Abstract. Euler Equations Solutions for Incompressible Fluid Flow. Copyright A. A. Frempong

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1 Copright A. A. Frempong Euler Equations Solutions for Incompressible Fluid Flow Abstract The simplest solution is usuall the best solution---albert Einstein This paper covers the solutions of the Euler equations in -D and 4-D for incompressible fluid flow.the solutions are the spin-offs of the author's previous analtic solutions of the Navier-Stokes equations vira: of 014. However, some of the solutions contained implicit terms. In this paper, the implicit terms have been epressed eplicitl in terms of,, and t. The author applied a new law, the law of definite ratio for fluid flow. This law states that in incompressible fluid flow, the other terms of the fluid flow equation divide the gravit term in a definite ratio, and each term utilies gravit to function. The sum of the terms of the ratio is alwas unit. This law evolved from the author's earlier solutions of the Navier-Stokes equations. In addition to the usual approach of solving these equations, the Euler equations have also been solved b a second method in which the three equations in the sstem are added to produce a single equation which is then integrated. The solutions b the two approaches are identical, ecept for the constants involved. From the eperience gained in solving the linearied Navier-Stokes equations, onl the equation with the gravit term as the subject of the equation was integrated. The eperience was that when each of the terms of the Navier-Stokes equation was used as the subject of the equation, onl the equation with the gravit term as the subject of the equation produced a solution. Ratios were used to split-up the direction Euler equation with the gravit term as the subject of the equation. The resulting five sub-equations were readil integrable, and even, the nonlinear sub-equations were readil integrated. The integration results were combined. The combined results satisfied the corresponding equation. This equation which satisfied its corresponding equation would be defined as the driver equation; and each of the other equations which would not satisf its corresponding equation would be called a supporter equation. A supporter equation does not satisf its corresponding equation completel, but provides useful information which is not apparent in the solution of the driver equation. The solutions and relations revealed the role of each term of the Euler equations in fluid flow. The gravit term is the indispensable term in fluid flow, and it is involved in the forward motion of fluids. The pressure gradient term is also involved in the forward motion. The variable acceleration term is also involved in the forward motion. The fluid flow behavior in the Euler solution ma be characteried as follows. The direction solution consists of linear, parabolic, and hperbolic terms. If one assumes that in laminar flow, the ais of smmetr of the parabola for horiontal velocit flow profile is in the direction of fluid flow, then in turbulent flow, the ais of smmetr of the parabola would have been rotated 90 degrees from that for laminar flow. The characteristic curve for the nonlinearterm is such a parabola whose ais of smmetr has been rotated 90 degrees from that of laminar flow. The nonlinear term is similar parabolicall to the nonlinear term. The characteristic curve for the nonlinear term is a combination of two similar parabolas and a hperbola. If the above direction flow is repeated simultaneousl in the and directions, the flow is chaotic and consequentl turbulent. 1

2 Euler Equations Solutions Method 1 Solutions of the Euler Equations of Fluid flow Method 1 In the Navier-Stokes equation, if µ 0, one obtains the Euler equation. From µ p g t, one obtains p Euler equation : µ 0 g t or p t g <---driver equation. Euler equation µ 0 : 1 p t g <---driver equation Split the equation using the ratio terms f, h, n, q, d,, and solve. f h n q d 1 1. fg t. hg 4. ng 4. qg. 1 p dg fgt 4 d hg fgt d d ng d d qg 1 p d dg d hgd d ngd d qgd; p hg or ng ψ qg ψ dg ng hg ψ qg ψ p dg C 0 ± hg 0 ng qg t fg t hg ψ ψ,,, ± C P dg f h n q d 1 0, 0 Find the test derivatives to check in the original equation. 1. fg. hg t ; hg ;. ng 4. hg, 0 0 qg. p dg 0 p ψ ψ 1 g t Above, and are arbitrar functions fg hg ng qg d 1? g g fg hg ng qg dg? g g f h n q d? g g 1? g f h n q d 1? g g Yes

3 Euler Equations Solutions Method 1 The relation obtained satisfies the Euler equation. Therefore the solution to the Euler equation : p t g is ng qg ψ ψ,,, t fgt ± hg ; P dg direction arbitrar functions 0, 0; d f h n q 1 Similarl, the equations and solutions for the other two directions are respectivel For, p g t λg λ8g ψ t g t g ψ,,, λ ± λ ; P λg 4 0, 0; λ λ λ λ λ For : p g t β g g β ψ t g t g ψ,,, β ± β8 ; P 0, 0; β β β β β β g 4 -direction -direction One will net solve the above sstem of solutions for,, in order to epress ng and qg e in terms of,,, and t. Solving for,, ng, and qg e ψ ψ. A. B ng qg fgt ± hg λg λ8g ψ ψ λgt ± λg βg β g ψ βgt ± β8g ψ. C Let, and., and are being used for simplicit. The will be changed back to,, and later, and the do not represent the variables, and in the sstem of solutions

4 Euler Equations Solutions Method 1 Step 1 From the above sstem of solutions, let A fgt hg ; D qg ; E ng B λ g t λ g ; F λ g ; G λ g 8 C β g t β g ; J β g ; L β g 8 Step : Then the solutions to the Euler sstem of equations become ignoring the arbitrar functions A D E B F G M C J L Step A D E B F G C J L 1 N Step 4 0 A D E B F G 0 A D E C J L 0 B F G C J L 4 P Maples software was used to solve sstem P to obtain Step Note: L FCD FCJ JLA JCE None of the popular academic programs C BLD BLJ GLA GCE could solve the sstem in M. L FCD FCJ JLA JCE Maples solved sstem P step 4 above for back to C BLD BLJ GLA GCE,,and in terms of A, B, C, D. E. F, G. L J. and L. ; C Note also that, and are not the same as L changing back to C as agreed to the, and in the sstem of equations.. LD J The were used for convenience and ; LA CE simplicit. LD J changing back to LA CE as agrred to Step : Appl and substitute from in steps -8 below A fgt ± hg ; B λgt ± λg ; C βgt ± β8g ; D qg ;, E ng ; F λ g ; G λ g J β g ; L β g 8 Step L β g C βgt ± β8g ng β g ng β gt± β g ng ng[ β g t g ± β8 ] β g 8 ; 0 Step ng [ β gt ng g ng ] ± [ β8 ] βg ng n g t g ng β β 8 ± β β g ng nβ gt β g ng 8 ± β β g 4

5 Euler Equations Solutions Method 1 Step 8: LD J LA CE β g qg g [ β ] βg fgt ± hg βgt ± β8g ng qg β g qg g qg [ β ] β g fg t ± hg β g t ± β g ng qg βg fgt ± hg βgt ± β8g ng βg [ qg βg ] qg βgfgt ± hgβg βgtng ± β8g ng βg [ qg βg ] qggβfgt ± hgβggq βggnqt ± β8g ggnq βg [ qg βg ] qg qggβfgt ± hgβggq βggnqt ± β8g ggnq βg[ qg βg] Dividing out the " " in the numerator and the denominator 8 qg qg g fg t hg g g q g g nqt g g g nq β ± β β ± β8 β g [ qg β g ] or qg qg g fg t hg g g q g g nqt g g g nq β ± β β ± β8 ββgg β qgg qg βfggg q βgg nq t ± hg βggq ± β8g ggnq ββgg β qgg Summar for the fractional terms of the -direction solution ng and qg in terms of,, and t ng gt g ng nβ β 8 ng kg gk ± β β g B kgt 1 ± g n k1 β β ; k β 8; k n qg ; β βfgggq e β ggnqt e ± hg ggq ± gggnq β β8 ββgg βqgg C qg g g k g g k t g k g g k g k g k 4 ± ± 8 gk 10 ggk 11 k4 β fq ; k β nq;; k h ; k 8 β 8; k9 nq k 10 ββ k11 βq 9

