User talk:EGM6321.f11.team1.yizhao

HW2| = R2.9 State and proof the full theorem =

Given
Review calculus, and find the minimum degree of differentiability of the function such that (2) p.9-3 is satisfied. State the full theorem and provide a proof. p.9-3 (2): $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$

Clairaut's theorem
In mathematical analysis, Clairaut's theorem or Schwarz's theorem,[1] named after Alexis Clairaut and Hermann Schwarz, states that if $$f:\mathbb R \rightarrow \mathbb R$$ has continuous second partial derivatives at any given point in $$ \mathbb R^n $$,say $$(a_1,...,a_n)$$, then for $$ 1<i,$$  $$ j<n,$$

$$\frac{\partial^2 f}{\partial x_i \partial x_j}(a_1,...,a_n)=\frac{\partial^2 f}{\partial x_j \partial x_i}(a_1,...,a_n)$$

In words, the partial derivations of this function are commutative at that point. One easy way to establish this theorem (in the case where n = 2, i = 1, and j = 2, which readily entails the result in general) is by applying Green's theorem to the gradient of f.

http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Clairaut.27s_theorem

Solution
The minimum degree of differentiability of the function such that (2) p.9-3 is satisfied is 3.

As Clairaut's theorem mentions, functions that satisfy (2) p.9-3 must have continuous second derivative.

For instance,

$$ z=f(x,y)=x^2+y^2 $$

$$\frac{\partial^2 f(x,y)}{\partial x\partial y}=\frac{\partial^2 f(x,y)}{\partial y\partial x}=2$$

The function

$$ z=f(x,y)=x^2+y^2 $$

can only be differentialed by 3 degrees

$$ \frac{\partial^2 f(x,y)}{\partial^2 x}=0 $$

And functions with degree of differentiability less than 3 can not satisfy (2) p. 9-3

For example f(x,y)=x+y

So The minimum degree of differentiability of the function such that (2) p.9-3 is satisfied is 3.

= R*2.14 Solve the general L1-ODE-VC=

Part 1
Solve the general L1-ODE-VC

Given
$$ a_1(x)\,y'+a_0(x)\,y=b(x)$$ $$\displaystyle (Equation\;14.1) $$

where

$$ a_1(x)=1,

a_0(x)=x,

b(x)=2x+3 $$

Solution
Substituting the conditions into 14.1

$$ y'+xy=2x+3$$ $$\displaystyle (Equation\;14.2) $$

Equation 14.2 is of the form

$$ y'+P(x)y=Q(x)$$

The integrating factor for this form is

$$ I(x)=\exp[P(x)dx]$$

For equation 14.2, the integrating factor is

$$ I(x)=\exp[\int xdx]=\exp[\frac{x^2}{2}] $$

Applying the integrating factor for equation 14.2:

$$ \exp[\frac{x^2}{2}]\frac{dy}{dx}+\exp[\frac{x^2}{2}]xy=2x\exp[\frac{x^2}{2}]+3\exp[\frac{x^2}{2}] $$ $$\displaystyle (Equation\;14.3) $$

(Eq.14.3) reduces to

$$ \frac{d}{dx}(y\exp(\frac{x^2}{2}))=2x\exp[\frac{x^2}{2}]+3\exp[\frac{x^2}{2}] $$

Integrating

$$ y\exp(\frac{x^2}{2})=\int 2x\exp[\frac{x^2}{2}]dx+\int3\exp[\frac{x^2}{2}]dx $$

Results in the following equation:

$$ y\exp(\frac{x^2}{2})=2\exp[\frac{x^2}{2}]+\int 3\exp [\frac{x^2}{2}]dx $$

There is no known anti-derivative of the last term, so using mathematical analysis tools the last therm results in the error-function. So that rearranging the equation we get for y(x):

$$ y=2+3\exp[\frac{x^2}{2}][\sqrt{\frac{\pi}{2}}erfi(\frac{x}{\sqrt2})] $$

Part 2
A general solution from the integrating factor method

Given
Assume $$ a(x)\ne0 $$ for all x, rearranging equationg 14.1 gives $$ y'+\frac{a_0(x)}{a_1(x)}y=\frac{b(x)}{a_1(x)} $$ $$\displaystyle (Equation\;14.4) $$

