User:Egm6321.f09.Team1.vasquez/Homework1

Differential Equations: Equations of Motion
Many physical, chemical and biological systems can be described using mathematical models (King et. al. 2003) . For instance, Vu-Quoc and Olson (1989) have formulated a model for the vehicle/structure interaction in high speed trains (e.g. Magnetic Levitaded Trains: Maglev). Figure 1 shows a basic model of the wheel and the deformed guideway from which the differential equation of motion can be obtained.



Figure 1 shows the following parameters


 * $$\displaystyle Y^1(t)$$ = nominal motion.
 * $$\displaystyle u^1(S,t)$$ = axial displacement of the guideway.
 * $$\displaystyle u^2(S,t)$$ = transverse displacement of the guideway.

The equation of motion can be obtained by finding the first and second time derivatives of the function $$\displaystyle f(Y^1 (t),t)$$. The total time derivative of $$\displaystyle f$$ is given by


 * {| style="width:100%" border="0"

$$ \frac{d}f\left( {Y^1 \left( t \right),t} \right) = \frac{\partial }f(Y^1 (t),t)\frac{d}Y^1 \left( t \right) + \frac{\partial }f(Y^1 (t),t) $$
 * style="width:95%" |
 * style="width:95%" |
 * style= | (1)
 * }


 * {| style="width:100%" border="0"

$$ \frac = \frac{\partial }f\left( {Y^1 ,t} \right)\dot Y^1 + \frac{\partial }f\left( {Y^1 ,t} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * style= | (2)
 * }

Changing the notation of the derivatives gives
 * {| style="width:30%" border="0"

$$ \frac = f_{,S} \left( {Y^1 ,t} \right)\dot Y^1 + f_{,t} \left( {Y^1 ,t} \right) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= | (3)
 * }

Now, we can take the second time derivative


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$$ \fracf\left( {Y^1 ,t} \right) = \frac{\partial }\left[ {\frac{\partial }f\left( {Y^1 ,t} \right)\dot Y^1 + \frac{\partial }f\left( {Y^1 ,t} \right)} \right]\frac{d}Y^1 + \frac{\partial }\left[ {\frac{\partial }f\left( {Y^1 ,t} \right)\dot Y^1 + \frac{\partial }f\left( {Y^1 ,t} \right)} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * style= | (4)
 * }

Taking the derivative of the products gives


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$$ \fracf\left( {Y^1 ,t} \right) = \left[ {\frac{\partial }\left( {\frac{\partial }f\left( {Y^1 ,t} \right)} \right)\dot Y^1  + \frac{\partial }f\left( {Y^1 ,t} \right)\cancelto{0}{\frac{\partial }\left( {\dot Y^1 } \right)}} \quad \right]\dot Y^1  + \frac{\partial }\left( {\frac{\partial }f\left( {Y^1 ,t} \right)} \right)\dot Y^1 + \frac{\partial }\left( {\frac{\partial }f\left( {Y^1 ,t} \right)} \right)\dot Y^1 + \frac{\partial }f\left( {Y^1 ,t} \right)\frac{\partial }\left( {\dot Y^1 } \right) + \frac{\partial }\left( {\frac{\partial }f\left( {Y^1 ,t} \right)} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * style= | (5)
 * }

Simplifying


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$$ \fracf\left( {Y^1 ,t} \right) = \fracf\left( {Y^1 ,t} \right)\left( {\dot Y^1 } \right)^2  + \fracf\left( {Y^1 ,t} \right)\dot Y^1  + \fracf\left( {Y^1 ,t} \right)\dot Y^1 + \frac{\partial }f\left( {Y^1 ,t} \right)\ddot Y^1  + \fracf\left( {Y^1 ,t} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * style= | (6)
 * }

Reorganizing yields


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$$ \fracf\left( {Y^1 ,t} \right) = \frac{\partial }f\left( {Y^1 ,t} \right)\ddot Y^1 + \fracf\left( {Y^1 ,t} \right)\left( {\dot Y^1 } \right)^2  + 2\fracf\left( {Y^1 ,t} \right)\dot Y^1  + \fracf\left( {Y^1 ,t} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * style= | (7)
 * }

