User:Egm6321.f12.team1.pb/report2

=Problem 2.4: Orthogonality of Legendre Functions=

Alternate Solution:
Let us prove the relation by contradiction. Let us assume $$\exists \alpha$$ such that 
 * $$y_{H}^{1}(\hat{x})= \alpha y_{H}^{2}(\hat{x})\qquad \qquad \forall x \in \mathbb R$$


 * $$\Rightarrow x = \alpha \frac{x}{2}\log \left (\frac{1+x}{1-x} \right )-\alpha \qquad \qquad \forall x \in \mathbb R$$

 Taking derivative with respect to x
 * $$1=\alpha \frac{d}{dx}\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-\frac{d}{dx}(\alpha)\qquad \qquad \forall x \in \mathbb R$$
 * $$\Rightarrow 1 = \frac{\alpha}{2}log\left(\frac{1+x}{1-x}\right)+\frac{\alpha x}{2}\left(\frac{1-x}{1+x}\right)\left(\frac{2}{(1-x)^2}\right) \qquad \qquad \forall x \in \mathbb R$$

 The Left Hand Side of the equation is a constant, but the Right Hand Side can never be reduced to a constant.  So the assumption that we made was incorrect.  So $$\exists x$$ such that $$y_{H}^{1}(\hat{x})\neq \alpha y_{H}^{2}(\hat{x})$$ for any given $$\alpha\quad \blacksquare$$

=Problem 2.6: Symmetry of second derivatives=

Find:
Find the minimum degree of differentiability of the function $$\phi(x,y)$$ such that the following equation
 * $$\frac {\partial^2 \phi (x,y)} {\partial x \partial y} = \frac {\partial^2 \phi (x,y)} {\partial y \partial x}$$

 is satisfied. State the full theorem and provide a proof.

Solution:
According to Mean Value Theorem, if $$f(x)$$ is continuous in $$[x,x+h]$$ and differentiable in $$(x,x+h)$$, then there exists a point $$x_1 \in (x,x+h)$$ such that


 * $$f(x+h)-f(x) = h \frac{\partial f(x)} {\partial x} \vert _{x=x_1}$$

Let us define


 * $$f(h,k):=[\phi(x+h,y+k)-\phi(x+h,y)]-[\phi(x,y+k)+\phi(x,y)]$$

 Applying Mean Value Theorem on the two terms in square braces
 * $$f(h,k) = k \frac{\partial \phi(x+h,y_1)} {\partial y} - k \frac{\partial \phi(x,y_2)} {\partial y}$$
 * $$\Rightarrow\frac{f(h,k)}{k} = \frac{\partial \phi(x+h,y_1)} {\partial y} - \frac{\partial \phi(x,y_2)} {\partial y}$$

 As $$k \rightarrow 0, y_1 \rightarrow y$$ and $$y_2 \rightarrow 2 $$


 * $$\Rightarrow \lim_{k \to 0}\frac{f(h,k)}{k} = \frac{\partial \phi(x+h,y_1)} {\partial y} - \frac{\partial \phi(x,y_2)} {\partial y} $$

 Applying Mean Value Theorem
 * $$\lim_{k \to 0}\frac{f(h,k)}{k} = h \frac{\partial \phi(x_1,y_1)} {\partial y}$$
 * $$\Rightarrow \lim_{k \to 0}\frac{f(h,k)}{hk} = \frac{\partial \phi(x_1,y_1)} {\partial y}$$



As $$h \rightarrow 0, x_1 \rightarrow x$$ 


 * $$\lim_{h \to 0} \lim_{k \to 0}\frac{f(h,k)}{hk} = \frac{\partial^2 \phi(x_1,y_1)}{\partial x \partial y}\qquad\qquad...7.1)$$

 Rearranging the terms of $$f(h,k)$$
 * $$f(h,k):=[\phi(x+h,y+k)-\phi(x,y+k)]-[\phi(x+h,y)+\phi(x,y)]$$

<br\> Following similar arguments, we can show that <br\>
 * $$\lim_{k \to 0} \lim_{h \to 0}\frac{f(h,k)}{hk} = \frac{\partial^2 \phi(x_1,y_1)}{\partial y \partial x} \qquad\qquad...7.2) $$

<br\> Equation 7.1 and 7.2 implies that
 * $$ \frac{\partial^2 \phi(x_1,y_1)}{\partial x \partial y} = \frac{\partial^2 \phi(x_1,y_1)}{\partial y \partial x} \qquad \qquad \qquad ...7.3)$$

<br\> To prove the relation, the only assumption we made was in the application of Mean Value Theorem, where we assumed that <br\><br\> <br\><br\> These are the three conditions necessary for equation 7.3 to hold. Since $$h \rightarrow 0, k \rightarrow 0$$, they can together be stated as <br\><br\>
 * 1) $$f(h,k)$$ is differentiable in $$(x,x+h)$$ and $$(y,y+k)$$
 * 2) $$\frac{\partial^2 \phi(x,y)}{\partial y}$$ is differntiable in $$(x,x+h)$$
 * 3) $$\frac{\partial^2 \phi(x,y)}{\partial x}$$ is differntiable in $$(y,y+k)$$
 * $$\phi(x,y), \frac{\partial^2 \phi(x,y)}{\partial x}, \frac{\partial^2 \phi(x,y)}{\partial y}$$ are required to be differentiable at $$(x,y)$$ for equation 7.3 to hold. That is $$\phi(x,y)$$ needs to be Double Differentiable.

=Problem 2.7=

Given:
Given the following equation
 * $$M(x,y)+N(y,x)y' = 0$$

<br\>
 * with $$M(x,y) = 75x^4, N(x,y) = cos(y)$$

and the proposed solution $$y(x) = sin^{-1}(k-15x^5)$$

Verify:
Verify that the proposed solution is indeed the solution of the given equation

Solution:
The proposed solution
 * $$y(x) = sin^{-1}(k-15x^5)$$

<br\>
 * $$\Rightarrow sin(y) = k-15x^5$$

<br\> Differentiating with respect to $$x$$
 * $$ cos(y)y' = -75x^4$$

<br\>
 * $$ 75x^4+cos(y)y' = 0$$

<br\> Which is the same as the given differential equation. <br\> So $$y(x) = sin^{-1}(k-15x^5)$$ is indeed the solution of the given equation.