Find
Find an expression for $$ y(x) $$ in terms of $$ a_0, a_1 $$ and b

Solution
Define $$P(x)=\frac{a_0(x)}{a_1(x)}$$ and $$Q(x)=\frac{b(x)}{a_1(x)}$$

Equation 14.4 is then in the form

$$y'+P(x)y=Q(x)$$

The integrating factor for this form is

$$I(x)=\exp[\int{P(x)}dx]$$

Introducing a dummy factor t and substituting in for P(x)

$$\exp[\int^x{\frac{a_0(t)}{a_1(t)}}dt]\frac{dy}{dx}+\exp[\int^x{\frac{a_0(t)}{a_1(t)}}dt]\frac{a_0(x)}{a_1(x)}y=\frac{b(x)}{a_1(x)}\exp[\int^x{\frac{a_0(t)}{a_1(t)}}dt] $$

Reducing

$$\frac{d}{dx}[\exp[\int^x\frac{a_0(t)}{a_1(t)}]y]=\frac{b(x)}{a_1(x)}\exp[\int^x\frac{a_0(t)}{a_1(t)}dt]$$

Integrating

$$y=A\exp[-\int\frac{a_0(t)}{a_1(t)}dt]+\exp[-\int^x\frac{a_0(t)}{a_1(t)}dt]\int^x\frac{b(s)}{a_1(s)}ds\exp[-\int^s\frac{a_0(t)}{a_1(t)}dt]ds$$

Part 3
Solution to linear, first order, ordinary differential equation with varying coefficients using Euler integrating factor method.

Solution
Given an equation in the form of 14.1, where

$$a_1(x)=x^2+1, a_0(x)=x,$$ and $$b(x)=2x,$$

Equation 14.1 becomes

$$(x^2+1)y'+xy=2x$$

Rearranging

$$y'+\frac{x}{x^2+1}y=\frac{2x}{x^2+1}$$$$\displaystyle (Equation\;14.5) $$

Equation 14.5 is in the form

$$y'+P(x)y=Q(x)$$|| $$\displaystyle (Equation\;14.6) $$

The integrating for this form is

$$I(x)=\exp[\int{\frac{x}{x^2+1}}dx]=\exp[\frac{1}{2}\ln(x^2+1)]=\sqrt {x^2+1} $$ $$\displaystyle (Equation\;14.7) $$

Multiply 14.6 by 14.7 and simplify

$$\frac{d}{dx}(y\sqrt {x^2+1})=\frac{2y}{\sqrt {x^2+1}}$$ $$\displaystyle (Equation\;14.8) $$

Integrating

$$\int^x\frac{d}{dx}(y\sqrt {x^2+1})=\int^x \frac{2y}{\sqrt {x^2+1}}dy$$$$\displaystyle (Equation\;14.9) $$

Gives

$$Y=2$$

=R*2.15 Show the necessity of constant k1 and k2=

Given
Since (2)-(3) p.11-3 is an L1-ODE-VC, there should be only one integration constant, not two.

k1 is contained in (3) p11.4

$$ h(x)= \exp[ \int a_0(s)ds +k_1]$$

k2 is contained in (1) p11.5

$$y(x)=\frac{1}{h(x)}[\int^xh(x)b(s)ds+k_2]$$

Find
Show that the integration constant k1 in (3 ) p.11-4 is not necessary, i.e, only k2 in (1) p.11-5 is necessary.

Solution
To show this we will use (6)p10.3 and show that k1 cancels out of the equation, but we must find the contants k1 and k2 from equations(eq8.1) and (eq8.2)

$$y(x)=\frac{1}{h(x)}\int^xh(s)b(s)ds$$

When we perform the integration in the RHS of (eq8.1) it becomes

$$\exp[\int^x a_0(s)ds]=\exp[\int a_0(x)dx+k]$$

Using the properties for exponential functions

$$\exp[a+b]=\exp[a]exp[b]$$

Now (eq8.3) becomes

$$\exp[\int a_0(x)dx]\underbrace{\exp[k]}_{=:k_1}$$

Plugging this back into (eq8.1) gives us

$$h(x)= k_1\exp[\int a_0(x)dx]$$

To find k2 we will evaluate the integral in (eq8.2)

$$\int^x h(s)b(s)ds=\int h(x)b(x)dx+k_2$$

Finally plug equations (eq8.5) and (eq8.6) into (6) p.10-3

$$y(x)=\frac{1}{\cancelto{}{k_1}\exp[\int a_0(x)dx]}[\int \cancelto{}{k_1}\exp[\int a_0(x)dx]b(x)dx+k_2]$$

This shows k1 is not necessary

$$y(x)=\frac{1}{\exp[\int a_0(x)dx]}[\int \exp[\int a_0(x)dx]b(x)dx+k_2]$$

= R3.1 =

Given
Show that the N1-ODE (1) p.13-2

$$ \underbrace{\bar b(x,y) c(y)}_{\displaystyle\color{blue}{N(x,y)}} \, y' + \underbrace{a(x) \bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0 $$

satisfies the condition (2) p,11-2 :

$$\frac{h_x}{h}=-\frac{1}{N}(N_x-M_y) \, =: \color{blue}{n(x)}$$

that an integrating factor h(x) can be found to render it exact, only if $$k_1(y)=d_1$$

Show that (1) p.13-2 includes (1) p.12-4

$$ M+Ny'=[a(x)\, y+k_2(x)] + \bar b(x) \,y'=0 $$

as a particular case.

Solution
=*R3.3 =

Given
$$ a(x)=sinx^3 $$

$$ b(x)=cosx $$

$$ c(y)= \exp(2y) $$

1.Find an N1-ODE of the form (1) p.13-2

$$ \underbrace{\bar b(x,y) c(y)}_{\displaystyle\color{blue}{N(x,y)}} \, y' + \underbrace{a(x) \bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0 $$

$$\,(Eq.3.1)$$

that is either exact or can be made exact by IFM. (See R3.1)

2.Find the first integral

$$\phi(x,y)=k $$

Solution
Part 1

$$\bar b(x)= \int\limits^x b(s)ds= \int\limits^x cos(s)ds= sin(x) $$

$$\bar c(y)= \int\limits^y c(s)ds= \int\limits^y \exp^{2s}ds= \frac{1}{2}\exp^{2y}$$

Now Eq 3.1 becomes:

$$sin(x)e^{2y}y'+\frac{1}{2}sin(x^3)e^{2y}=0$$ $$\,(Eq.3.2)$$

Eq 3.2 is the form of

$$M(x,y)+N(x,y)\,y'=0$$ and the first condition of Exactness is satisfied. $$\,(Eq.3.3)$$

The condition below is not only necessary but also sufficient for Eq 3.2 to be an exact differencial equation.

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

$$\frac{\partial M}{\partial y}=\frac{\partial(\frac{1}{2}sin(x^3)\exp^2y)}{\partial y}=sin(x^3)\exp^2y$$ $$\,(Eq.3.4)$$

$$\frac{\partial N}{\partial x}=\frac{\partial(sin(x)\exp^2y)}{\partial x}=cos(x)\exp^2y$$ $$\,(Eq.3.5)$$

Eq 3.4 and Eq 3.5 reveals that

$$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$

Hence the Eq.3.2 is not exact.

