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
Euler Equations and Functional Extremals
The question asks about the nature of the curve in the \((x, y)\)-plane that satisfies the Euler equations for the given functional.
The functional is given by:
\(J(y)=\int_0^5\left[y^2+x^3 y^{\prime}\right] d x\)
This is a standard functional in the calculus of variations, of the form \(J(y)=\int_{x_1}^{x_2} F(x, y, y') d x\), where \(F(x, y, y') = y^2 + x^3 y'\).
The extremals of such a functional satisfy the Euler-Lagrange equation:
\(\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0\)
We need to find the partial derivatives of \(F\) with respect to \(y\) and \(y'\), and then substitute them into the Euler-Lagrange equation.
Calculating Partial Derivatives
First, find the partial derivative of \(F\) with respect to \(y\):
\(\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(y^2 + x^3 y')\)
Since \(x^3 y'\) does not contain \(y\), its partial derivative with respect to \(y\) is zero. The partial derivative of \(y^2\) with respect to \(y\) is \(2y\).
So, \(\frac{\partial F}{\partial y} = 2y\).
Next, find the partial derivative of \(F\) with respect to \(y'\):
\(\frac{\partial F}{\partial y'} = \frac{\partial}{\partial y'}(y^2 + x^3 y')\)
Since \(y^2\) does not contain \(y'\), its partial derivative with respect to \(y'\) is zero. The partial derivative of \(x^3 y'\) with respect to \(y'\) is \(x^3\) (treating \(x\) as a constant).
So, \(\frac{\partial F}{\partial y'} = x^3\).
Applying the Euler-Lagrange Equation
Now, we need to calculate the total derivative with respect to \(x\) of \(\frac{\partial F}{\partial y'}\):
\(\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = \frac{d}{dx}(x^3)\)
The derivative of \(x^3\) with respect to \(x\) is \(3x^2\).
So, \(\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 3x^2\).
Now, substitute these results into the Euler-Lagrange equation:
\(\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0\)
\(2y - 3x^2 = 0\)
Identifying the Solution Curve
The equation satisfied by the extremals is \(2y - 3x^2 = 0\), which can be rewritten as:
\(2y = 3x^2\)
\(y = \frac{3}{2}x^2\)
This equation is of the form \(y = ax^2\), which represents a parabola in the \((x, y)\)-plane. A parabola is a type of quadratic curve.
Let's compare this form to the given options:
- linear: An equation of the form \(y = mx + c\). Our equation is not linear.
- quadratic: An equation of the form \(y = ax^2 + bx + c\). Our equation \(y = \frac{3}{2}x^2\) fits this form with \(a = \frac{3}{2}\), \(b = 0\), and \(c = 0\).
- cubic: An equation involving \(x^3\), like \(y = ax^3 + ...\). Our equation does not involve \(x^3\).
- trigonometric: An equation involving trigonometric functions like \(\sin(x)\) or \(\cos(x)\). Our equation does not involve trigonometric functions.
The solution curve defined by the Euler equations is \(y = \frac{3}{2}x^2\), which is a quadratic curve.
The final answer is \(\boxed{quadratic}\).
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
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
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?