Changing the notation of the derivatives gives


 * {| style="width:60%" border="0"

$$ \frac = f_{,S} \left( {Y^1 ,t} \right)\ddot Y^1 + f_{,SS} \left( {Y^1 ,t} \right)\left( {\dot Y^1 } \right)^2  + 2f_{,St} \left( {Y^1 ,t} \right)\dot Y^1  + f_{,tt} \left( {Y^1 ,t} \right) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= | (8)
 * }

Equation (8) is the differential equation that represents the nominal motion. It is a nonlinear second order partial differential equation with varying coefficients.

Linearity of differential equations
A differential equation can be interpreted as an operator and it is linear when its associated operator is linear, i.e. it satisfies additivity and homogeneity.

$$\displaystyle \forall u,v$$ functions of $$\displaystyle x$$ and $$\displaystyle \forall \alpha, \beta \in \mathbb{R} $$

The linear operator $$\displaystyle L(\cdot)$$ (associated with a linear differential equation) satisfies:


 * {| style="width:100%" border="0"

$$ L \left ( \alpha u + \beta v \right ) = \alpha L (u) + \beta L(v) $$
 * style="width:95%" |
 * style="width:95%" |
 * style= | (9)
 * }

Examples
Check the linearity of the following differential equations


 * $$\displaystyle 1) $$ $$ \quad 2 x^2 + \sqrt{y} + x^5 y^3 y' = 0$$

We can define the operator $$\displaystyle D(\cdot)$$ associated to the differential equation as follows:



D(\cdot)= 2 x^2 + \sqrt{(\cdot)} + x^5 (\cdot)^3 \frac{d( \cdot )}{dx} $$

We just need to check if it meets the linearity property given by (9). Using the left side gives



\begin{align} D(\alpha u + \beta v ) &= 2 x^2 + \sqrt{(\alpha u + \beta v)} + x^5 (\alpha u + \beta v)^3 \frac{d(\alpha u + \beta v)}{dx} \\ &=2 x^2 + \sqrt{(\alpha u + \beta v)} + x^5 (\alpha u + \beta v)^3 \left[ \alpha \frac{d( u )}{dx} + \beta \frac{d( v )}{dx} \right] \end{align} $$

Now, using the right side of (9)



\begin{align} \alpha D(u) + \beta D(v) &= \alpha \left [ 2 x^2 + \sqrt{(u)} + x^5 (u)^3 \frac{d(u)}{dx} \right ]+ \beta \left [ 2 x^2 + \sqrt{(v)} + x^5 (v)^3 \frac{d(v)}{dx} \right ]\\ &= \alpha \left [ 2 x^2 + \sqrt{(u)} \right ]+ \beta \left [ 2 x^2 + \sqrt{(v)} \right ] + x^5 \left [ \alpha (u)^3 \frac{d(u)}{dx} + \beta (v)^3 \frac{d(v)}{dx} \right ] \end{align} $$

Since $$\displaystyle D(\alpha u + \beta v ) \neq \alpha D(u) + \beta D(v)$$ we conclude that the given differential equation is nonlinear.


 * $$\displaystyle 2) $$ $$ \quad x^2 y^5+6 \left[ \frac{dy}{dx} \right]^2=0$$

We can define the operator $$\displaystyle D(\cdot)$$ associated to the differential equation as follows:



D(\cdot)=x^2 (\cdot)^5+6 \left[ \frac{d(\cdot)}{dx} \right]^2 $$

We just need to check if it meets the linearity property given by (9). Using the left side gives



\begin{align} D(\alpha u + \beta v ) &= x^2 (\alpha u + \beta v )^5+6 \left[ \frac{d(\alpha u + \beta v )}{dx} \right]^2\\ &=x^2 (\alpha u + \beta v )^5+6 \left[ \alpha \frac{d( u )}{dx} + \beta \frac{d( v )}{dx} \right]^2 \end{align} $$

