I want to determine the Green’s functions of the following BVP:
Let $y''=f(x)$ with the boundary conditions: $y(0)=0$, $y'(1)=0$
So the homogeneous solution for this is: $y_h(x)=C_1+C_2x$, has the fundamental system $ \{u_1(x),u_2(x)\}=\{1,x\}$
Now I'm stuck here, I want to determine the new FS $\{v_1(x),v_2(x)\}$:
the first boundary conditions gives us:
$y(0)=0 \Rightarrow C_1+C_2.0=0\Rightarrow C_1=0 \Rightarrow v_1(x)=x$
the second boundary conditions gives us:
$y'(1)=0 \Rightarrow C_2=0 \Rightarrow v_2(x)=1$
Hence $\{v_1(x),v_2(x)\}=\{x,1\}$ and furthermore satisfy the boudary conditions $v_1(0)=0,v'_2(1)=0 $ and we can easly show the Wronskian determinant $W(x)=-1$ is that correct? I think it should be different from $ \{u_1(x),u_2(x)\}=\{1,x\}$.
And I don't know how to continue this to find the green's function.
Thanks in advance for any help!
Taking it up on my comment, the Green's function associated to the problem $L(y) = y'' = f$, $y(0)=y'(0)=0$ is given by the solution to $L(G)=G''=\delta(x-\xi)$, $G(0) = G'(1) =0$, with $0<\xi<1$. One can apply integration by parts to show that $y(\xi) = \int^1_0 f(x) G(x;\xi)\,\mathrm{d}x$.
The task at hand is then to integrate $G''=\delta(x-\xi)$. Integrate once with respect to $x$ to have:
$$\int^{0}_{G'} \mathrm{d}G' = \int^1_x \delta(x-\xi)\,\mathrm{d}x \implies G' = H(x-\xi) - 1,$$
where $H$ is the Heaviside (or unit step) function. Note that the BC $G'(1)=0$ has been imposed. Integrate once more to find
$$\int^G_0 \,\mathrm{d}G = \int^x_0 \left[ H(x-\xi) -1 \right]\,\mathrm{d}x \implies G(x;\xi) = (x-\xi)H(x-\xi) - x,$$
where the BC $G(0)=0$ has been imposed.
I have checked that indeed this is the Green's function to the problem for the particular case $f(x) = x$, with the help of Mathematica.
I hope you find this useful.