User talk:Simmgibo

=Report 2=

Problem Statement
The minimum degree of differentiability of the function $$\displaystyle \phi(x,y) $$, and state the full theorem and provide a proof.

Given

 * Review calculus such that


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

$$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$
 * style="width:95%" |
 * style="width:95%" |
 * (6.1)
 * }

Nomenclature

 * $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\phi _{,xy}''$$

Solution

 * We have to know Clairaut's Theorem first.
 * If $$\displaystyle f:R^{n}\rightarrow R $$
 * {| style="width:100%" border="0"

$$\displaystyle f_{x_{i},y_{j}}(a_1, \dots, a_n) $$ = $$\displaystyle f_{y_{j},x_{i}}(a_1, \dots, a_n) $$ ,suppose that $$\displaystyle 1\leq i,j\leq n $$
 * style="width:95%" |
 * style="width:95%" |
 * (6.2)
 * }
 * {| style="width:100%" border="0"|-

According to Clairaut's Theorem 6.2, the minimum degree of differentiability of the function $$\displaystyle \phi(x,y) $$ must be three.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }


 * Proof of Clairaut's Theorem:
 * According to Mean Value Theorem, if $$ \displaystyle f:[a,b]\rightarrow\mathbb{R} $$ is continuous on $$ \displaystyle [a,b]$$ and differentiable on $$ \displaystyle (a,b)$$, and there exist c which is $$\displaystyle a(6.3)
 * }
 * Let K is the near the point which exist $$\displaystyle x,y\in K $$
 * {| style="width:100%" border="0"

$$f_{,y}' (a+n,b)-f_{,y} ' (a,b)=\lim_{k\rightarrow 0}\frac{f(a+h,b+k)-{f(a+h,b)}}{k}-\lim_{k\rightarrow 0}\frac{f(a,b+k)-{f(a,b)}}{k}$$
 * style="width:95%" |
 * style="width:95%" |
 * (6.4)
 * }
 * We can define the function after fixed k
 * $$\displaystyle g_{k}(t)=f(a+t,b+k)-f(a+t,b)$$
 * and after derivate $$g_{k}(t)$$
 * $$\displaystyle g_{k}'(t)=f_{,x}'(a+t,b+k)-f_{,x}'(a+t,b)$$
 * According to Mean Value theorem,
 * $$f_{,y}' (a+h,b)-f_{,y} ' (a,b)=\lim_{k\rightarrow 0}\frac{g_{k}(h)-{g_{k}(0)}}{k}=\lim_{k\rightarrow 0}\frac{hg_{k}'(\bar{h})}{k}$$
 * $$\frac{f_{,y}'(a+h,b)-f_{,y}'(a,b)}{h}=\lim_{k\rightarrow 0}\displaystyle \frac{f_{,x}'(a+\bar{h},b+k)-f_{,x}'(a+\bar{h},b)}{k}=f_{,xy}''(a+\bar{h},b)$$
 * if $$h\rightarrow 0$$, then $$\bar{h}\rightarrow 0$$
 * if $$h\rightarrow 0$$, then $$\bar{h}\rightarrow 0$$

So we can derive the following equation: $$f_{,yx}(a,b)=\lim_{h\rightarrow 0} \frac{f_{,y}'(a+h,b)-f_{,y}'(a,b)}{h}=f_{,xy}(a,b)$$

Problem Statement

 * 1. Derive Eq.(12.1),i.e.,the 2nd relation in the 2nd exactness condition, by differentiating the definition of $$ \displaystyle g(x,y,y^{'})$$ in (3)p.16-4 with respect to $$ \displaystyle p:=y^{'}$$ defined in (2)p.7-3.
 * 2. Derive Eq.(12.2),i.e., the 1st relation in the 2nd exactness condition.
 * 3. Verify that Eq.(12.3) satisfies the 2nd exactness condition.

