User:EGM6321.f11.team1.pan/HW1

= Problem 5 =

Given
Consider the variables seperated in 3_D curvilinear coordinates $$X\left(\xi\right)=\left(X(\xi_1),X(\xi_2),X(\xi_3)\right)$$ the separated equations for $$\displaystyle i=1,2,3$$ is

Find
These seperated equations can be expressed by

$$\displaystyle y''+a_1(x)y'+a_0(x)y=0$$

in which,

$$a_1(x)=\frac{g'(x)}{g(x)}$$

Solution
According to the differential of product, $$(Eq.5.1)$$ is

$$\frac{1}{g_i(\xi_i)}\left[\frac{dg_i(\xi_i)}{d\xi_i}\frac{dx_i(\xi_i)}{d\xi_i}+g_i(\xi_i)\frac{d^2x_i(\xi_i)}{d\xi_i^2}\right]+f_i(\xi_i)x_i(\xi_i)=0$$

For $$\displaystyle i=1,2,3$$ , set $$\displaystyle x$$ is the varible $$\displaystyle \xi_i$$, $$\displaystyle y$$ is its value $$\displaystyle x_i$$.

As for a specific $$\displaystyle i=1,2,3$$ , $$(Eq.5.2)$$ can be written as

$$\displaystyle a_2(x)y''+a_1(x)y'+a_0(x)y=0$$

in which,

$$\displaystyle a_2(x)=1;$$ $$\displaystyle a_1(x)=\frac{g'(x)}{g(x)}$$

= Problem 6 =

Given
In the Equation of Motion, the first term is

Find
Prove $$\displaystyle C_3(Y^1,t)\ddot{Y}^1$$  is nonlinear respect to $$\displaystyle Y^1$$.

Solution
If a function $$\displaystyle F(\cdot)$$ is linear, it must be satisfied by the following for any arbitrary number $$\displaystyle \alpha,\beta$$ :

As for the first term of EOM, we set $$\displaystyle F(\cdot)=C_3(\cdot,t)\ddot{(\cdot)}$$

Therefore, the left side of $$(Eq.6.2)$$ can be written as $$\displaystyle F(\alpha u+\beta v)=M\left[1-\overline{R}u^2_,ss(\alpha u+\beta v,t)\right]\frac{d^2(\alpha u+\beta v)}{dt^2}$$.

The right hand side of $$(Eq.6.2)$$ is

$$\displaystyle \alpha F(u)+\beta F(v)=\alpha M[1-\overline{R}u^2_,ss(u,t)]\frac{d^2 u}{dt^2}+\beta M[1-\overline{R}u^2_,ss(v,t)]\frac{d^2 v}{dt^2}$$.

It is obvious that $$\displaystyle F(\alpha u+\beta v)\neq \alpha F(u)+\beta F(v)$$, $$\displaystyle \forall \alpha ,\beta \in \mathbb R$$ In all, $$ (Eq.6.1) $$ is nonlinear with respect to $$\displaystyle Y^1$$.