Knowledge card

Momentum ODE

Mathematical Modelling — Fluid Dynamics & Heat TransferContinuum mechanics; boundary-layer similarity theory

Symbol

f ‴

The third-order similarity-reduced streamwise momentum equation; its formula is composed on demand from the Powell–Eyring, unsteady, magnetic, and buoyancy terms.

Main equation

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(f^{'''} S3.5 = (\frac{( ( ( A ( f ^{'} + ( \frac{η}{2} ) f ^{''} ) S3.1 + f ^{' 2} - f f ^{''} ) S3.2 - ( \frac{φ _{β}}{φ _{ρ}} ) λ θ cos α ) S3.3 + ( \frac{φ _{σ}}{φ _{ρ}} ) M f ^{'} )}{\frac{φ _{μ} + ε}{φ _{ρ} - ( \frac{ε δ}{φ _{ρ}} ) f ^{'' 2}}})_{[S3.4]})

Dimension:1Maths:𝔹 equation(·) — A statement that two sides are equal, depending on another quantityDimensionally verifiedMathematically verifiedFully reduced to base quantities

In words:equals(plusminusplus)divided by.

Open full derivation chain: 5 stepsDerivation

1
Open Step 1: Given / definition · The derived Momentum PDE (equation 3.24) as the given definition for reduction: unsteady accele…
The derived Momentum PDE (equation 3.24) as the given definition for reduction: unsteady acceleration plus convective acceleration equals the Powell–Eyring stress terms plus thermal buoyancy minus the Lorentz body force — five relational blocks.Given / definition
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About levels
Why the levels load one at a time
Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$((\frac{\partial u}{\partial t} S5.1 + u \cdot \nabla u) S5.4 = ((τ S5.2 + ρ g β) S5.3 - J \times B))$
2
Open Step 2: Similarity transformation · Substitute the similarity maps into EVERY block [eq. 3.40]: u=U_wf′, v=−√(cν_f/(1−γt))f, and T−…
Substitute the similarity maps into EVERY block [eq. 3.40]: u=U_wf′, v=−√(cν_f/(1−γt))f, and T−T∞=(T_r−T∞)θ with T_r−T∞=ΔT₀x/(1−γt)².Similarity transformation Substitution · η
Open operand record: Substitution η
Substitution:L0 $η$ L2 $(\frac{c}{ν _{f} ( 1 - γ t )}) \sqrt ? y$
Open term-change ledger: 5 records
Open record 1: ∂u/∂t → γ(cx/(1−γt)²)(f′+f″(η/2))
Consumes:L0 $\partial u / \partial t$ L1 $\frac{\partial U _{w} f ^{'} ( η )}{\partial t}_{?}$ Produces:L0 $γ (c x / (1 - γ t)^{2}) (f' + f ″ (η /2))$ L1 $((γ i \frac{c x}{( 1 - γ t ) ^{2}}) i f^{'} + f^{''} (\frac{η}{2}))$ The similarity transform inserts u = U_w f′(η) into the local acceleration ∂u/∂t; carrying out the time derivative (detailed in the Substituted Unsteady Acceleration node) turns the block into the growth scale cx/(1−γt)² times the transient velocity group f′+f″(η/2).γ(cx/(1−γt)²)(f′+f″(η/2))
Open record 2: u·∇u → (c²x/(1−γt)²)(f′²−ff″)