Now, using the right side of (9)



\begin{align} \alpha D(u) + \beta D(v) &= \alpha \left [ x^2 (u)^5+ 6 \left[ \frac{d(u)}{dx} \right]^2 \right ]+ \beta \left [ x^2 (v)^5+ 6 \left[ \frac{d(v)}{dx} \right]^2 \right ]\\ &= \alpha x^2 (u)^5+ \beta x^2 (v)^5 + 6 \left [ \alpha \left[ \frac{d(u)}{dx} \right]^2 + \beta \left[ \frac{d(v)}{dx} \right]^2 \right ] \end{align} $$

Since $$\displaystyle D(\alpha u + \beta v ) \neq \alpha D(u) + \beta D(v)$$ we conclude that the given differential equation is nonlinear.


 * $$\displaystyle 3) $$ $$ \quad y''+y'+y=0 $$

We can define the operator $$\displaystyle D(\cdot)$$ associated to the differential equation as follows:



D(\cdot)=\frac{d^2(\cdot)}{dx^2}+\frac{d(\cdot)}{dx}+(\cdot) $$

We just need to check if it meets the linearity property given by (9). Using the left side gives



\begin{align} D(\alpha u + \beta v ) &= \frac{d^2(\alpha u + \beta v )}{dx^2}+\frac{d(\alpha u + \beta v )}{dx}+(\alpha u + \beta v )\\ &= \alpha \frac{d^2(u)}{dx^2}+\beta \frac{d^2(v)}{dx^2}+\alpha \frac{d(u)}{dx}+\beta \frac{d(v)}{dx}+\alpha u + \beta v \end{align} $$

Now, using the right side of (9)



\begin{align} \alpha D(u) + \beta D(v) &= \alpha \left [ \frac{d^2(u)}{dx^2}+\frac{d(u)}{dx}+(u) \right ]+ \beta \left [ \frac{d^2(v)}{dx^2}+\frac{d(v)}{dx}+(v) \right ]\\ &= \alpha \frac{d^2(u)}{dx^2}+\alpha \frac{d(u)}{dx}+\alpha u+\beta \frac{d^2(v)}{dx^2}+\beta \frac{d(v)}{dx} + \beta v\\ &= \alpha \frac{d^2(u)}{dx^2}+\beta \frac{d^2(v)}{dx^2}+\alpha \frac{d(u)}{dx}+\beta \frac{d(v)}{dx}+\alpha u + \beta v \end{align} $$

Since $$\displaystyle D(\alpha u + \beta v ) = \alpha D(u) + \beta D(v)$$ we conclude that the given differential equation is linear.

Exact ODEs. Integrating factors
A first-order ODE


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$


 * style= | (10)
 * }

is said to be an exact differential equation if the differential form


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle M(x,y)dx+N(x,y)dy=0$$
 * $$\displaystyle M(x,y)dx+N(x,y)dy=0$$


 * style= | (11)
 * }

is exact, (Kreyszig 2006). The not only necessary but also sufficient condition for (10) to be and exact differential equation is given by


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$


 * style= | (12)
 * }

The implicit general solution for (10) is given by $$\displaystyle \phi(x,y)=c$$.

Integrating factors
If the ODE given by (10) is not exact, i.e. $$\displaystyle \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$, we can find $$\displaystyle h(x,y)$$ such that the differential equation


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \left[ h(x,y)M(x,y) \right] + \left[ h(x,y)N(x,y)\right]y'=0$$
 * $$\displaystyle \left[ h(x,y)M(x,y) \right] + \left[ h(x,y)N(x,y)\right]y'=0$$


 * style= | (13)
 * }

is exact. $$\displaystyle h(x,y)$$ is called the Euler integrating factor and it makes the non-exact differential equation integrable. Equation (13) satisfies the exactness condition if


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle h_xN - h_yM + h \left( N_x-M_y\right)=0$$
 * $$\displaystyle h_xN - h_yM + h \left( N_x-M_y\right)=0$$


