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Let $q, f \in C[0, 1]$ with $f < 0$ on $[0, 1]$. We need to find solution $y: [0, 1] \to \mathbb{R}$ to the BVP $$y'' + q y' = f$$ with $y(0) = y(1) = 0$ and to show $y'(0) > 0$ and $y'(1) < 0$.

My idea is to let $x = y'$ and re-write the ODE into $$x'= -qx + f$$ Using the general solution for $1^{st}$ order ODE, we have $$x = \frac{1}{e^{\int q(t) \mathrm{d}t}}\left(\int e^{\int q(t) \mathrm{d}t}f(t) \mathrm{d}t + c_1\right)$$ for some $c_1 \in \mathbb{R}$

Integrate again, we get $$y = \int \frac{1}{e^{\int q(t) \mathrm{d}t}}\left(\int e^{\int q(t) \mathrm{d}t}f(t) \mathrm{d}t + c_1\right) \mathrm{d}t + c_2$$ for some $c_1, c_2 \in \mathbb{R}$

I don't how to proceed from here. How am I suppose to plug in the boundary conditions and to show the inequalities? Thank you.

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  • $\begingroup$ $y(0) = 0$ and $y(0) > 0$? $\endgroup$ – Mattos Dec 6 '17 at 15:45
  • $\begingroup$ @Mattos fixed!! $\endgroup$ – Alex Vong Dec 6 '17 at 15:47
  • $\begingroup$ you can determine the constants $c_1$ and $c_2$ by using BCs: $y(0)=y(1)=0$. It may be helpful to write $\int_0^x$ instead $\int$. $\endgroup$ – daulomb Dec 6 '17 at 16:11
  • $\begingroup$ @daulomb What is "BCs"? $\endgroup$ – Alex Vong Dec 6 '17 at 16:13
  • $\begingroup$ $y(0)=y(1)=0$ are the boundary conditions and do not confuse $x$ as I mentioned with your $x$ in the substitution case. $\endgroup$ – daulomb Dec 6 '17 at 16:14
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Multiply the equation by the function $e^{\int_0^tq(s)ds}$ (integration factor) to get $$\frac{d}{dt}\bigg(y'e^{\int_0^tq(s)ds}\bigg)=f(t)e^{\int_0^tq(s)ds}.$$ Integrate both sides w.r.t. to $t$ we get $$y'e^{\int_0^tq(s)ds}=\int_0^tf(u)e^{\int_0^uq(s)du}+c_1,$$ and from which we have $$y'=e^{-\int_0^tq(s)ds}\int_0^tf(u)e^{\int_0^uq(s)ds}du+c_1e^{-\int_0^tq(s)ds}.\quad (1)$$

Integrating once more we find $$y=\int_0^t\bigg[\bigg(e^{-\int_0^r q(s)ds}\int_0^rf(u)e^{\int_0^uq(s)}du\bigg)+c_1e^{-\int_0^r q(s)ds}\bigg]dr+c_2.$$ Since $y(0)=0$, $c_2=0$ and hence we have $$y=\int_0^t\bigg[\bigg(e^{-\int_0^rq(s)ds}\int_0^rf(u)e^{\int_0^uq(s)}du\bigg)+c_1\displaystyle e^{-\int_0^r q(s)ds}\bigg]dr.\qquad (2)$$(This is the solution with $c_1$ given below). Also $y(1)=0$ implies that $$c_1=-\bigg(\int_0^1\displaystyle e^{\int_0^r q(s)ds}dr\bigg)\int_0^1\bigg[\int_0^r\bigg(e^{-\int_0^rq(s)ds}\int_0^rf(u)e^{\int_0^uq(s)}du\bigg)\bigg]dr,$$ which is a positive number since $f<0$. Substituting this into (1) we see that $y'(0)=c_1>0.$ A$$c_1=-\bigg(\int_0^1\displaystyle e^{\int_0^r q(s)ds}dr\bigg)\int_0^1\bigg[\int_0^r\bigg(e^{-\int_0^rq(s)ds}\int_0^rf(u)e^{\int_0^uq(s)}du\bigg)\bigg]dr,$$ which is a positive number since $f<0$. Substituting this into (1) we see that $y'(0)=c_1>0$. However I couldn't see why $y'(1)$ should be positive. It is not easy to deduce it from (1) directly. In this case I can only say that since $y'(t)$ is continuous and $y'(0)>0$ (means it increases as $t$ increases to $1$, but since $y(1)=0$ after some $t$ it starts to decreasing till to $1$, so $y'(1)<0$.(This could also be concluded if we know that the graph is concave,i.e., $y^{''}<0$.)(requires a combursome calculation). May be some doyens of the forum make contribution to clarify this point for me too.

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