6 Euler Equations Solutions Method 1 Analsis of Solutions ng qg ψ ψ,,, t fgt ± hg ; P dg direction arbitrar functions 0, 0; d f h n q 1 ng nβ gt β g ng 8 ± β β g B qg β fg q g g nq t hg g g q g nq g β ± ± β β8 g ββ βqgg C g d f h n q 1; λ4 λ λ λ λ8 1; β4 β β β β8 1 One observes above that the most important insight of the above solution is the indispensabilit of the gravit term in incompressible fluid flow. Observe that if gravit, g were ero, the first four terms of the velocit solution and P would all be ero. This result can be stated emphaticall that without gravit forces on earth, there would be no incompressible fluid flow on earth as is known. More Observations: Comparison of the Euler solutions with equations of motion under gravit and liquid pressure of elementar phsics Motion under gravit equations: B : gt; C : g ; Liquid Pressure, P at the bottom of a liquid of depth h units is given b P gh Observe the following similarities above: 1. Observe the " gt " in gt of B of the motion equations and the fg t in the Euler solution.. Observe the " g " in g of C and the hg in the Euler solution.. Observe the P gh of the liquid pressure and the P d g of the Euler solution. There are five main terms ignoring the arbitrar functions in the Euler solution. Of these five terms, three terms, namel, fg t, hg, d g are the same ecept for the constants involved as the terms in the equations of motion under gravit. This similarit means that the approach used in solving the Euler equation is sound. One should also note that to obtain these three terms simultaneousl, onl the equation with the gravit term as the subject of the equation will ield these three terms. The author suggests that this form of the equation with the gravit term as the subject of the equation be called the standard form of the Euler equation, since in this form, one can immediatel split-up the equations using ratios, and integrate. elocit profile The direction solution consists of linear, parabolic, and hperbolic terms. If one assumes that in laminar flow, the ais of smmetr of the parabola for horiontal velocit flow profile is in the direction of fluid flow, then in turbulent flow, the ais of smmetr of the parabola would have been rotated 90 degrees from that for laminar flow. The characteristic curve for the nonlinearterm is such a parabola whose ais of smmetr has been rotated 90 degrees from that of laminar flow. The nonlinear term is similar parabolicall to the nonlinear term. The characteristic curve for the nonlinear term is a combination of two similar parabolas and a hperbola. If the above direction flow is repeated simultaneousl in the and directions, the flow is chaotic and consequentl turbulent

7 Euler Equations Solutions Method 1 Standard form of the -direction Euler equation for incompressible fluid flow One will call the Euler equation with the gravit term as the subject of equation in A, the standard form of the Euler equation for the ratio method of solving these equations, since this form produces a solution on integration. None of the other forms in B, C, D, E, or F, produces a solution. That is, the integration results of each of the other five equations do not satisf the corresponding equation. t p g A < standard form p t g B p g C t p g t p g t p g t F Uniqueness of the solution of the Euler equation When each term of the linearied Navier-Stokes equation was made subject of the N-S equation, onl the equation with the gravit term as the subject of the equation produced a solution. vira: of 014. Similarl. the solution of the Euler solution is unique. Etra: Linearied Euler Equation: If one linearies the Euler equation as was done in the lineariation of the Navier-Stokes equation, one obtains 4 1 p g t ; whose solution is fg t C ; P d g. 4 Note: B comparison with Navier-Stokes equation and its relation, a relation to Euler equation can be found b deleting the Navier-Stokes relation resulting from the µ -terms. D E Also note: The liquid pressure, P at the bottom of a liquid of depth h units is given b P gh. From the Euler solution in this paper, P d g from integrating dp d g where d is ratio d term. Each of the other terms in the Euler equation must also be set equal to the product of a ratio term and g. This result implies that the approach used in solving the Euler equations is valid.

