Wave and Elasticity Equations

Transcription

1 1 Wave and lasticity quations Now let us consider the vibrating string problem which is modeled by the one-dimensional wave equation. Suppose that a taut string is suspended by its extremes at the points (0g 0 ) and (1g 1 ) (Figure 0.0.1) and that a vertical force of f(xt) per unit length is applied to it at (xu(xt)) where u(xt) represents the vertical position of the string. Figure We will assume that u x is small for all x and t. This assures that the string is never stretched much from its unloaded length and therefore the string tension can be taken to be a constant T. Figure shows the forces acting on a small segment of the string between x and x+ x. If the angle between the string tangent and the horizontal is called α(xt) then setting the sum of the vertical components of the forces acting on the string segment equal to the product of the mass and vertical acceleration of that segment gives ρ x u tt = T sinα(x+ xt) T sinα(xt)+ xf(xt) where ρ is the linear density (mass per unit length) of the string. Figure Since u x is assumed to be small we can take and thus sinα(xt) tanα(xt) = u x (xt) ρu tt = T u x(x+ xt) u x (xt) x +f(xt).

2 2 Finally taking the limit as x 0 gives the wave equation ρu tt = Tu xx +f. (0.0.1) There may also be a frictional force acting on this string segment that is proportional to the velocity and in an opposing direction. This introduces a term au t intotheright-handside andwehavethe damped waveequation: ρu tt = au t +Tu xx +f. (0.0.2) The multidimensional version of this equation is ρu tt = au t +T 2 u+f. (0.0.3) In two dimensions it models the displacement u(xyt) of an elastic membrane. Theboundaryconditionsfor0.0.1or0.0.2maybeu(0t) = g 0 u(1t) = g 1 where g 0 and g 1 may possibly be functions of t. For u may be specified around the entire boundary but other boundary conditions are possible. Since the wave equation is second order in time both the initial position and initial velocity of each portion of the string (or membrane) must be specified so the initial conditions are u = h at t = 0 u t = q at t = 0. The equations that model displacements in an elastic body such as a metal object can be similarly derived using Newton s second law. If we focus on an arbitrary small volume of this body and set the mass times the acceleration

3 3 equal to the sum of the internal and external forces acting on we get ρu tt dx dy dz = (σ xx n x +σ xy n y +σ xz n z ) da + f 1 dx dy dz ρv tt dx dy dz = (σ yx n x +σ yy n y +σ yz n z ) da + f 2 dx dy dz ρw tt dx dy dz = (σ zx n x +σ zy n y +σ zz n z ) da + f 3 dx dy dz. (0.0.4) Here ρ(xyz) is the density and (u(xyzt) v(xyzt) w(xyzt)) is the displacement vector. That is the body element that is at (xyz) when the elastic body is unloaded and in equilibrium is displaced to (x+u y+v z+w) at time t. Thus (u tt v tt w tt ) is the acceleration vector. The stresses σ pq are the elements of a stress tensor (matrix) defined so that σ xx σ xy σ xz σ yx σ yy σ yz σ zx σ zy σ zz n x n y n z gives the force per unit area that the rest of the body exerts on a boundary element of that has unit outward normal (n x n y n z ). Thus σ pq (pq = xy orz)canbeinterpretedasthecomponentinthep-axisdirectionoftheinternal force experienced by a boundary element of unit area that is perpendicular to the q-axis. The vector (f 1 (xyzt) f 2 (xyzt) f 3 (xyzt)) represents the force per unit volume attributable to external sources such as gravity. After the divergence theorem is applied to the boundary integrals in 0.0.4

4 4 we get ρu tt dx dy dz = ρv tt dx dy dz = ρw tt dx dy dz = (σ xx ) x +(σ xy ) y +(σ xz ) z +f 1 dx dy dz (σ yx ) x +(σ yy ) y +(σ yz ) z +f 2 dx dy dz (σ zx ) x +(σ zy ) y +(σ zz ) z +f 3 dx dy dz. Since is an arbitrary volume it can be made so small that the integrands are almost constant so that at any point ρu tt = (σ xx ) x +(σ xy ) y +(σ xz ) z +f 1 ρv tt = (σ yx ) x +(σ yy ) y +(σ yz ) z +f 2 (0.0.5) ρw tt = (σ zx ) x +(σ zy ) y +(σ zz ) z +f 3. For a linear isotropic (direction independent) elastic body it has been determined experimentally that σ xx = [(1 µ)u x +µv y +µw z ] σ yy = [µu x +(1 µ)v y +µw z ] σ zz = [µu x +µv y +(1 µ)w z ] σ xy = σ yx = [u y +v x ] 2(1+µ) σ xz = σ zx = [u z +w x ] 2(1+µ) σ yz = σ zy = [v z +w y ] 2(1+µ) where and µ are material properties called the elastic modulus and the Poisson ratio respectively. Therefore for such a body represents a system of second-order partial differential equations whose properties are qualitatively similar to those of the wave equation On the boundary normally either the displacement vector (u v w) is specified or else an applied external boundary force (per unit area) vector (g 1 (xyzt) g 2 (xyzt) g 3 (xyzt)) is given. The boundary conditions that model the latter situation are obtained by balancing the internal and

5 5 external forces on the boundary: σ xx σ xy σ xz n x g 1 σ yx σ yy σ yz n y = g 2 σ zx σ zy σ zz n z g 3 At t = 0 the initial displacements (uvw) and the initial displacement velocities (u t v t w t ) must be specified throughout the body.