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Let's consider a boundary value problem $$ u''(x) = f(u(x)) + g(x)$$ with boundary conditions $u(0) = u(1) = 0$. We assume that the functions $g \in \mathscr C[0,1]$ and $f \in \mathscr C(\mathbb R)$ are given, $u(x)$ remains unknown.

How can we transform the equation to the integral form $$u(x) = \int_0^1 K(x,y) [f(u(y)) + g(y)] \,\textrm{d}y,$$ where the function $K$ depends on $f$ and $g$? I have tried to found an implicit formula for $K$ and failed - that's a technique I have discovered quite recently and haven't mastered yet.

What additional assumption do I need in order to say that the solution $u(x)$ of my equation is unique? Maybe the Lipschitz condition for $f$ or its derivative or some type of regularity of $u$, like being twice differentiable ($\mathscr C^2$)?

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  • $\begingroup$ Thinking out loud. Taking 2 derivatives seems to imply $$u''(x) = \int_0^1 K_{xx}(x,y) [f(u(y))+g(y)]dy$$ can we use this somehow? $\endgroup$ – gt6989b Oct 28 '15 at 18:27
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Look up Greens kernel, and indeed one condition can be written as $$K_{xx}(x,y)=\delta(x-y),$$ thus \begin{align} K_x&=H(x-y)+C,\\ K(x,y)&=(x-y)_++Cx+D \end{align} and evaluating for the boundary conditions $$0=K(0,y)=0+C·0+D,\qquad 0=K(1,y)=(1-y)+C$$ implies $$ K(x,y)=(x-y)_+-(1-y)·x=\begin{cases}-x·(1-y),&x\le y\\ -(1-x)·y,&x\ge y\end{cases} $$


Which falls in the general formula for Green kernels of Sturm-Liouville BVP $$ K(x,y)=u(\min(x,y))·v(\max(x,y)) $$ where $u,v$ are solution of the homogeneous linear operator with $u(0)=v(1)=0$, $u'(0)=v'(1)=1$.

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  • $\begingroup$ excuse me, but how can I look up Greens kernel? I have never heard of that. $\endgroup$ – user207868 Feb 23 '16 at 18:32
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    $\begingroup$ google.de/search?q=sturm+liouville+green+function $\endgroup$ – LutzL Feb 23 '16 at 22:17
  • $\begingroup$ The problem is that my equation involves the nonlinear part $f(u(x))$, does the Green's kernel trick still applies? $\endgroup$ – user207868 Feb 24 '16 at 10:37
  • $\begingroup$ Yes, the $K$ only depends on the linear differential operator part that you want to invert. The integral operator is now a map from $C^0$ to $C^2$, to get a fixed point operator you have to embed $C^2$ in $C^0$, this part is compact and makes in composition the integral operator compact. $\endgroup$ – LutzL Feb 24 '16 at 10:49

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