But it can be made exact by using Integrating factor method.Refer to Eq(2), page 11-2 of the lecture notes, and let's multyply Eq 3.2 by h(x) in order to have an exact differential equation.

$$h(x)=exp[\int^x(-\frac{1}{N}(N_x-M_y))dx]$$

We can rewrite Eq 3.2 by cancelling $$e^{2y}$$from both term as:

$$\underbrace{sin(x)}_{N(x,y)}{y'}+ \underbrace{\frac{1}{2}sin(x^3)}_{M(x,y)}=0 $$

Now:

$$ h(x)=exp[\int^x(-\frac{1}{sin(x)}(cos(x)-0))dx]=exp[\int^x(-\frac{cos(x)}{sin(x)})dx]=exp[-ln(sin(x))]=\frac{1}{sin(x)}$$

$$\,(Eq.3.6)$$

Now Exact differential equation is given by

$$h(x)[sin(x)e^{2y}y'+\frac{1}{2}sin(x^3)\frac{1}{sin(x)}=0]$$

That is

$$sin(x)\frac{1}{sin(x)}y'+\frac{1}{2}sin(x^3)\frac{1}{sin(x)}=0$$

$$y'+\frac{1}{2}{sin(x^3)}{sin(x)}=0$$ $$\,(Eq.3.7)$$

Part 2 We know that

$$M(x,y)=\phi_x(x,y)$$ <p style="text-align:right">$$\,(Eq.3.8)$$

$$N(x,y)=\phi_y(x,y)$$ <p style="text-align:right">$$\,(Eq.3.9)$$

Also,

$$\phi_x(x,y)=\frac{\partial\phi(x,y)}{\partial x}$$ <p style="text-align:right">$$\,(Eq.3.10)$$

$$\phi_y(x,y)=\frac{\partial\phi(x,y)}{\partial y}$$ <p style="text-align:right">$$\,(Eq.3.11)$$

By integrating Eq.3.10 and 3.11 with respect to x, y, respectively, we can have equations as given below.

$$\int \phi_x(x,y)dx=\phi(x,y)=\int M(x,y)dx +k(y)$$ <p style="text-align:right">$$\,(Eq.3.12)$$

$$\int \phi_y(x,y)dy=\phi(x,y)=\int N(x,y)dy +k(x)$$ <p style="text-align:right">$$\,(Eq.3.13)$$

Now we can calculate $$k(y)$$ by differentiating Eq 3.12 with respect to y and equating to Eq 3.9=>

$$k'(y)=N(x,y)=1$$

Now integrate the above equation =>

$$k(y)=y+k1$$

Then$$\phi(x,y) $$ is obtained by substituting

$$k(y)$$ in Eq 3.12

$$\phi(x,y)=\int M(x,y)dx+y+k1$$

That is

$$\phi(x,y)=\frac{1}{2}\int \frac{sin(x^3)}{sinx}dx +y=k$$

=*R3.11 SC_L1_ODE_CC=

Given
Use (2)-(3) p.15-4 together with (1) p.14-4 to show (3) p.15-3. (2) p.15-4


 * $$\displaystyle \frac{d}{dt} \mathbf{\Phi}(t,t_0)= \mathbf{A} \, \mathbf{\Phi}(t,t_0)$$

<p style="text-align:right">$$\,(Eq.11.1)$$

(3) p.15-4


 * $$\mathbf \Phi (t_0,t_0)= \mathbf I$$

<p style="text-align:right">$$\,(Eq.11.2)$$

(1) p.14-4


 * $$\mathbf{\dot{x}}(t)= \mathbf{A}(t) \, \mathbf{x}(t) + \mathbf{B}(t) \, \mathbf{u}(t)$$

<p style="text-align:right">$$\,(Eq.11.3)$$

(3)p.15-3


 * $$ \mathbf{x}(t)=\mathbf{\Phi}(t,t_0) \mathbf{x}(t_0) + \int^t_{t_0} \mathbf{\Phi}(t, \tau) \, \mathbf{B}(\tau) \, \mathbf{u}{\tau} \, d\tau $$

