Example 1. Reduce the boundary value problem

\[
\begin{gathered}
y^{\prime \prime}+\lambda y=x \\
y(0)=y\left(\frac{\pi}{2}\right)=0
\end{gathered}
\]

to an integral equation.
Solution. First, we find the Green's function $G(x, \xi)$ for the corresponding homogeneous problem

\[
\begin{gathered}
y^{\prime \prime}(x)=0 \\
y(0)=y\left(\frac{\pi}{2}\right)=0
\end{gathered}
\]

Since the linearly independent solutions of the equation $y^{\prime \prime}(x)=0$, satisfying the conditions $y(0)=0$ and $y\left(\frac{\pi}{2}\right)=0$, are the functions $y_{1}(x)=x$ and $y_{2}=x-\frac{\pi}{2}$, we seek the Green's function in the form

\[
G(x, \xi)= \begin{cases}\frac{y_{1}(x) y_{2}(\xi)}{W(\xi)}, & 0 \leqslant x \leqslant \xi \\ \frac{y_{1}(\xi) y_{2}(x)}{W(\xi)}, & \xi \leqslant x \leqslant \frac{\pi}{2}\end{cases}
\]

where

\[
W(\xi)=\left|\begin{array}{cc}
\xi & \xi-\frac{\pi}{2} \\
1 & 1
\end{array}\right|=\frac{\pi}{2}
\]

Thus,

\[
G(x, \xi)= \begin{cases}\left(\frac{2}{\pi} \xi-1\right) x, & 0 \leqslant x \leqslant \xi \\ \left(\frac{2}{\pi} x-1\right) \xi, \quad \xi \leqslant x \leqslant \frac{\pi}{2}\end{cases}
\]

Next, using the Green's function (8) as the kernel of the integral equation, we obtain for $y(x)$ the following integral equation:

\[
y(x)=f(x)-\lambda \int_{0}^{x / 2} G(x, \xi) y(\xi) d \xi
\]

where

\[
f(x)=\int_{0}^{\pi / 2} G(x, \xi) \xi d \xi=\int_{0}^{\pi}\left(\frac{2 x}{\pi}-1\right) \xi^{2} d \xi+\int_{0}^{\pi / 2}\left(\frac{2 \xi}{\pi}-1\right) x \xi d \xi=\frac{1}{6} x^{3}-\frac{\pi^{2}}{24} x
\]

Thus, the boundary value problem (6)-(7) has been reduced to the integral equation

\[
y(x)+\lambda \int_{0}^{\pi / 2} G(x, \xi) y(\xi) d \xi=\frac{1}{6} x^{3}-\frac{\pi^{2}}{24} x
\]

## Problems for Independent Solution

Reduce the following boundary value problems to integral equations:

219. $y^{\prime \prime}=\lambda y+x^{2} ; y(0)=y\left(\frac{\pi}{2}\right)=0$.
220. $y^{\prime \prime}=\lambda y+e^{x} ; \quad y(0)=y(1)=0$.
221. $y^{\prime \prime}+\frac{\pi^{2}}{4} y=\lambda y+\cos \frac{\pi x}{2} ; y(-1)=y(1), y^{\prime}(-1)=y^{\prime}(1)$.
222. $y^{\prime \prime}+\lambda y=2 x+1 ; y(0)=y^{\prime}(1), y^{\prime}(0)=y(1)$.
223. $y^{\text {Iv }}=\lambda y+1 ; \quad y(0)=y^{\prime}(0)=0, y^{\prime \prime}(1)=y^{\prime \prime \prime \prime}(1)=0$.
224. $y^{\prime \prime \prime}+\lambda y=2 x ; \quad y(0)=y(1)=0, y^{\prime}(0)=y^{\prime}(1)$.
225. $y^{\prime \prime}+\lambda y=e^{x} ; \quad y(0)=y^{\prime}(0), y(1)=y^{\prime}(1)$.
226. Transform the differential equation (obtained from the Schrödinger wave equation with a meson potential)

\[
\frac{d^{2} y}{d r^{2}}-k^{2} y(r)+v_{0} \frac{e^{-r}}{r} y(r)=0
\]

( $V_{0}$ and $\boldsymbol{k}^{2}$ - constants), the solution of which satisfies the boundary conditions

\[
y(0)=y(+\infty)=0
\]

into the integral equation

\[
y(r)=V_{0} \int_{0}^{+\infty} G(r, \xi) \frac{e^{-\xi}}{\xi} y(\xi) d \xi
\]

where

\[
G(r, \xi)= \begin{cases}\frac{1}{k} e^{-k \xi} \operatorname{sh} k r, & 0 \leqslant r \leqslant \xi \\ \frac{1}{k} e^{-k r} \operatorname{sh} k \xi, & \xi \leqslant r \leqslant+\infty\end{cases}
\]

## Chapter II Application of Integral Transforms to the Solution of Integral Equations

## § 17. Application of Fourier Transform to the Solution of Some Integral Equations

Let the Fredholm integral equation of the second kind with a kernel depending on the difference of the arguments be given:

\[
\varphi(x)=f(x)+\int_{-\infty}^{+\infty} K(x-t) \varphi(t) d t
\]

where $f(x) \in L_{2}(-\infty,+\infty)$ and $K(x) \in L_{2}(-\infty,+\infty)$.

Applying the Fourier transform and using the convolution theorem, we obtain

\[
\Phi(\omega)=F(\omega)+\sqrt{2 \pi} \Phi(\omega) \tilde{K}(\omega)
\]

where $\Phi(\omega), F(\omega), \tilde{K}(\omega)$ are the Fourier transforms of the functions $\varphi(x), f(x), K(x)$, respectively. From the equality (2), under the condition $1-\sqrt{2 \pi} \tilde{K}(\omega) \neq 0$, we find

\[
\Phi(\omega)=\frac{F(\omega)}{1-\sqrt{2 \pi} \tilde{K}(\omega)}
\]

Applying the inverse Fourier transform formula, we obtain the solution of the equation (1):

\[
\varphi(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \frac{F(\omega)}{1-\sqrt{2 \pi} \widetilde{K}(\omega)} e^{i z \omega} d \omega
\]

If $1-\sqrt{2 \pi} \tilde{K}(\omega)=0$ for some real values of $\omega$, then the equation (1), generally speaking, does not have an absolutely integrable solution on the entire axis $O x$.

Similarly, the Fredholm integral equation of the first kind with a kernel depending on the difference of the arguments:

\[
\int_{-\infty}^{+\infty} K(x-t) \varphi(t) d t=f(x)
\]

has a solution of the form

\[
\varphi(x)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} \frac{F(\omega)}{\widetilde{K}(\omega)} e^{i x \omega} d \omega
\]