Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $y(x)$ be a solution to $y''+e^xy=0$. Prove that there exists $t$ such that $0\le t\le T$ and $$\int_0^Te^{-x} y'y'' \, dx=\int_0^ty'y''\,dx.$$

share|cite|improve this question

The statement is trivial for the zero solution, so we'll ignore it for the rest of the answer.

Note that $ -\int_0^T e^{-x} y'y''dx = \int_0^T yy'dx = \frac{1}{2}\big(y(T)\big)^2 - \frac{1}{2}\big(y(0)\big)^2$ and $ -\int_0^t y'y''dx = \int_0^t e^x yy'dx. $


Per the above computations, it suffices to show that for any $T \geq 0$, there exists a $t \in [0, T]$ such that

$$\int_0^t e^x yy' \, dx = \int_0^T yy' \, dx \equiv \frac{1}{2}\bigl(y(T)\bigr)^2 - \frac{1}{2}\bigl(y(0)\bigr)^2.$$

We do this (Corollary 4) in a suitably abstract setting, namely where $y \colon [0,\infty) \to \mathbb R$ is a nontrivial solution to $y'' + q(x) y = 0$ with $q \colon [0,\infty) \to \mathbb R$ positive, nondecreasing, and $C^1$.

The key for our method is the second positive zero of $yy'$, which we denote by $x_2$. Corollary 4 follows from the following two results, to be proven:

  • The function $T \mapsto \int_0^T yy'\,dx$ (which is essentially $T \mapsto \bigl(y(T)\bigr)^2$) attains its global maximum and minimum, and hence every value in its image, somewhere on the interval $[0, x_2]$ (Corollary 2). This is due to the fact that $y$ is oscillatory and that the absolute value of the relative extrema of $y$ are nonincreasing (Proposition 1).
  • For $\tau \in [0, x_2]$, we show the existence of a $t \in [0, \tau]$ satisfying $\int_0^t e^x yy'\,dx = \int_0^\tau yy'\,dx$ (Proposition 3). Intuitively, if we let the upper limit of the integrals be variable, the integral on the left will reach a given value earlier than the one on the right due to the extra factor $e^x$ (which is, crucially, greater than or equal to $1$ and nondecreasing).

The details

Proposition 1. Let $q \colon [0,\infty) \to \mathbb R$ be positive, nondecreasing, and $C^1$. Let $y\colon [0,\infty) \to \mathbb R$ be a nontrivial solution to $y'' + q(x) y = 0$. Then:

  1. The set of zeros and local extrema of $y$ is countably infinite and has no limit point (in the whole domain $[0, \infty)$). Thus, they can be enumerated in increasing order. We denote the zeros on $(0,\infty)$ as $a_1 < a_2 < \dotsb$ and the extrema on $(0,\infty)$ as $b_1 < b_2 < \dotsb$.

  2. $y$ is oscillatory, i.e. $\lim_{n \to \infty} a_n = \infty$.

  3. On $(0,\infty)$, the zeros and the extrema are interlaced. That is, either $a_1 < b_1 < a_2 < b_2 < \dotsb$ or $b_1 < a_1 < b_2 < a_2 < \dotsb$.

  4. $\bigl(y(b_1)\bigr)^2 \geq \bigl(y(b_2)\bigr)^2 \geq \dotsb$.

Proof. Using Sturm's comparison theorem on $y$ and the solution to $\tilde y'' + q(0) \tilde y = 0$ yields an unbounded set of zeros of $y$. Now if the set $A$ of all zeros of $y$ were to have a limit point $a^* \in [0,\infty)$, then after choosing a sequence $a^{(n)} \to a^*$ in $A$, we would find that $y(a^*) = \lim y(a^{(n)}) = 0$ and $$ y'(a^*) = \lim_{x \to a^*} \frac{y(x) - y(a^*)}{x - a^*} = \lim_{n \to \infty} \frac{y(a^{(n)}) - y(a^*)}{a^{(n)} - a^*} = \lim_{n \to \infty} \frac{0 - 0}{a^{(n)} - a^*} = 0. $$ It follows by uniqueness of solution that $y$ is identically zero, a contradiction. Hence, $A$ does not have a limit point. We have proven 1 for the zeros, along with 2.

Next, Rolle's theorem tells us that there exists at least one extremum between each successive pair of zeros $a_n$ and $a_{n+1}$. On the other hand, $y'' = -qy$ has no zeros on the intervals $(a_n, a_{n+1})$ as well as $(0, a_1)$, so $y'$ is strictly monotonic on each of those intervals. In particular, $y'$ has at most one zero on each of those intervals. And by the uniqueness of solutions, zeros and interior extrema do not coincide for nontrivial solutions. This completes 1 and 3.

Finally for 4, a (possibly) standard trick (cf. [1], p. 232) is to observe that $q' \geq 0$ implies $$ \left(y^2 + \frac{(y')^2}{q} \right)' = \frac{2y'(qy+y'')}{q} - \frac{q'(y')^2}{q^2} = -\frac{q'(y')^2}{q^2} \leq 0, $$ meaning that the sum being differentiated is nonincreasing. This leads immediately to 4 because $y'(b_n) = 0$ for all $n$.

Corollary 2. Let $a_n$, and $b_n$ be defined as in the above proposition, and let $x_2 = \max\{a_1, b_1\}$, which by item 3 of the proposition is the second smallest positive zero of $yy'$. Define $L \colon [0, \infty) \to \mathbb R$ by $L(T) = \frac{1}{2}\bigl(y(T)\bigr)^2 - \frac{1}{2}\bigl(y(0)\bigr)^2 = \int_0^T yy' \, dx$. Then $L([0,x_2]) = L([0,\infty))$.

Proof. The global maximum of $y^2$ occurs at either $0$ or $b_1$, while the global minimum of $y^2$ occurs at any zero of $y$, including $a_1$. Use the intermediate value theorem.

Proposition 3. For $0 \leq x_1 \leq x_2$, let $f \colon [0, x_2] \to \mathbb R$ be an integrable function satisfying either

  1. $\left.f\right|_{(0,x_1)} \geq 0$ and $\left.f\right|_{(x_1,x_2)} \leq 0$; or
  2. $\left.f\right|_{(0,x_1)} \leq 0$ and $\left.f\right|_{(x_1,x_2)} \geq 0$.

Let $g \colon [0, x_2] \to [1, \infty)$ be a nondecreasing, integrable function. Then for any $\tau \in [0, x_2]$, there exists a $t \in [0, \tau]$ such that $\int_0^\tau f \,dx = \int_0^t gf \, dx$.

Proof. It suffices to prove the first case. We break it down into three smaller cases:

  • If $\tau \in [0, x_1]$, then $\int_0^\tau gf \, dx \geq \int_0^\tau f \, dx$ so the conclusion follows by the intermediate value theorem.
  • If $\tau \in (x_1, x_2]$ but $\int_0^\tau f \, dx \geq 0$, then there exists a $\tilde \tau \in [0, x_1]$ such that $\int_0^\tau f \, dx = \int_0^{\tilde\tau} f \, dx$ and we reduce to the previous case.
  • If $\tau \in (x_1, x_2]$ and $\int_0^\tau f \, dx < 0$, then $\int_0^\tau g(x)f(x) \, dx \leq \int_0^\tau g(x_1)f(x) \, dx \leq \int_0^\tau f(x) \, dx$, whereupon we use the intermediate value theorem.

Corollary 4. Given any $T \in [0,\infty)$, there exists a $t \in [0, T]$ such that $$\int_0^t e^x yy' \, dx = \int_0^T yy' \, dx.$$

Proof. By Corollary 2, there exists a $\tau \in [0, \min\{x_2, T\}]$ such that $\int_0^\tau yy' \, dx = \int_0^T yy' \, dx$. Apply Proposition 3 with $f = yy'$ and $g = \exp$.


[1]: Agarwal and O'Regan, An Introduction to Ordinary Differential Equations (1st Ed., 2008)

P.S. The exact form of the solutions can be given in terms of the Bessel functions: $$ y(x) = \alpha J_0(2e^{x/2}) + \beta Y_0(2e^{x/2}), $$ where $J_\nu$ and $Y_\nu$ are the Bessel functions of $\nu$-th order of the first and second kind, respectively, and $\alpha, \beta \in \mathbb R$. The Wolfram Alpha output for the ODE has a plot of sample solutions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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