Reynolds transport theorem: Difference between revisions

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revert incorrect change: Editor OptimizedShape101 incorrectly identified t as the variable of integration in the General Form of Reynolds transport theorem. The integrals are over the volume and its surface, so x is the variable of integration
 
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Latest revision as of 05:42, 22 September 2024

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}}In differential calculus, the Reynolds transport theorem (also known as the Leibniz–Reynolds transport theorem), or simply the Reynolds theorem, named after Osborne Reynolds (1842–1912), is a three-dimensional generalization of the Leibniz integral rule. It is used to recast time derivatives of integrated quantities and is useful in formulating the basic equations of continuum mechanics.

Consider integrating Template:Math over the time-dependent region Template:Math that has boundary Template:Math, then taking the derivative with respect to time: ddtΩ(t)𝐟dV. If we wish to move the derivative into the integral, there are two issues: the time dependence of Template:Math, and the introduction of and removal of space from Template:Math due to its dynamic boundary. Reynolds transport theorem provides the necessary framework.

General form

Reynolds transport theorem can be expressed as follows:[1][2][3] ddtΩ(t)𝐟dV=Ω(t)𝐟tdV+Ω(t)(𝐯b𝐧)𝐟dA in which Template:Math is the outward-pointing unit normal vector, Template:Math is a point in the region and is the variable of integration, Template:Math and Template:Math are volume and surface elements at Template:Math, and Template:Math is the velocity of the area element (not the flow velocity). The function Template:Math may be tensor-, vector- or scalar-valued.[4] Note that the integral on the left hand side is a function solely of time, and so the total derivative has been used.

Form for a material element

In continuum mechanics, this theorem is often used for material elements. These are parcels of fluids or solids which no material enters or leaves. If Template:Math is a material element then there is a velocity function Template:Math, and the boundary elements obey 𝐯b𝐧=𝐯𝐧. This condition may be substituted to obtain:[5] ddt(Ω(t)𝐟dV)=Ω(t)𝐟tdV+Ω(t)(𝐯𝐧)𝐟dA.

Template:Math proof

A special case

If we take Template:Math to be constant with respect to time, then Template:Math and the identity reduces to ddtΩfdV=ΩftdV. as expected. (This simplification is not possible if the flow velocity is incorrectly used in place of the velocity of an area element.)

Interpretation and reduction to one dimension

The theorem is the higher-dimensional extension of differentiation under the integral sign and reduces to that expression in some cases. Suppose Template:Mvar is independent of Template:Mvar and Template:Mvar, and that Template:Math is a unit square in the Template:Mvar-plane and has Template:Mvar limits Template:Math and Template:Math. Then Reynolds transport theorem reduces to ddta(t)b(t)f(x,t)dx=a(t)b(t)ftdx+b(t)tf(b(t),t)a(t)tf(a(t),t), which, up to swapping Template:Mvar and Template:Mvar, is the standard expression for differentiation under the integral sign.

See also

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References

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