## 数学代写|应用数学代写applied mathematics代考|MATH101

2023年3月22日
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## 数学代写|应用数学代写applied mathematics代考|The Euler-Lagrange equation

The Euler-Lagrange equation for (3.12) is
$$\frac{d}{d r}\left{\frac{r u^{\prime}}{\left[1+\left(u^{\prime}\right)^2\right]^2}\right}=0 .$$
Since the Lagrangian is independent of $u$, this has an immediate first integral,
$$r u^{\prime}=-c\left[1+\left(u^{\prime}\right)^2\right]^2$$
where $c \geq 0$ is a constant of integration.
If $c=0$ in (3.14), then we get $u^{\prime}=0$, or $u=$ constant. This solution corresponds to the cylinder with maximum resistance $1 / 2$. The maximum is not attained, however, within the class absolutely continuous functions $u \in X_M$, since for such functions if $u^{\prime}$ is zero almost everywhere with respect to Lebesgue measure, then $u$ is constant, and it cannot satisfy both boundary conditions $u(0)=M, u(1)=0$.
If $c>0$ in (3.14), then it is convenient to parametrize the solution curve by $p=u^{\prime}<0$. From (3.14), the radial coordinate $r$ is given in terms of $p$ by
$$r=-\frac{c\left(1+p^2\right)^2}{p}$$
Using this equation to express $d r$ in terms of $d p$ in the integral
$$u=\int p d r$$
and evaluating the result, we get
$$u=u_0-c\left(-\log |p|+p^2+\frac{3}{4} p^4\right),$$
where $u_0$ is a constant of integration.
From (3.15), we see that the minimum value of $r(p)$ for $p<0$ is
$$r_0=\frac{16 \sqrt{3} c}{9}$$ at $p=-1 / \sqrt{3}$. Thus, although this solution minimizes the resistance, we cannot use it over the whole interval $0 \leq r \leq 1$, only for $r_0 \leq r \leq 1$. In the remaining part of the interval, we use $u=$ constant, and we obtain the lowest global resistance by placing the blunt part of the body around the nose $r=0$, where it contributes least to the area and resistance.

## 数学代写|应用数学代写applied mathematics代考|Constrained variational principles

It often occurs that we want to minimize a functional subject to a constraint. Constraints can take many forms. First, consider the minimization of a functional
$$\mathcal{F}(u)=\int_a^b F\left(x, u, u^{\prime}\right) d x$$
over functions such that $u(a)=0, u(b)=0$, subject to an integral constraint of the form
$$\mathcal{G}=\int_a^b G\left(x, u, u^{\prime}\right) d x$$
Variational problems with integral constraints are called isoperimetric problems after the prototypical problem of finding the curve (a circle) that encloses the maximum area subject to the constraint that its length is fixed. ${ }^3$

We may solve this problem by introducing a Lagrange multiplier $\lambda \in \mathbb{R}$ and seeking stationary points of the unconstrained functional
$$\mathcal{F}(u)-\lambda \mathcal{G}(u)=\int_a^b\left{F\left(x, u, u^{\prime}\right)-\lambda G\left(x, u, u^{\prime}\right)\right} d x$$
The condition that this functional is stationary with respect to $\lambda$ implies that $\mathcal{G}(u)=0$, so a stationary point satisfies the constraint.

The Euler-Lagrange equation for stationarity of the functional with respect to variations in $u$ is
$$-\frac{d}{d x} F_{u^{\prime}}\left(x, u, u^{\prime}\right)+F_u\left(x, u, u^{\prime}\right)=\lambda\left[-\frac{d}{d x} G_{u^{\prime}}\left(x, u, u^{\prime}\right)+G_u\left(x, u, u^{\prime}\right)\right]$$

# 应用数学代考

## 数学代写|应用数学代写applied mathematics代考|The Euler-Lagrange equation

(3.12) 的欧拉-拉格朗日方程是
Ifrac ${d}{d$ r $} \backslash$ left $\left{\mid\right.$ frac $\left{r u^{\wedge}{\backslash\right.$ prime $\left.}\right}{\backslash$ left[ $[1+$ \left(u^ ${\backslash$ prim

$$r u^{\prime}=-c\left[1+\left(u^{\prime}\right)^2\right]^2$$

$$r=-\frac{c\left(1+p^2\right)^2}{p}$$

$$u=\int p d r$$

$$u=u_0-c\left(-\log |p|+p^2+\frac{3}{4} p^4\right),$$

$$r_0=\frac{16 \sqrt{3} c}{9}$$

## 数学代写|应用数学代写applied mathematics代考|Constrained variational principles

$$\mathcal{F}(u)=\int_a^b F\left(x, u, u^{\prime}\right) d x$$

$$\mathcal{G}=\int_a^b G\left(x, u, u^{\prime}\right) d x$$

Imathcal ${F}(u)-$-lambda Imathcal ${G}(u)=$ int_a $a^{\wedge} b \backslash l e f t{F \backslash l e f t(x, u$,

$$-\frac{d}{d x} F_{u^{\prime}}\left(x, u, u^{\prime}\right)+F_u\left(x, u, u^{\prime}\right)=\lambda\left[-\frac{d}{d x} G_{u^{\prime}}(x\right.$$

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## MATLAB代写

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