User:Egm6321.f12.team2/Report2

=Introduction=

Summary
In this report, we explore various topics in engineering analysis. This ranges from proofs of linearity in various 1st and 2nd order ordinary differential equations, to solving for integrating factors and proving whether a differential equation is exact or not. 

Team 2 Members
Yitien Yan Zhe Wang Ranjit Alexander Karl Van Wyk Anubhav Mohan

=R*2.1:Prove two functions are solutions to the Legendre differential equation=

Given
The following two functions,

and the Legendre differential equation,

Find
Prove that equations 2.1.1 and 2.1.2 are solutions to 2.1.3. That is, show that,

Solution for L2(y1)=0
The first and second derivatives of $$ \displaystyle y_H^1$$:

From equation 2.1.3, we get,

Thus, 2.1.1 is indeed a solution to 2.1.3.

Solution for L2(y2)=0
The process for showing that $$L_2(y^2_H)=0$$ is the same as above. First we take the first two derivatives of $$y_H^2$$:

For the second derivative, we can easily differentiate the first term by using the results from taking the first derivative.

From equation 2.1.3, we get,

Therefore, equation 2.1.2 is indeed a solution to 2.1.3. In conclusion, equations 2.1.1 and 2.1.2 are both solutions to 2.1.3.

We solved this by ourselves.

Given
The following equation,

and the differential equation,

Find
Verify that 2.2.1 is indeed a solution to 2.2.2.

Solution
The first derivative of 2.2.1,

Substituting 2.2.1 and 2.2.3 into 2.2.2, we obtain,

In conclusion, equation 2.2.1 is indeed a solution to 2.2.2

We solved this by ourselves.

Find
Show that, 2.3.1 is affine in $$y'$$, that it is in general an N1-ODE, but is not the most general N1-ODE. Provide an example of a more general N1-ODE.

Solution
Since the differential equation is affine,the function is linear in $$y'$$ but does not pass through zero. The function is affine due to the presence of $$M(x,y)$$.Hence, to prove the function is linear we can remove $$M(x,y)$$ reducing the equation to

Applying the linearity condition to check if equation 2.3.2 is linear with respect to $$y'$$,

and $$G(y',y,x)$$ does not pass through the origin; adding $$M(x,y)$$ makes $$G(y',y,x)$$ affine with repect to $$y'$$. A more general N1-ODE is of the form:

Where $$F(y')$$ is some function of $$y'$$. For example,

Which yields, We solved this by ourselves.

Given
The following two equations:

Find
Show that 2.4.1 and 2.4.2 are linearly independent. That is, for any given $$ c \in \mathbb R$$, show that $$ y_H^1(x) \ne c y_H^2(x)$$

Solution
For $$y_H^1(x)$$ & $$y_H^2(x)$$ to be linearly independent,they have to follow the condition $$y_H^1(x) \neq c y_H^2(x)$$ where $$c$$ is any scalar. Since $$y_H^1(x)$$ is a polynomial function and $$y_H^2(x)$$ is a log function, multipying any one of the two function will not yield the other function. For example, if $$\hat x = 0$$ then,

Therefore, $$y_H^1$$ and $$y_H^2$$ are linearly independent.

Figure 1 visually demonstrates on a smaller domain how the two functions 2.4.1 and 2.4.2 are linearly independent. We solved this by ourselves.

Given
The following function,

Find
The following function,

and show that it is a N1-ODE.

Solution
From chain-rule:

$$\frac{d\phi}{dx} = \frac{\partial\phi}{\partial x} + \frac{\partial\phi}{\partial y}\frac{\partial y}{\partial x}$$

Applying chain-rule to (2.5.1) yields:

$$ \begin{align} \frac{\partial\phi}{\partial x} &= 2xy^{\frac{3}{2}} + 3x^{-1} \\ \frac{\partial\phi}{\partial y} &= \frac{3}{2}x^2y^{\frac{1}{2}} + 2y^{-1} \end{align} $$

where $$ \displaystyle y' := \frac{\partial y}{\partial x}. $$

Thus, 2.5.3 satisfies the first exactness condition for N1-ODE's. Take partial derivative of $$ \displaystyle M $$ and $$ \displaystyle N $$ with respect to $$ x $$ and $$ y $$ respectively:

Based on equations 2.5.4 and 2.5.5, the second exactness condition for N1-ODE's is satisfied. An operator, $$F$$ is linear if $$F(\alpha u + \beta v) = \alpha F(u) + \beta F(v)$$ where $$\alpha,\beta$$ are some real constants. Therefore, to show that 2.5.3 is a nonlinear equation:

Since 2.5.6 and 2.5.7 are not equal, equation 2.5.3 is nonlinear. Also, the highest derivative in this equation is 1, therefore it is of first order, and $$x$$ is the only independent variable in this equation which makes 2.5.3 an ODE. Since the first and second exactness conditions are satisfied as well, 2.5.3 is an exact N1-ODE. We solved this by ourselves.

Given

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$$ M=\frac{\partial \phi(x,y)}{\partial x}\displaystyle $$ (2.6.1)
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$$ N=\frac{\partial \phi(x,y) }{\partial y}\displaystyle $$ (2.6.2)
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x} \displaystyle $$ (2.6.3)
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Find
Review calculus, and find the minimum degree of differentiability of the function $$\displaystyle \phi(x,y) $$. State the full theorem and provide a proof.

Mixed Derivative Theorem
If $$\displaystyle f(x,y)$$ and its partial derivatives $$\displaystyle f_{x}, f_{y}, f_{xy}$$ and $$\displaystyle f_{yx}$$ are defined in a neighborhood of $$\displaystyle (a,b)$$ and all are continuous at $$\displaystyle (a,b)$$, then $$\displaystyle f_{xy}(a,b)=f_{yx}(a,b)$$.

Mean Value Theorem
Assume $$\displaystyle f:\mathbb {R}^2\rightarrow \mathbb {R}$$ is differentiable. Define $$\displaystyle X_{0} = (x_{0}, y_{0})$$ and $$\displaystyle X = (x_{0} + u, y_{0} + v)$$.

Then there exists $$\displaystyle C$$ which lies on the line joining $$\displaystyle X_{0}$$ and $$\displaystyle X$$ such that
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$$f(X) = f(X_{0}) + f'(C)(X-X_{0}) \displaystyle $$ i.e, there exists $$ c\in (0,1) \displaystyle $$ such that
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$$f(x_{0} + u, y_{0} + v) - f(x_{0}, y_{0}) = uf_{x}(C) + vf_{y}(C) \displaystyle $$ where $$\displaystyle C = (x_{0} + cu, y_{0} + cv)$$
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Proof
Suppose $$\displaystyle f:\mathbb [a,b]\rightarrow \mathbb R$$
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$$F_{1}(x,y) = f(x+u,y+v) - f(x+u,y)-f(x,y+v)+f(x,y) \displaystyle $$ (2.6.4) and
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$$F_{2}(x,y) = f(x,y+v) - f(x,y) \displaystyle $$ (2.6.5) From Mean Value theory: Because
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$$\displaystyle {F_{2}(a+u,b)-F_{2}(a,b)}=f(a+u,b+v) - f(a+u,b)-f(a,b+v)+f(a,b)=F_{1}(a,b) \displaystyle $$ (2.6.7) Replace $$F_{2}(a+u,b)-F_{2}(a,b)$$ in (2.6.6) with $$ F_{1}(a,b)$$
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$$\displaystyle \frac{F_{1}(a,b)}{u}=\frac{\partial F_{2}}{\partial x}(a+cu,b) $$ (2.6.8) Referring to (2.6.5), the right hand side of (2.6.8) can be expressed as:
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$$\displaystyle \frac{\partial F_{2}}{\partial x}(a+cu,b)=\frac{\partial f}{\partial x}(a+cu,b+v)-\frac{\partial f}{\partial x}(a+cu,b) $$ (2.6.9) Applying mean value theory to (2.6.9):
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$$\displaystyle \frac{\frac{\partial f}{\partial x}(a+cu,b+v)-\frac{\partial f}{\partial x}(a+cu,b)}{v}=\frac{\partial^{2}}{\partial x\partial y}f(a+cu,b+cv) $$ (2.6.10) From (2.6.8), we know that $$\displaystyle \frac{F_{1}(a,b)}{u} =[ \frac{\partial f}{\partial x}(a+cu,b+v)-\frac{\partial f}{\partial x}(a+cu,b) ]u $$ The (2.6.10) can then be written as: When $$\displaystyle u\rightarrow 0$$, $$\displaystyle v\rightarrow 0$$, the limit of (2.6.11) can be expressed as: When we exchange $$u$$ with $$v$$ : As we define $$\displaystyle [a,b]\rightarrow \mathbb {R}$$, we can conclude that:
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We solved this by ourselves.

