Solving the boundary value problem?

Question

Let $G(x,y)$ be the Green's function of the boundary value problem $$[(1+x)u']'+(\sin{x})u=0,~x\in[0,1],~u(0)=u(1)=0.$$

Then, the function $g$ defined by $$g(x)=G(x,\dfrac{1}{2}),~x\in[0,1].$$

1. is continuous
2. is discontinuous at $x=1/2$
3. is differentiable
4. doesnot have the left derivative at $x=1/2$

I have tried so many times to find the function but failed. How can I do this?

Answer 1

This question is not asking you to find the function. It wants you to remember the properties of the Green's function of a linear operator $L$, namely

• $ G(0,y)=G(1,y)=0 $ (satisfies the boundary conditions)
• $ L[G](x,y) = \delta(x-y) $ (solves the differential equation forced with a point impulse on the right-hand side).

The second of these means in particular that $$ [(1+x)g']' + (\sin{x}) g = \delta(x-1/2). $$ Therefore $g$ is a solution to the differential equation on $[0,1/2)$ and $(1/2,1]$ (with boundary conditions at $1/2$ that we can determine), and therefore it is continuously differentiable for every point in $[0,1]\setminus\{1/2\}$.

Now we have to examine conditions at $x=1/2$. The only term that can be discontinuous is the second derivative, because if locally $g'=\delta+h$ where $u$ is discontinuous, then $g'' = \delta'+$ less singular terms. Integrating over a small interval around the discontinuity gives $$ [(1+x)g']_{1/2-\epsilon}^{1/2+\epsilon} = 1, $$ from integrating the delta function and using that $g$ must be continuous at $1/2$ (if it isn't, $g'$ looks like a $\delta$, and we noted above we have problems then). $(1+x) \approx 3/2$ on this interval, so $g'$ has a discontinuity: it jumps by $3/2$ as we cross $x=1/2$.

So, to recap, $g$ is continuous everywhere, and the first derivative is continuous except where it jumps at $1/2$. Therefore the only correct statement is that $g$ is continuous.