KIRAN SAJJAN

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MaplePrimes Activity


These are replies submitted by KIRAN SAJJAN

@tomleslie  i need the same figures  MHD_dissipative_flow_stream_lines
 

@tomleslie main aim is to draw the stream lines by the given ode and Bc

I m not having 100% clarity how to plot the thing in maple for that reason i am refering the demo code related to that

Which type of plots i need i will give a demo image for that from other research paper.

@tomleslie

I will write the code according to my knowledge by seeing that reference if found errors or mistaks then can you rectify it sir.

@tomleslie By Referring below work seat how can i plot similar graphs for  my equation i want to plot U'(Y) versus Y  and Theta (Y) versus Y with differnt values of M = 0.5,1, 1.5

And table values  of Cf and Nu with Different values of M=0.5,1,1.5

@tomleslie  can we get similar table values and graphs in HPM method above given file need to match the answer for both the method 

but i have observed in the hpm code table values and Q as well as Theta_b values also not matching single degit for different values of parameter like M=3, Pr=7.2 and N=0.1

for any value not matching with numerical solution

thank you

@tomleslie  why you used in Q integration of U(Y) limit is 0 to 1 but u have given U(Y)(Y) what it means sir similarly in theta B also

@tomleslie By Referring below work seat how can i plot similar graphs for  my equation i want to plot U'(Y) versus Y  and Theta (Y) versus Y with differnt values of M = 0.5,1, 1.5

And table values  of Cf and Nu with Different values of M=0.5,1,1.5

nanofluid_BVP_Sheikholeslami.mw

@tomleslie  can we plot the stream lines for Nussel number and skinfriction with given parameter values sir

@tomleslie i have given mine governing equation sir.. Below i m given same equation from the base paper

@tomleslie

Which answer needed sir.

According to that i will follow the things

@tomleslie any demo code already ploted is there share me according to that i will try to do the same

NULL

eqat := {M.(D(theta))(0)+2.*Pr.f(0) = 0, diff(phi(eta), eta, eta)+2.*Sc.f(eta).(diff(phi(eta), eta))-(1/2)*S.Sc.eta.(diff(phi(eta), eta))+N[t]/N[b].(diff(theta(eta), eta, eta)) = 0, diff(g(eta), eta, eta)-2.*(diff(f(eta), eta)).g(eta)+2.*f(eta).(diff(g(eta), eta))-S.(g(eta)+(1/2)*eta.(diff(g(eta), eta)))-1/(sigma.Re[r]).((1+LinearAlgebra:-HermitianTranspose(d).exp(-eta))/(1+d.exp(-eta))).g(eta)-LinearAlgebra:-HermitianTranspose(beta).((1+LinearAlgebra:-HermitianTranspose(d).exp(-eta))^2/sqrt(1+d.exp(-eta))).g(eta).sqrt((diff(f(eta), eta))^2+g(eta)^2) = 0, diff(theta(eta), eta, eta)+2.*Pr.f(eta).(diff(theta(eta), eta))-(1/2)*S.Pr.eta.(diff(theta(eta), eta))+N[b].Pr.((diff(theta(eta), eta)).(diff(phi(eta), eta)))+N[t].Pr.((diff(theta(eta), eta))^2)+4/3.N.(diff((C[T]+theta(eta))^3.(diff(theta(eta), eta)), eta)) = 0, diff(f(eta), eta, eta, eta)-(diff(f(eta), eta))^2+2.*f(eta).(diff(f(eta), eta))+g(eta)^2-S.(diff(f(eta), eta)+(1/2)*eta.(diff(f(eta), eta, eta)))-1/(sigma.Re[r]).((1+LinearAlgebra:-HermitianTranspose(d).exp(-eta))/(1+d.exp(-eta))).(diff(f(eta), eta))-LinearAlgebra:-HermitianTranspose(beta).((1+LinearAlgebra:-HermitianTranspose(d).exp(-eta))^2/sqrt(1+d.exp(-eta))).(diff(f(eta), eta)).sqrt((diff(f(eta), eta))^2+g(eta)^2) = 0, g(0) = 1, g(6) = 0, phi(0) = 1, phi(6) = 0, theta(0) = 1, theta(6) = 0, (D(f))(0) = 1, (D(f))(6) = 0}

sys1 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .2, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys2 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .4, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys3 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .6, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys4 := eval(eqat, {M = 0, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .8, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys5 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .2, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys6 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .4, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys7 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .6, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys8 := eval(eqat, {M = .5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .8, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys9 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .2, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys10 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .4, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys11 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .6, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys12 := eval(eqat, {M = 1, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .8, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys13 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .2, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys14 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .4, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys15 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .6, LinearAlgebra:-HermitianTranspose(d) = 1.5})

sys16 := eval(eqat, {M = 1.5, N = 2, Pr = .8, S = -2.5, Sc = .5, d = 2, sigma = .2, C[T] = .5, N[b] = .4, N[t] = .4, Re[r] = 1.1, LinearAlgebra:-HermitianTranspose(beta) = .8, LinearAlgebra:-HermitianTranspose(d) = 1.5})

Download Nur(_Beta_star_Vs.M).mw

@tomleslie

These are the equation and Bc for the plot with different parameters here i kept demo figures from other research paper i want plot the similar

@tomleslie

Those plots are demo plots from different paper i need to plot similar for the given parameter values in the equation with different values of M(M=1, M=2,M=3) and other parameters kept constant mentioned in the above equation 

@nm 

restart; N := 0; g := 1; A := 1; B := 0; M := 1; lambda := 1; Ec := 1; OdeSys := diff(U(Y), Y, Y)+Theta(Y)+N*(Theta(Y)*Theta(Y))-(M*M)*U(Y), diff(Theta(Y), Y, Y)+Ec*(diff(U(Y), Y))^2; Cond := U(0) = lambda*(D(U))(0), Theta(0) = A+g*(D(Theta))(0), U(1) = 0, Theta(1) = B; Ans := dsolve([OdeSys, Cond], numeric, output = listprocedure);
NULL;
NUMERIC := [(eval(diff(U(Y), Y), Ans))(.5), (eval(U(Y), Ans))(.5)];

I m getting the numericala ans but i want analytical solution expression with same boundary condition is it possible?

@Carl Love  reference fig i kept sir.

 I have given the all parameter ranges in my code

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