<p style="text-align:right">$$\,(Eq.11.4)$$

Solution
Substituting$$\mathbf{x}(t)$$ in Eq11.1 by $$\mathbf{\phi}(t,t_0)$$ <p style="text-align:right">$$\,(Eq.11.5)$$

to obtain the differential equation of the state transition matrix, we get:


 * $$ \dot\phi (t,t_0)=A \mathbf{\phi}(t,t_0) + B u(t)$$

Re-arranging equation 11.5 in the following form:


 * $$\underbrace{1}_{N}\dot\phi - \underbrace{A \phi(t,t_0)}_{M}=Bu(t)$$

<p style="text-align:right">$$\,(Eq.11.6)$$

It is clear that this equation complies with the first condition of exactness.

So to make this equation exact we will apply the Euler Integrating Factor Method.

It has been previously shown that in order to meet the second condition of exactness we can multiply equation 11.6 by a function h to obtain the following:


 * $$\underbrace{h1}_{N}\dot\phi - \underbrace{hA \phi}_{M}=hBu$$

Now the second condition of exactness may be applied, where x is the dependent variable and t is the independent variable.


 * $$\underbrace{(h \frac{\partial M}{\partial{\phi}}+M \frac{\partial h}{\partial{\phi}})}_{\frac{\partial M}{\partial{\phi}}}=\underbrace{(h \frac{\partial N}{\partial{t}}+N \frac{\partial h}{\partial{t}})}_{\frac{\partial N}{\partial{t}}}$$

Now for the special case where $$h_x$$ equals zero we obtain the following:


 * $$hM_\phi=hN_t+Nh_t$$

Which may be rearranged to obtain:


 * $$hN_t+Nh_t-hM_\phi=0$$

We can now solve for h as a function of t.


 * $$h(t)=e^{-\int^t_{t_0}\frac{1}{N}(N_t-M_\phi)dt}$$

Where:
 * $$N=1$$


 * $$N_t=\frac{\partial 1}{\partial t}=0$$


 * $$M=-A \phi$$


 * $$M_\phi=-A$$


 * $$h(t)=e^{\int^t-Ads}=e^{-At}$$

<p style="text-align:right">$$\,(Eq.11.7)$$

Substituting 11.7 in the equation 11.6 we get


 * $$\underbrace{e^{-At} \dot\phi(t,t_0)}_{f'(x)\cdot g(x)}-\underbrace{e^{-At} \dot\phi(t,t_0)}_{f(x)\cdot g'(x)}=e^{-At} Bu(t)

$$ <p style="text-align:right">$$\,(Eq.11.8)$$

We should recognize equation 11.8 as the derivative of $$ (h\phi)$$ , by using the relation given in equation 11.2:


 * $$\frac{d(h\phi)}{dt}=hBu$$

Therefore, to find the solution we integrate both sides in the interval
 * $$t_0 \le \tau \le t$$


 * $$\int^t_{t_0} \frac{d(e^{-A\tau}\phi)}{d\tau}d\tau=\int^t_{t_0} e^{-A\tau}Bu(\tau)d\tau

$$ <p style="text-align:right">$$\,(Eq.11.9)$$

Equation 11.9 yields:


 * $$e^{-A(t-t_0)}\phi(t,t_0)-e^{-A(t_0-t_0)}\phi(t_0,t_0)+\int^t_{t_0}e^{-A\tau}Bu(\tau)d\tau$$

<p style="text-align:right">$$\,(Eq.11.10)$$

Re-arranging terms in equation 11.10 we obtain:


 * $$\phi(t,t_0)=e^{A(t-t_0)} \phi(t_0,t_0)+\int^t_{t_0} e^{A(t-\tau)} Bu(\tau)d\tau$$

<p style="text-align:right">$$\,(Eq.11.11)$$

Substituting $$\phi(t,t_0)$$ in the equation 11.11 by $$x(t)$$ we obtain the differential equation in the form of equation 11.4:


 * $$x(t)=e^{A(t-t_0)}x(t_0)+\int^t_{t_0}e^{A(t-\tau)}bu(\tau)d\tau$$