Strauss PDEs 2e: Section Exercise 1 Page 1 of 6

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1 Strauss PDEs 2e: Section 3 - Exercise Page of 6 Exercise Carefully derive the equation of a string in a medium in which the resistance is proportional to the velocity Solution There are two ways (among others) to go about this problem One is the integral formulation, where we consider the forces acting over a finite portion of the string The other is the differential formulation, where we consider the forces acting on an infinitesimal element of the string In both cases we come to the same governing PDE, so use whichever you prefer The Integral Formulation Figure : Schematic of the string (integral formulation) In order to derive the equation of motion, we will invoke Newton s second law, which states that the sum of the forces acting on a body is equal to its mass times its acceleration Mathematically this is written as F = ma Note that this is a vector equation; in other words, there is a separate equation for each component of force and corresponding component of acceleration For this problem we will choose the coordinate system as shown in the figure, so there are two equations of significance Fx = ma x Fu = ma u There are two forces acting on this string, T at x = and T at x = x, in addition to the resistive force of the medium The motion of the string is entirely vertical, which means there is no horizontal component of acceleration (a x = 0) the resistive force of the medium will only act vertically The tensions, on the other hand, have components in both the x-direction and u-direction and have to be resolved using cosine and sine, respectively Horizontal component of T at x = : T cos θ 0 Horizontal component of T at x = x : +T cos θ Vertical component of T at x = : T sin θ 0 Vertical component of T at x = x : +T sin θ,

2 Strauss PDEs 2e: Section 3 - Exercise Page 2 of 6 where θ 0 and θ are the angles between the vectors and the x-axis at and x, respectively To determine θ it is necessary to note that tan θ is equal to rise over run, the slope If the height of the string is u(x, t), the slope is given by u/x = u x As shown in Figure, the hypotenuse can be determined using Pythagorean s theorem And now cos θ can be written in terms of u Newton s second law in the x-direction is thus Fx = T cos θ 0 cos θ = ma x = 0 T + u 2 x x=x0 + u 2 x x=x = 0 T + u 2 x x=x0 = T + u 2 x x=x small for all x and t The binomial theorem tells us that + u 2 x = + ( ) u 4 Compared to, u 2 x and all higher powers of u x can be considered negligible Approximating the square root terms as, the equation of motion simplifies to T x=x0 T x=x This equation tells us that the magnitude of the tension at is equal to the magnitude of the tension at x ; that is, T is constant Let s move on to Newton s second law in the u-direction If the height of the string is u(x, t), the rate of change of the height with respect to time, the velocity, is given by u/t = u t Consequently, the rate of change of velocity with respect to time, the acceleration, is 2 u/t 2 = u tt As said in the problem statement, the resistive force is proportional to velocity Mathematically this can be expressed as F r resistive force proportional to u t velocity To change this proportionality to an equation we can use, we have to introduce a constant of proportionality Let s call it r The minus sign indicates that the force acts in the direction that opposes the motion F r = ru t Understand that this force is applied at a specific point x at time t on the string To get the total force acting on the string from x = to x = x, we have to sum (integrate) the resistive forces (F r ) from to x Think of F r as a distributed force that we have to integrate over x to get the total of Note, too, that u tt is the acceleration at a specific point on the string, x, at time t To get the force we therefore have to multiply u tt by a tiny bit of mass dm Mass, of course, is density times length, so this can be written in terms of arc length, s, as dm = ρ ds The total force is obtained by integrating u tt dm over the mass of the string Newton s second law in the u-direction is thus ˆ x ˆ Fu = T sin θ 0 sin θ + F r dx = u tt dm mass of string

3 Strauss PDEs 2e: Section 3 - Exercise Page 3 of 6 Write θ in terms of u using the right triangle in Figure and substitute dm = ρ ds T u x + u 2 x x=x0 ˆ u x x ru t dx = + u 2 x x=x ˆ length of string ρu tt ds = ˆ x ρu tt + u 2 x dx small for all x and t The binomial theorem tells us that + u 2 x = + ( ) u 4 Compared to, u 2 x and all higher powers of u x can be considered negligible Approximating the square root terms as, the equation of motion simplifies to T u x (x, t) T u x (, t) According to the fundamental theorem of calculus, so the left side can be written as ˆ b a ˆ x ru t dx f(x) dx = F (b) F (a), T u x (x, t) T u x (, t) = ˆ x ˆ x ˆ x x (T u x) dx [ x (T u x) ru t Thus, the integrands must be equal to each other ˆ x ru t dx = ] dx = ˆ x x (T u x) dx ˆ x ˆ x ρu tt dx ρu tt dx ρu tt dx The tension is constant, so Therefore, where 2k = r/ρ and c 2 = T/ρ x (T u x) ru t = ρu tt T u xx ru t = ρu tt u tt + 2ku t = c 2 u xx,

