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数学代写|数学建模代写Mathematical Modeling代考|Euler-Lagrange Equation

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数学代写|数学建模代写Mathematical Modeling代考|Euler-Lagrange Equation

Let
$$I=\iint_S F\left(x, y, u, u_x, u_y\right) d x d y$$
where $F()$ is a known function, then the value of $I$ depends on $u(x, y)$ and our object is to choose $u(x, y)$ so that the integral $I$ has a maximum or minimum value. Such a function is given by the Euler-Lagrange equation of calculus of variations viz.
$$\frac{\partial F}{\partial u}-\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial u_x}\right)-\frac{\partial}{\partial y}\left(\frac{\partial F}{\partial u_y}\right)=0$$
Since $F$ is a known function of $x, y, u, u_x, u_{y^{\prime}}$ therefore $\partial F / \partial u, \partial F / \partial u_x \partial F / \partial u_y$ are also known functions of $x, y, u, u_x, u_y$. As such the left-hand side of Eqn. (72) is a known function of $x, y$, $u, u_x, u_y, u_{x x}, u_{x y}, u_{y y}$, so that Eqn. (72) gives a partial differential equation of second order for determining $u(x, y)$.
To illustrate the use of Eqn. (72), we consider the problem of finding the surface with minimum area out of all those surfaces which are bounded by a given skew curve. The surface area is given by
$$I=\iint_S \sqrt{1+p^2+q^2} d x d y ; p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}$$
so that Eqn. (72) gives
$$0-\frac{\partial}{\partial x}\left(\frac{p}{\sqrt{1+p^2+q^2}}\right)-\frac{\partial}{\partial y}\left(\frac{q}{\sqrt{1+p^2+q^2}}\right)=0$$
or
$$\left(1+q^2\right) r+\left(1+p^2\right) t-2 p q s=0 ; r=\frac{\partial^2 z}{\partial x^2}, s=\frac{\partial^2 z}{\partial x \partial y}, t=\frac{\partial^2 z}{\partial y^2}$$
Now if Eqn. (75) is satisfied, then the sum of the principal radii of curvature at every point of the surface is zero, i.e., the mean curvature is zero at every point. A surface for which the mean curvature is zero at every point is called a minimal surface and the previous discussion explains the reason for this.
It can be shown that the only ruled surface which is a minimal surface is a right helicoid. It can also be shown that the catenoid obtained by rotating a catenary about its directrix is a minimal surface.

数学代写|数学建模代写Mathematical Modeling代考|Minimal Surfaces

To illustrate the use of Eqn. (72), we consider the problem of finding the surface with minimum area out of all those surfaces which are bounded by a given skew curve. The surface area is given by
$$I=\iint_S \sqrt{1+p^2+q^2} d x d y ; p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}$$
so that Eqn. (72) gives
$$0-\frac{\partial}{\partial x}\left(\frac{p}{\sqrt{1+p^2+q^2}}\right)-\frac{\partial}{\partial y}\left(\frac{q}{\sqrt{1+p^2+q^2}}\right)=0$$
or
$$\left(1+q^2\right) r+\left(1+p^2\right) t-2 p q s=0 ; r=\frac{\partial^2 z}{\partial x^2}, s=\frac{\partial^2 z}{\partial x \partial y}, t=\frac{\partial^2 z}{\partial y^2}$$
Now if Eqn. (75) is satisfied, then the sum of the principal radii of curvature at every point of the surface is zero, i.e., the mean curvature is zero at every point. A surface for which the mean curvature is zero at every point is called a minimal surface and the previous discussion explains the reason for this.
It can be shown that the only ruled surface which is a minimal surface is a right helicoid. It can also be shown that the catenoid obtained by rotating a catenary about its directrix is a minimal surface.

数学代写|数学建模代写Mathematical Modeling代考|Euler-Lagrange Equation

$$I=\iint_S F\left(x, y, u, u_x, u_y\right) d x d y$$

$$\frac{\partial F}{\partial u}-\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial u_x}\right)-\frac{\partial}{\partial y}\left(\frac{\partial F}{\partial u_y}\right)=0$$

$$I=\iint_S \sqrt{1+p^2+q^2} d x d y ; p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}$$

$$0-\frac{\partial}{\partial x}\left(\frac{p}{\sqrt{1+p^2+q^2}}\right)-\frac{\partial}{\partial y}\left(\frac{q}{\sqrt{1+p^2+q^2}}\right)=0$$

$$\left(1+q^2\right) r+\left(1+p^2\right) t-2 p q s=0 ; r=\frac{\partial^2 z}{\partial x^2}, s=\frac{\partial^2 z}{\partial x \partial y}, t=\frac{\partial^2 z}{\partial y^2}$$

数学代写|数学建模代写Mathematical Modeling代考|Minimal Surfaces

$$I=\iint_S \sqrt{1+p^2+q^2} d x d y ; p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}$$

$$0-\frac{\partial}{\partial x}\left(\frac{p}{\sqrt{1+p^2+q^2}}\right)-\frac{\partial}{\partial y}\left(\frac{q}{\sqrt{1+p^2+q^2}}\right)=0$$

$$\left(1+q^2\right) r+\left(1+p^2\right) t-2 p q s=0 ; r=\frac{\partial^2 z}{\partial x^2}, s=\frac{\partial^2 z}{\partial x \partial y}, t=\frac{\partial^2 z}{\partial y^2}$$

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。