Consumes:L0 $u \cdot \nabla u$ L1 $((u i \frac{\partial u}{\partial x}) i + (v i \frac{\partial u}{\partial y}))$ Produces:L0 $(c^{2} x / (1 - γ t)^{2}) (f'^{2} - f f ″)$ L1 $((\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) f^{' 2} i - (\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) f f^{''})$ The similarity transform inserts u = U_w f′(η) and v = −√(cν_f/(1−γt))f(η) into the convective terms u∂u/∂x and v∂u/∂y, turning the block into the convective scale c²x/(1−γt)² times f′²−ff″ (detailed per-term in the Substituted Convective Acceleration node).(c²x/(1−γt)²)(f′²−ff″)
Open record 3: τ → τ_PE(η)
Consumes:L0 $τ$ L2 $(c_{1} ? \frac{\partial ^{2} u}{\partial y ^{2}}) P E^{1} ? - (c_{2} ? (\frac{\partial u}{\partial y})^{2} ? \frac{\partial ^{2} u}{\partial y ^{2}}) P E^{2}$ Produces:L0 $τ_{P} E (η)$ L1 $((\frac{ν _{hn f} + 1}{ρ _{hn f} β ς}) U_{w} (\frac{c}{ν _{f} ( 1 - γ t )}) f^{'''} i - (\frac{1}{2 ρ _{hn f} β ς ^{3}}) U_{w}^{3} (\frac{c}{ν _{f} ( 1 - γ t )})^{2} f^{'' 2} f^{'''})$ The similarity transform inserts the shear-rate and curvature maps ∂u/∂y = U_w(c/(ν_f(1−γt)))^{1/2}f″(η), ∂²u/∂y² = U_w(c/(ν_f(1−γt)))f‴(η) into the Powell–Eyring diffusion and retardation terms (detailed step-by-step in the Substituted Powell–Eyring Stress Terms node), exposing their dimensional coefficients explicitly. Consumed at DEPTH 2 so the second-order velocity gradients (∂²u/∂y², (∂u/∂y)²) that the maps act on are visible, not the opaque τ symbol.τ_PE(η)
Open record 4: ρgβ → gβ_hnf(Tr−T∞)θcosα
Consumes:L0 $ρ g β$ L1 $g ? β_{hn f} ? (T r - T \infty) θ ? cos α$ Produces:L0 $g β_{h} n f (T r - T \infty) θ cos α$ L1 $((((g i β_{hn f}) i T r - T \infty) i θ) i cos α)$ The similarity transform inserts T−T∞=(T_r−T∞)θ with T_r−T∞=ΔT₀x/(1−γt)², preserving a temperature scale distinct from the velocity scale.gβ_hnf(Tr−T∞)θcosα
Open record 5: J×B → (σ_hnf B²/ρ_hnf)U_w f′
Consumes:L0 $J \times B$ L1 $\frac{σ B ^{2}}{ρ} ? U_{w} f^{'} (η)$ Produces:L0 $(σ_{h} n f B^{2} / ρ_{h} n f) U_{w} f'$ L1 $((\frac{σ B ^{2}}{ρ} i U w) i f^{'})$ The similarity transform inserts u = U_w f′(η) into the Lorentz body force, turning it into the magnetic coefficient times U_w f′ (detailed in the Substituted Lorentz Braking node); B(t) remains time-dependent at this stage.(σ_hnf B²/ρ_hnf)U_w f′
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About levels
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Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$((γ (\frac{c x}{( 1 - γ t ) ^{2}}) (f^{'} + f^{''} (\frac{η}{2})) S1.1 + (\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) (f^{' 2} - f f^{''})) S1.4 = ((τ_{P E} (η) S1.2 + g β_{hn f} (T r - T \infty) θ cos α) S1.3 - (\frac{σ _{hn f} B ^{2}}{ρ _{hn f}}) U_{w} f^{'}))$
3
Open Step 3: Rearrange terms · Move every term to one side and set the equation to zero [eq. 3.42], the standard homogeneous f…