Given

 * 2nd exactness condition for N2-ODEs
 * {| style="width:100%" border="0"

$$\displaystyle f_{xp} + pf_{yp} + 2 f_y = g_{pp}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.1)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx} + 2pf_{xy} + p^2 f_{fyy} = g_{xp} + p g_{yp} - g_y $$
 * style="width:95%" |
 * style="width:95%" |
 * (12.2)
 * }
 * The N2-ODE
 * {| style="width:100%" border="0"

$$\displaystyle x\left( y^{\prime} \right)^2 + yy^{\prime} + \left( xy \right) y^{''} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * (12.3)
 * }

Nomenclature

 * $$ \displaystyle p := y^{'}=\frac{dy}{dx}$$
 * $$ \displaystyle y^{''}=\frac{d^{2}y}{dx^{2}}$$

Solution

 * 1. Derive Eq.(12.1),i.e.,the 2nd relation in the 2nd exactness condition
 * According to (3)p.16-4
 * {| style="width:100%" border="0"

$$\displaystyle g(x,y,p):=\phi_{x}(x,y,p)+\phi_{y}(x,y,p)p$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.4)
 * }
 * and, (4)p.16-4
 * {| style="width:100%" border="0"

$$\displaystyle f(x,y,p):=\phi_{p}(x,y,p)$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.5)
 * }
 * Differentiate the Eq.(12.4) with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{p}=\phi_{xp}+\phi_{yp}p+\phi_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.6)
 * }
 * Differentiate the Eq.(12.6) with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{pp}=\phi_{xpp}+\phi_{ypp}p+\phi_{yp}+\phi_{yp}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.7)
 * }
 * According to Eq.(12.5)
 * {| style="width:100%" border="0"

$$\displaystyle g_{pp}=f_{xpp}+f_{yp}p+2f_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.8)
 * }
 * {| style="width:100%" border="0"|-

This Eq.(12.8) is equal to the Eq.(12.1)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * 2.Derive Eq.(12.2),i.e., the 1st relation in the 2nd exactness condition
 * According to (1)p.17-3, we can get these equations,
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{xy}=\phi_{yx}, \phi_{yp}=\phi_{py}, \phi_{px}=\phi_{xp}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.9)
 * }
 * First, from Eq.(12.9)$$\displaystyle \phi_{px}=\phi_{xp}$$
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$$\displaystyle \left( \phi_{p} \right)_{x} = \left( \phi_{x} \right)_{p}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.10)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle (g - \phi_y y^{\prime})_{p} =(f)_{x}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.11)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle g_{p} -\phi_{yp} y^{\prime} - \phi_y{} = f_x$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.12)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle g_{p}-(\phi_{p})_{y}p-\phi_y = f_x$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.13)
 * }
 * So,
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{y}=g_{p} - f_{y}p - f_{x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.14)
 * }
 * Second, $$\displaystyle (\phi_{y})_{p}=(\phi_{p})_{y}$$
 * and, $$\displaystyle g:=\phi_{x}+\phi_{y}p$$
 * {| style="width:100%" border="0"

$$\displaystyle (\frac{g-\phi_{x}}{p})_{p}=f_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.15)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle -\frac{g-\phi_{x}}{p^2} + \frac{g_{p} - f_{x}}{p}=f_y$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.16)
 * }
 * So,
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{x}=g-p(g_{p}-f_{x})+p^{2}f_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.17)
 * }
 * Then, differentiate the Eq.(12.14) with respect to x
 * {| style="width:100%" border="0"

$$\displaystyle (\phi_{y})_{x} = g_{px}-f_{yx}p-f_{xx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.18)
 * }
 * differentiate the Eq.(12.17) with respect to y
 * {| style="width:100%" border="0"

$$\displaystyle (\phi_{x})_{y}=g_{y}-p(g_{py}-f_{xy})+p^{2} f_{yy}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.19)
 * }
 * According to Eq.(12.9)$$\displaystyle \phi_{xy}=\phi_{yx}$$ we can know,
 * {| style="width:100%" border="0"

$$\displaystyle g_{px}-f_{yx}p-f_{xx}=g_{y}-p(g_{py}-f_{xy})+p^{2} f_{yy}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.20)
 * }
 * After rearrange, we can get equation as follow,
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.21)
 * }
 * {| style="width:100%" border="0"|-

