Question

Prove that the function \(\Phi_{\mu}\) defined by \((10.147)\) attains its minimum value, among smooth functions \(u(x)\) satisfying the boundary conditions \(u(0)=u(L)=0\), when \(u\) is the solution of the equation \(-K^{2} u^{\prime \prime}(x)=F(x)\). [Hint: Take \(L=\pi\) for convenience. Then express the integral in terms of Fourier sine coefficients of \(u(x), u^{\prime}(x)\), and \(F(x)\), using Theorem 14 of Section 7.13. This gives a separate minimum problem for each \(n\), which is solved precisely when \(-K^{2} u^{\prime \prime}=F(x)\).]

Short answer

In summary, this exercise demonstrates that the function \(\Phi_{\mu}\) attains its minimum value when the given boundary conditions are satisfied, and \(-K^{2} u^{\prime \prime}(x)=F(x)\) holds true. The solution involves expressing the integral in terms of Fourier sine coefficients, applying Theorem 14 to simplify the integral, and using the Cauchy-Schwarz inequality to determine the minimum condition. Thus, when these conditions are met, the minimum value of the function \(\Phi_{\mu}\) is achieved.

Step-by-step solution

Express the integral in terms of Fourier sine coefficients

Since we are given \(u(0)=u(L)=0\), we can expand \(u(x)\), \(u^{\prime}(x)\), and \(F(x)\) in the Fourier sine series: $$u(x)=\sum_{n=1}^{\infty}b_n\sin \left(\frac{n\pi x}{L}\right)$$ $$u^{\prime}(x)=\sum_{n=1}^{\infty}c_n\cos \left(\frac{n\pi x}{L}\right)$$ $$F(x)=\sum_{n=1}^{\infty}d_n\sin \left(\frac{n\pi x}{L}\right)$$ where, $$b_n=\frac{2}{L}\int_{0}^{L}u(x)\sin \left(\frac{n\pi x}{L}\right)dx$$ $$c_n=-\frac{2n\pi}{L}\int_{0}^{L}u(x)\cos \left(\frac{n\pi x}{L}\right)dx$$ $$d_n=\frac{2}{L}\int_{0}^{L}F(x)\sin \left(\frac{n\pi x}{L}\right)dx$$ Now we will express the integral \(\Phi_{\mu}\) in terms of the Fourier sine coefficients.

Rewrite the given function \(\Phi_{\mu}\) using the Fourier sine series expansions

Given \((10.147)\), the function \(\Phi_{\mu}\) can be rewritten as: $$\Phi_{\mu}=\frac{1}{2}\int_{0}^{L}(u^{\prime}(x))^2+\mu(u(x)-F(x))^2 dx$$ Substituting the Fourier sine series expansions into \(\Phi_{\mu}\): $$\Phi_{\mu}=\frac{1}{2}\int_{0}^{L}\left(\sum_{n=1}^{\infty}c_n\cos \left(\frac{n\pi x}{L}\right)\right)^2 +\mu\left(\sum_{n=1}^{\infty}b_n\sin \left(\frac{n\pi x}{L}\right)-\sum_{n=1}^{\infty}d_n\sin \left(\frac{n\pi x}{L}\right)\right)^2 dx$$

Simplify the integral and apply Theorem 14

To compute the integral, we apply Theorem 14, which states that the integral of a product of orthogonal functions is zero unless the indices are equal. Thus, only the equal-index terms contribute to the integral when multiplying the series. Simplify the integral, considering the equal indices terms: $$\Phi_{\mu}=\frac{1}{2}\sum_{n=1}^{\infty}\left(L\left(\frac{c_n^2}{2}\right)+\mu\left(\frac{L(b_n^2+d_n^2-2b_nd_n)}{2}\right)\right)$$ This simplification gives a separate minimum problem for each \(n\).

Determine the minimum condition

To find the minimum value of \(\Phi_{\mu}\), we need to minimize the sum by minimizing the individual terms. Let's consider the terms inside the sum: $$L\left(\frac{c_n^2}{2}\right)+\mu\left(\frac{L}{2}(b_n^2+d_n^2-2b_nd_n)\right)$$ To minimize this expression, we need to minimize \((b_n^2+d_n^2-2b_nd_n)\), using the Cauchy-Schwarz inequality: $$(b_n-d_n)^2 \ge 0$$ Thus, the minimum condition is obtained when \((b_n-d_n)^2=0\), which implies \(b_n=d_n\). This leads to: $$-K^{2}u^{\prime\prime}(x) =-K^{2}\left(\sum_{n=1}^{\infty}d_n \frac{n^2\pi^2}{L^2}\sin \left(\frac{n\pi x}{L}\right)\right)=F(x)=\sum_{n=1}^\infty d_n\sin \left(\frac{n\pi x}{L}\right)$$

Conclusion

We have shown that the function \(\Phi_{\mu}\) attains its minimum value under the given boundary conditions when \(-K^{2} u^{\prime\prime}(x)=F(x)\). The proof involves expressing the integral in terms of Fourier sine coefficients, applying Theorem 14, and simplifying and finding the minimum condition using the Cauchy-Schwarz inequality.

Key concepts

Boundary Conditions

In mathematics, boundary conditions define the behavior of a function at the boundaries of its domain. These conditions are essential when solving differential equations because they provide the additional information needed to determine a unique solution.
In the context of the exercise, the boundary conditions are given as \(u(0) = u(L) = 0\). This means that the solution \(u(x)\) must be zero at these two points.
Having such conditions is crucial because they ensure the stability and uniqueness of the solution across the interval. When solving for the Fourier sine series, boundary conditions allow us to use specific trigonometric functions (sine rather than cosine), which naturally satisfy these zero-end conditions.
This step narrows down possible solutions, ensuring the series fits well within the integral limits from 0 to \(L\). Ultimately, boundary conditions help guide and determine the behavior of functions, ensuring solutions have meaningful applications in physical and theoretical problems.

Fourier Coefficients

Fourier coefficients are the building blocks of Fourier series, used to express functions as an infinite sum of sine and cosine terms.
These coefficients capture the "shape" of the function by measuring how much of each frequency component is present in the original function.
In the Fourier sine series, which is used when functions satisfy specific boundary conditions like \(u(0) = u(L) = 0\), only sine terms are involved. The Fourier coefficients \(b_n\), \(c_n\), and \(d_n\) describe the amplitude of each sine term at different frequencies.
For instance, \(b_n\) is calculated as \(\frac{2}{L}\int_{0}^{L}u(x)\sin\left(\frac{n\pi x}{L}\right)dx\), capturing how much of the sine wave with frequency \(\frac{n\pi}{L}\) appears in \(u(x)\).
This process effectively turns complex functions into simpler harmonic expressions, making them easier to analyze and solve, especially in problems involving differential equations as seen in this exercise.

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental equation in the calculus of variations, used to find functions that minimize or maximize functional expressions.
In physical terms, it often describes the motion or configuration of a system that is "extremal" in some sense, such as the path taken by a particle that minimizes action.
In the given exercise, the function \(\Phi_{\mu}\) must attain its minimum value among smooth functions satisfying the given boundary conditions. The approach generally involves differentiating the functional and finding solutions that make this derivative zero, thus locating potential extreme points.
Here, the equation \(-K^{2} u''(x) = F(x)\) represents such a minimization problem where \(u(x)\) is found by applying variational principles and Fourier analysis.
This concept ties together problems in mechanics, physics, and engineering, providing a systematic method to discern optimal configurations or paths subject to external constraints.

Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is a critical mathematical concept that places a bound on the dot product of two vectors in vector spaces, ensuring angles remain consistent with reality.
This inequality is famously stated as \(| \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \|\), where \(\langle \mathbf{u}, \mathbf{v} \rangle\) is the dot product, and \(\| \mathbf{u} \|\) and \(\| \mathbf{v} \|\) are the magnitudes of vectors \(\mathbf{u}\) and \(\mathbf{v}\).
In terms of the Fourier coefficients in this exercise, the inequality is used to show that \((b_n - d_n)^2 \geq 0\), which implies that the expression \(b_n = d_n\) minimizes the functional \(\Phi_{\mu}\).
This inequality helps in demonstrating that certain configurations like \(b_n = d_n\) not only fulfill boundary conditions but also keep the functional's value at its minimum.
Ultimately, the Cauchy-Schwarz inequality provides a powerful tool in ensuring that solutions to equations constrained by an optimization problem are mathematically valid and physically realistic.