The cardinality of the set of extremals of \(\displaystyle J[y]=\int_0^1\left(y^{\prime}\right)^2 d x, \) subject to y(0) = 1, y(1) = 6, \(\displaystyle \int_0^1 y d x=3\) is
1
Understanding the Problem
We are asked to find the cardinality of the set of extremals for a given functional subject to boundary conditions and an integral constraint. This is a classic problem in the calculus of variations, specifically an isoperimetric problem.
The functional is given by \(\displaystyle J[y]=\int_0^1\left(y^{\prime}\right)^2 d x\).
The boundary conditions are \(y(0) = 1\) and \(y(1) = 6\).
The integral constraint is \(\displaystyle \int_0^1 y d x=3\).
Extremals for isoperimetric problems satisfy the Euler-Lagrange equation modified by a Lagrange multiplier associated with the integral constraint.
Applying the Euler-Lagrange Equation
For an isoperimetric problem with functional \(\displaystyle \int_a^b L(x, y, y') dx\) and constraint \(\displaystyle \int_a^b g(x, y, y') dx = C\), we introduce a Lagrange multiplier \(\lambda\) and consider the augmented Lagrangian \(L^*(x, y, y', \lambda) = L(x, y, y') + \lambda g(x, y, y')\). The extremals satisfy the Euler-Lagrange equation for \(L^*\):
\[ \frac{\partial L^*}{\partial y} - \frac{d}{dx}\left(\frac{\partial L^*}{\partial y'}\right) = 0 \]In our case, \(L(x, y, y') = (y')^2\) and \(g(x, y, y') = y\). So, the augmented Lagrangian is:
\[ L^*(x, y, y', \lambda) = (y')^2 + \lambda y \]Now, we compute the partial derivatives:
- \(\displaystyle \frac{\partial L^*}{\partial y} = \frac{\partial}{\partial y}((y')^2 + \lambda y) = \lambda\)
- \(\displaystyle \frac{\partial L^*}{\partial y'} = \frac{\partial}{\partial y'}((y')^2 + \lambda y) = 2y'\)
Next, we compute the total derivative with respect to \(x\) for the second term:
\[ \frac{d}{dx}\left(\frac{\partial L^*}{\partial y'}\right) = \frac{d}{dx}(2y') = 2y'' \]Substitute these into the Euler-Lagrange equation:
\[ \lambda - 2y'' = 0 \] \[ 2y'' = \lambda \] \[ y'' = \frac{\lambda}{2} \]This is a simple second-order linear differential equation.
Solving the Differential Equation
We integrate the equation \(y'' = \frac{\lambda}{2}\) twice with respect to \(x\).
Integrating once gives the expression for \(y'\):
\[ y' = \int \frac{\lambda}{2} dx = \frac{\lambda}{2}x + C_1 \]Integrating again gives the expression for \(y\):
\[ y = \int \left(\frac{\lambda}{2}x + C_1\right) dx = \frac{\lambda}{4}x^2 + C_1x + C_2 \]Here, \(C_1\) and \(C_2\) are integration constants.
Applying Boundary Conditions
We use the given boundary conditions \(y(0) = 1\) and \(y(1) = 6\) to determine the constants \(C_1\) and \(C_2\) in terms of \(\lambda\).
Using \(y(0) = 1\):
\[ y(0) = \frac{\lambda}{4}(0)^2 + C_1(0) + C_2 = 1 \] \[ C_2 = 1 \]Using \(y(1) = 6\):
\[ y(1) = \frac{\lambda}{4}(1)^2 + C_1(1) + C_2 = 6 \] \[ \frac{\lambda}{4} + C_1 + C_2 = 6 \]Substitute \(C_2 = 1\) into the second equation:
\[ \frac{\lambda}{4} + C_1 + 1 = 6 \] \[ C_1 = 5 - \frac{\lambda}{4} \]Now substitute the expressions for \(C_1\) and \(C_2\) back into the solution for \(y(x)\):
\[ y(x) = \frac{\lambda}{4}x^2 + \left(5 - \frac{\lambda}{4}\right)x + 1 \] \[ y(x) = \frac{\lambda}{4}x^2 + 5x - \frac{\lambda}{4}x + 1 \] \[ y(x) = \frac{\lambda}{4}(x^2 - x) + 5x + 1 \]Applying the Integral Constraint
Finally, we use the integral constraint \(\displaystyle \int_0^1 y d x=3\) to find the value(s) of \(\lambda\).
\[ \int_0^1 \left(\frac{\lambda}{4}(x^2 - x) + 5x + 1\right) dx = 3 \]We can split the integral:
\[ \frac{\lambda}{4} \int_0^1 (x^2 - x) dx + \int_0^1 (5x + 1) dx = 3 \]Evaluate the integrals:
\[ \int_0^1 (x^2 - x) dx = \left[\frac{x^3}{3} - \frac{x^2}{2}\right]_0^1 = \left(\frac{1^3}{3} - \frac{1^2}{2}\right) - \left(\frac{0^3}{3} - \frac{0^2}{2}\right) = \frac{1}{3} - \frac{1}{2} = \frac{2-3}{6} = -\frac{1}{6} \] \[ \int_0^1 (5x + 1) dx = \left[\frac{5x^2}{2} + x\right]_0^1 = \left(\frac{5(1)^2}{2} + 1\right) - \left(\frac{5(0)^2}{2} + 0\right) = \frac{5}{2} + 1 = \frac{5+2}{2} = \frac{7}{2} \]Substitute these values back into the constraint equation:
\[ \frac{\lambda}{4} \left(-\frac{1}{6}\right) + \frac{7}{2} = 3 \] \[ -\frac{\lambda}{24} + \frac{7}{2} = 3 \]Solve for \(\lambda\):
\[ -\frac{\lambda}{24} = 3 - \frac{7}{2} \] \[ -\frac{\lambda}{24} = \frac{6}{2} - \frac{7}{2} \] \[ -\frac{\lambda}{24} = -\frac{1}{2} \] \[ \frac{\lambda}{24} = \frac{1}{2} \] \[ \lambda = \frac{24}{2} = 12 \]Cardinality of Extremals
We found a unique value for the Lagrange multiplier, \(\lambda = 12\). This means there is only one function \(y(x)\) of the form \(y(x) = \frac{\lambda}{4}(x^2 - x) + 5x + 1\) that satisfies the integral constraint along with the boundary conditions. Each valid value of \(\lambda\) corresponds to exactly one extremal function.
Since we found exactly one value for \(\lambda\), there is exactly one extremal function that satisfies all given conditions. The cardinality of the set of extremals is the number of such functions.
The unique extremal function is obtained by substituting \(\lambda=12\):
\[ y(x) = \frac{12}{4}(x^2 - x) + 5x + 1 = 3(x^2 - x) + 5x + 1 = 3x^2 - 3x + 5x + 1 = 3x^2 + 2x + 1 \]This function satisfies \(y(0)=1\), \(y(1)=6\), and \(\int_0^1 (3x^2 + 2x + 1) dx = [x^3+x^2+x]_0^1 = 1+1+1 = 3\).
Therefore, the set of extremals contains only this single function.
Conclusion
The number of extremals satisfying the given conditions is 1. The cardinality of the set of extremals is 1.
The correct option is the one indicating a cardinality of 1.
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
