User:Egm6321.f12.team5.yi/report1

= Problem R* 1.7 = Show that $$L_{2}(\cdot )$$is linear.

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

Solution
A function can be said to be linear if and only if:

$$\displaystyle F(\alpha u+\beta v)\equiv \alpha F(u)+\beta F(v), \forall \alpha, \beta\in\mathbb R$$

In this problem, we can proof:

$$L_{2}(\alpha (\cdot)_1+\beta (\cdot)_2)=\frac{d^2(\alpha (\cdot)_1+\beta (\cdot)_2)}{dx^2}+a_1(x)\frac{d(\alpha (\cdot)_1+\beta (\cdot)_2)}{dx}+a_0(x)(\alpha (\cdot)_1+\beta (\cdot)_2)$$

$$=\alpha \frac{d^2(\cdot )_1}{dx^2}+\beta \frac{d^2(\cdot )_2}{dx^2}+\alpha a_1(x)\frac{d(\cdot )_1}{dx}+\beta a_1(x)\frac{d(\cdot )_2}{dx}+\alpha a_0(x)(\cdot )_1+\beta a_0(x)(\cdot )_2$$

$$=\alpha (\frac{d^2(\cdot )_1}{dx^2}+a_1(x)\frac{d(\cdot )_1}{dx}+a_0(x)(\cdot )_1) +\beta (\frac{d^2(\cdot )_2}{dx^2}+a_1(x)\frac{d(\cdot )_2}{dx}+a_0(x)(\cdot )_2)$$

$$=\alpha L_{2}(\cdot)_1 + \beta L_{2}(\cdot)_2$$

So we can say $$L_{2}(\cdot )$$ is linear.