Move every term to one side and set the equation to zero [eq. 3.42], the standard homogeneous form: subtracting the stress and buoyancy blocks and adding the Lorentz block collects the whole balance against zero — the SAME five blocks, now equated to nought.Rearrange terms
Level 0 · more available
Brief — named quantities
About levels
Why the levels load one at a time
Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$(((((γ (\frac{c x}{( 1 - γ t ) ^{2}}) (f^{'} + f^{''} (\frac{η}{2})) S2.1 + (\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) (f^{' 2} - f f^{''})) S2.2 - τ_{P E} (η)) S2.3 - g β_{hn f} (T r - T \infty) θ cos α) S2.4 + (\frac{σ _{hn f} B ^{2}}{ρ _{hn f}}) U_{w} f^{'}) S2.5 = 0)$
4
Open Step 4: Non-dimensionalisation · Divide by c²x/(1−γt)² and identify each raw coefficient: γ/c=A; (ν_hnf+1/(ρ_hnfβς))/ν_f=(φ_μ+ε)…
Divide by c²x/(1−γt)² and identify each raw coefficient: γ/c=A; (ν_hnf+1/(ρ_hnfβς))/ν_f=(φ_μ+ε)/φ_ρ; retardation=εδ/φ_ρ; gβ_hnfΔT₀/c²=(φ_{ρβ}/φ_ρ)λ with λ=Gr_x/Re_x²=gβ_fΔT₀/c²; and magnetic damping=(φ_σ/φ_ρ)M. Every explicit x and t cancels.Non-dimensionalisation Divisor · c²x/(1−γt)²
Open operand record: Divisor c²x/(1−γt)²
Divisor:L0 $c^{2} x / (1 - γ t)^{2}$ L1 $\frac{c ^{2} ? x}{( 1 - γ t ) ^{2}}_{?}$
Open term-change ledger: 5 records
Open record 1: γ(cx/(1−γt)²)(f′+f″(η/2)) → A(f′+(η/2)f″)
Consumes:L0 $γ (c x / (1 - γ t)^{2}) (f' + f ″ (η /2))$ L1 $((γ i \frac{c x}{( 1 - γ t ) ^{2}}) i f^{'} + f^{''} (\frac{η}{2}))$ Produces:L0 $A (f' + (η /2) f ″)$ L1 $(A i f^{'} + f^{''} (\frac{η}{2}))$ Dividing the unsteady block by the convective scale c²x/(1−γt)² leaves the raw ratio γ/c, identified BY DEFINITION as the unsteadiness parameter A — giving the unsteadiness term A(f′+(η/2)f″).A(f′+(η/2)f″)
Open record 2: (c²x/(1−γt)²)(f′²−ff″) → f′²−ff″
Consumes:L0 $(c^{2} x / (1 - γ t)^{2}) (f'^{2} - f f ″)$ L1 $((\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) f^{' 2} i - (\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) f f^{''})$ Produces:L0 $f'^{2} - f f ″$ L1 $(f^{' 2} i - f f^{''})$ Dividing the convective block by the same convective scale c²x/(1−γt)² cancels its common dimensional factor exactly — that cancellation IS the self-similarity — leaving the dimensionless inertia f′²−ff″.f′²−ff″
Open record 3: τ_PE(η) → ((φ_μ + ε)/φ_ρ) f‴ − ((εδ/φ_ρ))f″²f‴
Consumes:L0 $τ_{P} E (η)$ L1 $((\frac{ν _{hn f} + 1}{ρ _{hn f} β ς}) U_{w} (\frac{c}{ν _{f} ( 1 - γ t )}) f^{'''} i - (\frac{1}{2 ρ _{hn f} β ς ^{3}}) U_{w}^{3} (\frac{c}{ν _{f} ( 1 - γ t )})^{2} f^{'' 2} f^{'''})$ Produces:L0 $((φ_{μ} + ε) / φ_{ρ}) f ‴ - ((ε δ / φ_{ρ})) f ″^{2} f ‴$ L1 $(c_{1} f^{'''} i - (\frac{ε δ}{φ _{ρ}}) f^{'' 2} f^{'''})$ Dividing the Powell–Eyring block by the convective scale exposes the raw diffusion and retardation coefficients, identified BY DEFINITION as (φ_μ+ε)/φ_ρ and εδ/φ_ρ (the verified grouping carrying 1/φ_ρ onto BOTH stress groups).((φ_μ + ε)/φ_ρ) f‴ − ((εδ/φ_ρ))f″²f‴