 * style= | (14)
 * }

Case 1
$$\displaystyle h_yM =0$$

In this case $$\displaystyle h(x,y)=h(x)$$ and (14) becomes


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \frac{h_x}{h}=\frac{1}{N} \left( M_y - N_x \right) = f(x)$$
 * $$\displaystyle \frac{h_x}{h}=\frac{1}{N} \left( M_y - N_x \right) = f(x)$$


 * style= | (15)
 * }

and the integrating factor is found by integrating both sides


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle h(x)=\exp \int^x f(s)ds$$
 * $$\displaystyle h(x)=\exp \int^x f(s)ds$$


 * style= | (16)
 * }

Case 2
$$\displaystyle h_xN =0$$

In this case $$\displaystyle h(x,y)=h(y)$$ and (14) becomes


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \frac{h_y}{h}=\frac{1}{M} \left( N_x - M_y \right) = g(y)$$
 * $$\displaystyle \frac{h_y}{h}=\frac{1}{M} \left( N_x - M_y \right) = g(y)$$


 * style= | (17)
 * }

and the integrating factor is found by integrating both sides


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle h(y)=\exp \int^y g(s)ds$$
 * $$\displaystyle h(y)=\exp \int^y g(s)ds$$


 * style= | (18)
 * }

Example
Find the solution of the first-order ODE


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle p' + p = x$$
 * $$\displaystyle p' + p = x$$


 * style= | (18)
 * }

To find the solution, we need to perform the exactness test which requires two steps: 1. Writing the ODE in the form of (10) and 2. Checking the condition given by (12).

1. Re-writing (18) in the for of (10) gives


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle (p-x)+ (1)p' = 0$$
 * $$\displaystyle (p-x)+ (1)p' = 0$$


 * style= | (19)
 * }

where $$\displaystyle M=(p-x)$$ and $$\displaystyle N=1$$.

2. Now, we have to test if $$\displaystyle M_p=N_x$$


 * $$\displaystyle M_p = 1$$ and $$\displaystyle N_x = 0$$

Therefore, (18)) is not an exact differential equation. Then, we need to look for an integrating factor (Case 1) in order to the ODE be exact.


 * $$\displaystyle f(x)=\frac{1}{N}\left[ M_p - N_x \right ]= \frac{1}{1}\left[ 1 - 0 \right ] = 1 $$

Now, we can use (16) to get the integrating factor


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle h(x)=\exp \int^x 1ds = e^x$$
 * $$\displaystyle h(x)=\exp \int^x 1ds = e^x$$


 * style= | (20)
 * }

Multiplying both sides of (19) by (20) gives


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle e^x(p-x) + e^x p' = 0$$
 * $$\displaystyle e^x(p-x) + e^x p' = 0$$


 * style= | (21)
 * }

which is now an exact differential equation ($$\displaystyle M_p = N_x = e^x$$).

To solve the differential equation we can use the following equation


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \phi(x,p)=\int^p{N(s)}ds+ l(x)$$
 * $$\displaystyle \phi(x,p)=\int^p{N(s)}ds+ l(x)$$


 * style= | (22)
 * }

Substituting $$\displaystyle N = e^x$$ into (22) gives


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \phi(x,p)=\int^p(e^x) ds+ l(x)$$
 * $$\displaystyle \phi(x,p)=\int^p(e^x) ds+ l(x)$$


 * style= | (23)
 * }

Integrating with respect to $$\displaystyle p $$


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \phi(x,p)=pe^x + l(x)$$
 * $$\displaystyle \phi(x,p)=pe^x + l(x)$$


 * style= | (24)
 * }

Now we can take the derivative of $$\displaystyle \phi(x,p)$$ to find $$\displaystyle l(x)$$


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \frac{\partial \phi(x,p)}{\partial x}= e^x p + \frac{\partial l(x)}{\partial x} = M = e^x (p - x)$$
 * $$\displaystyle \frac{\partial \phi(x,p)}{\partial x}= e^x p + \frac{\partial l(x)}{\partial x} = M = e^x (p - x)$$