8 Euler Equations Solutions Method 1 Conclusion The direction Euler equation with the gravit term as the subject of the equation was split-up into sub-equations using ratios. The sub-equations were solved and combined. The relation obtained from the integration satisfied the corresponding Euler equation. Similarl the direction and the direction equations were solved. B solving algebraicall and simultaneousl for, and, the ng and qg terms have been epressed eplicitl in terms of,, and t. One ma note that in checking the relations obtained for integrating the equations for possible solutions, one needs not have eplicit epressions for,, and, since these behave as constants in the identit checking process. The above solution is the solution everone has been waiting for, for nearl 10 ears. It was obtained in two simple steps, namel, splitting the equation using ratios, integrating and successfull checking for identit in the original equation. The fluid flow behavior in the Euler solution ma be characteried as follows. The direction solution consists of linear, parabolic, and hperbolic terms. If one assumes that in laminar flow, the ais of smmetr of the parabola for horiontal velocit flow profile is in the direction of fluid flow, then in turbulent flow, the ais of smmetr of the parabola would have been rotated 90 degrees from that for laminar flow. The characteristic curve for the nonlinearterm is such a parabola whose ais of smmetr has been rotated 90 degrees from that of laminar flow. The nonlinear term is similar parabolicall to the nonlinear term. The characteristic curve for the nonlinear term is a combination of two similar parabolas and a hperbola. If the above direction flow is repeated simultaneousl in the and directions, the flow is chaotic and consequentl turbulent. The impediment to solving the Euler equations has been due to how to obtain sub-equations from the si-term equation. The above solution was made possible after pairing the terms of the equation using ratios ratio terms. The author was encouraged b Lagrange's use of ratios and proportion in solving differential equations. One advantage of the pairing approach is that the above solution can easil be etended to an number of dimensions. Finall, to solve a partial differential equation, one integrates properl, and check the integration relations for identit in the original equation. Since the above procedure has been followed and completed successfull, the Euler equations for incompressible fluid flow have been solved. P.S. The liquid pressure, P at the bottom of a liquid of depth h units is given b P gh. From the Euler solution in this paper, P d g from integrating dp d g where d is ratio d term. Each of the other terms in the Euler equation must also be set equal to the product of a ratio term and g. This result implies that the approach used in solving the Euler equations is valid. and perhaps ma be the onl approach to produce such a result b integrating the Euler equation. 8

9 Euler Equations Solutions Method Solutions of -D Euler Equations Method Here, the three equations below, will be added together; and a single equation will be integrated p t g 1 p g t p t g Step 1: Appl the aiom, if a b and c d, then a c b d; and therefore, add the left sides and add the right sides of the above equations. That is, 1 g g g p p p t t t g g g Two lines per equation Let g g g G, where G g g g to obtain p p p t t t G Step : Solve the above 1-term equation using the ratio method. 1 ratio terms The ratio terms to be used are respectivel the following: Sum of the ratio terms 1 β1, β, β, β4, β, β, λ1, λ, λ, λ4, λ, λ, λ, λ8, λ9 p p p β ; β β 1 G G ; G ; β β G G βg λg t 4 ; t ; t ; 1 ; λ G ; G G λ λ ; 4 ; λ G ; G λ λg λg ; ; 8 ; λ 9 G 1. p β 1 G dp βg d 1 P βg C 1 19 p β G dp βg d P βg C 0 p β G dp βg d P βg C 1 9

