1208. Find the particular solution $u(x, t)$ of the differential equation $\frac{\partial^{2} u}{\partial t^{2}}-a^{2} \frac{\partial^{2} u}{\partial x^{2}}=0$, satisfying the boundary conditions:

$\begin{array}{ll}\text { 1) } u(0, t)=0, & \text { 2) } u(l, t)=0,\end{array}$
3) $u(x, 0)=\varphi_{1}(x)$
4) $u_{t}(x, 0)=\varphi_{2}(x)$.
Solution. By the Fourier method, we first look for particular solutions of the given equation in the form of the product of two functions, one of which depends only on $x$, and the other only on $t$:

$$
u(x, t)=X(x) T(t)
$$

Finding the derivatives $u_{x x}^{\prime \prime}=T X_{x x}^{\prime \prime}, u_{t t}^{\prime \prime}=X T_{t t}^{\prime \prime}$ and substituting them into the given equation, we get

$$
X T^{\prime \prime}-a^{2} T X^{\prime \prime}=0, \quad \text { or } \quad \frac{X^{\prime \prime}}{X}=\frac{T^{\prime \prime}}{a^{2} T}
$$

In the last equality, the variables are separated. The left part does not depend on $t$, and the right part does not depend on $x$. This is possible only if both parts of the equality do not depend on $\ell$ or $x$, i.e., they represent the same constant. Denoting this constant by $-\lambda^{2}$, we obtain two ordinary differential equations:

$$
\frac{X^{\prime \prime}}{X}=-\lambda^{2} \quad \text { and } \quad \frac{T^{\prime \prime}}{a^{2} T}=-\lambda^{2}
$$

or

$$
X^{\prime \prime}+\lambda^{2} X=0 \quad \text { and } \quad T^{\prime \prime}+\lambda^{2} a^{2} T=0
$$

Solving these as linear homogeneous equations with constant coefficients (§ 7), we find

$$
X=A \cos \lambda x+B \sin \lambda x ; \quad T=C \cos a \lambda t+D \sin a \lambda t
$$

where $A, B, C, D$ are arbitrary constants.

Substituting these expressions for $X$ and $T$ into equation (1), we get

$$
u(x, t)=(A \cos \lambda x+B \sin \lambda x)(C \cos u \lambda t+D \sin a \lambda t)
$$

Next, based on the given boundary conditions, we determine the values of the constants. Substituting into equation (2) the given values $x=0, u=0$ (first condition) and $x=-l, u=0$ (second condition) and canceling the factor $T(t) \neq 0$, we obtain

$$
0=A \cos 0+B \sin 0, \quad 0=A \cos \lambda l+B \sin \lambda l
$$

From the first equation, we find $A=0$, and from the second, $\sin \lambda l=0$ (since $B \neq 0$ when $A=0$), from which the parameter
$\lambda=\frac{n \pi}{l}, n=1,2,3, \ldots$, which was also arbitrary *.

Each value of $\lambda$ (or $n$) corresponds to a particular solution of the form

$$
u_{n}=X_{n} T_{n}=\left(x_{1:} \cos \frac{a n \pi t}{l}+\beta_{n} \sin \frac{a n \pi t}{l}\right) \sin \frac{n \pi x}{l}
$$

where $\alpha_{n}=B_{n} C_{n}, \beta_{n}=B_{n} D_{n}$ are arbitrary constants.

Due to the linearity and homogeneity of the given equation, the sum of its solutions is also a solution. Therefore, the sum of the series

$$
u(x, t)=\sum_{n=1}^{+\infty} u_{n}=\sum_{n=1}^{+\infty}\left(\alpha_{n} \cos \frac{a n \pi t}{l}+\beta_{n} \sin \frac{a n \pi t}{l}\right) \sin \frac{n \pi x}{l}
$$

is a solution of the given equation, satisfying conditions 1) and 2).

To determine the constants $\alpha_{n}$ and $\beta_{n}$, we use the last two boundary conditions. Substituting into equation (3) $t=0$, $u=\varphi_{1}(x)$ (third condition), we get

$$
P_{1}(x)=\sum_{n=1}^{+\infty} \alpha_{n} \sin \frac{n \pi x}{l}
$$

Differentiating equation (3) with respect to $t$

$$
\frac{\partial u}{\partial l}=\sum_{n=1}^{+\infty} \frac{a n \pi}{l}\left(\beta_{n} \cos \frac{a n \pi t}{l}-\alpha_{n} \sin \frac{a n \pi t}{l}\right) \sin \frac{n \pi x}{l}
$$

and substituting into the result $t=0, \frac{\partial u}{\partial t}=\varphi_{2}(x)$ (fourth condition), we get

$$
\Gamma_{2}(x)=\sum_{n=1}^{+\infty} \frac{a n \pi}{l} \beta_{n} \sin \frac{n \pi x}{l}
$$

Equalities (4) and (5) represent the expansions of the given functions $\varphi_{1}(x)$ and $\varphi_{2}(x)$ in the interval $(0, l)$ into incomplete Fourier series containing only sines. The coefficients of such expansions are determined by the known formula, Chapter IX, § 7:

$$
\text { If } f(x)=\sum_{n=1}^{+\infty} b_{n} \sin \frac{n \pi:}{l}, \text { then } b_{n}=\frac{2}{l} \int_{0}^{l} f(x) \sin \frac{n \pi x}{l} d x
$$[^52]

According to this formula

$$
\alpha_{n}=\frac{2}{l} \int_{0}^{l} \psi_{1}(x) \sin \frac{n \pi x}{l} d x, \quad \beta_{n}=\frac{2}{n a \pi} \int_{i}^{i} \digamma_{2}(x) \sin \frac{n \pi x}{l} d x
$$

Therefore, the sought particular solution of the given equation, satisfying the specified boundary conditions, is the function (3), where the constants $\alpha_{n}$ and $\beta_{n}$ are determined by formulas (6).

Obviously, for different initial data $a, l, \varphi_{1}(x)$, $\varphi_{2}(x)$, different values for $\alpha_{n}$ and $\beta_{n}$ will be obtained by formulas (6), and consequently, different series (3) for the function $u(x, t)$, satisfying the given differential equation and boundary conditions.

A physical interpretation of the solution to this problem, for example, can be as follows. A stretched string, fixed at the points $x=0$ and $x=l$ on the $O x$ axis, at the initial moment of time $t=0$ had the shape of the curve $u=\varphi_{1}(x)$, and each of its points with abscissa $x$ had the velocity $u_{t}^{\prime}=\varphi_{2}(x)$; then this string, left to itself, oscillates, remaining in the plane $x O u$. The given equation is the differential equation of transverse oscillations of the string. (The parameter $a^{2}=\frac{h}{\beta}$, where $h$ is the tension, $\rho$ is the density of the string.) The found solution $u(x, t)$ determines the shape of the string at any moment of time $t$.