# Two-point boundary value problem: $u'' = -\lambda \cos u$

Suppose that we have the following

$$u'' = -\lambda \cos u$$

with $u(0) = u(1) = 0$. How can it be shown that if $|\lambda|$ is sufficiently small then the problem above has a unique solution?

The hint is to reformulate the problem as a nonlinear integral equation. So here is my attempt:

$$u'(t) = C -\int_{0}^{t} \lambda \cos(u(s)) ds$$

If we define the map $T$ as

$$T(u(t)) = C -\int_{0}^{t} \lambda \cos(u(s)) ds$$

we can write

$$u' = T(u)$$

With the contraction mapping theorem I could show that if $T(x) = x$ and $T$ is a contraction mapping then $T(x) = x$ has a unique solution.

I suppose that if I integrate again:

$$u(x) = C_2 + C_1 x - \int_{0}^{x} \int_{0}^{t} \lambda \cos(u(s)) ds dt$$

but that is where I'm running out of ideas.

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## 3 Answers

I think you have the right idea, but you can calculate expressions for $C_1, C_2$ by using the boundary conditions. Integrating twice yields $$u(t) = u(0) + tu'(0) - \lambda \int_0^t \int_0^{\tau} \cos(u(s)) \, ds \, d\tau$$ We immediately eliminate $u(0) = 0$. If we evaluate at $t = 1$, we get $$0 = u(1) = u'(0) - \lambda \int_0^1 \int_0^{\tau} \cos(u(s)) \, ds \, d\tau$$ Substituting everything, we get $$u(t) = t\lambda \int_0^1 \int_0^{\tau} \cos(u(s)) \, ds \, d\tau - \lambda \int_0^t \int_0^{\tau} \cos(u(s)) \, ds \, d\tau$$ Try to work with this integral equation.

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I'm not quite sure $C_1 = u'(0)$ and $C_2 = u(0)$. In any case, this still gives us an integral in terms of $u$ so this doesn't solve anything. – Stan Dec 18 '12 at 0:55
I'm simply iterating the fundamental theorem of calculus to calculate the constants. You're right, this isn't an explicit formula for $u$. However, you now have a nonlinear fixed point problem of the form $I(u) = u$, where $I$ is the integral operator from above, so could try to use the fixed point method. – Christopher A. Wong Dec 18 '12 at 1:04

While this is not the intended solution method, you may find another approach of interest. I'll assume $\lambda > 0$ (the case $\lambda = 0$ is trivial, and $\lambda < 0$ can be obtained from $\lambda > 0$ by $u \to -u$). The differential equation describes a nonlinear pendulum. It is a Hamiltonian system, where the total energy $E = (u')^2/2 + \lambda \sin(u)$ is constant. A part of the phase portrait is shown here:

The solutions in which $u$ is not monotonic are those for $-\lambda < E < \lambda$, and in order for $u$ to ever reach $0$ we need $E \ge 0$. For $0 \le E < \lambda$ there is a trajectory that goes from $(u,u') = (0,\sqrt{2E})$ to $(0,-\sqrt{2E})$ to the right of $u=0$. There are also trajectories that go from $(0,-\sqrt{2E})$ to $(0,\sqrt{2E})$ to the left of $u=0$, but these cover more than half a period, which is longer than $1$ if $\lambda$ is sufficiently small. For $E$ near $0$ the time to get from $(0,\sqrt{2E})$ to $(0,-\sqrt{2E})$ is near $0$, while as $E \to \lambda$ the time goes to $+\infty$ because the trajectory passes near the fixed point $(\pi/2, 0)$. By the Intermediate Value Theorem there is some $E \in (0,\lambda)$ for which the time is $1$, corresponding to a solution with $u(0) = u(1) = 0$. It seems somewhat tricky to show that the time is an increasing function of $E$, which you would need for uniqueness.

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This is not an answer, but a suggestion. Your equations actually can be written in the form $$\frac{(\dot u)^2}{2}+\lambda \sin u=0,$$ from which the implicit solution is given by $$\pm\frac {1}{\sqrt 2}\int_{u_0}^u\frac{du}{\sqrt{E-\lambda \sin u}}=t-t_0,$$ where $E$ is a constant that determines the orbit.

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