Open record 4: gβ_hnf(Tr−T∞)θcosα → (φ_β/φ_ρ) λ θ cos α
Consumes:L0 $g β_{h} n f (T r - T \infty) θ cos α$ L1 $((((g i β_{hn f}) i T r - T \infty) i θ) i cos α)$ Produces:L0 $(φ_{β} / φ_{ρ}) λ θ cos α$ L1 $((((\frac{φ _{ρβ}}{φ _{ρ}})_{[i]} i λ) i θ) i cos α)$ Using T_r−T∞=ΔT₀x/(1−γt)², division leaves gβ_hnfΔT₀/c². The explicit Gr_x and Re_x formulas identify this as (φ_{ρβ}/φ_ρ)λ with λ=gβ_fΔT₀/c².(φ_β/φ_ρ) λ θ cos α
Open record 5: (σ_hnf B²/ρ_hnf)U_w f′ → (φ_σ/φ_ρ) M f′
Consumes:L0 $(σ_{h} n f B^{2} / ρ_{h} n f) U_{w} f'$ L1 $((\frac{σ B ^{2}}{ρ} i U w) i f^{'})$ Produces:L0 $(φ_{σ} / φ_{ρ}) M f'$ L1 $(((\frac{φ _{σ}}{φ _{ρ}})_{[i]} i M) i f^{'})$ Dividing the Lorentz block by the convective scale and inserting B(t)² = B₀²/(1−γt) cancels the residual (1−γt) exactly, identifying σ_hnf B₀²/(ρ_hnf c) BY DEFINITION as (φ_σ/φ_ρ)M.(φ_σ/φ_ρ) M f′
Level 0 · more available
Brief — named quantities
About levels
Why the levels load one at a time
Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$(((((A (f^{'} + (\frac{η}{2}) f^{''}) S6.1 + f^{' 2} - f f^{''}) S6.2 - (\frac{φ _{μ} + ε}{φ _{ρ}}) f^{'''} - ((\frac{ε δ}{φ _{ρ}})) f^{'' 2} f^{'''}) S6.3 - (\frac{φ _{β}}{φ _{ρ}}) λ θ cos α) S6.4 + (\frac{φ _{σ}}{φ _{ρ}}) M f^{'}) S6.5 = 0)$
★
Open Step 5: Separation of variables · Isolate the highest derivative: factor f‴ out of the Powell–Eyring stress and divide the balanc…
Isolate the highest derivative: factor f‴ out of the Powell–Eyring stress and divide the balance by its coefficient, giving the explicit normal form f‴ = … that the numerical solver marches.Separation of variables Divisor · (φ_μ+ε)/φ_ρ−(εδ/φ_ρ)f″²
Open term-change ledger: 1 records
Open record 1: ((φ_μ + ε)/φ_ρ) f‴ − ((εδ/φ_ρ))f″²f‴ → f‴ + (φ_μ+ε)/φ_ρ−(εδ/φ_ρ)f″²
Consumes:L0 $((φ_{μ} + ε) / φ_{ρ}) f ‴ - ((ε δ / φ_{ρ})) f ″^{2} f ‴$ Produces:L0 $f ‴$ L0 $(φ_{μ} + ε) / φ_{ρ} - (ε δ / φ_{ρ}) f ″^{2}$ The Powell–Eyring stress ((φ_μ+ε)/φ_ρ)f‴ − (εδ/φ_ρ)f″²f‴ = f‴·[(φ_μ+ε)/φ_ρ − (εδ/φ_ρ)f″²] separates into the common factor f‴ (isolated on the left as the highest derivative) and the effective-diffusion denominator K = (φ_μ+ε)/φ_ρ − (εδ/φ_ρ)f″² that the balance is divided by.
Level 0 · more available
Brief — named quantities
About levels
Why the levels load one at a time
Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$(f^{'''} S3.5 = (\frac{( ( ( A ( f ^{'} + ( \frac{η}{2} ) f ^{''} ) S3.1 + f ^{' 2} - f f ^{''} ) S3.2 - ( \frac{φ _{β}}{φ _{ρ}} ) λ θ cos α ) S3.3 + ( \frac{φ _{σ}}{φ _{ρ}} ) M f ^{'} )}{\frac{φ _{μ} + ε}{φ _{ρ} - ( \frac{ε δ}{φ _{ρ}} ) f ^{'' 2}}})_{[S3.4]})$

