# Linear Two-Point Boundary Value Problem (BVP)

I ran across the following. In the text I was reading, it was left unproved. Can anyone help me see why it's true?

If the linear two-point boundary value problem \begin{cases} y''=p(x)y'+q(x)y+r(x)\\ y(a)=A\\ y(b)=B \end{cases} satisfies

1. $p(x),q(x),r(x)$ are continuous on $[a,b]$
2. $q(x)>0$ on $[a,b]$

then the problem as a unique solution.

Thanks!

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There are already nice answers here. An alternative would be to apply more general results from elliptic PDE theory, or from Sturm-Liouville theory. –  timur Aug 15 '12 at 0:10

Let $z$ be the difference of two solutions. Then $z$ solves the homogeneous equation $z'' =pz' +qz$ with homogeneous boundary conditions.

If $z$ is not identically zero, then it's either strictly positive or strictly negative somewhere. Let's take on the first case, since we can switch to $-z$ anyway. At the point of absolute maximum (which is not an endpoint) we have $z' =0$, $z''\le 0$ and $qz>0$. Contradiction.

(Existence). I assume the existence for IVP is known. Let $y_0$ be the solution of the IVP $y_0(a)=A$, $y_0'(a)=0$ (and same DE as in the problem). Also, let $y_1$ be the solution of the homogeneous equation $y_1''=py_1'+qy$ with the initial conditions $y_1(a)=0$ and $y_1'(a)=1$. Let us observe that for any number $k$ the combination $y_0+ky_1$ solves your DE and satisfies the left-endpoint boundary condition.

It remains to choose $k$ so that $y_0(b)+ky_1(b)=B$. Clearly, this is possible as long as $y_1(b)\ne 0$. And how do we know that $y_1(b)\ne 0$? Because if $y_1(b)=0$, then $y_1$ and $\equiv 0$ both solve the homogeneous BVP for $y''=py'+qy$, contradicting uniqueness.

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Also, you may be interested in this book which despite its title covers linear BVP as well. –  user31373 Aug 15 '12 at 0:00
I think the approach in the existence proof is called shooting method. It can be used also to solve some nonlinear problems. –  timur Aug 15 '12 at 0:06
@timur Yes, that's what it's called. First time I saw it I could not believe that this is also a reasonable way to solve BVP numerically, but turns out it is. Scholarpedia has a nice article on BVP. –  user31373 Aug 15 '12 at 0:09
Yes, this converts the problem into finding a correct initial angle to shoot. For nonlinear problems there can be many possible angles. –  timur Aug 15 '12 at 0:14

Let $y_1$ and $y_2$ be two solutions of the given BVP, and let $(a_+,b_+)$ be a connected component of the open set $\{x \in (a,b): y_2(x)<y_1(x)\}$. Setting $y=y_2-y_1$ we have $$y''=p(x)y'+q(x)y \ \text{ in } (a_+,b_+),\quad y(a_+)=0=y(b_+).$$ Thanks to Rolle's Theorem, there is some $c \in (a_+,b_+)$ such that $y'(c)=0$. Hence $y''(c)=q(c)y(c)>0$, i.e. $c$ is a minimum for $y$. By the Maximum Principle we have $c \in \{a_+,b_+\}$, contradicting the fact that $c \in (a_+,b_+)$. Theorefore $(a_+,b_+)$ is empty. Similarly, one shows that any connected component of $\{x \in (a,b):\ y_2(x)>y_1(x)\}$ is empty. Thus $y_2(x)=y_1(x)$ for every $x \in (a,b)$.

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Thanks! Both of these prove the uniqueness part, but what about existence? How do you know any solutions exist? –  Fortunato Aug 14 '12 at 15:01
@Fortunato As you found out, you don't need any rep points to comment within your own thread. –  user31373 Aug 14 '12 at 23:06
@Fortunato: You'll notice a few things: firstly, I changed your answer to a comment. Secondly, I merged the unregistered account that you used to ask the question with your current account. –  mixedmath Aug 15 '12 at 4:10