Example 4. Find the influence function $G(x, y)$ for a beam supported at both ends $x=0$ and $x=1$. (Here $G(x, y)$ is the displacement parallel to the $O z$ axis of the cross-section at the point $x=y$, caused by the action of a unit load concentrated at the point $x=y$ and acting parallel to the $O z$ axis.
Solution. Let $R_{0}$ and $R_{1}$ be the unknown reactions at the support points caused by the action of a unit load at point $x=y$ (Fig. 5). Then the bending moment $M$ at point $x$ of the beam will be

$$
M=\left\{\begin{array}{lll}
-R_{0} x, & \text { if } & 0 \leqslant x \leqslant y, \\
-R_{1}(1-x), & \text { if } & y \leqslant x \leqslant 1 .
\end{array}\right.
$$

For the equilibrium of the system of three forces $R_{0}, R_{1}$, and $Q=1$, we must have

$$
1-R_{0}-R_{1}=0, \quad R_{1} \cdot 1=1 \cdot y
$$

From this, $R_{1}=y, R_{0}=1-y$. Then the bending moment will be

$$
M=M(x, y)=\left\{\begin{array}{lll}
-x(1-y), & \text { if } \quad 0 \leqslant x \leqslant y \\
-y(1-x), & \text { if } y \leqslant x \leqslant 1
\end{array}\right.
$$

Therefore, the influence function $G(x, y)$ must satisfy the differential equation of the bending moment

$$
E_{x} I_{x} \frac{\partial^{2} G}{\partial x^{2}}=M
$$

and the boundary conditions

$$
G(0, y) \equiv 0, \quad G(1, y) \equiv 0 .
$$

Here $\boldsymbol{E}_{x}$ is the Young's modulus, $\boldsymbol{I}_{\boldsymbol{x}}$ is the moment of inertia of the cross-section of the beam relative to the neutral axis perpendicular to the $x O z$ plane at point $x$.

To solve the boundary value problem (29)-(30), we will use the following integral:

$$
M^{*}(x, y)=\int_{0}^{1} M(x, z) M(z, y) F(z) d z
$$

where

$$
F(x)=\frac{1}{E_{x} I_{x}}
$$

Indeed, $M^{*}$ satisfies the boundary conditions (30), since

$$
M^{*}(0, y)=M^{*}(1, y) \equiv 0
$$

$M^{*}(x, y)$ satisfies the differential equation (29), since it is a symmetric function of $\boldsymbol{x}$ and $\boldsymbol{y}$, having the following explicit expression for $\boldsymbol{x} \leqslant \boldsymbol{y}$:

$$
\begin{array}{r}
M^{*}(x, y)=x y \int_{0}^{1}(1-z)^{2} F(z) d z-x \int_{0}^{y}(1-y)(y-z) F(z) d z- \\
-(1-y) \int_{0}^{x} z(x-z) F(z) d z
\end{array}
$$

from which it follows that

$$
\frac{\partial M^{*}}{\partial x}=y \int_{0}^{1}(1-z)^{2} F(z) d z-\int_{0}^{y}(1-y)(y-z) F(z) d z-(1-y) \int_{0}^{x} z F(z) d z
$$

and

$$
\frac{\partial^{2} M^{*}}{\partial x^{2}}=-(1-y) x F(x)=\frac{M(x, y)}{E_{x} I_{x}}, \quad \text { if } \quad 0 \leqslant x \leqslant y
$$

Therefore, for the case of a beam supported at the ends, we obtain

$$
G(x, y)=\int_{0}^{1} M(x, z) M(z, y) F(z) d z
$$

## Problems for Independent Solution

In the following examples, determine whether a Green's function exists for the given boundary value problem, and if it exists, construct it.

190. $y^{\prime \prime}=0$

$y(0)=y^{\prime}(1), y^{\prime}(0)=y(1)$.

191. $y^{\prime \prime}=0$

$y(0)=y(1), y^{\prime}(0)=y^{\prime}(1)$.

192. $y^{\prime \prime}+y=0$

$y(0)=y(\pi)=0$.

193. $y^{\mathrm{N}}=0$;

$y(0)=y^{\prime}(0)=y^{\prime \prime}(1)=y^{\prime \prime \prime}(1)=0$.

194. $y^{\prime \prime \prime}=0$;

$y(0)=y^{\prime}(1)=0, y^{\prime}(0)=y(1)$.

195. $y^{\prime \prime \prime}=0$;

$y(0)=y(1)=0, y^{\prime}(0)=y^{\prime}(1)$.

196. $y^{\prime \prime}=0 ; \quad y(0)=0, y(1)=y^{\prime}(1)$.
197. $y^{\prime \prime}+y^{\prime}=0 ; \quad y(0)=y(1), y^{\prime}(0)=y^{\prime}(1)$.
198. $y^{\prime \prime}-k^{2} y=0 ;(k \neq 0) y(0)=y(1)=0$.
199. $y^{\prime \prime}+y=0 ; \quad y(0)=y(1), y^{\prime}(0)=y^{\prime}(1)$.
200. $y^{\prime \prime \prime}=0 ; \quad y(0)=y(1)=0, y^{\prime}(0)+y^{\prime}(1)=0$.
201. $y^{\prime \prime}=0 ; \quad y^{\prime}(0)=h y(0), y^{\prime}(1)=-H y(1)$.
202. $x^{2} y^{\prime \prime}+2 x y^{\prime}=0 ; \quad y(x)$ is bounded as $x \rightarrow 0, y(1)=\alpha y^{\prime}(1)$.
203. $x^{3} y^{\prime V}+6 x^{2} y^{\prime \prime \prime}+6 x y^{\prime \prime}=0 ; y(x)$ is bounded as $x \rightarrow 0, y(1)=y^{\prime}(1)=0$.
204. $x^{2} y^{\prime \prime}+x y^{\prime}-y=0 ; \quad y(x)$ is bounded as $x \rightarrow 0, y(1)=0$.
205. $x y^{\prime \prime}+y^{\prime}-\frac{1}{x} y=0 ; \quad y(0)$ is finite, $y(1)=0$.
206. $x^{2} y^{\prime \prime}+x y^{\prime}-n^{2} y=0 ; \quad y(0)$ is finite, $y(1)=0$.
207. $x^{2}(\ln x-1) y^{\prime \prime}-x y^{\prime}+y=0 ; \quad y(0)$ is finite, $y(1)=0$.
208. $\frac{d}{d x}\left[\left(1-x^{2}\right) \frac{d y}{d x}\right]=0 ; \quad y(0)=0, y(1)$ is finite.
209. $x y^{\prime \prime}+y^{\prime}=0 ; y(0)$ is bounded, $y(l)=0$.
210. $y^{\prime \prime}-y=0 ; \quad y(0)=y^{\prime}(0), y(l)+\lambda y^{\prime}(l)=0$.

(Consider the cases: $\lambda=1, \lambda=-1,|\lambda| \neq 1$.)

211. Find the Green's function for the equation

$$
\begin{gathered}
\frac{d}{d x}\left[(1+\alpha x)^{4} \frac{d y}{d x}\right]=0, \quad \alpha>0 \\
y(0)=y(l)=0
\end{gathered}
$$

## § 15. Application of the Green's Function to Solving Boundary Value Problems

Let the differential equation with a right-hand side be given by

$$
L[y] \equiv p_{0}(x) y^{(n)}(x)+p_{1}(x) y^{(n-1)}(x)+\ldots+p_{n}(x) y(x)=f(x)
$$

and the boundary conditions

$$
V_{1}(y)=0, \quad V_{2}(y)=0, \quad \ldots, \quad V_{n}(y)=0
$$

where, as in $\S 14$, we assume that the linear forms $V_{1}, V_{2}, \ldots, V_{n}$ of $y(a), y^{\prime}(a), \ldots, y^{(n-1)}(a), y(b), y^{\prime}(b), \ldots, y^{(n-1)}(b)$ are linearly independent.

Theorem. If $G(x, \xi)$ is the Green's function of the homogeneous boundary value problem

$$
L[y]=0, \quad V_{k}(y)=0 \quad(k=1,2, \ldots, n)
$$

then the solution of the boundary value problem (1)-(2) is given by the formula

$$
y(x)=\int_{a}^{b} G(x, \xi) f(\xi) d \xi
$$