IS_Math-2013-02

题目来源：Problem 2 日期：2024-08-09 题目主题：Math-Calculus-Variational Calculus

解题思路

这道题目涉及函数泛函的最小化问题，需要使用变分法来求解。我们要研究函数 $I[f]$，它由函数 $f(x)$ 和它的导数 $\frac{\mathrm{d}f(x)}{\mathrm{d}x}$ 通过一个积分定义。为了找到使得该泛函最小的函数，我们需要：

1. 证明泛函的凸性：我们需要证明给定的泛函满足某种形式的凸性不等式。
2. 导出满足欧拉-拉格朗日方程的微分方程：使用变分法，通过对泛函取变分并设为零，可以导出一个对应的常微分方程。这个微分方程描述了使泛函最小的函数的必要条件。
3. 分析解的最优性：证明通过求解欧拉-拉格朗日方程获得的解确实是使泛函 $I[f]$ 最小的解。这通常涉及计算二阶变分并验证其非负性。
4. 求解微分方程：根据边界条件，求解从欧拉-拉格朗日方程得到的微分方程，找出具体的函数形式。

Solution

1. Convexity of the functional $I[f]$

To prove that $I[(1-t)f+tg]\le (1-t)I[f]+tI[g]$ for any $f,g\in \mathcal{F}$ and $t\in [0,1]$, we proceed as follows:

$$\begin{array}{rl}I[(1-t)f+tg]& ={\int }_0^1\left\{((1-t)f+tg{)}^2+{\left(\frac{\mathrm{d}}{\mathrm{d}x}((1-t)f+tg)\right)}^2\right\}\mathrm{d}x\\ & ={\int }_0^1\left\{((1-t)f+tg{)}^2+((1-t)f^{\prime }+t{g}^{\prime }{)}^2\right\}\mathrm{d}x\\ & \le {\int }_0^1\left\{(1-t)f^2+t{g}^2+(1-t)(f^{\prime }{)}^2+t(g^{\prime }{)}^2\right\}\mathrm{d}x\text{(by Cauchy-Schwarz inequality)}\\ & =(1-t){\int }_0^1(f^2+(f^{\prime }{)}^2)\mathrm{d}x+t{\int }_0^1(g^2+(g^{\prime }{)}^2)\mathrm{d}x\\ & =(1-t)I[f]+tI[g]\end{array}$$

The inequality in the third line follows from a variation of the Cauchy-Schwarz inequality applied to both the function values and their derivatives separately.

Equality Condition for the Convexity Inequality

Recall the convexity inequality we proved in part 1:

For any $f,g\in \mathcal{F}$ and $t\in [0,1]$,

$I[(1-t)f+tg]\le (1-t)I[f]+tI[g]$

To determine when equality holds, we need to examine the steps in our proof closely:

1. The key step in our proof used the Cauchy-Schwarz inequality:

   $((1-t)a+tb{)}^2\le (1-t)a^2+t{b}^2$

2. For the Cauchy-Schwarz inequality, equality holds if and only if $a$ and $b$ are linearly dependent, i.e., there exists a constant $k$ such that $a=kb$.

3. In our case, this means equality in the convexity inequality holds if and only if:

   a) $f(x)=k\cdot g(x)$ for some constant $k$ and for all $x\in [0,1]$, and

   b) $f^{\prime }(x)=k\cdot g^{\prime }(x)$ for the same constant $k$ and for all $x\in [0,1]$

4. However, we also need to consider the constraints of our function space $\mathcal{F}$:

   • $f(0)=g(0)=0$
   • $f(1)=g(1)=1$

5. Given these constraints, the only way for $f$ and $g$ to be linearly dependent and satisfy the boundary conditions is if $f=g$.

Therefore, we can conclude:

The equality in $I[(1-t)f+tg]\le (1-t)I[f]+tI[g]$ holds if and only if $f=g$ almost everywhere in $[0,1]$.

This result has an important implication:

• The strict inequality $I[(1-t)f+tg]<(1-t)I[f]+tI[g]$ holds for all $t\in (0,1)$ whenever $f$ and $g$ are distinct functions in $\mathcal{F}$.

This strict inequality is crucial because it implies that $I[f]$ is strictly convex, which in turn guarantees the uniqueness of the minimizer we found later in (3) using the Euler-Lagrange equation.

2. Deriving the Euler-Lagrange equation

Let $f\in \mathcal{F}$ satisfy ${\frac{\mathrm{d}}{\mathrm{d}t}I[(1-t)f+tg]|}_{t=0}=0$ for any $g\in \mathcal{F}$. We have:

$$\begin{array}{rl}0& ={\frac{\mathrm{d}}{\mathrm{d}t}I[(1-t)f+tg]|}_{t=0}\\ & ={\frac{\mathrm{d}}{\mathrm{d}t}{\int }_0^1\left\{((1-t)f+tg{)}^2+((1-t)f^{\prime }+t{g}^{\prime }{)}^2\right\}\mathrm{d}x|}_{t=0}\\ & ={\int }_0^1\left\{2((1-t)f+tg)(-f+g)+2((1-t)f^{\prime }+t{g}^{\prime })(-f^{\prime }+g^{\prime })\right\}\mathrm{d}x{|}_{t=0}\\ & ={\int }_0^1\left\{2f(g-f)+2f^{\prime }(g^{\prime }-f^{\prime })\right\}\mathrm{d}x\end{array}$$

Now, let’s integrate by parts the term with $f^{\prime }(g^{\prime }-f^{\prime })$:

$$\begin{array}{rl}{\int }_0^{1}2f^{\prime }(g^{\prime }-f^{\prime })\mathrm{d}x& ={2f^{\prime }(g-f)|}_0^1-{\int }_0^{1}2f^{\prime \prime }(g-f)\mathrm{d}x\\ & =2f^{\prime }(1)(1-1)-2f^{\prime }(0)(0-0)-{\int }_0^{1}2f^{\prime \prime }(g-f)\mathrm{d}x\\ & =-{\int }_0^{1}2f^{\prime \prime }(g-f)\mathrm{d}x\end{array}$$

The boundary terms vanish because $f(0)=g(0)=0$ and $f(1)=g(1)=1$ for $f,g\in \mathcal{F}$.

Substituting back:

$$\begin{array}{rl}0& ={\int }_0^1\left\{2f(g-f)-2f^{\prime \prime }(g-f)\right\}\mathrm{d}x\\ & ={\int }_0^{1}2(f-f^{\prime \prime })(g-f)\mathrm{d}x\end{array}$$

For this equation to hold for all $g\in \mathcal{F}$, we must have:

$${\int }_0^1(f-f^{\prime \prime })(g-f)\mathrm{d}x=0$$

Using the given property, we conclude that:

$$f-f^{\prime \prime }=0$$

Therefore, the ordinary differential equation that $f$ should satisfy is:

$$f^{\prime \prime }=f$$

This differential equation, along with the boundary conditions $f(0)=0$ and $f(1)=1$, fully specifies the boundary value problem for $f$.

3. Minimization of $I[f]$

The solution of the Euler-Lagrange equation minimizes $I[f]$ due to the following reasons:

1. Convexity of $I[f]$: We proved in part 1 that for any $f,g\in \mathcal{F}$ and $t\in [0,1]$,

   $I[(1-t)f+tg]\le (1-t)I[f]+tI[g]$

   This inequality is the definition of a convex functional. Convexity is crucial because it ensures that any local minimum of the functional is also a global minimum.

2. Stationary point from the Euler-Lagrange equation: The Euler-Lagrange equation we derived,

   $f^{\prime \prime }=f$

   along with the boundary conditions $f(0)=0$ and $f(1)=1$, gives us a stationary point of $I[f]$. To see this, recall that we showed:

   ${\frac{\mathrm{d}}{\mathrm{d}t}I[(1-t)f+tg]|}_{t=0}=0$

   for all $g\in \mathcal{F}$. This is precisely the definition of a stationary point in the calculus of variations.

3. Global minimum property: For a convex functional, any stationary point is a global minimum. We can prove this as follows:

   Let $f^{\ast }$ be the solution to the Euler-Lagrange equation, and let $h$ be any function in $\mathcal{F}$. Consider the function:

   $\varphi (t)=I[(1-t)f^{\ast }+th]$

   We know that ${\varphi }^{\prime }(0)=0$ because $f^{\ast }$ is a stationary point. By the convexity of $I[f]$, we have:

   $\varphi (t)\le (1-t)I[f^{\ast }]+tI[h]$

   Subtracting $I[f^{\ast }]$ from both sides:

   $\varphi (t)-I[f^{\ast }]\le t(I[h]-I[f^{\ast }])$

   Dividing by $t>0$ and taking the limit as $t\to 0^{+}$:

   $0={\varphi }^{\prime }(0)\le I[h]-I[f^{\ast }]$

   Therefore, $I[f^{\ast }]\le I[h]$ for all $h\in \mathcal{F}$, proving that $f^{\ast }$ is indeed a global minimum.

4. Uniqueness of the solution: The strict convexity of $I[f]$ (which can be shown from the strict inequality in the Cauchy-Schwarz inequality) implies that this global minimum is unique.

In conclusion, the solution to the Euler-Lagrange equation $f^{\prime \prime }=f$ with boundary conditions $f(0)=0$ and $f(1)=1$ provides the unique global minimizer of the functional $I[f]$ over the set $\mathcal{F}$. This result beautifully illustrates the power of variational methods in finding optimal solutions to complex problems.

Question 4

The ordinary differential equation obtained is:

$$\frac{{\mathrm{d}}^{2}f(x)}{\mathrm{d}{x}^2}=f(x).$$

The general solution to this differential equation is:

$$f(x)=A{e}^x+B{e}^{-x}$$

where $A$ and $B$ are constants. Using the boundary conditions:

1. $f(0)=0$: $A+B=0$
2. $f(1)=1$: $Ae+B{e}^{-1}=1$

Solving these equations:

$$\begin{array}{rl}A& =\frac{1}{e-e^{-1}}\\ B& =-\frac{1}{e-e^{-1}}\end{array}$$

Therefore, the solution that minimizes $I[f]$ is:

$$f(x)=\frac{e^x-e^{-x}}{e-e^{-1}}$$

知识点

泛函积分不等式Euler-Lagrange方程常微分方程凸函数变分法

难点思路

这道题目的难点在于理解变分法的基本思想，以及如何从变分导出 Euler-Lagrange 方程。关键是要理解为什么对任意 $g\in \mathcal{F}$，导数为零意味着积分中的被积函数恒为零。

解题技巧和信息

1. 在处理变分问题时，首先证明泛函的凸性是个好习惯，这有助于确保我们找到的是全局最小值。
2. 在导出 Euler-Lagrange 方程时，注意处理边界条件，特别是在分部积分时。
3. 解决常微分方程时，注意使用题目给定的边界条件来确定具体解。

重点词汇

• Functional 泛函
• Variational calculus 变分法
• Euler-Lagrange equation 欧拉-拉格朗日方程
• Convexity 凸性
• Stationary point 驻点
• Ordinary differential equation (ODE) 常微分方程

参考资料

1. Gelfand, I. M., & Fomin, S. V. (2000). Calculus of variations. Dover Publications. Chapter 1-2.
2. Weinstock, R. (1974). Calculus of variations: with applications to physics and engineering. Dover Publications. Chapter 3.