7
$\begingroup$

Considering homogeneous, linear, second order differential equation $$y'' + p(x)y' + q(x)y = r(x), a<x<b$$ $p, q, r$ 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?

$\endgroup$

1 Answer 1

5
$\begingroup$

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 $[a, b]$.

$\endgroup$
5
  • 1
    $\begingroup$ why does $z'(a)=0$ lead to $z(x)=0$ and what does "by the uniqueness of solutions" mean? $\endgroup$
    – XT Chen
    Dec 24, 2018 at 6:17
  • $\begingroup$ @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 $\endgroup$ Dec 24, 2018 at 6:20
  • $\begingroup$ @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 $z(a) = z'(a) = 0$. $\endgroup$ Dec 24, 2018 at 6:23
  • $\begingroup$ picard theorem only gives the situation of order one, does it hold for order two? and what is the $z(b)=0$ used for? $\endgroup$
    – XT Chen
    Dec 24, 2018 at 6:39
  • $\begingroup$ @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. $\endgroup$ Dec 24, 2018 at 6:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.