4 Strauss PDEs 2e: Section 3 - Exercise Page 4 of 6 The Differential Formulation Figure 2: Schematic of the string (differential formulation) In order to derive the equation of motion, we will invoke Newton s second law, which states that the sum of the forces acting on a body is equal to its mass times its acceleration Mathematically this is written as F = ma Note that this is a vector equation; in other words, there is a separate equation for each component of force and corresponding component of acceleration For this problem we will choose the coordinate system as shown in the figure, so there are two equations of significance Fx = ma x Fu = ma u There are two forces acting on this string, T at x and T at x, in addition to the resistive force of the medium The motion of the string is entirely vertical, which means there is no horizontal component of acceleration (a x = 0) the resistive force of the medium will only act vertically The tensions, on the other hand, have components in both the x-direction and u-direction and have to be resolved using cosine and sine, respectively Horizontal component of T at x: T cos θ x Horizontal component of T at x: +T cos θ x+ x Vertical component of T at x: T sin θ x Vertical component of T at x: +T sin θ x+ x, where θ x and θ x+ x are the angles between the vectors and the x-axis at x and x, respectively To determine θ it is necessary to note that tan θ is equal to rise over run, the slope If the height of the string is u(x, t), the slope is given by u/x = u x As shown in Figure 2, the hypotenuse can be determined using Pythagorean s theorem And now cos θ can be written in terms of u Newton s second law in the x-direction is thus Fx = T cos θ x cos θ x+ x = ma x = 0 T + u 2 x x + u 2 x x+ x = 0

5 Strauss PDEs 2e: Section 3 - Exercise Page 5 of 6 T + u 2 x x = T + u 2 x x+ x small for all x and t The binomial theorem tells us that + u 2 x = + ( ) u 4 Compared to, u 2 x and all higher powers of u x can be considered negligible Approximating the square root terms as, the equation of motion simplifies to T x T x+ x This equation tells us that the magnitude of the tension at x is equal to the magnitude of the tension at x; that is, T is constant Let s move on to Newton s second law in the u-direction If the height of the string is u(x, t), the rate of change of the height with respect to time, the velocity, is given by u/t = u t Consequently, the rate of change of velocity with respect to time, the acceleration, is 2 u/t 2 = u tt As said in the problem statement, the resistive force is proportional to velocity Mathematically this can be expressed as F r resistive force proportional to u t velocity To change this proportionality to an equation we can use, we have to introduce a constant of proportionality Let s call it r The minus sign indicates that the force acts in the direction that opposes the motion F r = ru t Understand that this force is applied at a specific point x at time t on the string To get the total force acting on the string from x to x, we have to sum (integrate) the resistive forces (F r ) from x to x Think of F r as a distributed force that we have to integrate over x to get the total of Note, too, that u tt is the acceleration at a specific point on the string, x, at time t To get the force we therefore have to multiply u tt by a tiny bit of mass m Mass, of course, is density times length, so this can be written in terms of arc length, s, as m = ρ s Newton s second law in the u-direction is thus Fu = T sin θ x sin θ x+ F r dx = u tt m Write θ in terms of u using the right triangle in Figure 2 and substitute m = ρ s T u x + u 2 x x u x + u 2 x x+ x x x ru t dx = ρu tt s = ρu tt + u 2 x x small for all x and t The binomial theorem tells us that + u 2 x = + ( ) u 4

6 Strauss PDEs 2e: Section 3 - Exercise Page 6 of 6 Compared to, u 2 x and all higher powers of u x can be considered negligible Approximating the square root terms as, the equation of motion becomes T u x ( x, t) T u x (x, t) x ru t dx ρu tt x Now take the limit of both sides as x 0 The integrand is constant from x to x in the limit, so it can be pulled out in front of the integral So lim [T u x( x, t) T u x (x, t)] lim ru t dx = lim ρu tt x x Now divide both sides by x lim [T u x( x, t) T u x (x, t)] ru t lim x = ρu tt According to the definition of the derivative, T u x ( x, t) T u x (x, t) lim ru t = ρu tt x lim x The tension is constant, so Therefore, where 2k = r/ρ and c 2 = T/ρ df dx = lim f( x) f(x) x x (T u x) ru t = ρu tt T u xx ru t = ρu tt u tt + 2ku t = c 2 u xx, PS The solution to this PDE is quite involved but very interesting It can be found in the solution to exercise 2 of section 33 in Asmar s Partial Differential Equations with Fourier Series and Boundary Value Problems