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


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .