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Simple Differential Equation Modeling

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data
    I am trying to model using differential equations rain falling from the sky, hitting the pavement, and then running off the pavement into the ground. I'm a little rusty on differential equations and I'm just wondering if my answer is correct. In this model I'm assuming that the rain is only landing on the pavement, not on the ground around the pavement. I'm also assuming that the source of the rain will never run out.


    3. The attempt at a solution
    [itex]\frac{dS}{dt}[/itex]=c-P-G

    [itex]\frac{dP}{dt}[/itex]=P-G

    [itex]\frac{dG}{dt}[/itex]=-S-P

    [itex]\frac{dS}{dt}[/itex] is the amount of rain in the sky per unit time

    [itex]\frac{dP}{dt}[/itex] is the amount of rain on the pavement per unit time

    [itex]\frac{dG}{dt}[/itex] is the amount of rain on the ground per unit time

    c is the constant source of rain
    P is the rain on the pavement
    G is the rain on the ground
    S is the rain in the sky

    Does this look correct or is it completely off? Thanks!
     
  2. jcsd
  3. Apr 11, 2014 #2

    Simon Bridge

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    I think you need to be more clear about the model you are using - it looks like you are modelling rainfall as if there is a reservoir of water in the sky that drizzles out somehow - falls to the ground and runs away to some unspecified place.

    You can check the model by putting the DEs into normal language and see if they say what you hoped.

    S(t) is the rain in the sky, then dS/dt is the rate of change of the rain in the sky ... so it would be how hard it is raining?

    The 1st DE seems to be saying that the rate of rainfall increases as "the source of rain" which you don't define (it looks like a rate the S(t) increases ... so it would be water flowing into the reservoir? What?)
    It also decreases as "the amount of water on the pavement" and "the amount of water on the ground".
    So the more rain that is on the ground and on the pavement the slower it falls from the sky?

    Does this make sense?
     
  4. Apr 12, 2014 #3
    I see what you're saying. I think I want dS/dt to be how hard it is raining, which would be constant. Taking that into account, does this look right?

    dS/dt= c

    dP/dt= c - G

    dG/dt= P

    dS/dt is how hard it is raining, for example inches per minute
    dP/dt is how much rain is on the pavement per unit time
    dG/dt is how much rain is on the ground per unit time
    c is constant
    G how much rain is on the ground
    P is how much rain is on the pavement

    I'm still not really sure if this is right...
     
  5. Apr 12, 2014 #4

    Simon Bridge

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    You said that P is the amount of rain on the pavement, and G is the rain on the ground ... then:
    ... dP/dt would be the rate of change of the amount of rain on the pavement.
    ... dG/dt would be the rate of change of the amount of rain on the ground.

    dP/dt= c - G

    ... is saying that the rate of change of rain on the pavement increases by rate that it falls but decreases the more water is on the ground? (This may be valid if you think the ground becomes saturated over time, so it takes water at a slower rate.)

    dG/dt = P
    ... is saying that the rate of change of rain in the ground increases with the amount of rain on the pavement (which is probably reasonable - the more water on the pavement, the faster it flows from the pavement ... but not all that physical. No matter - you are still getting the concepts ironed out and fluid flow is complicated.)

    Go back to what you are modelling. Off your description so far, you want something like:

    Rain falls from the sky at a constant rate ... c (what does c mean? how is c measured?)

    (1) Rain accumulates on the pavement, therefore, by the rate it arrives minus the rate it leaves.
    $$\frac{dP}{dt} = c - \frac{dG}{dt}$$
    (2) The rate it leaves is proportional to the amount currently on the ground.
    $$\frac{dG}{dt}=kP $$... introducing constant of proportionality ##k##.

    Subbing (2) into (1): $$\frac{dP}{dt} = c-kP$$ ... which you can solve by hand.

    But I still think you need to look more at the details of your model.
    Define everything - leave nothing implied.
     
    Last edited: Apr 12, 2014
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