Navier Stokes in Higher Order Dual Numbers and Other Algebraic Systems

Navier Stokes in Higher Order Dual Numbers and Other Algebraic Systems Mohammed Farhaan, BMSCE Bengaluru December 7, 2025 Abstract The Navier-Stokes Equation needs no introduction. In this paper we define an Infinitely Adjoined Algebra and we rewrite the Navier-Stokes Equation in this new Algebra and observe that the Equation is well preserved except for a new constant. The paper was written as an encouragement to draw similarities hyperreals and dual numbers ,and as evidence that studying infinitely adjoined algebras could be useful The Navier-Stokes equation is a notoriously nonlinear PDE. We aim to explore higher-dimensional algebras, particularly to observe what happens as we move towards infinite-dimensional algebras. Introduction A PDE is defined as an equation of the form: F(x1, x2, . . . , xn, u, ∂u ∂x1 ,∂u ∂xn ,∂2u ∂xi∂xj , . . . )=0 Some examples include: •Heat Equation: ∂u ∂t =α∇2u •Wave Equation: ∂2u ∂t2=c2∇2u •Navier-Stokes Equation: ∂u ∂t + (u · ∇)u =−∇p+v∇2u 1 Algebras and Adjoined Algebras An algebra Aover a field Kis a vector space with a bilinear operation: A×A→A Axioms: •(a+b)c=ac +bc •a(b+c) = ab +ac •λ(ab) = (λa)b=a(λb) We form adjoined algebras as follows: F[x]/(p(x)) = F[m],where p(m)=0 These are standard Some examples of Adjoined Algebras include: •R[x]/(x2+ 1) - Complex Numbers •R[x]/(x2) - Dual Numbers •R[x]/(x2−1) - Split Complex Numbers Infinite Order Algebras We define infinite order algebras as limits: lim n→∞ R[x]/(xn) (Infinite Order Dual Numbers) lim n→∞ R[x]/(xn+ 1) (Infinite Order Complex Numbers) lim n→∞ R[x]/(xn−1) (Infinite Order Split Complex Numbers) Function Definitions For functions in infinite order algebras: f:R(∞)→R(∞) f=f0+f1ϵ+f2ϵ2+f3ϵ3+. . . , fj→f∞ For complex infinite order algebras: f=f0+f1i+f2i2+f3i3+. . . , fj→f∞ For split complex infinite order algebras: f=f0+f1j+f2j2+f3j3+. . . , fj→f∞ 2 Results a) Dual Number Series S= ∞ X n=0 ϵn=1 1−ϵ b) Complex Number Series S= ∞ X n=0 in=1 + i 1−i c) Split-Complex Number Series S= ∞ X n=0 jn= 1 Looking for differences in PDEs in infinite dimensional algebras. so an interesting thing to observe is what happens to these PDEs in D∞as the non linearity in the PDE is completely removed transformed into something algebraic? We also explore what happens in C∞and Cs ∞ Heat Equation in D∞, C∞, Cs ∞ let the function u=u0+u1ϵ+u2ϵ2· · · − u∞ϵ∞then in the Heat equation ∂u ∂t =∂u0 ∂t +∂u1 ∂t ϵ+∂u2 ∂t ϵ2· · · − ∂u∞ ∂t ϵ∞ and ∂2u ∂x2=∂2u0 ∂x2+ϵ∂2u1 ∂x2+ϵ2∂2u2 ∂x2· · · − ϵ∞∂2u∞ ∂x2 by equating the ϵkparts, we get heat equation for each component of u u0to u∞ ∂u0 ∂t =α∂2u0 ∂x2 ∂u1 ∂t =α∂2u1 ∂x2 . . . 