This Eq.(12.21) is equal to the Eq.(12.2)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }
 * 3.Verify that Eq.(12.3) satisfies the 2nd exactness condition
 * Suppose
 * {| style="width:100%" border="0"

$$\displaystyle f=xy$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.22)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle g=yy^{'}+x(y^{'})^{2}=yp+xp^{2}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.23)
 * }
 * Differentiate the Eq.(12.23)with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{p}=y+2xp$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.24)
 * }
 * Differentiate the Eq.(12.24)with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{pp}=2x$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.25)
 * }
 * Differentiate the Eq.(12.23)with respect to y:
 * {| style="width:100%" border="0"

$$\displaystyle g_{y}=p$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.26)
 * }
 * Differentiate the Eq.(12.26)with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{yp}=1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.27)
 * }
 * Differentiate the Eq.(12.22)with respect to x and y:
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx}=f_{yy}=f_{yp}=f_{xp}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.28)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle f_{y}=x $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.29)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle f_{xy}=1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.30)
 * }
 * The 2nd exactness condition satisfies:
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=2p$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.31)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle g_{xp}+pg_{yp}-g_{y}=2p$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.32)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle f_{xp}+2f_{y}+f_{yp}p=2x=g_{pp}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.33)
 * }

Problem Statement

 * Show that these three relations for the symmetry of mixed 2nd partial derivatives are true
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{xy}=\phi_{yx}, \ \ \phi_{yp}=\phi_{py}, \ \ \phi_{px}=\phi_{xp}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.1)
 * }

Given

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

$$\displaystyle g_{o}-\frac{d g_{1}}{dx}+\frac{d^2 g_{2}}{dx^{2}}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.2)
 * }

Nomenclature

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

$$\displaystyle p=y'$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.3)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2} g_{2}}{dx^{2}}=\frac{d}{dx}(\frac{d g_{2}}{dx})$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.4)
 * }

Solution

 * According to the definition,
 * {| style="width:100%" border="0"

$$\displaystyle g_{0}:=\frac{\partial}{\partial y}(\frac{d \phi}{dx})$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.5)
 * }
 * And then
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx} g_{1}=\frac{d}{dx}[\phi_{xy'}+\phi_{yy'}y'+\phi_y+\phi_{y'y'} y'']$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.6)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}}{dx^{2}} g_{2}=\frac{d}{dx}[\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y'']$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.7)
 * }
 * By Substituting Eq. (4.5), (4.6), and (4.7) into Eq. (4.2) we get
 * {| style="width:100%" border="0"

$$\frac{\partial}{\partial y} (\frac{d \phi}{dx})-\frac{d}{dx} [ \phi_{xy'}+\phi_{yy'}y'+\phi_{y}+\phi_{y'y'}y ]+\frac{d}{dx}[ \phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y]=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.8)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\frac{\partial}{\partial y} (\frac{d \phi}{dx})-\frac{d}{dx}(\phi_{xp})+\frac{d}{dx}( \phi_{yp}p)+\frac{d}{dx}(\phi_y)+\frac{d}{dx}(\phi_{pp}y)+\frac{d}{dx}(\phi_{px})+\frac{d}{dx}(\phi_{py}p)+\frac{d}{dx}(\phi_{pp}y)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.9)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle [\frac{\partial}{\partial y}(\frac{d \phi}{dx})-\frac{d}{dx}\phi_{y}]+[\frac{d}{dx}\phi_{px}-\frac{d}{dx}\phi_{xp}]+[\frac{d}{dx}(\phi_{py}p)-\frac{d}{dx}(\phi_{yp}p)]+[\frac{d}{dx}(\phi_{pp}y)-\frac{d}{dx}(\phi_{pp}y)]=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.10)
 * }
 * From Eq.(4.10), we can get
 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx})-\frac{d}{dx}\phi_{y}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.11)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx})=\frac{d}{dx}\phi_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.12)
 * }
 * And
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}\phi_{px}-\frac{d}{dx}\phi_{xp}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.13)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}\phi_{px}=\frac{d}{dx}\phi_{xp}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.14)
 * }
 * And
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}(\phi_{py}p)-\frac{d}{dx}(\phi_{yp}p)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.15)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}(\phi_{py}p)=\frac{d}{dx}(\phi_{yp}p)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.16)
 * }
 * From Eq.(4.12) ,(4.14) and (4.16), we can get
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{xy}=\phi_{yx}, \ \ \phi_{px}=\phi_{xp}, \ \ \phi_{yp}=\phi_{py}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.17)
 * }
 * {| style="width:100%" border="0"|-

