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Homework Help: System of Partial Differential Equations

  1. May 17, 2014 #1
    System of PDEs--Heat Equation For Two Objects

    Hello everyone,

    Before is a system of partial differential equations; to be specific, it is this system:

    [itex]\frac{\partial U_A }{\partial t} = - \frac{k_B}{k_A} \alpha_A \left( \frac{\partial^2 U_B}{\partial x^2} + \frac{\partial^2 U_B}{\partial y^2} + \frac{\partial^2 U_B}{\partial z^2} \right)[/itex]


    [itex]\frac{\partial U_B }{\partial t} = \alpha_B \left( \frac{\partial^2 U_B}{\partial x^2} + \frac{\partial^2 U_B}{\partial y^2} + \frac{\partial^2 U_B}{\partial z^2} \right)[/itex]

    I am not very certain as to how to solve this--as a matter of fact, I do not even know if it is possible to solve this. So, does this system have a solution [itex]U_A(x,y,z,t)[/itex] and [itex]U_B(x,y,z,t)[/itex]? And if it does, could someone help me with solving it, such as providing hints or suitable reading materials? I would certainly appreciate it.
    Last edited: May 17, 2014
  2. jcsd
  3. May 17, 2014 #2
    I think I might have figured it out: would I solve the second partial differential equation, and then use the function that I would find during the process, and then use that to solve the first partial differential equation?
  4. May 17, 2014 #3


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    The second equation is the heat equation.

    The second equation gives [itex]\nabla^2 U_B = \frac{1}{\alpha_B} \frac{\partial U_B}{\partial t}[/itex], which on substitution into the first equation gives [tex]
    \frac{\partial U_A}{\partial t} = -\frac{k_B \alpha_A}{k_A \alpha_B} \frac{\partial U_B}{\partial t}[/tex] from which it follows that (assuming [itex]\frac{k_B \alpha_A}{k_A \alpha_B}[/itex] to be independent of time)
    \frac{\partial }{\partial t} \left(U_A + \frac{k_B \alpha_A}{k_A \alpha_B} U_B\right) = 0.
    The quantity in the brackets is thus a function of space only, and is therefore determined by the initial conditions.
  5. May 17, 2014 #4
    So, the functions [itex]U_A[/itex] and [itex]U_B[/itex] are both independent of time? Really? Hmmm....That is not what I expected. Perhaps the way in which I am approaching this problem is wrong.
  6. May 17, 2014 #5
    What I am trying to do is calculate the time it takes for two objects call them A and B, which are initially at different temperatures, to reach thermal equilibrium. My starting assumption was that the rate at which heat flows from object B is equal to negative the rate at which heat leaves A. This idea could also be expressed in the heat flux equation (Fourier's heat conduction equation, I think):

    [itex]\dot{\mathbf{q}}_B = - \dot{\mathbf{q}}_A [/itex]

    From this I could get

    [itex]\frac{\partial U_A}{\partial x} = - \frac{k_B}{k_A} \frac{\partial U_B}{\partial x}[/itex]

    [itex]\frac{\partial U_A}{\partial y} = - \frac{k_B}{k_A} \frac{\partial U_B}{\partial y}[/itex]

    [itex]\frac{\partial U_A}{\partial z} = - \frac{k_B}{k_A} \frac{\partial U_B}{\partial z}[/itex]

    Differentiating each one with their respective spatial variable, we get

    [itex]\frac{\partial^2 U_A}{\partial x^2} = - \frac{k_B}{k_A} \frac{\partial^2 U_B}{\partial x^2}[/itex]

    and so forth.

    From these premises, I set up the heat equation for both objects, A and B, and made the substitutions using the above relationships.
  7. May 17, 2014 #6
    In light of this new information I have provided, should this thread be moved into, say, the classical physics thread?
  8. May 17, 2014 #7


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    No. It just says that If the temperature of ##U_A## changes, then ##U_B## changes in proportion, but in the opposite direction.

    With the benefit of hindsight (a wonderful exact science!), that should be obvious from what you are modeling:
    But you probably need to think a bit more about your PDEs, because if A and B are separate objects, the two equations apply to different regions in space corresponding to the two objects, and it's not obvious how they describe how heat gets from one object to the other. That could be conduction if they have a common boundary, or radiation, or convection involving something else apart from A and B.
  9. May 17, 2014 #8
    So, would you say that I am approaching this problem incorrectly?

  10. May 17, 2014 #9
    What if I consider the two objects as one 'composite object', and use one heat equation. So, if object A is defined by the equation [itex]S_1 = g_1(x,y,z)[/itex], and object B by the equation [itex]S_2 = g_2(x,y,z)[/itex], then we can consider the composite object [itex]S_1 \cap S_2[/itex]. For the initial conditions, we can constrain the temperature of the [itex]S_1[/itex] to be [itex]T_1[/itex], and [itex]S_2[/itex] to be [itex]T_2[/itex], where [itex]T_1 > T_2[/itex]; that is, [itex]u(x,y,z,0) = T_1 ~\forall x,y,z \in S_1 [/itex] and [itex]u(x,y,z,0) = T_2 \forall x,y,z \in S_2[/itex]. Then, at time [itex]t = \tau[/itex], the time at which [itex]S_1[/itex] and [itex]S_2[/itex] come to thermal equilibrium, [itex]u(x,y,z,\tau) = T_{EQ} ~ \forall x,y,z \in S_1 \cap S_2[/itex].