3 ∂u∞ ∂t =α∂2u∞ ∂x2 We get the same implications by comparing and equating the ikparts and jkparts in C∞and Cs ∞respectively. Burger PDE in D∞, C∞, Cs ∞ a)Burgers PDE in D∞ Given by ∂u ∂t +u∂u ∂x =ν∂2u ∂x2 u=u0+u1ϵ+u2ϵ2. . . u∞ϵ∞ ∂u ∂t =∂u0 ∂t +∂u1 ∂t ϵ+∂u2 ∂t ϵ2. . . ∂u∞ ∂t ϵ∞ u∂u ∂x = (u0+u1ϵ+u2ϵ2. . . u∞ϵ∞)(∂u0 ∂x +∂u1 ∂x ϵ+∂u2 ∂x ϵ2. . . ) ∂u ∂x = limn→∞(Pn i=0 uiϵi)(Pn j=0 ∂uj ∂x ϵj) and we know that ( n X i=0 aixi)( n X j=0 bjxj) = 2n X k=0 ckxk = n X k=0 ckxk+ 2n X k=n+1 ckxk we know cp k=Pk i=0 aibk−iu∂u ∂x = limn→∞(Pn k=0 ckϵk+P2n k=n+1 ckϵk) = lim n→∞ n X k=0 ( k X i=0 ui ∂uk−i ∂x )ϵk if ui=u0,∀ithen u∂u ∂x = lim n→∞ n X k=0 ( k X i=0 u0 ∂u0 ∂x )ϵk = lim n→∞ n X k=0 (k+ 1)u0 ∂u0 ∂x ϵk =u0 ∂u0 ∂x lim n→∞ n X k=0 (k+ 1)ϵk 4 what is limn→∞ Pn k=0 kϵk? S= 0 + ϵ+ 2ϵ2+ 3ϵ3. . . ∞ϵ∞ Sϵ = 0 + ϵ2+ 2ϵ3. . . (∞ − 1)ϵ∞+ϵ∞+1 (S−Sϵ) = 0 + ϵ+ϵ2+ϵ3. . . ∞+ 1 = S S(1 −ϵ) = 1 1−ϵ−1 = 1−(1 −ϵ) 1−ϵ=ϵ 1−ϵ S=ϵ (1 −ϵ)2 Giving us u∂u ∂x =ϵ2 (1 −ϵ)2u0 ∂u0 ∂x Now lets do the same substitution ∂u ∂t = (1 + ϵ+ϵ2−ϵ∞)∂u0 ∂t ν∂2u ∂x2= (1 + ϵ+ϵ2−ϵ∞)ν∂2u0 ∂x2 Burgers Eqn in D∞is given by the following. let u= (1 + ϵ+ϵ2−ϵ∞) and letu0=1 1−ϵu0 ∂u ∂t +u∂u ∂x =ν∂2u ∂x2 converts to (1 + ϵ+ϵ2−ϵ∞)∂u0 ∂t +ϵ2 (1 −ϵ)2u0 ∂u0 ∂x = (1 + ϵ+ϵ2−ϵ∞)ν∂2u0 ∂x2 giving 1 1−ϵ ∂u0 ∂t +ϵ2 (1 −ϵ)2u0 ∂u0 ∂x =1 1−ϵν∂2u0 ∂x2 ∂u0 ∂t +ϵ2 1−ϵu0 ∂u0 ∂x =ν∂2u0 ∂x2 5 0.1 b) Doing the same in C∞ Letting u=u0+iu1+i2u2+i3u3· · · +i∞u∞ ∂u ∂t =∂u0 ∂t +i∂u1 ∂t +i2∂u2 ∂t · · · +i∞∂u∞ ∂t ∂2u ∂x2=∂2u0 ∂x2+i∂2u1 ∂x2+i2∂2u2 ∂x2. . . u∂u ∂x = lim n→∞( n X i=0 uiii)( n X j=0 ∂uj ∂x ij) = lim n→∞( n X k=0 ( k X i=0 ui ∂uk−i ∂x )ik) = lim n→∞( n−1 X k=0 ( k X i=0 ui ∂uk−i ∂x )ik+ ( n X i=0 ui ∂un−i ∂x )in) = lim n→∞( n X k=0 ( k X i=0 ui ∂uk−i ∂x )ik− n−1 X k=0 ( k X i=0 ui ∂uk−i ∂x )ik+n) if we say ui=u, ∀ithen u∂u ∂x = lim n→∞( n X k=0 (k+ 1)u0 ∂u0 ∂x ik) =u0 ∂u0 ∂x lim n→∞ n X k=0 (k+ 1)ik =u0 ∂u0 ∂x (1 ·0+2·i+ 3 ·i2+ 4 ·i3· · · + (∞+ 1)i∞) ⇒u0 ∂u0 ∂x (0 + i−2−3i· · · + (∞+ 1) + i∞) ⇒u0 ∂u0 ∂x (0 + i(1 − ∞)−2(1 − ∞). . . ) =u0 ∂u0 ∂x (∞ × i+ 1 1−i) and from Cs ∞and im sure if evaluated as Cs ∞would also have some unbounded nature in the form of ∞or 1 (1−ϵ)2 Now Finally lets see what happens when we take Navier Stokes in D∞ 6 Navier Stokes in D∞ Starting with simpler terms ∂u ∂t + (u · ∇)u =−∇p+v∆uu =u0+ϵu1+ϵ2u2+ϵ3u3........