It is same as the Eq.(4.1)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Problem Statement

 * Use matlab to plot $$ \displaystyle F(5,-10;1;x) $$ near $$ X=0 $$ to display the local maximum (or maxima)in this region.
 * show that
 * {| style="width:100%" border="0"

$$ \displaystyle F(5,-10;1;x)=(1-x)^{6}(1001x^{4}-1144x^{3}+396x^{2}-44x+1) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(10.1)
 * }

Nomenclature

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

$$ \displaystyle F(a,b;c;x):=\sum_{k=0}^\infty\frac{(a)_k\,(b)_k}{(c)_k}\,\frac{x^k}{k!} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(10.2)
 * }

Solution

 * First part:$$ \displaystyle F(5,-10;1,x) $$
 * [[File:Problem 10.5 1.jpg]]
 * Second part:$$ \displaystyle F(x)=(1-x)^{6}(1001x^{4}-1144x^{3}+396x^{2}-44x+1) $$
 * [[File:Problem 10.5 2.jpg]]
 * {| style="width:100%" border="0"|-

Obviously, the graph of the first part is same as the graph of the second part.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Problem Statement
Find the expressions for$$ \displaystyle X(x)$$in terms of $$ \displaystyle \cos K x, \ \sin K x, \ \cosh K x, \ \sinh K x $$

Given

 * L4-ODE-CC
 * {| style="width:100%" border="0"

$$ \displaystyle X^{(4)}-K^4X=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(14.1)
 * }
 * Assume
 * {| style="width:100%" border="0"

$$ \displaystyle X(x)=e^{(rx)} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(14.2)
 * }
 * Solution
 * {| style="width:100%" border="0"

$$ \displaystyle r_{1,2}=\pm K $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(14.3)
 * }
 * and
 * {| style="width:100%" border="0"

$$ \displaystyle r_{3,4}=\pm i\,K$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(14.4)
 * }

Nomenclature
$$ \displaystyle i:=\sqrt{-1} $$

Solution

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

$$ \displaystyle X{(x)}=c_1e^{Kx}+c_2e^{-Kx}+c_3e^{iKx}+c_4e^{-iKx} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(14.5)
 * }
 * Eqn.(14.5) can be separate for two parts
 * First,
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} c_1e^{Kx}+c_2e^{-Kx} &= c_1(\frac{e^{Kx}+e^{-Kx}}{2}+\frac{e^{Kx}-e^{-Kx}}{2}) + c_2(\frac{e^{Kx}+e^{-Kx}}{2}-\frac{e^{Kx}-e^{-Kx}}{2}) \\&= c_1(\cosh Kx+ \sinh Kx) + c_2(\cosh Kx- \sinh Kx)\\ &=(c_1+c_2)\cosh Kx + (c_1-c_2)\sinh Kx \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(14.6)
 * }
 * Second,
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} c_3e^{iKx}+c_4e^{-iKx} &=c_3(\cos Kx +i\sin Kx)+c_4(\cos Kx -i\sin Kx) \\ &= (c_3+c_4)\cos Kx + (c_3-c_4)i\sin Kx \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(14.7)
 * }
 * Combine with two parts:
 * {| style="width:100%" border="0"|-

$$ \displaystyle X{(x)}=(c_1+c_2)\cosh Kx + (c_1-c_2)\sinh Kx + (c_3+c_4)\cos Kx + (c_3-c_4)i\sin Kx $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">(14.8)
 * }

Problem Statement

 * Find the particular solution $$ \displaystyle y_P(x)$$ after knowing the homogeneous solution $$ \displaystyle y_H(x)$$

Given

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

$$ \displaystyle y'(x)+P(x)y(x)=Q(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.1)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle y_H(x)=exp[-\int P(x)dx] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.2)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle y_P(x)=A(x)y_H(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.3)
 * }

Nomenclature

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

$$ \displaystyle y'(x)= \frac{d}{dx}y$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution

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

$$ \displaystyle y_P'(x)=A'(x)y_H(x)+A(x)y_H'(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.4)
 * }
 * Because,
 * {| style="width:100%" border="0"

$$ \displaystyle y_H'(x)=-P(x)exp[-\int P(x)dx]=-P(x)y_H(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.5)
 * }
 * We can get Eq.(5.4) as,
 * {| style="width:100%" border="0"

$$ \displaystyle y_P'(x)=A'(x)exp[-\int P(x)dx]-A(x)P(x)exp[-\int P(x)dx] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.6)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ \displaystyle y_P'(x)=exp[-\int P(x)dx][A'(x)-A(x)P(x)] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.7)
 * }
 * Using Eq.(5.3) and Eq.(5.7) substitute for Eq.(5.1)
 * {| style="width:100%" border="0"

$$ \displaystyle exp[-\int P(x)dx][A'(x)-A(x)P(x)]+P(x)A(x)exp[-\int P(x)dx]=Q(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.8)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ \displaystyle exp[-\int P(x)dx][A'(x)-A(x)P(x)+P(x)A(x)]=Q(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.9)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ \displaystyle A'(x)exp[-\int P(x)dx]=Q(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.10)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ \displaystyle A'(x)=\frac{Q(x)}{exp[-\int P(x)dx]}=Q(x)exp[\int P(x)dx] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.11)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ \displaystyle A(x)=\int Q(x)exp[\int P(x)dx]dx $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.12)
 * }
 * Using Eq.(5.2) and Eq.(5.12) substitute for Eq.(5.3)
 * {| style="width:100%" border="0"|-

$$ \displaystyle y_P(x)=\int Q(x)exp[\int P(x)dx]dx(exp[-\int P(x)dx]) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">(5.13)
 * }

Problem Statement

 * Explain why $$ \displaystyle r_2(x)$$ is not a valid root, i.e.,$$\displaystyle u_2(x)=e^{xr_2(x)} $$ is not a valid solution.

Given

 * Application: King 2003 p.28 (Pb.1.1 ab)
 * {| style="width:100%" border="0"

$$ \displaystyle (x-1)y''-xy'+y=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.1)
 * }
 * lst homogeneous solution; pretend not knowing:
 * {| style="width:100%" border="0"

$$ \displaystyle u_1(x)=e^x $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.2)
 * }
 * Trial solution: $$\displaystyle y=e^{rx}$$, r=constant
 * Characteristic equation:
 * {| style="width:100%" border="0"

$$ \displaystyle (x-1)r^2-xr+1=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.3)
 * }
 * WA:
 * {| style="width:100%" border="0"

$$ \displaystyle r_1=1 ==> constant$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.4)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle r_2(x)=\frac{1}{x-1} ==> non-constant $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.5)
 * }

Nomenclature

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

$$ \displaystyle y'= \frac{d}{dx}y$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle y''= \frac{d}{dx}(\frac{d}{dx}y)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution

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

$$ \displaystyle y=e^{xr_2(x)}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.6)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle y'=-r_2^{-2}(x)e^{xr_2(x)}=-\frac{1}{(x-1)^2}e^{\frac{x}{x-1}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.7)
 * }
 * And
 * {| style="width:100%" border="0"

$$ \displaystyle y''=2(x-1)^{-3}e^{xr_2(x)}+\frac{1}{(x-1)^2}\frac{1}{(x-1)^2}e^{xr_2(x)}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.8)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ \displaystyle y''=\frac{2x-1}{(x-1)^4}e^{\frac{x}{x-1}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.9)
 * }
 * Using Eq.(8.3), Eq.(8.7) and Eq.(8.9) substitute for Eq.(8.1)
 * {| style="width:100%" border="0"

$$ \displaystyle (x-1)\frac{2x-1}{(x-1)^4}e^{\frac{x}{x-1}}+x\frac{1}{(x-1)^2}e^{\frac{x}{x-1}}+e^{\frac{x}{x-1}}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.10)
 * }
 * Consider the left side,
 * {| style="width:100%" border="0"

$$ \displaystyle [\frac{2x-1}{(x-1)^3}+\frac{x}{(x-1)^2}+1]e^{\frac{x}{x-1}}=[\frac{x^3-2x^2+4x-2}{(x-1)^3}]e^{\frac{x}{x-1}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.11)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle [\frac{x^3-2x^2+4x-2}{(x-1)^3}]e^{\frac{x}{x-1}} $$ is not sure to be zero, so $$ \displaystyle r_2(x) $$ is not a valid root.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Problem Statement

