Problem 4. In this problem, we aim to connect chain rule and the divergence theorem through a problem in physics. Assume that p(t, x, y, z) denote the density of a fluid at the point (x, y, z) ER3 at time t. Assume that this fluid moves with the velocity V = (Vx, Vy, Vz). We would like to explore the change of density along its flow line which is induced by the velocity field V. Let's consider a control volume E. As we know from our previous lectures, the total mass in E at time t is derived by the integral: M (t ) = ]] p (t, x, y, z ) aV . a) The rate of change of M(t) is M (t ) - at // p(t, x, y, 2 ) av - JJ p (t, x, y, z ) dV. The conservation of mass indicates that this change is equal to the net amount of mass passing through the boundary bud(E), that is, d M (t) = - fond(E) Pond ( B P (t , x, y, z ) Vinds, where n is the exterior unit vector on the boundary bud(E). For example, the rate of change of the students in a class is equal to the rate of students that enter the class minus the number who leave the class. Use the divergence theorem and obtain the equation (/ { pitdiv ( PV) } dV= 0. It turns out that the equality holds for the following expression: pt +div (p V) =0. Conclude the following equation at P (t, 20 ( t ) , y (t ) , z ( t ) ) = - pdiv (V ), where y(t) = (x(t), y(t), z(t)) is the trajectory of the control volume E induced by the velocity field V. The left hand side is the change of mass in E along the trajectory y(t).b) Conclude: .) if div (V) =0, then p(t, x(t), y(t), z(t)) = p(0, x(0), y(0), z(0)) .) if div (V) > 0, then p(t, a(t), y(t), z(t)) p(0, x(0), y(0), z (0)). Thus the rate of change of p along the flow lines depends on the divergence of the velocity. Consider the following 2D velocity filed V =(-y-0.1x, x - 0.ly). The filed is shown below: 05 -0.5 -0.5 0.5 Run the following code in Matlab and explain your observation. In particular, explain the change of density, and explain any type of circulation or rotation you observe. a=0. 1; [r, th] =meshgrid (0:0. 01:0. 1,0:pi/10:2*pi) ; x=0. 5+r. *cos (th) ; y=0. 5+r. *sin(th) ; t=0:pi/4:4*pi; phi=@(t) exp(-a*t)*[cos(t), -sin(t) ; sin(t) , cos(t)]; for i=1: length (t) S=phi (t (i) ); xt=S(1, 1)*x+S(1, 2)*y; yt=S(2, 1) *x+S(2, 2)*y; ut=exp (2*a*t (i) ) . *exp(-x. ~2-y . -2) ; surf (xt , yt, ut) ; view(2), colormap(flipud (hot) ); axis equal tight; shading interp, axis ([-1 1 -1 1]); clim([0 2.5]); grid off; hold on pause (0.2) ; end