# How to prove the existence and uniqueness of the solution of a second order linear ODE?

Considering homogeneous, linear, second order differential equation $$y'' + p(x)y' + q(x)y = r(x), a $$p, q, r$$ are given. We already know $$y(a) = A, y(b) = B$$ and $$q(x) < o$$. How to prove that this equation has unique solution in $$[a,b]$$ if it has a solution?

Suppose $$y_1(x)$$ and $$y_2(x)$$ both satisfy

$$y'' + p(x) y' + q(x)y = r(x) \tag 1$$

with

$$y_1(a) = y_2(a) = A, \; y_1(b) = y_2(b) = B; \tag 2$$

then we have

$$y_1'' + p(x) y_1' + q(x)y_1 = r(x) \tag 3$$

and

$$y_2'' + p(x) y_2' + q(x)y_2 = r(x), \tag 4$$

and, subtracting,

$$(y_1 - y_2)'' + p(x) (y_1 - y_2)' + q(x)(y_1 - y_2) = 0, \tag 5$$

and

$$(y_1 - y_2)(a) = y_1(a) - y_2(a) = 0 = y_1(b) - y_2(b) = (y_1 - y_2)(b); \tag 6$$

setting

$$z(x) = y_1(x) - y_2(x), \tag 7$$

we have

$$z'' + p(x) z' + q(x) z = 0, \; z(a) = z(b) = 0; \tag 8$$

now if $$z'(a) = 0$$, then by uniqueness of solutions we must have $$z(x) = 0$$, $$x \in [a, b]$$, since $$z(a) = 0$$, and thus $$y_1(x) = y_2(x)$$, and we are done; so we assume $$z'(a) \ne 0$$; in fact, since (8) is linear, we may take $$z'(a) > 0$$, and thus, by the continuity of $$z'(x)$$, there is some $$\delta > 0$$ such that

$$z'(x) > 0, \; x \in [a, a + \delta); \tag 9$$

we then have

$$z(x) = z(x) - z(a) = \displaystyle \int_0^x z'(s) \; ds > 0, x \in (a, a + \delta); \tag{10}$$

since $$z(x) > 0$$ somewhere on $$[a, b]$$, it must have a positive absolute maximum $$x_M \in (a, b)$$; at such a point

$$z'(x_M) = 0, \tag{11}$$

and thus by (8),

$$z''(x_M) = -q(x_M)z(x_M) > 0 \tag{12}$$

by virtue of the facts that $$z(x_M) > 0$$, $$q(x_M) < 0$$; but (12) is impossible at a maximum of $$z(x)$$, where we must have $$z''(x_M) \le 0$$; this contradiction forces $$z'(a) = 0$$ and thus also

$$z(x) = 0, \; x \in [a, b], \tag{13}$$

and hence

$$y_1(x) - y_2(x) = z(x) = 0, \; x \in [a, b]; \tag{14}$$

we conclude the equation (1) with boundary conditions (2) has at most one solution on $$[a, b]$$.

• why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean? Commented Dec 24, 2018 at 6:17
• @X.TChen: it's the standard result on uniqueness of solutions to ODEs with initial conditions. $z(x) =0$ is the only solution with $z(a) = z'(a) = 0$. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem Commented Dec 24, 2018 at 6:20
• @X.TChen: "uniqueness of solutions" means there is at most one solution for given initial conditions. So since solutions are unique, we have $z(x) = 0$ if $z(a) = z'(a) = 0$. Commented Dec 24, 2018 at 6:23
• picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for? Commented Dec 24, 2018 at 6:39
• @X.TChen: Picard-Lindeloef extends to all orders. $z(b) = 0$ is what forces a maximum of $z(x)$ to exist; it means that $z(x)$, once positive, must return to $0$ and not just keep growing. Commented Dec 24, 2018 at 6:44