Consider the problem of extremising the functional \(J(y)=\int_1^3 y(3 x-y) d x\) with boundary conditions y(1) = 1 and y(3) = 2. Which of the following statements are true?
Understanding the Functional and Extremals
The problem asks us to find extremals of the functional \(J(y)=\int_1^3 y(3 x-y) d x\) subject to the boundary conditions \(y(1) = 1\) and \(y(3) = 2\). In calculus of variations, finding extremals of a functional is analogous to finding critical points of a function in differential calculus. Extremals are functions \(y(x)\) that minimize or maximize the functional \(J(y)\).
To find the extremals, we use the Euler-Lagrange equation. The functional is given in the form \(J(y) = \int_{x_1}^{x_2} F(x, y, y') dx\), where in this case, \(F(x, y, y') = y(3x - y) = 3xy - y^2\). Note that \(F\) does not depend on \(y'\).
Applying the Euler-Lagrange Equation
The Euler-Lagrange equation is given by:
\(\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0\)
First, we calculate the partial derivatives of \(F\) with respect to \(y\) and \(y'\):
- \(\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(3xy - y^2) = 3x - 2y\)
- \(\frac{\partial F}{\partial y'} = \frac{\partial}{\partial y'}(3xy - y^2) = 0\)
Now, substitute these into the Euler-Lagrange equation:
\((3x - 2y) - \frac{d}{dx}(0) = 0\)
\(3x - 2y = 0\)
This gives us the equation that any potential extremal must satisfy:
\(2y = 3x\)
\(y(x) = \frac{3x}{2}\)
This function \(y(x) = \frac{3x}{2}\) is the only possible candidate for an extremal satisfying the Euler-Lagrange equation for this functional \(F(x, y, y') = 3xy - y^2\).
Checking Boundary Conditions
For \(y(x) = \frac{3x}{2}\) to be an actual extremal for the given problem, it must also satisfy the specified boundary conditions: \(y(1) = 1\) and \(y(3) = 2\).
- Check at \(x=1\):
\(y(1) = \frac{3(1)}{2} = \frac{3}{2}\). The required condition is \(y(1) = 1\). Since \(\frac{3}{2} \neq 1\), the boundary condition at \(x=1\) is not satisfied. - Check at \(x=3\):
\(y(3) = \frac{3(3)}{2} = \frac{9}{2}\). The required condition is \(y(3) = 2\). Since \(\frac{9}{2} \neq 2\), the boundary condition at \(x=3\) is not satisfied.
Since the function \(y(x) = \frac{3x}{2}\), which is derived from the Euler-Lagrange equation, does not satisfy the given boundary conditions, there is no function that can be an extremal for this functional under these specific boundary conditions.
Analyzing the Options
Let's review the given options in light of our findings:
- Option 1: There is a unique extremal
This is false because the only candidate function satisfying the Euler-Lagrange equation does not meet the boundary conditions. - Option 2: \(y(x)=\frac{3 x}{2}\) is an extremal
This is false. Although \(y(x)=\frac{3 x}{2}\) satisfies the Euler-Lagrange equation, it fails the boundary conditions \(y(1)=1\) and \(y(3)=2\). An extremal must satisfy both. - Option 3: \(y(x)=\frac{x}{2}+\frac{1}{2}\) is an extremal
Let's check if this function satisfies the Euler-Lagrange equation \(3x - 2y = 0\). Substitute \(y(x) = \frac{x}{2}+\frac{1}{2}\) into the equation: \(3x - 2\left(\frac{x}{2}+\frac{1}{2}\right) = 3x - (x+1) = 3x - x - 1 = 2x - 1\). Since \(2x-1\) is not identically zero for all \(x \in [1, 3]\), this function does not satisfy the Euler-Lagrange equation. Therefore, it cannot be an extremal. (We can also check boundary conditions for completeness: \(y(1) = \frac{1}{2} + \frac{1}{2} = 1\), which matches \(y(1)=1\). \(y(3) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2\), which matches \(y(3)=2\). It satisfies the boundary conditions but not the EL equation). - Option 4: There are no extremals
This is true. The only function that satisfies the necessary condition (Euler-Lagrange equation) fails to satisfy the required boundary conditions. Therefore, no such extremal function exists.
Based on the analysis, the correct statement is that there are no extremals for the given functional with the specified 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
