Extremal Curve using Calculus of Variations

The problem asks us to find the curve \(y(x)\) connecting the points (1, 2) and (2, 8) on which an extremal of the given functional can be attained. An extremal curve satisfies the Euler-Lagrange equation derived from the calculus of variations.

The given functional is \(J[y]=\displaystyle\int_1^2\left(1+x^3 y^{\prime}\right) y^{\prime} d x\).

The integrand is \(f(x, y, y') = (1+x^3 y') y' = y' + x^3 (y')^2\).

Euler-Lagrange Equation

For a functional of the form \(J[y] = \int_{x_1}^{x_2} f(x, y, y') dx\), the curve \(y(x)\) that is an extremal satisfies the Euler-Lagrange equation:

\(\frac{\partial f}{\partial y} - \frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right) = 0\)

First, we compute the partial derivatives of \(f\) with respect to \(y\) and \(y'\):

• \(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(y' + x^3 (y')^2) = 0\)
• \(\frac{\partial f}{\partial y'} = \frac{\partial}{\partial y'}(y' + x^3 (y')^2) = 1 + 2x^3 y'\)

Substitute these into the Euler-Lagrange equation:

\(0 - \frac{d}{dx}(1 + 2x^3 y') = 0\)

\(\frac{d}{dx}(1 + 2x^3 y') = 0\)

Solving the Differential Equation

Integrating the equation \(\frac{d}{dx}(1 + 2x^3 y') = 0\) with respect to \(x\), we find that the expression inside the derivative must be a constant:

\(1 + 2x^3 y' = C\)

where \(C\) is an integration constant.

Now, we solve for \(y'\):

\(2x^3 y' = C - 1\)

\(y' = \frac{C - 1}{2x^3}\)

Let \(K = \frac{C - 1}{2}\). The equation becomes:

\(y' = \frac{K}{x^3} = K x^{-3}\)

Integrate \(y'\) with respect to \(x\) to find the function \(y(x)\):

\(y(x) = \int K x^{-3} dx = K \frac{x^{-3+1}}{-3+1} + D\)

\(y(x) = K \frac{x^{-2}}{-2} + D\)

\(y(x) = -\frac{K}{2x^2} + D\)

Let \(A = -\frac{K}{2}\). The general solution for the extremal curve γ is:

\(y(x) = \frac{A}{x^2} + D\)

where \(A\) and \(D\) are constants.

Applying Boundary Conditions

The curve γ passes through the points (1, 2) and (2, 8). We use these points to determine the constants \(A\) and \(D\).

Using point (1, 2): Substitute \(x=1\) and \(y=2\) into \(y(x) = \frac{A}{x^2} + D\).

\(2 = \frac{A}{1^2} + D\)

\(2 = A + D\) (Equation 1)

Using point (2, 8): Substitute \(x=2\) and \(y=8\) into \(y(x) = \frac{A}{x^2} + D\).

\(8 = \frac{A}{2^2} + D\)

\(8 = \frac{A}{4} + D\) (Equation 2)

Now we solve the system of linear equations:

1) \(A + D = 2\)

2) \(\frac{A}{4} + D = 8\)

Subtract Equation 2 from Equation 1:

\((A + D) - \left(\frac{A}{4} + D\right) = 2 - 8\)

\(A - \frac{A}{4} = -6\)

\(\frac{3A}{4} = -6\)

\(3A = -24\)

\(A = -8\)

Substitute the value of \(A = -8\) into Equation 1:

\(-8 + D = 2\)

\(D = 10\)

Thus, the equation of the extremal curve γ is:

\(y(x) = \frac{-8}{x^2} + 10\)

This can be written as \(y(x) = 10 - \frac{8}{x^2}\).

Checking the Points on the Curve

We are given several points and need to determine which one lies on the curve γ defined by \(y(x) = 10 - \frac{8}{x^2}\).

Let's check the point \((\sqrt{2}, 6)\) by substituting \(x = \sqrt{2}\) into the equation for \(y(x)\):

\(y(\sqrt{2}) = 10 - \frac{8}{(\sqrt{2})^2}\)

\(y(\sqrt{2}) = 10 - \frac{8}{2}\)

\(y(\sqrt{2}) = 10 - 4\)

\(y(\sqrt{2}) = 6\)

The calculated y-coordinate for \(x = \sqrt{2}\) is 6, which matches the y-coordinate of the point \((\sqrt{2}, 6)\). Therefore, the point \((\sqrt{2}, 6)\) lies on the curve γ.

For completeness, let's quickly verify the other options.

• Point \((\sqrt{2}, 3)\): \(y(\sqrt{2}) = 6 \neq 3\). The point is not on the curve.
• Point \((\sqrt{3}, \frac{22}{3})\): \(y(\sqrt{3}) = 10 - \frac{8}{(\sqrt{3})^2} = 10 - \frac{8}{3} = \frac{30-8}{3} = \frac{22}{3}\). The point is on the curve.
• Point \((\sqrt{3}, \frac{23}{3})\): \(y(\sqrt{3}) = \frac{22}{3} \neq \frac{23}{3}\). The point is not on the curve.

The point \((\sqrt{2}, 6)\) lies on the extremal curve γ.