User:EGM6321.f12.team7.Marjanovic/report1

=Problem 7=

Given
$$\displaystyle L_2(\cdot ) = \frac{d^2 (\cdot)}{dx^2}+a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot)$$

Find
Show that $$\displaystyle L_2(\cdot) $$ is linear.

Solution
If $$\displaystyle L_2(\cdot )$$ is linear, then the following must be true: $$\displaystyle L_2(\alpha u +\beta v) = \alpha L_2(u) + \beta L_2(v) $$ $$\displaystyle \forall \alpha, \beta \in \mathbb{R} $$ $$\displaystyle L_2(\alpha u + \beta v) = \frac{d^2 (\alpha u + \beta v)}{dx^2}+a_1(x)\frac{d(\alpha u + \beta v)}{dx}+a_0(x)(\alpha u + \beta v) $$ $$\displaystyle L_2(\alpha u + \beta v)= \alpha u + \beta v +a_1(x)[\alpha u' + \beta v'] + a_0(x)[\alpha u + \beta v] $$ $$\displaystyle L_2(\alpha u + \beta v) = [\alpha u +a_1(x)\alpha u' + a_0(x)\alpha u] + [\beta v +a_1(x)\beta v' + a_0(x)\beta v] $$ $$\displaystyle L_2(\alpha u + \beta v) = \alpha[u +a_1(x)u' + a_0(x)u] + \beta[v +a_1(x)v' + a_0(x)v] $$ $$\displaystyle L_2(\alpha u + \beta v) = \alpha L_2(u) + \beta L_2(v) $$ Thus $$\displaystyle L_2(\cdot) $$ is a linear operator.