User:Egm6321.f09.Team6.dianafoster/HW3

Problem 4:
Statement: 

For the case N1-ODE $$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$$ show that

$$f_0-\frac{df_1}{dx}=0\Leftrightarrow\Phi{xy}=\Phi{yx}$$

Specifically, find $$f_0$$ and $$f_1$$ in terms of $$\Phi$$.

Solution: 

We know the first condition for exactness is $$F=\frac{d\Phi}{dx}(x,y)=\Phi_x+\Phi_yy'$$

We also know the second condition for exactness is $$f_i=\frac{\delta F}{\delta y^{(i)}}$$   for i=1,2,...,n

Thus, $$f_o=\frac{\delta F}{\delta y}$$ $$=\frac{\delta}{\delta y}(\Phi_x+\Phi_yy')$$ $$=\Phi_{xy}$$

and $$f_1=\frac{\delta F}{\delta y'}$$ $$=\frac{\delta}{\delta y'}(\Phi_x+\Phi_yy')$$ $$=\Phi_y$$

Therefore, $$\Phi_{xy}-\frac{d}{dx}\Phi_y=0$$ $$\Phi_{xy}-\Phi_{yx}=0$$ $$\Phi_{xy}=\Phi_{yx}$$

Problem 5:
Statement: 

For the case N2-ODE, show: $$\frac{d}{dx}(\Phi_y)=f_0$$

$$f_1=\frac{df_2}{dx}+\Phi_y$$

$$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

Relate this last equation to the following: $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

Solution: 

We know the first condition for exactness is $$F=\frac{d\Phi}{dx}(x,y,y')=\Phi_x+\Phi_yy'+\Phi_{y'}y^{''}$$

We also know the second condition for exactness is $$f_i=\frac{\delta F}{\delta y^{(i)}}$$   for i=1,2,...,n

Thus, we can show: $$f_o=\frac{\delta F}{\delta y}=\frac{\delta}{\delta y}(\Phi_x+\Phi_yy'+\Phi_{y'}y^{''})$$

$$=\Phi_{xy}$$

$$=\Phi_{yx}$$

$$=\frac{d}{dx}(\Phi_y)$$

Furthermore, we can show: $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

$$\frac{df_1}{dx}=f_0+\frac{d^2f_2}{dx^2}$$

$$\int{\frac{df_1}{dx}dx}=\int{f_0dx}+\int{\frac{d^2f_2}{dx^2}dx}$$

$$f_1=\int{\frac{d}{dx}\Phi_ydx}+\int{\frac{d^2f_2}{dx^2}dx}$$

$$f_1=\Phi_y+\frac{df_2}{dx}$$

By substituting the found values above into $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

we get: $$\frac{d}{dx}\Phi_y-\frac{d}{dx}(\Phi_y+\frac{df_2}{dx})+\frac{d^2f_2}{dx^2}=0$$

$$\frac{d}{dx}\Phi_y-\frac{d}{dx}\Phi_y-\frac{d^2f_2}{dx^2}+\frac{d^2f_2}{dx^2}=0$$

$$0=0$$

Problem 6:
Statement: 

For the following Legendre differential equation

$$F=(1-x^2)y^{''}-2xy'+n(n+1)y=0$$

verify exactness using two methods. Method 1: $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

Method 2:

$$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$

Solution: 

We know $$f=(1-x^2)$$ and $$g=-2xy'+n(n+1)y$$

Thus: $$f_x=-2x$$ → $$f_{xx}=-2$$ → $$f_{xy}=0$$ → $$f_{xy'}=0$$ $$f_y=0$$ → $$f_{yy}=0$$ → $$f_{yy'}=0$$ $$g_x=-2y'$$ → $$g_{xy'}=-2$$ $$g_y=n(n+1)$$ → $$g_{yy'}=0$$ $$g_{y'}=-2x$$ → $$g_{y'y'}=0$$

Plugging the above into method 1 equations, we get: $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$-2+2p(0)+p^2(0)=-2+p(0)-n(n+1)$$ $$-2=-2-n(n+1)$$ NOT TRUE!

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$ $$0+p(0)+2(0)=0$$ $$0=0$$ TRUE!

Method 1 has only one equation satisfied; therefore, the method as a whole is not satisfied and exactness cannot be shown.

Next, we use method 2. We know:

$$f_0=\frac{\delta F}{\delta y}=n(n+1)$$

$$f_1=\frac{\delta F}{\delta y'}=-2x$$ → $$\frac{df_1}{dx}=-2$$

$$f_2=\frac{\delta F}{\delta y''}=(1-x^2)$$ → $$\frac{df_2}{dx}=-2x$$ → $$\frac{d^2f_2}{dx^2}=-2$$

Thus: $$n(n+1)-(-2)-2=0$$ $$n(n+1)=0$$ NOT TRUE!

As the above equation is not true, method 2 cannot show exactness. We will need to use the integrating factor to show exactness. So, we must find $$x^my^n$$ such that:

$$F=x^my^n[(1-x^2)y''-2xy'+n(n+1)y]=0$$

For simplification, let's assume $$p=y'$$, and $$p'=y''$$. We know:

$$\Phi=h(x,y)+\int{f(x,y,p)dp}$$ $$=h(x,y)+\int{(1-x^2)dp}$$ $$=h(x,y)+(1-x^2)p$$ $$=h(x,y)+p-x^2p$$

and:

$$g=\Phi_x+\Phi_yp$$

Therefore:

$$2xp+n(n+1)y=(h_x-2xp)+(h_y+0)p$$ $$2xp+n(n+1)y=h_x-2xp+h_yp$$ $$4xp+n(n+1)y=h_x+h_yp$$

So: $$4xp=h_yp$$ $$4x=h_y$$ which leads to: $$h=4xy+k_x$$

Also: $$n(n+1)y=h_x$$ which leads to: $$h=n(n+1)xy+k_y$$

Therefore: $$4xy=n(n+1)xy$$ $$4=n(n+1)$$ $$4=n^2+n$$ $$n=1.56$$ or $$n=-2.56$$

That means $$h(x,y)=x^my^n$$ has two solutions for exactness: $$h(x,y)=1.56xy$$ or $$h(x,y)=-2.56xy$$

Problem 7:
Statement: 

Show that: $$\forall u,v$$ functions of $$x$$, $$L(u+v)=L(u)+L(v)$$ and $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\forall$$ functions of $$ x$$

are equivalent to: $$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$$

Solution: 

We know that the linearity operator L(.) allows use of superposition. Thus: $$L(u+v)=L(u)+L(v)$$ $$L(\lambda u)=\lambda L(u)$$

Therefore, we can use the above two relations to show: $$L(\alpha u+\beta v)= L(\alpha u)+ L(\beta v)=\alpha L(u)+\beta L(v)$$