User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework4

= Problem 2 - Verify Exactness of Bessel Equation =

Given

 * Bessel Equation is given by
 * {| style="width:100%" border="0"

$$
 * $${{x}^{2}}{{y}^{''}}+x{{y}^{'}}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y=0$$
 * $$\displaystyle (Eq. 2.1)
 * $$\displaystyle (Eq. 2.1)
 * }
 * }


 * Where, $$\displaystyle\upsilon \in \mathbb{R}$$

Find

 * $$\displaystyle\left( 1 \right)$$ Verify Exactness of $$\displaystyle Eq.2.1$$ using 2 methods


 * Method 1


 * Second conditions of Exactness:
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$$ $$
 * First Relation:$$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}
 * First Relation:$$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}
 * $$\displaystyle (Eq.2.2)
 * }
 * }


 * and


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$$ $$
 * Second relation:$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}
 * Second relation:$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}
 * $$\displaystyle (Eq.2.3)
 * }
 * }


 * Method 2


 * Another form of second Exactness condition for a $$\displaystyle N2-ODE$$:


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f_0 - \frac {df_1}{dx} + \frac {d^2f_2}{dx^2}=0 $$ $$.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq.2.4)
 * }
 * }
 * $$\displaystyle\left( 2 \right)$$ If $$\displaystyle Eq.2.1$$ not Exact, see whether it can be made Exact using Integrating Factor Method with:


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$$
 * $$h\left( x,y \right)={{x}^{m}}{{y}^{n}}$$
 * $$\displaystyle (Eq. 2.5)
 * $$\displaystyle (Eq. 2.5)
 * }
 * }


 * That is,find $$\displaystyle (m,n) $$ such that Eq 2.1 is exact.

Solution

 * PART 1: Verify Exactness of Bessel Equation


 * The first condition for exactness of $$\displaystyle N2-ODE$$ is that it is in the following form:
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F(x,y,p) = f(x,y,p)y'' + g(x,y,p)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2.6)
 * }
 * }


 * where


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p = y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2.7)
 * }
 * }


 * Now comparing Eq.2.1 with the form of Eq.2.6,we can see that:


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$$
 * $$\displaystyle f(x,y,p)={{x}^{2}}$$
 * $$\displaystyle (Eq.2.8)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.8)
 * }
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$$
 * $$\displaystyle g(x,y,p)=x{p}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.9)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.9)
 * }
 * }


 * Hence the first condition for exactness is satisfied.


 * Now verify second condition of Exactness:


 * (a) Using Method 1- Eq.2.2 and Eq.2.3


 * Taking derivative of Eq.2.8 yields:


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

\begin{array}{*{20}l} {f_x = 2x} & {f_{xx}  = 2} & {f_{xy}  = 0} & {f_{xp}  = 0 }  \\ {f_y = 0} & {f_{yy}  = 0} & {f_{yp}  = 0}\\ \end{array} $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * Similarly taking derivative of Eq.2.9 yields:


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\begin{array}{*{20}l} {g_x = p+2xy} & {g_{xp} = 1} & {{g}_{y}}={{x}^{2}}-{{\upsilon }^{2}} & {g_{yp}  = 0 }  \\ {g_p = x} & {g_{pp}  = 0} \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting the above values to Eq.2.3 (first relation) and Eq.2.4 (second relation):


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 * $$2+2p\times 0+{{p}^{2}}\times 0=1+p\times 0-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 * }
 * }
 * }


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$$
 * $$\Rightarrow 2=1-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.10)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.10)
 * }
 * }


 * So first relation for second exactness condition is not satisfied.


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$$
 * $$0+p\times 0+2\times 0=0\Rightarrow 0=0$$, second relation is satisfied
 * <p style="text-align:right;">$$\displaystyle (Eq.2.11)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.11)
 * }
 * }


 * Since the first relation is not satisfied, the Eq.2.1 is not Exact


 * (b) Using Method 2- Eq.2.4


 * Differentiating $$\displaystyle Eq.2.1$$, partially with respect to $$\displaystyle y,{{y}^{'}},{{y}^{''}}$$,


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 * $${{f}_{0}}=\frac{\partial F}{\partial y}={{x}^{2}}-{{\upsilon }^{2}}$$
 * }
 * }
 * }


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 * $${{f}_{1}}=\frac{\partial F}{\partial {{y}^{'}}}=x$$
 * }
 * }
 * }


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 * $${{f}_{2}}=\frac{\partial F}{\partial {{y}^{''}}}={{x}^{2}}$$
 * }
 * }
 * }


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 * Now, $$\frac{\partial {{f}_{1}}}{\partial x}=1$$, and $$\frac{\partial {{x}^{2}}}=2$$
 * }
 * }
 * }


 * Substittuting above values in $$\displaystyle Eq.2.4$$ yields,


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


 * $${{x}^{2}}-{{\upsilon }^{2}}-1+2\ne 0\Rightarrow $$not satisfied and hence the Eq.2.1 is not Exact.
 * }
 * }
 * }


 * <H4>PART 2: Make Bessel Equation exact using Integrating Factor Method</H4>


 * Multiply $$\displaystyle Eq.2.1$$ by Integrating Factor ($$\displaystyle Eq.2.5$$)$$\Rightarrow $$


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$$
 * $$\underbrace_{f}{{y}^{''}}+\underbrace{{{x}^{m+1}}{{y}^{n}}{{y}^{'}}+{{x}^{m+2}}{{y}^{n+1}}-{{x}^{m}}{{y}^{n+1}}{{\upsilon }^{2}}}_{g}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.12)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.12)
 * }
 * }


 * New $$\displaystyle f$$ and $$\displaystyle g$$ are:


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$$
 * $$f\left( x,y,p \right)={{x}^{m+2}}{{y}^{n}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.13)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.13)
 * }
 * }


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$$
 * $$g\left( x,y,p \right)={{x}^{m+1}}{{y}^{n}}p+{{x}^{m+2}}{{y}^{n+1}}-{{x}^{m}}{{y}^{n+1}}{{\upsilon }^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.14)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.14)
 * }
 * }


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Analyzing the second relation of second exactness condition ($$\displaystyle Eq.2.3 $$):$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}$$, it can be noted that most of the terms drop out since $$\displaystyle f$$ is independent of $$\displaystyle p$$. Consequently all derivatives of $$\displaystyle f$$ with respect to $$\displaystyle p$$ become zero. Moreover $$\displaystyle {{g}_{pp}}=0$$ and hence $$\displaystyle Eq.2.3 $$ is simplified to,
 * }
 * }


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$$
 * $$\displaystyle 2{{f}_{y}}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.15)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.15)
 * }
 * }


 * Taking derivative of $$\displaystyle (Eq.2.13)$$ with repect to $$\displaystyle y$$ and substituting to$$\displaystyle (Eq.2.15)$$ yields,


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 * $$\displaystyle n{{x}^{m+2}}{{y}^{n-1}}=0$$
 * }
 * }
 * }


 * The above equation is true only when,


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$$\displaystyle n=0$$ $$
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 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.16)
 * }
 * }


 * As $$\displaystyle n=0 $$, the function $$\displaystyle f$$ and $$\displaystyle g$$ can be rewritten as:


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$$
 * $$f\left( x,y,p \right)={{x}^{m+2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.17)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.17)
 * }
 * }


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$$
 * $$g\left( x,y,p \right)={{x}^{m+1}}p+{{x}^{m+2}}y-{{x}^{m}}y{{\upsilon }^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.18)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.18)
 * }
 * }


 * Taking derivative of $$\displaystyle Eq.2.18$$ yields,


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\begin{array}{*{20}l} {{f}_{x}}=\left( m+2 \right){{x}^{m+1}}, & {{f}_{xx}}=\left( m+2 \right)\left( m+1 \right){{x}^{m}}, & {f_{xy} = 0}, & {f_{xp}  = 0 }  \\ {f_y = 0}, & {f_{yy}  = 0}, & {f_{yp}  = 0}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Similarly taking derivative of $$\displaystyle Eq.2.17$$ yields,


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\begin{array}{*{20}l} {{g}_{x}}=\left( m+1 \right){{x}^{m}}p+\left( m+2 \right){{x}^{m+1}}y-m{{x}^{m-1}}y{{\upsilon }^{2}}\\ {{g}_{xp}}=\left( m+1 \right){{x}^{m}}\\ {{g}_{y}}={{x}^{m+2}}-{{x}^{m}}{{\upsilon }^{2}}\\ {g_{yp} = 0} \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting above differentials in $$\displaystyle Eq.2.17$$ gives,


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


 * $$\left( m+2 \right)\left( m+1 \right){{x}^{m}}+0+0=\left( m+1 \right){{x}^{m}}+0-\left( {{x}^{m+2}}-{{x}^{m}}{{\upsilon }^{2}} \right)$$
 * }
 * }
 * }


 * Simplifying,


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 * $$\left( m+2 \right)\left( m+1 \right)=m+1+{{\upsilon }^{2}}-{{x}^{2}}$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{m}^{2}}+2m+\left( {{x}^{2}}-{{\upsilon }^{2}}+1 \right)=0$$
 * }
 * }
 * }


 * $$\displaystyle m$$ is obtained by solving the above quadratic equation:


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$$m=-1\pm \sqrt{\left( {{\upsilon }^{2}}-{{x}^{2}} \right)}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.19)
 * }
 * }


 * So the integration factor is obtained by substituting $$\displaystyle Eq. 2.16 $$ and $$\displaystyle Eq. 2.19 $$ in $$\displaystyle Eq. 2.5 $$:


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$$h={{x}^{-1\pm \sqrt{\left( {{\upsilon }^{2}}-{{x}^{2}} \right)}}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.20)
 * }
 * }


 * Thus Bessel Equation can be made exact using the above integrating factor.