Consider the following ODE on the interval $[a,b]$

$ y''+p(x)y'+q(x)y=0 $

such that $p$ and $q$ are continuous functions in $[a,b]$.

Show that if a solution of the above ODE is tangent to the $x$ axis, then is null in $[a,b]$

What I did say was that if $y_{1}$ is a solution of the ODE and it is tangent to the $x$ axis at a point, then $y_{1} (x_{0}) = 0$, $y'_{1} (x_{0}) = 0$, $a <x_{0} <b$. Then replacing it in the differential equation can be deduced that $y_{1}'' (x_{0})$ = 0, then use a theorem of existence and uniqueness. Can anyone give me a hand

thank you very much


If $p,q$ are continuous then the initial value problem for $y''+p(x)y'+q(x)y=0$ has a unique solution for each initial data set in $[a,b]$. Consider then, the zero solution $y=f(x)=0$ for all $x \in [a,b]$ is a solution since it is clear that $f''(x)+p(x)f'(x)+q(x)f(x)=0$ for all $x \in [a,b]$. Moreover, if $x_o \in (a,b)$ such that $y(x_o)=0$ and $y'(x_o)=0$ as the solution must be a horizontal tangent since the $x$-axis is horizontal. But, $y(x_o)=0, y'(x_o)=0$ is a set of initial values hence it uniquely determines a solution on the interval $[a,b]$. It follows that this solution must be the null solution by uniqueness of solution to the initial value problem. This theorem I'm using is usually called the "fundamental existence and uniqueness theorem for ODEs". It ought to be in whatever text you're using.

  • $\begingroup$ I think I was very close to the answer, but you're right uniqueness makes grace. thx $\endgroup$ – user119144 May 7 '14 at 3:20
  • $\begingroup$ @user119144 yes. You were very very very close. Glad to help. $\endgroup$ – James S. Cook May 7 '14 at 4:57

The solution the OP gave is basically correct, and the reasoning not fundamentally flawed, though introduction of the fact that $y_1''(x_0) = 0$ is perhaps a trifle redundant. This problem indeed illustrates an important principle in the theory of linear ordinary differential equations, that being as follows: let $\vec r(t)$ satisfy the linear, time dependent system

$\dot{\vec r} = A(t) \vec r, \tag{1}$

where $A(t)$ is a continuous $n \times n$ matrix function of $t$; then the only solution which takes the value $\vec r(t_0) = 0$ at some $t_0 \in \Bbb R$ the identically zero solution $r(t) = 0$ for all $t \in D \subset \Bbb R$, where $D$ is the domain (interval) on which the system is defined; this assertion indeed follows from the uniqueness of solutions which always applies to a linear system with continuous c0efficients, since then $A(t)\vec r$ is easily seen to Lipschitz continuous in $\vec r$ and continuous in $t$, which are the necessary hypotheses to insure existence and uniqueness of solutions. So, if $\vec r(t_0) = 0$, $\vec r(t_0)$ agrees with the zero solution at $t_0$, hence everywhere in $D$ by uniqueness.

This concept applies to the OP's problem in the following manner: with

$y''+p(x)y'+q(x)y=0 \tag{2}$

we can set

$z = y' \tag{3}$

so that (2) becomes

$z' = -p(x)z - q(x) y, \tag{4}$

and then setting

$\vec r = \begin{pmatrix} y \\ z \end{pmatrix} \tag{5}$

we then write (2) as the two-dimensional system

$\vec r'(x) = A(x)r(x) \tag{6}$


$A(x) = \begin{bmatrix} 0 & 1 \\ -q(x) & -p(x) \end{bmatrix}. \tag{7}$

The condition that the graph of $y_1(x)$ be tangent to the $x$-axis at $x_0$ translates to $y_1(x_0) = 0$ (since the graph must meet the axis) and $z_1(x_0) = y_1'(x_0) = 0$ (since the graph is tangent to the axis at the point of meeting). Thus we have

$\vec r_1(x_0) = \begin{pmatrix} y_1(x_0) \\ z_1(x_0) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \tag{8}$

and hence $\vec r_1(x) = 0$ for all $x \in [a, b]$ in light of the general situation described in the previous paragraph.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

  • $\begingroup$ another way to look, thank you very much $\endgroup$ – user119144 May 8 '14 at 3:51
  • $\begingroup$ Glad to help out! $\endgroup$ – Robert Lewis May 8 '14 at 3:58

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