User:Egm6321.f10.team2.zou/HW1

= Problem 5-4: Boundary Value Problem =

Given
From the note 5-3 and 5-4 we know,


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y(x)=cy_{H}^{1}(x)+dy_{H}^{2}(x)+{{y}_{P}}(x),$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }


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y(a)=\alpha, $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }


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y(b)=\beta. $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
$$\displaystyle c, d$$ in terms of $$\displaystyle \alpha, \beta. $$

Solution
By substituting $$\displaystyle a $$ and $$\displaystyle b $$ into $$\displaystyle Eq. 1 $$ and combining with $$\displaystyle Eq. 2 $$ and $$\displaystyle Eq. 3 $$, we derive,


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cy_{H}^{1}(a)+dy_{H}^{2}(a)+{{y}_{P}}(a)=\alpha, $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }


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cy_{H}^{1}(b)+dy_{H}^{2}(b)+{{y}_{P}}(b)=\beta. $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

By multipling $$\displaystyle y_{H}^{2}(b) $$ in both sides of $$\displaystyle Eq. 4 $$ and $$\displaystyle y_{H}^{2}(a) $$ of $$\displaystyle Eq. 5 $$, we obtain,


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y_{H}^{1}(a)y_{H}^{2}(b)*c+y_{H}^{2}(a)y_{H}^{2}(b)*d+{{y}_{P}}(a)y_{H}^{2}(b)=\alpha y_{H}^{2}(b),$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }


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y_{H}^{1}(b)y_{H}^{2}(a)*c+y_{H}^{2}(a)y_{H}^{2}(b)*d+{{y}_{P}}(b)y_{H}^{2}(a)=\beta y_{H}^{2}(a).$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

Let $$\displaystyle Eq. 6 $$ minus $$\displaystyle Eq. 7 $$, then $$\displaystyle d $$ is cancelled, so we get,


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c=\frac{\left[ \alpha -{{y}_{P}}(a) \right]y_{H}^{2}(b)-\left[ \beta -{{y}_{P}}(b) \right]y_{H}^{2}(a)}{y_{H}^{1}(a)y_{H}^{2}(b)-y_{H}^{1}(b)y_{H}^{2}(a)}. $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

Note the structrue of $$\displaystyle Eq.4 $$ and $$\displaystyle Eq.5 $$, we find that $$\displaystyle c $$ is bound to $$\displaystyle y_{H}^{1} $$ while $$\displaystyle d $$ is bound to $$\displaystyle y_{H}^{2} $$, that is, when we switch the superscript between $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$, $$\displaystyle d $$ will become the former $$\displaystyle c $$ thus the same result of $$\displaystyle Eq. 8 $$. After that we switch back the superscript between $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$ then we get the right answer of $$\displaystyle d $$. By doing this, simply exchange the superscript of right side of $$\displaystyle Eq. 8 $$, we have,


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d=\frac{\left[ \alpha -{{y}_{P}}(a) \right]y_{H}^{1}(b)-\left[ \beta -{{y}_{P}}(b) \right]y_{H}^{1}(a)}{y_{H}^{2}(a)y_{H}^{1}(b)-y_{H}^{2}(b)y_{H}^{1}(a)}. $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

= Problem 6-1: Verifying 2 homogeneous solutions =

Given

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{{L}_{2}}(y)=(1-{{x}^{2}})\ddot{y}-2x\dot{y}+2y$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

where $$\displaystyle {{L}_{2}} $$ is an operator of functions.


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y_{H}^{1}(x)=x,$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }


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y_{H}^{2}(x)=\frac{x}{2}\log (\frac{1+x}{1-x})-1,$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

where $$\displaystyle y_{H}^{1}(x) $$ and $$\displaystyle y_{H}^{2}(x) $$ are homogeneous solutions of the equation $$\displaystyle {{L}_{2}}(y)=0$$.

Find
Verify:
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{{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Solution
From $$\displaystyle Eq.2 $$, we have,


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\dot{y}_{H}^{1}(x)=1,$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }


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\ddot{y}_{H}^{1}(x)=0.$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

Then by sustituting $$\displaystyle \dot{y}_{H}^{1}(x)$$ and $$\displaystyle \ddot{y}_{H}^{1}(x)$$ into $$\displaystyle {{L}_{2}}(y)$$ we have,


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\begin{align} {{L}_{2}}(y_{H}^{1})& =0-2x+2x \\ & =0. \end{align}$$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

From $$\displaystyle Eq.3 $$, we derive,


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\begin{align} \dot{y}_{H}^{2}(x)& ={{\left[ \frac{x}{2}\log \left( \frac{2}{1-x}-1 \right)-1 \right]}^{\prime }} \\ & ={{\left( \frac{x}{2} \right)}^{\prime }}*\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{2}*{{\left[ \log \left( \frac{2}{1-x}-1 \right) \right]}^{\prime }} \\ & =\frac{1}{2}\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{1-{{x}^{2}}}, \end{align}$$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }


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\begin{align} \ddot{y}_{H}^{2}(x) & ={{\left[ \frac{1}{2}\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{1-{{x}^{2}}} \right]}^{\prime }} \\ & ={{\left[ \frac{1}{2}\log \left( \frac{2}{1-x}-1 \right) \right]}^{\prime }}+{{\left( \frac{x}{1-{{x}^{2}}} \right)}^{\prime }} \\ & =\frac{1}{1-{{x}^{2}}}+\frac{1+{{x}^{2}}}. \end{align}$$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Then by substituting $$\displaystyle \dot{y}_{H}^{2}(x)$$ and $$\displaystyle \ddot{y}_{H}^{2}(x)$$ into $$\displaystyle {{L}_{2}}(y)$$ we have,


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\begin{align} {{L}_{2}}(y_{H}^{2}) & =(1-{{x}^{2}})\left( \frac{1}{1-{{x}^{2}}}+\frac{1+{{x}^{2}}} \right)-2x\left( \frac{1}{2}\log M+\frac{x}{1-{{x}^{2}}} \right)+2\left( \frac{x}{2}\log M-1 \right) \\ & =1+\frac{1+{{x}^{2}}}{1-{{x}^{2}}}-x\log M-\frac{2{{x}^{2}}}{1-{{x}^{2}}}+x\log M-2 \\ & =0, \end{align}$$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

where $$\displaystyle M=\frac{2}{1-x}-1$$.

By $$\displaystyle Eq. 6$$ and $$\displaystyle Eq. 9$$, we have proved $$\displaystyle {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$.