 * Show that the infinitesimal length $$ ds $$ in Eq.3.2 can be written in spherical coord. as follows
 * {| style="width:100%" border="0"

$$ \displaystyle ds^2=1\cdot dr^2+r^2 d\theta^2+r^2(\cos\theta)^2\,d\varphi^2 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.1)
 * }

Given

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

$$ \displaystyle ds^2=dx_i dx_i=\sum_{i=1}^3(dx_i)^2 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.2)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle x_1=x=r\cos\theta\cos\varphi=\xi_1\cos\xi_2\,\cos\xi_3 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.3)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle x_2=y=r\cos\theta\sin\varphi=\xi_1\cos\xi_2\,\sin\xi_3 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.4)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle x_3=z=r\sin\theta=\xi_1\sin\xi_2 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.5)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \Delta u=\frac{1}{h_1h_2h_3}\sum^3_{i=1}\frac{\partial}{\partial\xi_i}\left[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}\right] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.6)
 * }

Nomenclature

 * $$ \displaystyle dr=\frac{d}{dr} $$
 * $$ \displaystyle d\theta=\frac{d}{d\theta} $$
 * $$ \displaystyle d\varphi=\frac{d}{d\varphi} $$
 * $$ \displaystyle dr^2=dr\,dr $$
 * $$ \displaystyle d\theta^2=d\theta\,d\theta $$
 * $$ \displaystyle d\varphi^2=d\varphi\,d\varphi $$

Solution

 * Differentiate the Eq.3.3
 * {| style="width:100%" border="0"

$$ \displaystyle dx_1=\cos\theta\,cos\varphi\,dr-r\sin\theta\,\cos\varphi\,d\theta-r\cos\theta\,\sin\varphi\,d\varphi $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.7)
 * }
 * And,
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} dx_1dx_1=&(\cos\theta)^2\,(\cos\varphi)^2\,drdr-r\sin\theta\,\cos\theta\,(\cos\varphi)^2\,drd\theta-r(\cos\theta)^2\,\sin\varphi\,\cos\varphi\,drd\varphi\\&-r\cos\theta\,\sin\theta\,(\cos\varphi)^2\,d\theta\,dr+r^2(\sin\theta)^2\,(\cos\varphi)^2\,d\theta\,d\theta+r^2\cos\theta\,\sin\theta\,\cos\varphi\,\sin\varphi\,d\theta\,d\varphi\\&-r(\cos\theta)^2\,\cos\varphi\,\sin\varphi\,d\varphi\,dr+r^2\sin\theta\,cos\theta\,\cos\varphi\,\sin\varphi\,d\varphi\,d\theta+r^2(\cos\theta)^2\,(\sin\varphi)^2\,d\varphi\,d\varphi \end{align}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.8)
 * }
 * Differentiate the Eq.3.4
 * {| style="width:100%" border="0"

$$ \displaystyle dx_2=\cos\theta\,sin\varphi\,dr-r\sin\theta\,\sin\varphi\,d\theta+r\cos\theta\,\cos\varphi\,d\varphi $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.9)
 * }
 * And,
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} dx_2dx_2=&(\cos\theta)^2\,(\sin\varphi)^2\,drdr-r\sin\theta\,\cos\theta\,(\sin\varphi)^2\,drd\theta+r(\cos\theta)^2\,\sin\varphi\,\cos\varphi\,drd\varphi\\&-r\sin\theta\,\cos\theta\,(\sin\varphi)^2\,d\theta\,dr+r^2(\sin\theta)^2\,(\sin\varphi)^2\,d\theta\,d\theta-r^2\sin\theta\,\cos\theta\,\sin\varphi\,\cos\varphi\,d\theta\,d\varphi\\&+r(\cos\theta)^2\,\sin\varphi\,\cos\varphi\,d\varphi\,dr-r^2\sin\theta\,cos\theta\,\sin\varphi\,\cos\varphi\,d\varphi\,d\theta+r^2(\cos\theta)^2\,(\cos\varphi)^2\,d\varphi\,d\varphi \end{align}  $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.10)
 * }
 * Differentiate the Eq.3.5
 * {| style="width:100%" border="0"

