# About Second-order Linear Homogenous ODE

One way to solve second-order linear homogeneous ode with constant coefficients is to do the following things: $$a\left(\frac{\mathrm d^2}{\mathrm dx^2}\right)f+b\left(\frac{\mathrm d}{\mathrm dx}\right)f+cf=0$$ $$aD^2f+bDf+cf=0$$ $$(D-\lambda_1I)(D-\lambda_2I)f=0$$ $$\Longrightarrow(D-\lambda_1I)f=0\textrm{ or }(D-\lambda_2I)f=0$$ What's the theoretical basis of the last step? This is equivalent to prove that

$$\ker(T+\lambda I)\oplus\ker(T-\lambda I)=\ker(T^2-\lambda^2I).$$

However that's not generally true, at least when $\lambda=0$, since it becomes $\ker T=\ker T^2$. Is it due to the particularity of $D$ or does this formula hold unless $\lambda=0$?

## 2 Answers

Difference equations also share similar properties:

$au_{n+2}+bu_{n+1}+cu_{n}=0$

$u_{n}=\left \{ \begin{array}{ccc} k_{1} \alpha^{n}+k_{2} \beta^{n} & , & \alpha \neq \beta \\ (k_{1}+nk_{2})\alpha^{n} & , & \alpha = \beta \end{array} \right.$

where $\alpha, \beta$ are the roots of $ax^{2}+bx+c=0$.

Linear superposition is crucial (proof omitted).

For $a\neq 0$,

$\displaystyle u_{n+2}+\frac{b}{a}u_{n+1}+\frac{c}{a}u_{n}=0$

$u_{n+2}-(\alpha+\beta) u_{n+1}+\alpha \beta u_{n}=0$

$u_{n+2}-\alpha u_{n+1}=\beta (u_{n+1}-\alpha u_{n})$

Let $v_{n}=u_{n+1}-\alpha u_{n}$, the two independent solutions can be written as:

$u_{n+1}=\alpha u_{n} \quad$ or $\quad v_{n+1}=\beta v_{n}$

$u_{n}=u_{0}\alpha^{n} \quad$ or $\quad v_{n}=v_{0}\beta^{n}$

$u_{n}=u_{0}\alpha^{n} \quad$ or $\quad \displaystyle u_{n}= u_{1}\left( \frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} \right)- u_{0}\alpha \beta \left( \frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta} \right)$

In particular, $u_{n}$ reduces to $u_{0} \beta^{n}$ when $u_{1}=\beta u_{0}$.

Take $\; \beta \to \alpha \;$ for double roots.

ODE can be treated in similar manner with $f''-\alpha f'=\beta (f'-\alpha f)$,

therefore $f(x)=e^{\alpha x}$ or $f'-\alpha f=e^{\beta x}$ and so on.

Let $p(X) = aX^2 + bX + c$ with $a \neq 0$. Assuming we work over the complex numbers, the second degree polynomial $p(X)$ splits as a product of linear factors $p(X) = ap_1(X) p_2(X)$ where $p_i(X) = X - \lambda_i$ for some $\lambda_i \in \mathbb{C}$. In order for the method to work, we need to have

$$\ker(p(D)) = \ker(p_1(D)) \oplus \ker(p_2(D)) = \ker(D - \lambda_1 I) \oplus \ker(D - \lambda_2 I)$$

which means that a general solution of the original ODE can be written as $Ae^{\lambda_1 x} + B e^{\lambda_2 x}$ for $A,B \in \mathbb{C}$.

The solution space $\ker(p(D))$ of the original ODE is always two dimensional while the solution spaces $\ker(D - \lambda_i I) = \mathrm{span} \{ e^{\lambda_i x} \}$ are one-dimensional and contained in $\ker(p(D))$ so we always have

$$\ker(p(D)) \supseteq \ker(D - \lambda_1 I) \oplus \ker(D - \lambda_2 I).$$

If $\lambda_1 \neq \lambda_2$, then both sides are two dimensional (as $e^{\lambda_1 x}$ and $e^{\lambda_2 x}$ are linearly independent, being eigenvectors that correspond to different eigenvalues) and we have equality. If $\lambda_1 = \lambda_2 = \lambda$, then the right hand side is one dimensional and we have strict inclusion

$$\ker(p(D)) = \ker((D - \lambda I)^2) \supsetneq \ker(D - \lambda I).$$

In this case, the solutions for the second order ODE are of the form $Axe^{\lambda x} + B e^{\lambda x}$ for $A,B \in \mathbb{C}$.

• Well, what I want to ask is why ker(p(D)) is always two-dimensional. And a piece of advice: the second paragraph may be simplified by Cayley-Hamilton theorem. – Han Li Feb 15 '16 at 15:33
• I wanted to keep the discussion as elementary as possible. The space $\ker(p(D))$ is two dimensional by the uniqueness and existence theorem for differential equations. – levap Feb 15 '16 at 15:41
• In fact, I don't see what it has to do with the Cayley-Hamilton theorem. If $p(X) = p_1(X) \dots p_k(X)$ then $\ker(p(T)) \supseteq \ker(p_1(T)) \oplus \dots \oplus \ker(p_k(T))$ follows immediately from the definition. – levap Feb 15 '16 at 15:52