Deriving a diffusion process's main PDE starts from a flux-balance argument. That argument takes exactly one tool: the divergence theorem. It turns "the flux crossing the boundary" into "the divergence of the flux, at every point inside." Green's identities build on that same idea. They return again, later in this course, as the argument that a Dirichlet or Neumann boundary value problem has only one solution.
The divergence theorem trades one kind of bookkeeping for another. Instead of tracking how much of a field pokes out through every patch of a region's outer skin, you switch views. You add up how much the field is "spreading out" at every point on the inside. The two totals always agree.
Green's first identity is just that same trade, applied to one specific field. That field is built out of one function times another function's gradient. A gradient is just the direction and rate that function is climbing fastest. This is exactly the combination that shows up once you multiply a PDE by a test function and integrate.
One special case of that trade is worth remembering on its own. Multiply a function by itself, instead of by a second function. Then the messy mixed term turns into a squared length, which can never come out negative.
Suppose the outer-skin side of the bookkeeping comes out to zero. That happens because the function, or its outward rate of change, is zero all along that skin. Then the inside total is forced to zero too. A total built entirely of non-negative pieces can only be zero if every piece in it is zero.
The divergence theorem makes a claim about a vector field \mathbf F. Say \mathbf F is smooth on a solid region E. That region E has boundary surface \partial E and outward unit normal \mathbf n. Then \iiint_E \operatorname{div}\mathbf F\,dV = \oiint_{\partial E}\mathbf F\cdot\mathbf n\,dS. This trades a flux integral over the closed surface for a volume integral. That volume integral is of the divergence, \operatorname{div}\mathbf F=\nabla\cdot\mathbf F, taken throughout the solid it encloses.
Green's first identity follows from one substitution. Pick the specific field \mathbf F=u\nabla v. It is built out of one scalar function times another's gradient. Expand its divergence with the product rule: \operatorname{div}(u\nabla v)=u\nabla^2v+\nabla u\cdot\nabla v. Drop that expansion into the divergence theorem's volume integral. On the boundary, \nabla v\cdot\mathbf n is exactly the directional derivative \partial v/\partial n, pointing straight out of the surface. That gives \iiint_E\left(u\nabla^2v+\nabla u\cdot\nabla v\right)dV = \oiint_{\partial E}u\,\frac{\partial v}{\partial n}\,dS.
One further substitution matters more downstream than the general identity itself. Set u=v=w. Then the mixed term \nabla u\cdot\nabla v turns into \nabla w\cdot\nabla w, which can never be negative. The identity becomes \iiint_E\|\nabla w\|^2\,dV = \oiint_{\partial E}w\,\frac{\partial w}{\partial n}\,dS.
Suppose the right side vanishes. That happens because w, or its normal derivative, is zero all along the boundary. Then the left side is forced to zero as well. A non-negative quantity that integrates to zero must be zero everywhere. So \nabla w is pinned down to \mathbf 0 throughout E.
A uniform source, and the sphere it flows through
Toggle the sphere and slide its radius. The divergence readout stays fixed no matter the radius -- that constant IS the source strength the divergence theorem is measuring.
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Worked example
Divergence of a radial source field
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1
Find \operatorname{div}\mathbf F for \mathbf F=\langle x,y,z\rangle.
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\operatorname{div}\mathbf F = \dfrac{\partial}{\partial x}(x)+\dfrac{\partial}{\partial y}(y)+\dfrac{\partial}{\partial z}(z) = 1+1+1.
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3\boxed{\operatorname{div}\mathbf F = 3}
By the divergence theorem, the flux of this field out through any closed surface enclosing volume V is 3V. That's the same number the widget above reports, no matter which sphere radius is chosen.
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Worked example
Setting up Green's first identity
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Write Green's first identity for u=x, v=x^2+y^2+z^2 on the unit ball.