    I am not sure if my mathematical statements are correct, but hopefully the idea is conveyed.

    One problem I see, however, is that will have different values of [itex]\alpha[/itex] for S1 and S2
    Last edited: May 17, 2014
  11. May 17, 2014 #10


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    We didn't know that when post #2 was made. For a system of equations (ODEs, PDEs or anything else) it's a reasonable assumption all the equations apply to the same region in space and time. Post #2 would be physically meaningful if for example you sprayed liquid droplets into a gas at a different temperature.

    I recommend you keep the physics clear in your mind, rather than getting lost in inventing mathematical notation. The equations in your OP don't really say how heat gets from object A to object B. They seem to be trying to say "the rate of temperature change of B at one point in space is somehow related to the heat flux in A at some other point". But if you have one equation containing ##U_A(x,y,z,t)## and ##U_B(x,y,z,t)##, then ##x##, ##y##, ##z## and ##t## represent the same point in space and the same moment in time.
  12. May 17, 2014 #11
    Hmm, I definitely see the difficulty, now. What would you suggest I do?
  13. May 17, 2014 #12
    Post #5 explains it all, and you are definitely on the right track.

    Let's take a specific example. Suppose you have two identical slabs of different materials, one at temperature TA0 and the other at TB0. Slab A runs from x = -L to x = 0, and slab B runs from x = 0 to x = +L. The slabs are insulated at x = -L and at x = +L, respectively. At time t equal to zero, the two slabs are brought into thermal contact, and heat is allowed to flow between them. You would like to find the temperature distribution in the slabs as a function of time and spatial position. Does this pretty much capture what you are looking for? If so, you might first be interested in considering the simpler problem in which the two slabs are made of the same material.

  14. May 17, 2014 #13
    Chestermiller: Yes, that appears to be what I am trying to do. My question is, in the scenario you have just proposed, is it possible to calculate the time at which the objects equilibrate? Also, would it still be wise to set up a heat equation for each slab?
  15. May 17, 2014 #14
    Yes, effectively. The mathematics would predict that they would never quite reach the exact same temperature, but, you can readily calculate the finite amount of time it would take for them to reach say 0.1% of the original temperature difference.
    Yes, definitely. That's what I would do (and have done many times). Let's see your set up of this problem, including the boundary conditions (particularly the boundary condition at the interface between the slabs).

    After you get the problem properly formulated, I can give you some thoughts an how to solve the equations.

  16. May 18, 2014 #15
    I can't quite figure this out. The heat equation for A would be [itex]\frac{\partial U_A}{\partial t} = \alpha \frac{\partial^2 U_A}{\partial x^2}[/itex], where the unknown function [itex]U_A(x,t)[/itex] must satisfy the initial condition [itex]U_A(x,0) = T_0~~\forall x \in [-L,0][/itex]. The object B would have a similar heat equation, and the initial condition for this would be [itex]U_B(x,0) = T_1 ~~\forall x \in [0,L][/itex]. At the boundary between both objects, [itex]x=0[/itex], heat flow will occur, the heat flow being described by the heat flux equation [itex]\dot{q}_{A,x} = - \dot{q}_{B,x}[/itex].

    I am not sure how to implement these ideas, though.
    Last edited: May 18, 2014
  17. May 18, 2014 #16
    Nice job so far. You don't have to use separate subscripts for the temperatures of A and B, because the temperature will be continuous at the interface. But you do have to use separate subscripts for the thermal diffusivities of A and B, αA and αB, because the thermal properties in the two regions are different.

    In addition to the temperature being constant at the interface, the heat flux must also be constant at the interface:
    where the k's are the thermal conductivities. Note that, although the heat flux is constant across the interface, the temperature gradient is not.

    There are also two other boundary conditions that need to be included in the formulation. These are the insulation boundary conditions at x = -L and x = +L:

    This completes the formulation of the problem. I'll give you a chance to look this over, and then respond if you have any questions. Then we can proceed to articulating what is happening, and to discussing how to go about solving the equations.

    (This analysis applies to the case where the physical properties of A and B are different. I'm assuming that you want to skip the case where the properties of the two slabs are the same. If you first want to solve the case where the properties of the two slabs are the same, please indicate that desire.)

  18. May 18, 2014 #17
    I do have one question, why does the temperature at the interface have to be constant? Is that just a simplifying assumption?
  19. May 18, 2014 #18
    It doesn't have to be constant at the interface. It has to be continuous at the interface. At finite times, temperature of A must match the temperature of B at the interface.

  20. May 18, 2014 #19
    Oh, okay. I am sorry, I misread what you said. That makes more sense.
  21. May 18, 2014 #20
    OK. So let's get started. But, first, which case do you want to consider?

    1. A and B have different thermal properties

    2. A and B have the same thermal properties

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