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Equation

Essence
S1The SAME balance after substitution: substituted unsteady plus substituted convective acceleration equals substituted Powell–Eyring stress plus substituted buoyancy minus substituted Lorentz braking.
S1.1essenceThe sum of γ(cx/(1−γt)²)(f′+f″(η/2)) and (c²x/(1−γt)²)(f′²−ff″). [1.1]
S1.2essenceThe sum of τ_PE(η) and gβ_hnf(Tr−T∞)θcosα. [1.1]
S1.3essence(τ_PE(η) + gβ_hnf(Tr−T∞)θcosα) reduced by (σ_hnf B²/ρ_hnf)U_w f′. [1.1]
S1.4essence(γ(cx/(1−γt)²)(f′+f″(η/2)) + (c²x/(1−γt)²)(f′²−ff″)) set equal to ((τ_PE(η) + gβ_hnf(Tr−T∞)θcosα) − (σ_hnf B²/ρ_hnf)U_w f′). [1.1]
S2The substituted balance in homogeneous form: substituted unsteady plus substituted convective acceleration minus substituted Powell–Eyring stress minus substituted buoyancy plus substituted Lorentz braking equals zero.
S2.1essenceThe sum of γ(cx/(1−γt)²)(f′+f″(η/2)) and (c²x/(1−γt)²)(f′²−ff″). [1.1]
S2.2essence(γ(cx/(1−γt)²)(f′+f″(η/2)) + (c²x/(1−γt)²)(f′²−ff″)) reduced by τ_PE(η). [1.1]
S2.3essence((γ(cx/(1−γt)²)(f′+f″(η/2)) + (c²x/(1−γt)²)(f′²−ff″)) − τ_PE(η)) reduced by gβ_hnf(Tr−T∞)θcosα. [1.1]
S2.4essenceThe sum of (((γ(cx/(1−γt)²)(f′+f″(η/2)) + (c²x/(1−γt)²)(f′²−ff″)) − τ_PE(η)) − gβ_hnf(Tr−T∞)θcosα) and (σ_hnf B²/ρ_hnf)U_w f′. [1.1]
S2.5essence((((γ(cx/(1−γt)²)(f′+f″(η/2)) + (c²x/(1−γt)²)(f′²−ff″)) − τ_PE(η)) − gβ_hnf(Tr−T∞)θcosα) + (σ_hnf B²/ρ_hnf)U_w f′) set equal to 0. [1.1]
S3Isolate the highest derivative: factor f‴ out of the Powell–Eyring stress and divide the balance by its coefficient, giving the explicit normal form f‴ = … that the numerical solver marches.
S3.1essenceThe sum of A(f′+(η/2)f″) and f′²−ff″.
S3.2essence(A(f′+(η/2)f″) + f′²−ff″) reduced by (φ_β/φ_ρ) λ θ cos α.
S3.3essenceThe sum of ((A(f′+(η/2)f″) + f′²−ff″) − (φ_β/φ_ρ) λ θ cos α) and (φ_σ/φ_ρ) M f′.
S3.4essence(((A(f′+(η/2)f″) + f′²−ff″) − (φ_β/φ_ρ) λ θ cos α) + (φ_σ/φ_ρ) M f′) divided by (φ_μ+ε)/φ_ρ−(εδ/φ_ρ)f″².
S3.5essencef‴ set equal to ((((A(f′+(η/2)f″) + f′²−ff″) − (φ_β/φ_ρ) λ θ cos α) + (φ_σ/φ_ρ) M f′) / (φ_μ+ε)/φ_ρ−(εδ/φ_ρ)f″²).

Dimension I

Definition

What it is — and what it is not

What it is: S4The similarity-reduced streamwise momentum equation.
S5The momentum balance: unsteady plus convective acceleration equals Powell–Eyring stress plus buoyancy minus the Lorentz force.
S5.1essenceThe sum of ∂u/∂t and u·∇u. [1.1]
S5.2essenceThe sum of τ and ρgβ. [1.1]
S5.3essence(τ + ρgβ) reduced by J×B. [1.1]
S5.4essence(∂u/∂t + u·∇u) set equal to ((τ + ρgβ) − J×B). [1.1]
Wisdoms: S6The reduced momentum ODE in named dimensionless parameters: the unsteadiness term plus the dimensionless convective inertia minus the Powell–Eyring diffusion/retardation group minus buoyancy plus magnetic damping equals zero.
S6.1essenceThe sum of A(f′+(η/2)f″) and f′²−ff″. [1.1]
S6.2essence(A(f′+(η/2)f″) + f′²−ff″) reduced by ((φ_μ + ε)/φ_ρ) f‴ − ((εδ/φ_ρ))f″²f‴. [1.1]
S6.3essence((A(f′+(η/2)f″) + f′²−ff″) − ((φ_μ + ε)/φ_ρ) f‴ − ((εδ/φ_ρ))f″²f‴) reduced by (φ_β/φ_ρ) λ θ cos α. [1.1]
S6.4essenceThe sum of (((A(f′+(η/2)f″) + f′²−ff″) − ((φ_μ + ε)/φ_ρ) f‴ − ((εδ/φ_ρ))f″²f‴) − (φ_β/φ_ρ) λ θ cos α) and (φ_σ/φ_ρ) M f′. [1.1]
S6.5essence((((A(f′+(η/2)f″) + f′²−ff″) − ((φ_μ + ε)/φ_ρ) f‴ − ((εδ/φ_ρ))f″²f‴) − (φ_β/φ_ρ) λ θ cos α) + (φ_σ/φ_ρ) M f′) set equal to 0. [1.1]
Attributes: S7Balances the Powell–Eyring diffusion and nonlinear retardation against convective inertia, unsteadiness, magnetic damping, and buoyancy; the equation is composed on demand. [1.1]

Dimension II

In practice

How to deal with it

Consequences: S8The buoyancy term carries θ, coupling it to the energy ODE.

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