ϵ∞u∞ ∂u ∂t =∂u0 ∂t +ϵ∂u1 ∂t +ϵ2∂u2 ∂t +. . . ϵ∞∂u∞ ∂t (u · ∇)u is complicated so we leave it for later . ∇p=∇p1+ϵ∇p1+ϵ2∇p2........ϵ∞∇p∞ ∆u = ∆u0+ϵ∆u1+ϵ2∆u2........ϵ∞∆u∞ now we move to (u · ∇)u Evaluating (u · ∇)u in D∞ to see what happens to the (u · ∇)u term in D∞lets start in D2 1) in D2 ϵ2= 0 u =u0+ϵu1 u · ∇u = (u0+ϵu1)· ∇(u0+ϵu1) = (u0· ∇)u0+ϵ[(u0· ∇)u1+ (u1· ∇)u0] Real Part = (u0· ∇)u0 Imaginary part or ϵpart = (u0· ∇)u1+ (u1· ∇)u0 2) in D3 ϵ3= 0 u =u0+ϵu1+ϵ2u2 (u · ∇)u = (u0+ϵu1+ϵ2u2)· ∇(u0+ϵu1+ϵ2u2) = (u0· ∇)u0+ϵ[(u0· ∇)u1+ (u1· ∇)u0] + ϵ2[(u0· ∇)u2+ (u1· ∇)u1+ (u2· ∇)u0] What can be noticed was the sum of the indices was constant which which were the number of terms for the ϵipart N(ϵ0) = 1 7 N(ϵ1) = 2 N(ϵ2) = 3 N(ϵi) =? This is because the (u · ∇)u term expands in the following way (u · ∇)u = 2n−2 X i=0 ϵi(X j+k=i 0≤j,k≤n−1 (uj· ∇)uk) now since ϵi>0 ϵi= 0 iff i∈[0,2n−2] ∩N (u · ∇)u = n−1 X i=0 ϵi(X j+k=i 0≤j,k≤n−1 (uj· ∇)uk) When u0=uj∀j∈N The ϵipart will be = N(ϵi)(u0· ∇)u0 (u · ∇)u = ( n−1 X i=0 N(ϵi)·ϵi)(u0· ∇)u0 or (u · ∇)u = (u0· ∇)u0(Pn−1 i=0 N(ϵi)×ϵi) now what is N(ϵi)? N(ϵi) = f(i) f:N−→ N Such that f(i) is the number of ways ∃j, k ∈Nand j, k ≤n−1 and j+k=i now if i, j, k ≤n−1 then f(i) = i+ 1 if ui=uj∀i, j then in D∞ (u · ∇)u = (u0· ∇)u0×( n−1 X i=0 (i+ 1)ϵi) 8 Now plugging this all to get a Navier Stokes in D∞gives us (if ui=uj) 1 1−ϵ ∂u0 ∂t + (u0· ∇)u0(1 + 2ϵ+ 3ϵ2. . . ) = −1 1−ϵ∇p0+ν∇2u0 or ∂u0 ∂t + (u0· ∇)u0(1 + 2ϵ+ 3ϵ2. . . )(1 −ϵ) = −∇p0+ν∇2u0 now what is (1 −ϵ)(1 + 2ϵ+ 3ϵ2. . . ) 1+2ϵ+ 3ϵ2+ 4ϵ3+ 5ϵ4+ 6ϵ5· · · +∞ϵ∞−1+∞ϵ∞ −ϵ−2ϵ2−3ϵ3−4ϵ4−5ϵ5−6ϵ6· · · − ∞ϵ∞ = 1 + ϵ+ϵ2+ϵ3· · · +∞ϵ∞−1 =1 1−ϵ in a strange sense (1 + 2ϵ+ 3ϵ2. . . ) = 1 (1 −ϵ)2 so in a weird way we have a scalar version of Navier Stokes with a coefficient. ∂u0 ∂t + (u0· ∇)u0 1 1−ϵ=−∇p0+ν∇u0 or ∂u0 ∂t + (u0· ∇)u0F=−∇p0+ν∇u0 in this if Fis a constant then this is probably a 1D version of Burgers Equation. References [1] Partial differential equation. Wikipedia. https://en.wikipedia.org/wiki/Partial_differential_equation [2] Heat equation. Wikipedia. https://en.wikipedia.org/wiki/Heat_equation [3] Wave equation. Wikipedia. https://en.wikipedia.org/wiki/Wave_equation [4] Navier–Stokes equations. Wikipedia. https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations [5] Algebra over a field. Wikipedia. https://en.wikipedia.org/wiki/Algebra_over_a_field 9