Determine of the wave equation in the task of electrical oscillations

One of the basic equations of mathematical physics (for instance function of two independent variables) is the differential equation with partial derivatives of the second order (3). This equation is called the wave equation, and is provided when considering the process of transverse oscillations of wire, longitudinal oscillations of rod, electrical oscillations in a conductor, torsional vibration at waves, etc... The paper shows how to form the equation (3) which is the equation of motion of each point of wire with abscissa x in time t during its oscillation. It is also shown how to determine the equation (3) in the task of electrical oscillations in a conductor. Then equation (3) is determined, and this solution satisfies the boundary and initial conditions. 1 Formation of a wave equation In mathematical physics, implied under a wire is a thin elastic thread. Let the wire of the length l at the initial moment match with O from O to l. Assume that the ends of the wire are fixed at points x=0 and x=1. If the wire is taken from its original position and then released, or without moving the wire, we provide a speed at its initial points, then the wire points will move the wire starts to oscillate, flickers (oscillates). The task is to determine the shape of the wire at any time and to determine the law of motion of each point of the wire with the abscissa x at the moment t. When the wire oscillates the deviations of the wire points from the initial position are small, on the Ox axis and are in one plane. The oscillations are determined by the function in u(x,t), which gives the displacement value of the wire points from the abscissa x at the moment t (Fig.1). Fig. 1. Displacement value of the wire points When we observe small wire deviations in the plane (x, u), we will assume that the length of the wire element is equal to its projection on the axis Ox, i.e. 2 1M M =x2x1. We also assume that the tension at all points of the wire is the same and is denoted by T. Consider the wire element MM' (Fig. 2). Fig. 2. Wire element MM' At the ends of this element, at the tangents, the force T acts on the wire. Let the tangents form with Ox axes the angles and     . Then the projection on the axis of the Ou of the forces acting on the element MM', will be equal to    sin ) sin( T T    .. When the angle  is small, we can put that   sin  tg , so we have:               Ttg Ttg T T ) ( sin ) sin(                    x x t x x u T x t x u x t x x u T 2 2 ) , ( ) , ( ) , (  , ) , ( 2 2 x x t x u T     1 0   (1) © The Authors, published by EDP Sciences. This is an open access article distributed under the terms of the Creative Commons Attribution License 4.0 (http://creativecommons.org/licenses/by/4.0/). MATEC Web of Conferences 184, 01023 (2018) https://doi.org/10.1051/matecconf/201818401023 Annual Session of Scientific Papers IMT ORADEA 2018


Formation of a wave equation
In mathematical physics, implied under a wire is a thin elastic thread. Let the wire of the length l at the initial moment match with O from O to l. Assume that the ends of the wire are fixed at points x=0 and x=1. If the wire is taken from its original position and then released, or without moving the wire, we provide a speed at its initial points, then the wire points will move -the wire starts to oscillate, flickers (oscillates). The task is to determine the shape of the wire at any time and to determine the law of motion of each point of the wire with the abscissa x at the moment t.
When the wire oscillates the deviations of the wire points from the initial position are small, on the Ox axis and are in one plane. The oscillations are determined by the function in u(x,t), which gives the displacement value of the wire points from the abscissa x at the moment t ( Fig.1).

Fig. 1. Displacement value of the wire points
When we observe small wire deviations in the plane (x, u), we will assume that the length of the wire element is equal to its projection on the axis Ox, i.e. 2 1 M M =x2-x1. We also assume that the tension at all points of the wire is the same and is denoted by T.
Where Lagrange's theorem is applied in the expression in square brackets.
In order to get the equation of motion, it is necessary to equalize the external forces acting on the element with the force of inertia. Let  be the specific weight of the wire. Then the mass of the wire element will be x. The element acceleration equals  2 u/t 2 . According to D'Alembert's principle, we obtain: ( If we shorten this equation with x  and put T/=a 2 , we get the following differential equation with partial derivatives of the second order [1-6]: It is a wave equation -the equation of oscillation of a wire. The equation (3) itself is insufficient to fully determine the motion of the wire. The required function in u(x, t) should satisfy even boundary conditions that determine what happens at the ends of the wire x=0 and x=1, and the initial conditions describing the position of the wire at the initial moment t=0.
Let, for example, as we have assumed, the ends of the wire in x=0 and x=l be immobile. Then for any t the equalities are exact: These equalities are boundary conditions in the task. In the initial moment t=0 the wire has a certain shape, which we have given it. Let this form be determined by the function f(x). Therefore, there must be: Further, the speed needs to be given in the starting point at each point of the wire, which is determined by the function The conditions (5) and (6) are the initial conditions. As indicated above, equation (3) is also obtained in the task of electrical oscillations in the conductor. This can be shown in the following way. The electrical current in the conductor is characterized by the size i(x,t) and the tension v(x,t), which depend on the coordinate of the conductor point h and on the time t. Observing the conductor element x  , we can write that the decrease in the tension on the element . This decrease in tension is equal to the sum of resistance where R and L are coefficients of resistance and induction on the unit of conductor length. The minus sign should be taken because the current flows in the opposite direction from growth of v. By shortening it with x  , we get the equality: (8) Furthermore, the difference in currents passing It is consumed on charging the element, it equals Cx(v/t)t, and on the outflowing through the side surface of the conductor due to the imperfection of the insulation, equaling t x Av   (here A -is the coefficient of outflow). By equating these expressions and shortening with t x  , we get: The equality (8) and (10) are called telegraphic equations.
Equations (8) and (10) can be obtained from the equation system, which contains only the required function i(x,t), and the equations, which contains only the required function v(x,t). Differential equality (10) by x; equation (8) is differentiated by t, multiplied by S and substract the two values. We obtain: . 0 (11) From (8) we express v/x and replace it in (11). We get: Analogously, we obtain the equation that determines v(x,t).
Differential equality (8) by x; equation (10) is differentiated by t, multiplied by L and substract the two values. We obtain: From the equation (10) we express i/x and replace it in (13). We get: If we can neglect the loss through insulation (A=0) and resistance (R=0), then the equations (12) We will seek the solution of the equation (16), which satisfies the boundary conditions (17) and (18) in the form of products of the two functions H(h) and T(t), the first of which depends on x, and the second on t: By substitution in the relation (10), we obtain: We divide the equation with XT a 2 and we obtain: and thus we have separated the variables.
On the left side of the equation is a function, which does not depend on x, and on the right the function which does not depend on t. The equation (20) From this equation we obtain two linear differential equations with constant coefficients: The general solutions of these equations are: where A, B, C, D are the production constants.
By putting the expressions X(x) and T(t) into equality (19), we obtain: Let's find the constants A and B so that they satisfy the conditions (17) and (18). Since T(t)  0 (otherwise it , which would be contrary to the set condition), then the function X(x) satisfies conditions (17) and (18), i.e. X(0)=0, X (l)= 0. Replacing the values x=0 and x=l in equality (23), on the basis of (17) we obtain: The first equality gives A=0. From the second equality follows: