Problem 2 - Calculus Of Variations, With Applications To Physics And Engineering

Question

(a) Regarding the left-hand member of the obvious inequality $$ \int_{x_{1}}^{x_{2}}[g(x)+t h(x)]^{3} d x \geqq 0 $$ as a quadratic function of \(t\), where \(t\) is arbitrary, prove the (Schwarz's) inequality $$ \int_{x_{1}}^{x_{2}} h^{2} d x \int_{x_{1}}^{x_{2}} g^{2} d x \geqq\left\\{\int_{x_{1}}^{x_{2}} g h d x\right\\}^{2} $$ where the equality sign holds if and only if \(g(x)=A h(x)\), where \(A\) is some constant. (b) Given that \(y\left(x_{1}\right)=y_{1}, y\left(x_{2}\right)=y_{2}\) and that \(p(x)\) is a known function, use (101) to prove that the absolute minimum of $$ I=\int_{x_{1}}^{* z} p^{2} y^{\prime 3} d x $$ is $$ \frac{\left(y_{2}-y_{1}\right)^{2}}{\int_{x_{1}}^{x_{2}}\left(d x / p^{2}\right)} $$ and that this minimum is attained if and only if $$ y^{\prime}=\frac{A}{p^{2}} $$ where \(A\) is an arbitrary constant. HINT: \(\int_{x_{1}}^{x_{2}} y^{\prime} d x=y_{2}-y_{1}\). (c) Show that (104) is a first integral of the Euler-Lagrange equation associated with the integral (102). Thus it is shown that the extremum of \((102)\) is an absolute minimum. Verify that \((103)\) is the value of \((102)\) when \((104)\) is substituted.

Step-by-Step Solution

Answer

The inequality can be derived using the dwarf's inequality principle and the Euler-Lagrange equation. The minimum function can be derived from this inequality, which is then confirmed by substituting it back into the integral to get the original function's value.

Step 1: Apply Schwarz's Inequality

The inequality stated in part (a) is known as Schwarz's inequality. Use the inequality to make the following calculation: \\[ \int_{x_{1}}^{x_{2}} h^{2} dx \int_{x_{1}}^{x_{2}} g^{2} dx - \left(\int_{x_{1}}^{x_{2}} g h dx\right)^{2} \geqq 0 \\] Because \( \int_{x_{1}}^{x_{2}}[g(x)+t h(x)]^{3} dx \) is a quadratic function of \(t\), the discriminant must be non-negative for \(t\) to be real, i.e., \\[4 \int_{x_{1}}^{x_{2}} h^{2} dx \int_{x_{1}}^{x_{2}} g^{2} dx - 4\left(\int_{x_{1}}^{x_{2}} g h dx\right)^{2} \geqq 0 \\] Dividing by 4 simplifies it to Schwarz's inequality.

Step 2: Apply the Given Condition

We are given that \(y(x_{1})=y_{1}, y(x_{2})=y_{2}\) and \(p(x)\) is a known function. Using the hint \(\int_{x_{1}}^{x_{2}} y' dx = y_{2} - y_{1}\), we can rewrite the integral \(I\) as follows: \\[I = \int_{x_{1}}^{x_{2}} p^{2} ( y_{2} - y_{1})^2 dx \\] The absolute minimum of I is given by: \\[I_{min} = \frac{( y_{2} - y_{1})^{2}}{\int_{x_{1}}^{x_{2}}\frac{dx}{p^{2}}}\\] which is derived from the inequality \\[\int_{x_{1}}^{x_{2}} h^{2} dx \int_{x_{1}}^{x_{2}} g^{2} dx \geqq (\int_{x_{1}}^{x_{2}} g h dx )^{2} \\] with \(h = p ( y_{2} - y_{1})\) and \( g = \frac{1}{p}\).

Step 3: Confirm the Euler-Lagrange Equation

Now to verify equation (104), let’s express it in proper form. As for any \(y\), the expression \(\delta I = \frac{\delta y}{p^2\delta x}\) is zero, equation (104) is effectively the Euler-Lagrange Equation. It satisfies the following condition: \\[0 = \frac{d}{dx}(p^2y') - 0 \\] which simplifies to \\[p^2y''=0 \\] which is an instance of the Euler-Lagrange equation.

Step 4: Verify the Integral Value

We can substitute the minimum function derived from equation (104) back into the function I to confirm the equivalence with the given value in equation (103): \\[I = \int_{x_{1}}^{x_{2}}p^2 \left(\frac{A}{p^{2}}\right)^2 dx = \frac{( y_{2} - y_{1})^{2}}{\int_{x_{1}}^{x_{2}}\frac{1}{p^{2}}dx}\\] which matches the value given in the problem.

Key Concepts

Schwarz's Inequality

Schwarz's Inequality is a fundamental result in the field of calculus of variations and analysis. It relates the integrals of two functions and their product. Given functions \( g(x) \) and \( h(x) \), Schwarz's Inequality states:

\[ \int_{x_{1}}^{x_{2}} h^{2} dx \int_{x_{1}}^{x_{2}} g^{2} dx \geqq \left(\int_{x_{1}}^{x_{2}} gh dx\right)^{2} \]The equality holds if and only if one function is a constant multiple of the other, i.e., \( g(x) = A h(x) \) for some constant \( A \). This suggests a geometric interpretation: the angle between two vector functions is zero when they are parallel. In the context of the exercise, by treating the expression \( \int_{x_{1}}^{x_{2}} [g(x) + t h(x)]^{3} dx \) as a quadratic in \( t \), it implies the discriminant must be non-negative for all real \( t \), reaffirming the Schwarz's Inequality.

Euler-Lagrange Equation

The Euler-Lagrange Equation is pivotal in solving extremum problems related to integrals. Consider a functional \( I[y] = \int_{x_{1}}^{x_{2}} L(x, y, y') dx \). The goal is to find the function \( y \) that makes \( I[y] \) extreme, usually a minimum or maximum. The Euler-Lagrange equation gives us the necessary condition:

\[ \frac{d}{dx} \frac{\partial L}{\partial y'} - \frac{\partial L}{\partial y} = 0 \]For the exercise at hand, the functional \( I \) involves \( p(x)\) and \( y'(x) \). The equation \( 0 = \frac{d}{dx}(p^2y') \) derived from the Euler-Lagrange equation ensures that the function \( y' \) that minimizes \( I \) satisfies this condition. This plays a crucial role in confirming that a derived function is indeed an extremum.

Extremum Problems

Extremum Problems are central to calculus of variations and are concerned with finding maxima or minima of functionals. A functional is an expression in the form \( I[y] = \int_{x_{1}}^{x_{2}} F(x, y, y') dx \). Such problems often occur in physics and engineering, where one seeks to optimize a specific quantity.In the original exercise, finding the minimum of the integral involves transformations and substitutions that satisfy boundary conditions. The extremum condition is achieved when the given differential condition \( y'(x) = \frac{A}{p^2(x)} \) is met, showing that the sought minimum value is \[ \frac{(y_{2} - y_{1})^{2}}{\int_{x_{1}}^{x_{2}} \frac{dx}{p^2(x)}} \].

Determining these extremums ensures the accurate and efficient solution to practical problems where resources or actions need to be optimized.