To check the sign and convexity of a transcendental function using naive approach

Question

I was solving this BVP: $$y''+y= \csc(x), y(0)=0=y(\pi/2), x\in[0,\pi/2]$$ By following the same lines of this solution using the variation of parameters method, I got the solution: $$y=-x\cos(x) + \sin(x)\log(\sin(x))$$ I now need to determine its sign and nature (concavity/convexity/a mix of both) in $[0,\pi/2]$.

My approach

1. To check the sign of y in $[0,\pi/2]$: Both the terms of $y$, viz. $-x\cos(x)$ and $\sin(x)\log(\sin(x))$ are negative and so is $y$.
2. To check the nature of y in $[0,\pi/2]$: On differentiating $y$ twice, I got: $$y''=x \cos(x)+\sin(x)+ \cot(x)\csc(c)-\sin(x)\log(\sin(x))$$ Interestingly, again on investigating the signs of all the four terms of $y''$ individually, they turned out to be each positive and thus so is $y''$ implying the convexity of $y$.

My question

Is there a nicer way to verify the sign and especially the nature of $y$? In particular can we do it without finding the solution, directly from the given data?

P.S. This question was supposed to be answered in an objective question paper, in a limited time frame without using any calculators or plotters.

Answer 1

If you use inspection, the first derivative cancels "close" to $\frac \pi 5$.

At this point, the value of the second derivative is $$\frac{1}{4} \left(1+\sqrt{5}\right) \left(\sqrt{1+\frac{2}{\sqrt{5}}}+\frac{\pi }{5}\right)-\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} \left(\frac{1}{2} \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)-1\right)$$ which is $2.52197$.

Answer 2

1. To check the sign of y in $[0,\pi/2]$: Both the terms of $y$, viz. $-x\cos(x)$ and $\sin(x)\log(\sin(x))$ are negative and so is $y$.
2. To check the nature of y in $[0,\pi/2]$: On differentiating $y$ twice, I got: $$y''=x \cos(x)+\sin(x)+ \cot(x)\csc(c)-\sin(x)\log(\sin(x))$$ Interestingly, again on investigating the signs of all the four terms of $y''$ individually, they turned out to be each positive and thus so is $y''$ implying the convexity of $y$. This implies $y$ to be negative and convex in $[0,\pi/2]$

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