Verify That
is indeed the solution for the N1-ODE,

Solution
Move individual variables to their respective sides of the equation,

Take the integral of both sides,

Resulting in,

And hence, We solved this by ourselves.

Given
The exact differential equation,

Which can be rewritten as,

Since 2.8.2 is exact, $$\bar M_y = \bar N_x$$. Thus, the following equation 2.8.3 can be obtained:

Find
Why solving equation (2.8.3) for the integration factor $$\displaystyle h(x,y) $$ is usually not easy ?

Solution
Solving equation 2.8.3 for the integrating factor $$\displaystyle h(x,y)$$ is usually not easy because equation(2.8.2) is an non-linear partial differential equation with multi-variable $$x$$ and $$y$$. In the case that both $$h_x \ne 0$$ and $$h_y \ne 0$$, the integration can be complicated due to the existence of both $$N$$ and $$M$$ which also contain $$x$$ and $$y$$. Differently, if either $$h_x = 0$$ or $$h_y = 0$$, we can integrate the equation with respect to only one variable which makes the processes much simpler. We solved this by ourselves.

Find
The Euler Integrating Factor, $$ h(x) $$.

Solution
Considering $$h_y$$ as the integration factor, and replacing $$\frac{1}{M}(N_x-M_y)$$ with $$m(y)$$,

Solving the integrals, leads to:

And then solving for $$h$$, we get:

We solved this by ourselves.

Find
and

Solution
Rewrite (2.10.1) as the following:

where, $$M(x,y)$$ and $$N(x,y)$$ are the variables in an ODE. In this case $$N(x,y)=1$$ Solving for the integrating factor,

And,

Equation (2.10.4) is not EXACT, because $$M_x = x^{-1} \neq N_y = 0$$ To make them EXACT, we multiply (2.10.4) by $$h(x)=x$$, resulting in,

Now, $$M_y = 1$$ and $$N_x = 1$$, so $$M_y=N_x$$ making the ODE EXACT and the integrating factor be $$h(x)=x$$ We can now integrate $$M(x,y)$$ and $$N(x,y)$$ from (2.10.4) individually,

Resulting in,

And solving for $$y$$ in (2.10.10) we get: We solved this by ourselves.

Given
The following general L1-ODE-VC's:

Find
The solution, y(x), for each equation (2.11.1, 2.11.2, 2.11.3).

The solution for equation 2.11.1
Checking case 1,

Where, $$N_x = \frac{\partial N}{\partial x} and M_y = \frac{\partial M}{\partial y} $$. Therefore, equation 2.11.6 is only a function of x, and case 1 is satisfied. Integrating 2.11.6 yields,

Equation 2.11.1 then becomes,

From 2.11.6,

Therefore,

Rearranging and integrating yields,

Solution for equation 2.11.2
Rearranging 2.11.2,

Checking case 1,

Equation 2.11.16 is only a function of x, and therefore case 1 is correct. Integrating 2.11.15 yields,

From 2.11.2,

Since from equation 2.11.16,

Equation 2.11.18 then becomes,

Solution for equation 2.11.3
Checking case 1,

Therefore, equation 2.11.6 is only a function of x, and case 1 is satisfied. Integrating 2.11.21 yields,

Following the formula derived in 2.11.19, We solved this by ourselves.