10 Euler Equations Solutions Method 4. βg t 4 d β G dt 4 β Gt C 4, λ 1 G λ1g d λ d 1G d λ1gd λ G 1 λ1g ± λ G C 1 βg t d β G dt β Gt C 8 λg d λ d G d λgd λg ψ λ G ψ βg t d β G dt β Gt C 9 4 λg d λ d G d λgd λg ψ λ G ψ 10 λ 4 G d λ d 4G d λ4gd λ4g ψ λ G ψ 4 1 λ G d λ G d d λgd λg ψ λ G ψ 11 λ G d λ G d d λ Gd λ G λ G ± λ G C 14 λg d λ d 8G d λ8gd λ8g ψ λ G ψ λg d λ d G d λgd λg ψ λ G ψ 1 λg d λ d 9G d λ9gd λ G 9 λ9g ± λ G C

11 Euler Equations Solutions Method Step : One Collects the integrals of the sub-equations, above, for,,, P, P, P For, P For, P For, P Sum of integrals from Sum of integrals from Sum of integrals from sub-equations #4, #, #8, sub-equations #, #10, sub-equations #, #1, #14, #9, #1, #11, #1, #. #1, #, β4gt C β Gt C 1 βgt C4 ± λ1g C λ G ψ λ G 4 ψ λ G ψ λ G ψ ± λg C 8 λ G ψ λ G ψ G C ± λ9 P β 1 G C19 P β G C P βg C 0 1 From above, For, Sum of integrals from sub-equations #4, #, #8, #9, #1, λ G λ G ψ t G G,,, β ± λ ψ 4 t 1 ; P βg C arbitrar functions For : Sum of integrals from sub-equations #, #10, #11, #1, #. λ G G t Gt G,,, β ± λ 4 λ ψ ψ ; P βg C arbitrar functions 0 For : Sum of integrals from sub-equations #, #1, #14, #1, #, λ G λ G ψ ψ Gt G β ± λ 8 P βg C arbitrar functions 9 1 Step 4: Simplif the sums of the integrals from above..method solutions of N-S equations λ G λ G ψ t Gt G,,, β ± λ ψ arbitrar functions P βg C 0, λ G λ G ψ t Gt G ψ,,, β ± λ arbitrar functions P βg C 0,

12 Euler Equations Solutions Method λ G λ G ψ t Gt G ψ,,, β ± λ arbitrar functions P βg C 0, 0 1 The above are solutions for, P, P, P.of the Euler Equations Comparison of Method 1 and Method of Solutions of Euler Equations Method 1: direction solution of Euler equation ng qg ψ t fg t hg ψ,,, ± C P dg; d f h n q 1 0, 0 arbitrar functions 44 A Method : direction solution of Eulerequation λ G λ G ψ t Gt G,,, β ± λ ψ B arbitrar functions P βg C 0, It is pleasantl surprising that the above solutions A and B are almost identical ecept for the constants, even though the were obtained b different approaches as in Method 1 and Method.. For the sstem of Euler equations, there are three driver equations, since each equation contains the gravit term. In Method 1, the gravit term was g. In Method, the gravit term was G, where G is the magnitude of the vector sum of the gravit terms. Note that in Method 1, the sum of the ratio terms ratio terms for each equation equals unit, but in Method, the sum of the ratio terms 1 ratio terms for the single driver equation solved equals unit. Note that in Method, onl a single "driver" equation was solved, but in Method 1, three "driver" equations were solved. In Method, one could sa that the sstem of Euler equations was "more simultaneousl" solved than in Method 1. About integrating onl a single equation If one asked for help in solving the Euler equations, and one was told to add the three equations together and then solve them, one would think that one was being given a nonsensical advice; but now, after studing Method above, one would appreciate such a suggestion. References: For paper edition of the above paper, see the book entitled "Power of Ratios" b A. A. Frempong, published b Yellowtetbooks.com. Without using ratios or proportion, the author would never be able to split-up the Euler equations into sub-equations which were readil integrable. The impediment to solving the Euler equations for over 10 ears has been due to finding a wa to split-up the equations. Since ratios were the ke to splitting the Euler equations and solving them, the solutions have also been published in the " Power of Ratios" book which covers definition of ratio and applications of ratio in mathematics, science, engineering, economics and business fields. 1