User:Egm6321.f12.team1.pb/report1

=Problem 4: Separation of Variables (Alternate Solution)=
 * Draw the polar coordinate lines $$(\xi _{1},\xi _{2})=(r,\theta )$$ in a 2-D plane emanating from a point, not at the origin

Solution:
Coordinate lines are lines that represent the direction of change of a particular coordinate, but orthogonal to other coordinates, i.e. other coordinates, than the one which is represented by the coordinate lines are constant at those coordinate lines.


 * The relation between Cartesian and polar coordinates are given by


 * $$x=r \cos \theta$$
 * and
 * $$y=r \sin \theta$$

According to the definition, the coordinate lines can be generated by eliminating corresponding coordinates from the equation.

At a point $$(r,\theta)$$

The coordinate lines corresponding to $$(\xi _{1})$$ or $$r$$ can be generated by eliminating $$r$$ from the equations. So in Cartesian coordinate, corresponding equation will be


 * $$\frac {y} {x} = tan \theta$$

Similarly the coordinate lines corresponding to $$(\xi _{2})$$ or $$\theta$$ is given by


 * $$ x^2+y^2 = r^2$$

=Problem 6: Proof of Non-linearity = We solved it by ourselves

Given:

 * $$c_3 (Y^1, t) = M [ 1 - \bar R u^2_{,ss}(Y^1 , t) ] $$

Proof That:

 * $$c_3 (Y^1, t) \ddot Y^1 $$ is non-linear with respect to $$Y^1$$

Solution:

 * A function $$F(x)$$ is non-linear with respect to $$x$$ if $$\exists \alpha, \beta \in \mathbb R $$ such that $$F(\alpha x + \beta y) \neq \alpha F(x)+\beta F(y)$$


 * Let us assume that $$c_3 (Y^1, t) \ddot Y^1 $$ is linear with respect to $$Y^1$$


 * $$\Rightarrow c_3(\alpha Y^1_1 + \beta Y^1_2, t)

(\alpha \ddot Y^1_1+\beta \ddot Y^1_2) = \alpha c_3(Y^1_1)\ddot Y^1_1+\beta c_3(Y^1_2)\ddot Y^1_2,\qquad \forall \alpha, \beta \in \mathbb R $$


 * $$\Rightarrow M[1-\bar R u^2_{,ss}(\alpha Y^1_1 + \beta Y^1_2, t)]

(\alpha \ddot Y^1_1+\beta \ddot Y^1_2) = \alpha M[1-\bar R u^2_{,ss}(Y^1_1,t)]\ddot Y^1_1+ \beta M[1-\bar R u^2_{,ss}(Y^1_2,t)]\ddot Y^1_2$$


 * Subtracting $$ M(\alpha \ddot Y^1_1+\beta \ddot Y^1_2)$$ from both sides


 * &#160;&#160;&#160;&#160;&#160;&#160;$$ M \bar R u^2_{,ss}(\alpha Y^1_1 + \beta Y^1_2, t)

(\alpha \ddot Y^1_1+\beta \ddot Y^1_2) = \alpha M\bar R u^2_{,ss}(Y^1_1,t) \ddot Y^1_1+ \beta M \bar R u^2_{,ss}(Y^1_2,t) \ddot Y^1_2$$


 * $$\Rightarrow \alpha u^2_{,ss}(\alpha Y^1_1 + \beta Y^1_2, t)\ddot Y^1_1+\beta u^2_{,ss}

(\alpha Y^1_1 + \beta Y^1_2, t)\ddot Y^1_1 = \alpha u^2_{,ss}(Y^1_1,t) \ddot Y^1_1+ \beta u^2_{,ss}(Y^1_2,t) \ddot Y^1_2,\qquad \forall \alpha, \beta \in \mathbb R $$


 * Which is clearly impossible whether $$u^2(Y^1_2,t)$$ is linear or non-linear with respect to $$Y^1_2$$


 * This implies the the only assumption we made was incorrect.


 * So $$c_3 (Y^1, t) \ddot Y^1 $$ is non-linear with respect to $$Y^1$$

=Problem 7: Proof of Linearity = We solved it by ourselves

Given:

 * $$L_2(y) = y''+a_1(x)y'+a_0(x)y$$

Proof That:

 * $$L_2(.)$$ is linear.

Solution:

 * A function $$F(.)$$ is linear if $$\forall \alpha, \beta \in \mathbb R, \quad F(\alpha y_1 + \beta y_2) = \alpha F(y_1)+\beta F(y_2)$$


 * $$L_2( y_1 ) = y_1+a_1(x)y'_1+a_0(x)y_1$$ and $$L_2( y_2 ) = y_2+a_1(x)y'_2+a_0(x)y_2$$


 * $$L_2( \alpha y_1 + \beta y_2) = \frac {d^2} {dx^2} ( \alpha y_1 + \beta y_2) +

a_1(x) \frac {d} {dx} ( \alpha y_1 + \beta y_2) +a_0(x)( \alpha y_1 + \beta y_2) $$


 * $$\;=\frac {d^2} {dx^2} ( \alpha y_1) +\frac {d^2} {dx^2} ( \beta y_2)+

a_1(x) \frac {d} {dx} ( \alpha y_1 )+ a_1(x) \frac {d} {dx} ( \beta y_2)+ a_0(x)( \alpha y_1 ) + a_0(x)( \beta y_2) $$


 * $$\;= \alpha y_1 + \beta y_2+ a_1(x) \alpha y'_1 + a_1(x) \beta y'_2+ a_0(x)\alpha y_1+a_0(x) \beta y_2 $$


 * $$\;= \alpha (y_1 + a_1(x) y'_1 + a_0(x) y_1)+\beta (y_2 + a_1(x) y'_2+ a_0(x) y_2) $$


 * $$\Rightarrow L_2( \alpha y_1 + \beta y_2) = \alpha L_2( y_1 ) + \beta L_2 (y_2) $$


 * So $$ L_2(.) $$ is a linear operator.