I don't understand part of Example 2. Vibrating String of the book PARTIAL DIFFERENTIAL EQUATIONS AN INTRODUCTION by Walter A. Strauss.
Consider a flexible, elastic homogenous string or thread of length $l$, which undergoes relatively small transverse vibrations. For instance, it could be a guitar string or a plucked violin string. Let $ u(x, t) $ be its displacement from equilibrium position at time $t$ and position $x$. Because the string is perfectly flexible, the tension (force) is directed tangentially along the string. Let $T(x, t)$ be the magnitude of this tension vector. Let $\rho$ be the density (mass per unit length) of the string. It is a constant because the string is homogeneous. We shall write down Newton’s law for the part of the string between any two points at $x = x_{0}$ and $x = x_{1}$. The slope of the string at $ x_{1} $ is $ u_{x}(x_{1}, t) $. Newton’s law $ F = ma $ in its transverse $(u)$ component is
Force acting on the string between the points $x_{0}$ and $x_{1} = T \sin \theta \vert _{x_{0}} ^{x_{1}} = T\dfrac{ u_{x}}{\sqrt{1 + u_{x}^{2}}} \vert _{x_{0}} ^{x_{1}} = \int_{x_{0}} ^{x_{1}} \rho u_{tt} dx $
The right side is the component of the mass times the acceleration integrated over the piece of string.
Why is the right side integrated over the piece of string if mass times the acceleration is just $\rho \, u_{tt} \Delta x$?
I appreciate any suggestion.
The force on the segment $(x_0,x_1)$ is a function of time only, $F(t)$. Hence, $ma$ should be the same kind of function. This is your first hint that $\rho \Delta x \cdot u_{tt}(x,t)$ is not the mass-times-acceleration for the entire segment, since it is dependent on $x$.
When dealing with an object like a string, we consider it as a collection of "point" masses, and we need to add up the effect of each point mass. Since the length of each of the "point" segments is infinitely small, we call it $dx$, and since their mass is density-times-length, $m=\rho dx$. Finally, acceleration is $u_{tt}(x,t)$, so for each "point" on the string, the effect of mass-times-acceleration that it has is $\rho u_{tt} dx$. Since we want to consider the total effect across the whole segment from $x_0$ to $x_1$, we add up all the pieces. Of course, when adding up infinitely many, infinitely small pieces, we call it integration, so we write $(ma)_{segment} = \int_{x_0}^{x_1}\rho u_{tt} dx$.