 * style= | (25)
 * }

Then, from (25)


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \frac{\partial l(x)}{\partial x} = \cancel{pe^x} - xe^x - \cancel{e^x p}$$
 * $$\displaystyle \frac{\partial l(x)}{\partial x} = \cancel{pe^x} - xe^x - \cancel{e^x p}$$


 * style= | (26)
 * }

Therefore,


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle l(x)= - \int xe^x dx$$
 * $$\displaystyle l(x)= - \int xe^x dx$$


 * style= | (27)
 * }

Integrating (27) by parts gives


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle l(x)= - \left ( x e^x- \int {e^x} dx \right ) $$
 * $$\displaystyle l(x)= - \left ( x e^x- \int {e^x} dx \right ) $$


 * style= | (28)
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle l(x)=- x e^x + e^x$$
 * $$\displaystyle l(x)=- x e^x + e^x$$


 * style= | (29)
 * }

Substituting (29) into (24)


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \phi(x,p)= e^x p - x e^x + e^x $$
 * $$\displaystyle \phi(x,p)= e^x p - x e^x + e^x $$


 * style= | (30)
 * }

Equation (30) represents is an implicit solution for the differential equation, and since $$\displaystyle \phi(x,y)=c$$,


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle e^x p - x e^x + e^x = A$$
 * $$\displaystyle e^x p - x e^x + e^x = A$$


 * style= | (31)
 * }

From (31), we can get an explicit solution $$\displaystyle p(x)$$


 * {| style="width:25%" border="0"


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle p(x)=A e^{-x}  + x - 1$$
 * $$\displaystyle p(x)=A e^{-x}  + x - 1$$


 * style= | (32)
 * }

Building exact nonlinear ODEs
To build an exact nonlinear differential equation, we just need to invent any $$\displaystyle \phi(x,y)$$ (nonlinear continuous function of $$\displaystyle x$$ and $$\displaystyle y$$), and find its total derivative with respect to $$\displaystyle x$$.


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle F=\frac{d}{dx} \phi(x,y)$$
 * $$\displaystyle F=\frac{d}{dx} \phi(x,y)$$


 * style= | (31)
 * }

Therefore


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle M=\frac{\partial \phi}{dx}= \phi_x$$
 * $$\displaystyle M=\frac{\partial \phi}{dx}= \phi_x$$


 * style= | (32)
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle N=\frac{\partial \phi}{dy}= \phi_y$$
 * $$\displaystyle N=\frac{\partial \phi}{dy}= \phi_y$$


 * style= | (33)
 * }

Examples

 * $$\displaystyle \phi(x,y)=6 x^4 + 2 y ^{\frac{3}{2}}$$


 * $$\displaystyle M=24 x^3$$


 * $$\displaystyle N=3 y ^{\frac{1}{2}}$$


 * $$\Rightarrow \displaystyle \left (24 x^3 \right) + \left( 3 y ^{\frac{1}{2}} \right) y'=0$$


 * $$\displaystyle \phi(x,y)=xy + x^2y^2$$


 * $$\displaystyle M=y + 2xy^2$$


 * $$\displaystyle N=x + 2yx^2$$


 * $$\Rightarrow \displaystyle \left (y + 2xy^2 \right) + \left (x + 2yx^2 \right) y'=0$$


 * $$\displaystyle \phi(x,y)=e^{xy}+x$$


 * $$\displaystyle M=ye^{xy}+1$$


 * $$\displaystyle N=xe^{xy}$$


 * $$\Rightarrow \displaystyle \left (ye^{xy}+1 \right) + \left (xe^{xy} \right) y'=0$$


 * $$\displaystyle \phi(x,y)=y \sin(x)$$


 * $$\displaystyle M=y \cos(x)$$


 * $$\displaystyle N=\sin(x)$$


 * $$\Rightarrow \displaystyle \left [y \cos(x) \right] + \left [\sin(x) \right] y'=0$$