# ODE : Existence and Uniqueness theorem

Just to make it explicit which theorem i am considering here, let me first state the Existence and Uniqueness theorem for a second order linear differential equation.

Theorem: Consider an equation of the form

$$a_0 y'' + a_1 y' + a_2 y = r(x),$$

where $a_0(x)$, $a_1(x)$, $a_2(x)$ and $r(x)$ are continuous functions on an interval $(a,b)$ and $a_0(x) \ne 0$ for each $x \in (a,b)$. Let $c_1$ and $c_2$ be arbitrary real numbers and $x_0 \in (a,b)$. Then there exists a unique solution $y(x)$ defined over $(a,b)$ for the above equation satisfying $y(x_0) = c_1$ and $y'(x_0) = c_2$.

Now the author also mentions below corollary to the above mentioned theorem.

Corollary: If $y(x)$ be a solution to

$$a_0 y'' + a_1 y' + a_2 y = 0$$

(note $r(x) = 0$ here) satisfying $y(x_0) = 0$ and $y'(x_0) = 0$ for some $x_0 \in (a,b)$, then $y(x)$ is identically zero on $(a,b)$.

I am not able to convince myself as to why this corollary is valid. Need help with an informal proof to understand why this corollary holds good.

• I edited your question to $\LaTeX$ify it. Cheers! – Robert Lewis May 7 '18 at 18:56
• I assume that you believe that the uniqueness part of the theorem holds true. The function identically equal to zero is a solution of the IVP $a_0y''+a_1y'+a_2y=0$, $y(x_0)=y'(x_0)=0$, so, by uniqueness, it is the only solution of the IVP. – user539887 May 7 '18 at 19:09

Given that the stated Theorem holds, the Corollary follows as follows:

Consider any solution $y(x)$ to

$$a_0 y''(x) + a_1 y'(x) + a_2 y(x) = 0 \tag 1$$

on $(a, b)$ such that

$y(x_0) = y'(x_0) = 0; \tag 2$

by our theorem, $y(x)$ the unique--the only--solution to (1) such that (2) binds.

Now consider the function

$z(x) = 0, \; x \in (a, b); \tag 3$

clearly

$z(x_0) = z'(x_0) = 0, \tag 4$

and obviously

$a_0 z''(x) + a_1 z'(x) + a_2 z(x) = a_0 \cdot 0 + a_1 \cdot 0 + a_2 \cdot 0 = 0; \tag 5$

so $z(x)$ satisfies (1) as well; furthermore, by (2) and (4), $z(x)$ and $z'(x)$ agree with $y(x)$ and $y'(x)$, respectively, at $x_0$; but $y(x)$ is the only solution satisfying (2), so we must have

$z(x) = y(x), \; \forall x \in (a, b); \tag 6$

the requisite conclusion now follows by virtue of (3).

Suppose otherwise that the solution is nonzero, i.e. there is a $z(x)\neq 0$ that satisfies the homogenous equation. Then this would contradict uniqueness in the original theorem, because if you have a nonzero solution $y$ to the original theorem, then $y+z$ is also such a solution, since $y(x_0)+z(x_0)=y(x_0)=c_1$ and $y'(x_0)+z'(x_0)=y'(x_0)=c_2$.