$$G(x,t) =\begin{cases} a+ b\log t & \text{if $0<x<t$ } \\[2ex] c+ d\log t & \text{if $t<x<1$ } \end{cases}$$

is a Greens function for $xy''+y'=0$ subject to $y$ being bounded as $x$ tends to $0$ and $y(1)=y'(1)$ if

The options are

1) $a=b=c=d=1$

2) $a=c=1$, $b=d=0$

3) $a=c=0$, $b=d=1$

4) $a=b=c=d=0$

I don't know how to tackle the problem. I've tried to find the Green's function. And I've got

$$ G(x,t) =\begin{cases} a+ b(\log x) & \text{if $0<x<t$ } \\[2ex] a+ b(\log x)-\log t+\log x & \text{if $t<x<1$ } \end{cases} $$

but this is not in the above given form. Also how to apply the boundary conditions? Please help me to solve the problem.

  • 3
    $\begingroup$ This is a CSIR QUESTION $\endgroup$
    – Tani
    Dec 14, 2015 at 6:24
  • $\begingroup$ is CSIR some ongoing contest? $\endgroup$
    – najayaz
    Dec 14, 2015 at 7:57
  • $\begingroup$ @G-man, a late reply, still: no, CSIR NET is a biannual 3-hour examination in India, for lectureship and scholarship for higher studies. $\endgroup$ May 9, 2016 at 6:31

1 Answer 1


I hope you meant $$G(x,t) =\begin{cases} a+ b\log t & \text{if $0<x\le t$ } \\[2ex] c+ d\log x & \text{if $t\le x\le1$ } \end{cases}$$

Now, using:

  1. $G(x,t)$ satisfies the boundary condition.

$$\implies G(1,t)=G'(1,t)\implies c+d\log 1=\frac d 1\implies c=d$$ That narrows our options to a and d.

  1. Derivative, $G'(x,t)$ jumps at $t$ by $1\over p(t)$=$1\over t$ $$\implies d=1$$

Hence 1.

  • $\begingroup$ I did not understand your $2$ point.Can you elaborate? $\endgroup$ Jun 6, 2016 at 9:46
  • $\begingroup$ @KushalBhuyan Follows #4 here: en.wikipedia.org/wiki/Green's_function#Theorem . Gives $\frac dt=\frac1t$ $\endgroup$ Jun 6, 2016 at 9:50
  • $\begingroup$ Okay. Then how you get $d=1$. Sorry haven't seen it very clearly. $\endgroup$ Jun 6, 2016 at 9:54
  • $\begingroup$ got it thanks.... $\endgroup$ Jun 6, 2016 at 10:09

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