Use Green's Theorem in the form of this equation to prove Green's first identity, where D and C satisfy the hypothesis of Green's Theorem and the appropriate partial derivatives of f and g exist and are continuous. (The quantity grad

Yasmin
2021-03-04
Answered

Use Green's Theorem in the form of this equation to prove Green's first identity, where D and C satisfy the hypothesis of Green's Theorem and the appropriate partial derivatives of f and g exist and are continuous. (The quantity grad

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tabuordy

Answered 2021-03-05
Author has **91** answers

Step 1

$\int {\int}_{D}f{\mathrm{\nabla}}^{2}gdA={\oint}_{C}f\left(\mathrm{\nabla}g\right).nds-\int {\int}_{D}\mathrm{\nabla}f.\mathrm{\nabla}gdA$

Add$\int {\int}_{D}\mathrm{\nabla}f.\mathrm{\nabla}gdA$ to both sides,

$\int {\int}_{D}f{\mathrm{\nabla}}^{2}gdA+\int {\int}_{D}\mathrm{\nabla}f.\mathrm{\nabla}gdA={\oint}_{C}f\left(\mathrm{\nabla}g\right).nds$

$\int \int (f{\mathrm{\nabla}}^{2}g+\mathrm{\nabla}f.\mathrm{\nabla}g)dA={\oint}_{C}f\left(\mathrm{\nabla}g\right).nds$

Step 2

Using the product rule for derivatives,

we can write$f{\mathrm{\nabla}}^{2}g+\mathrm{\nabla}f.\mathrm{\nabla}g=f\mathrm{\nabla}.\left(\mathrm{\nabla}g\right)+\mathrm{\nabla}f.\mathrm{\nabla}g=\mathrm{\nabla}.\left(f\mathrm{\nabla}g\right)$

$\int {\int}_{D}\mathrm{\nabla}.f\left(\mathrm{\nabla}g\right)dA={\oint}_{C}f\left(\mathrm{\nabla}g\right).nds$

$\int {\int}_{D}\xf7\left(f\mathrm{\nabla}g\right)dA={\oint}_{C}f\left(\mathrm{\nabla}g\right).nds$

Add

Step 2

Using the product rule for derivatives,

we can write

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Suppose I have a problem of the form

$mi{n}_{x\ge \u03f5,y\ge 0}\frac{f(x)+g(y)}{x+y}$

subject to some (convex) inequality constraints and some affine equality constraints, and where $f$ and $g$ are known to be convex, and $\u03f5>0$ is some known constant. We can also assume that the feasible set is compact (in addition to being convex).

The main challenge here is when $\frac{f(x)+g(y)}{x+y}$ is not convex (otherwise it is just a convex problem), and also we don't have an easy escape like $f$ and $g$ are log convex, for example.

In my specific situation, I also know that $f,g$ are of the form

$f(x)={\int}_{0}^{x}{h}_{f}(u)\text{}du\text{},$

and

$g(x)={\int}_{0}^{x}{h}_{g}(u)\text{}du\text{},$

for some (known) nondecreasing and integrable (but not necessarily continuous) functions ${h}_{f}:[0,{B}_{f}]\to \mathbb{R}$ and ${h}_{g}:[0,{B}_{g}]\to \mathbb{R}$ for some real numbers $0<{B}_{f},{B}_{g}<\mathrm{\infty}$.

Are there any easy ways to tackle this sort of problem? I am hoping to find ways that would be computationally not much more difficult than convex optimization, but I am not sure if this is possible... it seems like even the simpler problem of minimizing $f(x)/x$ is (surprisingly) difficult...

$mi{n}_{x\ge \u03f5,y\ge 0}\frac{f(x)+g(y)}{x+y}$

subject to some (convex) inequality constraints and some affine equality constraints, and where $f$ and $g$ are known to be convex, and $\u03f5>0$ is some known constant. We can also assume that the feasible set is compact (in addition to being convex).

The main challenge here is when $\frac{f(x)+g(y)}{x+y}$ is not convex (otherwise it is just a convex problem), and also we don't have an easy escape like $f$ and $g$ are log convex, for example.

In my specific situation, I also know that $f,g$ are of the form

$f(x)={\int}_{0}^{x}{h}_{f}(u)\text{}du\text{},$

and

$g(x)={\int}_{0}^{x}{h}_{g}(u)\text{}du\text{},$

for some (known) nondecreasing and integrable (but not necessarily continuous) functions ${h}_{f}:[0,{B}_{f}]\to \mathbb{R}$ and ${h}_{g}:[0,{B}_{g}]\to \mathbb{R}$ for some real numbers $0<{B}_{f},{B}_{g}<\mathrm{\infty}$.

Are there any easy ways to tackle this sort of problem? I am hoping to find ways that would be computationally not much more difficult than convex optimization, but I am not sure if this is possible... it seems like even the simpler problem of minimizing $f(x)/x$ is (surprisingly) difficult...

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Use Stokes' Theorem to compute ${\oint}_{C}\frac{1}{2}{z}^{2}dx+\left(xy\right)dy+2020dz$ , where C is the triangle with vertices at(1,0,0),(0,2,0), and (0,0,2) traversed in the order.

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multivariable functions works when the partial derivatives are continous but if the function is just differentiable does the chain rule work ? I mean if $z=f(x,y)$ is differentiable function and $x=g\left(t\right),y=h\left(t\right)$ are also differentiable functions can we write:

$\frac{dz}{dt}=\frac{\partial f}{\partial x}\times \frac{dx}{dt}+\frac{\partial f}{\partial y}\times \frac{dy}{dt}$

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Consider a minimization problem:

$\begin{array}{rl}& min\sum _{i=1}^{n}|{x}_{i}|,\\ & Ax=b,\end{array}$

where $A$ is an $m\times n$ matrix of rank $m$. I know that the minimum points contains one where at least $n-m$ components of $x$ is zero and I have proved this conclusion. My problem is that, I think this should be a developed proposition but I am not familiar with optimization theory, so I don't know where to find a reference. Could anybody help me find a reference like a book or article? Thanks!

$\begin{array}{rl}& min\sum _{i=1}^{n}|{x}_{i}|,\\ & Ax=b,\end{array}$

where $A$ is an $m\times n$ matrix of rank $m$. I know that the minimum points contains one where at least $n-m$ components of $x$ is zero and I have proved this conclusion. My problem is that, I think this should be a developed proposition but I am not familiar with optimization theory, so I don't know where to find a reference. Could anybody help me find a reference like a book or article? Thanks!

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Use Stokes' Theorem to evaluate $\int {\int}_{S}CURLf\cdot dS$ .

$F(x,y,z)={x}^{2}{y}^{3}zi+\mathrm{sin}\left(xyz\right)j+xyzk$ ,

S is the part of the cone$y}^{2}={x}^{2}+{z}^{2$ that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.

S is the part of the cone