$$ \displaystyle dx_3=\sin\theta\,dr+r\cos\theta\,d\theta $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.11)
 * }
 * And,
 * {| style="width:100%" border="0"

$$ \displaystyle dx_3dx_3=(\sin\theta)^2\,drdr+r\cos\theta\sin\theta\,drd\theta\,+r\sin\theta\cos\theta\,d\theta\,dr+r^2(\cos\theta)^2\,d\theta\,d\theta$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.12)
 * }
 * According to Eq.3.2
 * {| style="width:100%" border="0"

$$ \displaystyle ds^2=dx_1dx_1+dx_2dx_2+dx_3dx_3$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.13)
 * }
 * And $$ \displaystyle (\sin\theta)^2\,+(\cos\theta)^2\,=1$$
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} ds^2=&(\cos\theta)^2\,drdr-r\sin\theta\cos\theta\,drd\theta\,-r\cos\theta\sin\theta\,d\theta\,dr+r^2(\sin\theta)^2\,d\theta\,d\theta\,r^2(\cos\theta)^2\,d\varphi\,d\varphi\,\\&+(\sin\theta)^2\,drdr+r\cos\theta\sin\theta\,drd\theta\,+r\sin\theta\cos\theta\,d\theta\,dr+r^2(\cos\theta)^2\,d\theta\,d\theta \end{align}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.14)
 * }
 * {| style="width:100%" border="0"|-
 * {| style="width:100%" border="0"|-

$$ \displaystyle ds^2=1\cdot dr^2+r^2 d\theta^2+r^2(\cos\theta)^2\,d\varphi^2 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }
 * Because,
 * {| style="width:100%" border="0"

$$ \displaystyle \begin{align} &(h_2)^2=r^2 \\& d\theta^2=(d\xi_2)^2 \end{align}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.15)
 * }
 * So,
 * {| style="width:100%" border="0"

$$ \displaystyle h_2=r $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.16)
 * }


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

$$ \displaystyle \begin{align} &(h_3)^2=r^2(\cos\theta)^2 \\& d\varphi^2=(d\xi_3)^2\end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.17)
 * }
 * So,
 * {| style="width:100%" border="0"

$$ \displaystyle h_3=r\cos\theta $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.18)
 * }
 * And $$ \displaystyle h_1h_2h_3=r^2\cos\theta $$
 * According to Eq.3.6
 * $$ \displaystyle i=1 $$
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial}{\partial\xi_1}\left[\frac{h_1h_2h_3}{(h_1)^2}\frac{\partial u}{\partial\xi_1}\right]= \frac{\partial}{\partial r}\left[\frac{r^2\cos\theta}{(1)^2}\frac{\partial u}{\partial r}\right]$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.19)
 * }
 * $$ \displaystyle i=2 $$
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial}{\partial\xi_2}\left[\frac{h_1h_2h_3}{(h_2)^2}\frac{\partial u}{\partial\xi_2}\right]= \frac{\partial}{\partial \theta}\left[\frac{r^2\cos\theta}{(r)^2}\frac{\partial u}{\partial \theta}\right]$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.20)
 * }
 * $$ \displaystyle i=3 $$
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial}{\partial\xi_3}\left[\frac{h_1h_2h_3}{(h_3)^2}\frac{\partial u}{\partial\xi_3}\right]= \frac{\partial}{\partial \varphi}\left[\frac{r^2\cos\theta}{(r\cos\theta)^2}\frac{\partial u}{\partial \varphi}\right] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(3.21)
 * }


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

$$ \displaystyle \Delta u=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\cos^2\theta}\frac{\partial^2 u}{\partial \varphi^2}+\frac{1}{r^2\cos\theta}\frac{\partial}{\partial\theta}\left(\cos\theta\frac{\partial u}{\partial\theta}\right) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

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MaintenanceBot (discuss • contribs) 03:52, 11 November 2013 (UTC)