6. Conservation laws

Chapter status

Illustrative · compile-only. The local fluid balances target the current 0.1.0a7 surface; their integral interpretation is derived here.

Account for what can change

A conservation law is an accounting identity. For an extensive quantity $Q$ inside a fixed region $\Omega$,

$$\frac{\mathrm{d}}{\mathrm{d}t}\int_\Omega q\,\mathrm{d}V + \int_{\partial\Omega}\boldsymbol{j}\!\cdot\boldsymbol{n}\,\mathrm{d}A = \int_\Omega s\,\mathrm{d}V.$$

The three terms are storage, outward flux, and production. This sign convention makes positive $\boldsymbol{j}\!\cdot\boldsymbol{n}$ a loss from the region. Changing the orientation changes the flux sign, not the physics.

Learning outcomes

After this chapter, you should be able to:

• identify storage, flux, and source terms in a balance;
• use the divergence theorem to relate integral and local forms;
• distinguish conservation from a constitutive law that supplies a flux; and
• catch a sign error by checking a simple control-volume limit.

From a control volume to a local relation

For a sufficiently smooth flux, the divergence theorem gives

$$\int_{\partial\Omega}\boldsymbol{j}\!\cdot\boldsymbol{n}\,\mathrm{d}A = \int_\Omega \nabla\!\cdot\boldsymbol{j}\,\mathrm{d}V.$$

If the balance holds for every suitable subregion, its local form is

$$\partial_t q + \nabla\!\cdot\boldsymbol{j} - s = 0.$$

The integral form states the accounting principle directly. The local form is often convenient for authoring field relations. Moving between them requires regularity and a clear orientation; it is not a change of numerical method.

Two steady fluid balances

For constant-density incompressible flow, mass conservation reduces to

eqiora

relation incompressibility on body {
  div(velocity) = 0;
}

The zero divergence says that a small control volume has no net volumetric outflow. It does not determine the velocity by itself.

The same standard package writes a steady momentum balance as

eqiora

relation momentum on body {
  -div(
    2 * dynamic_viscosity * symmetric_part(grad(velocity))
    - isotropic_lift(pressure)
  ) - grad(force_potential) = 0;
}

Here the divergence acts on the stress-like flux. Viscosity and pressure specify how that flux depends on the fields; those are constitutive ingredients inside the conservation statement.

Read the standard fluid components.

Deliberate failure: reverse only one flux sign

Imagine a source-free, steady one-dimensional region with more flux leaving the right face than entering the left. Its contents cannot remain steady. A residual that declares this imbalance zero after reversing only the right-face sign has lost the chosen outward-normal convention.

Test the signs on this two-face control volume before manipulating the full field equation. A compiler can verify compatible types and dimensions, but it cannot infer which orientation you intended.

Exercises

1. Set $s=0$ and storage to zero. What must the total outward flux be?
2. Derive the local balance from the integral statement using the divergence theorem.
3. Integrate div(velocity) = 0 over a region and interpret the boundary integral.
4. In the momentum excerpt, identify the conserved balance and the constitutive ingredients separately.
5. Draw a one-dimensional control volume and label the outward normal at both ends. Use it to check the flux signs.

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