boundary conditions \(y(0) = 0\) and \(y(1) = 1\) to find the values of \(C_1\) and \(C_2\).
- Using \(y(0) = 0\):
- \(y(0) = \frac{0^3}{24} + C_1(0) + C_2 = 0\)
- \(0 + 0 + C_2 = 0 \implies C_2 = 0\)
- Using \(y(1) = 1\):
- \(y(1) = \frac{1^3}{24} + C_1(1) + C_2 = 1\)
- \(\frac{1}{24} + C_1 + 0 = 1\)
- \(C_1 = 1 - \frac{1}{24} = \frac{24 - 1}{24} = \frac{23}{24}\)
Extremal Function
Substitute the values of \(C_1\) and \(C_2\) back into the expression for \(y(x)\):
\(y(x) = \frac{x^3}{24} + \frac{23}{24} x + 0\)
\(y(x) = \frac{x^3}{24} + \frac{23 x}{24}\)
This function is the extremal of the given functional that satisfies the boundary conditions.
Comparing with Options
Let's compare our solution with the provided options:
- Option 1: \(y=\frac{x^2}{12}+\frac{11 x}{12}\) - Does not match.
- Option 2: \(y=\frac{x^3}{3}+\frac{2 x^2}{3}\) - Does not match.
- Option 3: \(y=\frac{x^2}{7}+\frac{6 x}{7}\) - Does not match.
- Option 4: \(y=\frac{x^3}{24}+\frac{23 x}{24}\) - Matches our derived solution.
Therefore, the extremal of the functional is \(y=\frac{x^3}{24}+\frac{23 x}{24}\).
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
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\(J(y)=\int_0^5\left[y^2+x^3 y^{\prime}\right] d x\)
define a solution curve in the (x, y) - plane which is
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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?