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\nabla u = (1,0,0), \nabla v = (2x,2y,2z), and \nabla^2v = 6. So u\nabla^2v+\nabla u\cdot\nabla v = 6x+2x = 8x.
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3\boxed{\iiint_E 8x\,dV = \oiint_{\partial E} x\,\frac{\partial v}{\partial n}\,dS}
Both sides equal 0 here by symmetry (x is odd on the ball and on its boundary sphere). That's a check worth running, not a coincidence to skip past.
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Where this goes next
Deriving the diffusion equation
Deriving how temperature spreads through a rod starts the same way. You multiply a flux-balance statement by an arbitrary test region. Then you apply the divergence theorem directly. That turns a flux-through-the-boundary statement into a pointwise PDE. Green's first identity returns even later, unabbreviated. There, it's the tool that proves a Dirichlet or Neumann boundary value problem has at most one solution, in the steady-state (no-source) case.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
The Divergence Theorem is the 3D analogue of which earlier result?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The tangential form's 3D analogue is Stokes' Theorem, which pairs with curl, not the Divergence Theorem.
That theorem is a 0-dimensional-boundary statement, further from this one in the family.
The conservative test is a different kind of statement -- an equivalence, not a boundary-to-interior integral trade.
Why
Both trade a boundary flux integral for an interior divergence integral -- the normal form of Green's Theorem in 2D, the Divergence Theorem in 3D.
Look for the earlier theorem that also trades a boundary flux integral for an interior divergence integral, just one dimension down.
From Calculus III — Multivariable Calculus
Using the Divergence Theorem, find \oiint_S \mathbf{F}\cdot d\mathbf{S} for \mathbf{F}(x,y,z)=(-y,\,x,\,0), any closed surface S.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\operatorname{div}\mathbf{F}=0+0+0=0 everywhere (a pure rotation has zero divergence), so the flux through any closed surface is 0.
Correct answer
Also accepted:
Why
Compute the divergence first -- if it's identically zero, the surface's exact shape never matters.
From Calculus III — Multivariable Calculus
Using the Divergence Theorem, find \oiint_S \mathbf{F}\cdot d\mathbf{S} for \mathbf{F}(x,y,z)=(2x,\,2y,\,2z), S the sphere of radius 3.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\operatorname{div}\mathbf{F}=2+2+2=6. Volume of the ball of radius 3 is \frac{4}{3}\pi(27)=36\pi. Flux =6\cdot36\pi=216\pi\approx678.6.
Correct answer
Also accepted:
Why
Compute the constant divergence, then multiply by the ball's volume \frac{4}{3}\pi r^3.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Advanced Multivariable Calculus
Green's first identity is derived by applying the Divergence Theorem to which vector field?
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This sum of gradients isn't the field the proof uses -- the product u times grad(v) is the key construction.
This is a sum of scalars, not a vector field at all -- the Divergence Theorem needs a genuine vector field input.
The Laplacian of u is a scalar, not the vector field the proof applies the Divergence Theorem to.
Why
Green's first identity comes from applying the Divergence Theorem to F = u*grad(v), whose divergence expands (via the product rule) into exactly the identity's left-hand integrand.
The Divergence Theorem needs an actual vector field as its input -- look for the option built by multiplying a scalar function by a gradient.
From Advanced Multivariable Calculus
How is Green's SECOND identity derived from the first?
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The second identity follows directly from the first, via a swap-and-subtract trick -- they are closely connected.
Only ONE additional application (with roles swapped) is needed, not three.
Setting u=v produces a different, simpler consequence (used in the uniqueness proof), not the second identity itself.
Why
Applying the first identity with u,v and then with v,u swapped, then subtracting, cancels the symmetric grad(u).grad(v) term, leaving exactly the antisymmetric combination in the second identity.
Start by writing the first identity down twice: once as originally stated, once with u and v swapped in both slots.
Now subtract the second version from the first, and look at what happens to the grad(u).grad(v) cross term, which appears identically in both.