Example 1. Construct the Green's function for the homogeneous boundary value problem

\[
\begin{gathered}
y^{\mathrm{IV}}(x)=0 \\
\left\{\begin{array}{l}
y(0)=y^{\prime}(0)=0 \\
y(1)=y^{\prime}(1)=0
\end{array}\right.
\end{gathered}
\]
Solution. First, we will show that the boundary value problem (15)-(16) has only a trivial solution. Indeed, the fundamental system of solutions for equation (15) is

$$
y_{1}(x)=1, \quad y_{2}(x)=x, \quad y_{3}(x)=x^{2}, \quad y_{4}(x)=x^{3}
$$

so that its general solution has the form

$$
y(x)=A+B x+C x^{2}+D x^{3}
$$

where $A, B, C, D$ are arbitrary constants. The boundary conditions (16) give us four relations to determine $A, B, C, D$:

$$
\begin{aligned}
y(0) & =A=0 \\
y^{\prime}(0) & =B=0 \\
y(1) & =A+B+C+D=0 \\
y^{\prime}(1) & =B+2 C+3 D=0
\end{aligned}
$$

From this, we have $A=B=C=D=0$.

Thus, the problem (15)-(16) has only the zero solution $y(x) \equiv 0$, which means that we can construct (and uniquely) the Green's function $G(x, \xi)$ for it.

Now let's construct the Green's function. Using the fundamental system of solutions (17), we represent the desired Green's function $G(x, \xi)$ in the form

$$
\begin{aligned}
& G(x, \xi)=a_{1} \cdot 1+a_{2} \cdot x+a_{3} \cdot x^{2}+a_{4} \cdot x^{3} \text { for } 0 \leqslant x \leqslant \xi, \\
& G(x, \xi)=b_{1} \cdot 1+b_{2} \cdot x+b_{3} \cdot x^{2}+b_{4} \cdot x^{3} \text { for } \xi \leqslant x \leqslant 1,
\end{aligned}
$$

where $a_{1}, a_{2}, a_{3}, a_{4}, b_{1}, b_{2}, b_{3}, b_{4}$ are currently unknown functions of $\xi$. Let $c_{k}(\xi)=b_{k}(\xi)-a_{k}(\xi)(k=1,2,3,4)$ and write down the system of linear equations for finding the functions $c_{k}(\xi)$ (see system (5)):

$$
\left\{\begin{aligned}
c_{1}+c_{2} \xi+c_{3} \xi^{2}+c_{4} \xi^{3} & =0 \\
c_{2}+c_{3} \cdot 2 \xi+c_{4} \cdot 3 \xi^{2} & =0 \\
c_{3} \cdot 2+c_{4} \cdot 6 \xi & =0 \\
c_{4} \cdot 6 & =1
\end{aligned}\right.
$$

Solving this system, we get

$$
c_{1}(\xi)=-\frac{1}{6} \xi^{3}, \quad c_{2}(\xi)=\frac{1}{2} \xi^{2}, \quad c_{3}(\xi)=-\frac{1}{2} \xi, \quad c_{4}(\xi)=\frac{1}{6}
$$

Next, we use property $4^{\circ}$ of the Green's function, namely that it must satisfy the boundary conditions (2), i.e.,

$$
\begin{array}{ll}
G(0, \xi)=0, & G_{x}^{\prime}(0, \xi)=0 \\
G(1, \xi)=0, & G_{x}^{\prime}(1, \xi)=0
\end{array}
$$

In our case, these relations take the form

$$
\left\{\begin{aligned}
a_{1} & =0 \\
a_{2} & =0 \\
b_{1}+b_{2}+b_{3}+b_{4} & =0 \\
b_{2}+2 b_{3}+3 b_{4} & =0
\end{aligned}\right.
$$

Using the fact that $c_{k}=b_{k}-a_{k}(k=1,2,3,4)$, from (20) and (21) we find:

$$
\begin{cases}a_{1}=0, & b_{1}=-\frac{1}{6} \xi^{3} \\ a_{2}=0, & b_{2}=\frac{1}{2} \xi^{2} \\ a_{3}=\frac{1}{2} \xi-\xi^{2}+\frac{1}{2} \xi^{3}, & b_{3}=\frac{1}{2} \xi^{3}-\xi^{2} \\ a_{4}=-\frac{1}{6}+\frac{1}{2} \xi^{2}-\frac{1}{3} \xi^{3}, & b_{4}=\frac{1}{2} \xi^{2}-\frac{1}{3} \xi^{3}\end{cases}
$$

Substituting the values of the coefficients $a_{1}, a_{2}, \ldots, b_{4}$ from (22) into (18) and (19), we obtain the desired Green's function:

$$
G(x, \xi)= \begin{cases}\left(\frac{1}{2} \xi-\xi^{2}+\frac{1}{2} \xi^{3}\right) x^{2}-\left(\frac{1}{6}-\frac{1}{2} \xi^{2}+\frac{1}{3} \xi^{3}\right) x^{3}, & 0 \leqslant x \leqslant \xi \\ -\frac{1}{6} \xi^{3}+\frac{1}{2} \xi^{2} x+\left(\frac{1}{2} \xi^{3}-\xi^{2}\right) x^{2}+\left(\frac{1}{2} \xi^{2}-\frac{1}{3} \xi^{3}\right) x^{3}, & \xi \leqslant x \leqslant 1\end{cases}
$$

The last expression can be easily transformed into the form

$$
G(x, \xi)=\left(\frac{1}{2} x-x^{2}+\frac{1}{2} x^{3}\right) \xi^{2}-\left(\frac{1}{6}-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}\right) \xi^{3} \quad \text { for } \quad \xi \leqslant x \leqslant 1
$$

so that $G(x, \xi)=G(\xi, x)$, i.e., the Green's function is symmetric. This could have been stated in advance, as the boundary value problem (15)-(16) is self-adjoint.

We recommend that the reader verify this independently. In addition, we advise checking that the Green's function we found satisfies all the requirements $1^{\circ}-4^{\circ}$ formulated in its definition.