The following example (Hubbard and West 1991, p.159) shows that in some physical

Question

The following example (Hubbard and West 1991, p.159) shows that in some physical situation, non-uniqueness is natural and obvious, not pathological. Consider a water bucket with a hole in the bottom. If you see an empty bucket, with a puddle beneath it, can you figure out when the bucket was full? No, of course not! It could have finished emptying a minute ago, ten minutes ago, or whatever. The solution to the corresponding differential equation must be non-unique when integrated backwards in time. Here's a crude model of the situation. Let h(t) = height of the water remaining in the bucket at time t; a = area of the hole; A = cross-sectional area of the bucket (assumed constant); v(t) = velocity of the water passing through the hole. a)Show that a nu(t) = Ah(t). What physical law are you invoking? b) To derive an additional equation, use conservation of energy First find the change in potential energy in the system, assuming that the height of the water in the bucket decreases by an amount Delta h and that the water has density rho. Then find the kinetic energy transported out of the bucket by the escaping water. Finally assuming all the potential energy is converted into kinetic energy, derive the equation v^2 = 2gh. c) Combining (b) and (c), show h = C Squareroot h, where C = Squareroot 2g (a/A). d) Given h(0) = 0 (bucket empty at t = 0) show that the solution for h(t) is non-unique in backwards time, i.e., for t

Explanation / Answer

part a:

as water is incompressible, volume of water leaked=volume of water reduced from the bucket

hence area of the hole*speed of water=cross section of water in the bucket*rate of change of height of water

==>a*v(t)=A*(dh/dt)...(1)

part b:

potential energy decreased =mass of water of decreased volume*g*decrease in height

increase in kinetic energy =0.5*mass *speed^2

so equating both,

0.5*mass*v^2=mass*g*h

==>v^2=2*g*h

==>v=-sqrt(2*g*h)

part c:

combining equation 1 and equation 2:

a*v=A*(dh/dt)

==>dh/dt=(a/A)*v=-(a/A)*sqrt(2*g*h)=-C*sqrt(h)

where C=sqrt(2*g)*(a/A)

part d:

dh/dt=-C*sqrt(h)

==>dh/sqrt(h)=-C*dt

integrating both sides:

2*sqrt(h)=-C*t+C1

where C1 is a constant

at t=0, h=0

==>C1=0

so equation becomes:

2*sqrt(h)=-C*t

==>h=0.25*C^2*t^2

as for both t<0 and t>0 , value of h is same, the solution is not unique in nature.

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