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Given positive scalars $k_1,k_2, x_1, u_1$ and continuous non-decreasing function $f: R\to R$ with $f(0) = 0$ how to prove the existence of unique $C^2([0,x_1])$ solution of the problem:

$$u'' = k_1u'+k_2f(u)\quad \hbox{ on }\ \langle0,x_0\rangle$$ $$u(0) = 0$$ $$u(x_1) = u_1.$$

Does it imply that $u\geq 0$?

I know this is standard result, but I have difficulties with references to this result.

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The tag differential-equations is intended for ordinary differential equations, there's no need to create a new tag. –  Martin Sleziak Jul 21 '12 at 14:42
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3 Answers

Uniqueness looks like a standard application of monotonicity. Suppose $u,v$ are two solutions. Use the ODE to show that the integral of $(u''-v'')(u-v)$ is nonnegative. On the other hand, integration by parts turns this into the integral of $-(u'-v')^2$. Hence $u\equiv v$.

For the nonnegativity, suppose $u$ attains negative minimum at some point $p$. Then $f(u(p))=0$, otherwise we already have a contradiction. But then $f(u)=0$ in a neighborhood of $p$, which reduces the equation to a linear one - with only the constant solution.

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If you want to proof the existence of the solution you need to use the Fixed point theorem. In your case you need to transform the differential equation to an integral equation first. I recommend you to look for a book on non linear differential equation. There you will find existence and uniqueness theorems.

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It's not clear to me how standard existence results apply to this boundary value problem. For instance, if you allow the constants to have arbitrary sign, the boundary value problem has completely different features. Can you please clarify what you had in mind? –  Sam Lisi Aug 20 '12 at 21:08
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LVK has already answered existence and the non-negativity of the solution. For existence, "shooting" works. For each parameter $\alpha \ge 0$, consider the solution to the IVP \[ \begin{split} &u'' = k_1 u' + k_2 f(u) \\ &u(0) = 0, \\ &u'(0) = \alpha. \end{split} \] Denote the unique solution to this IVP by $u_\alpha$. Then, $u_\alpha(x_1)$ is a continuous function of $\alpha$. Observe that $u_0$ is identically $0$, so $u_0(x_1) = 0$. Furthermore, by rewriting the ODE, we have \[ ( e^{-k_1 t} u' ) ' = e^{-k_1 t} k_2 f(u). \] Integrating both sides and using $f(u) \ge 0$, we get that $u_\alpha'(t) \ge \alpha$ for all $t \in [0, x_1]$. It then follows that $u_\alpha(x_1) \ge \alpha\, x_1$, so for $\alpha$ large enough, $u_\alpha(x_1) > u_1$.

The IVT gives the existence of an $\alpha$ for which a solution exists.

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