Dear Users,

I hope everyone here is fine. I want to extract dat file from the attached contour plot file. Please help me to resolve this issue. Many thanks in advance

Dat_file.mw

Dear Users,
I hope you are doing well. The following is the code to solve a nonlinear PD equation numerically and I plotted the graphs for T(y,t) sucessfully.

restart; with(plots); PDE1 := Pr*(diff(T(y, t), t)-Ree*(diff(T(y, t), y))) = (1+Nr*(T(y, t)+1)^3)*(diff(T(y, t), y, y))+3*Nr*(T(y, t)+1)^2*(diff(T(y, t), y))^2; ICandBC := {T(1, t) = 1, T(y, 0) = 1, (D[1](T))(0, t) = T(0, t)}; Ree := .1; Pr := 6.2; HA1 := [0, 1, 10]; AA := [red, green, blue, cyan, purple, black];
printlevel := 2; for i to nops(HA1) do Nr := op(i, HA1); print("Nr = ", %); PDE[i] := {PDE1}; pds[i] := pdsolve(PDE[i], ICandBC, numeric, spacestep = 1/200, timestep = 1/100); PlotsT[i] := pds[i]:-plot[display](T(y, t), t = 1, linestyle = "solid", labels = ["y", "u"], color = op(i, AA), numpoints = 800) end do;
display([`$`(PlotsT[j], j = 1 .. nops(HA1))], size = [1000, 600], axes = boxed, labels = [x, (convert("T", symbol))(x, T)], labelfont = ["Times", 14, Bold], labeldirections = [horizontal, vertical], axesfont = ["Arial", 14, Bold], thickness = 3)

I want to plot the graphs for (1+Nr*(T(y, t)+1)^3)*(diff(T(y, t), y)), at t = 1. Also want to plot diff(T(y, t), y) at y = 0 and y = 1 against Nr. Kindly help me in this matter.

Dear Users!
I hope are fine here. I got the following expression after a lot of computations

((1/2)*r*(r-1)+(1/6)*r*(r-1)*(r-2))*`Δy`[-1]^3+(1/2)*r*(r-1)*`Δy`[-1]^2+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4)*`Δy`[-2]^7+((1/6)*r*(r-1)*(r-2)+(1/12)*r*(r-1)*(r-2)*(r-3)+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4))*`Δy`[-2]^5+((1/6)*r*(r-1)*(r-2)+(1/24)*r*(r-1)*(r-2)*(r-3))*`Δy`[-2]^4+((1/24)*r*(r-1)*(r-2)*(r-3)+(1/60)*r*(r-1)*(r-2)*(r-3)*(r-4))*`Δy`[-3]^7+((1/24)*r*(r-1)*(r-2)*(r-3)+(1/60)*r*(r-1)*(r-2)*(r-3)*(r-4))*`Δy`[-3]^6+r*`Δy`[0]+y[0]

Actually, for the above, I want the factorization of each coefficient of `Δy`[0], `Δy`[-1], `Δy`[-2] etc and the above expression shoud be in descending order given as:

y[0]+r*`Δy`[0]+(1/2)*r*(r-1)*`Δy`[-1]^2+(1/6)*r*(r-1)*(1+r)*`Δy`[-1]^3+(1/24)*r*(r-1)*(r-2)*(1+r)*`Δy`[-2]^4+(1/120)*r*(r-1)*(r-2)*(r+2)*(1+r)*`Δy`[-2]^5+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4)*`Δy`[-2]^7+(1/120)*r*(r-1)*(r-2)*(r-3)*(-3+2*r)*`Δy`[-3]^6+(1/120)*r*(r-1)*(r-2)*(r-3)*(-3+2*r)*`Δy`[-3]^7

I am waiting for your positive response. Thanks

Dear Users! Hope everyone is fine here. I have some points x_0,x_1,..,x_N and corresponding to these number have values y_0, _1, ..., y_N as,

restart
a := 1; b := 5; h := 1; f := 1/x; N := (b-a)/h;
for i from 0 while i <= N do x[i] := h*i+a; y[i] := eval(f, x = x[i]) end do;

Now I want to develope the following difference table using the values of y_0, y_1, ..., y_N as,

where difference column generated using the following concept

Please help me in this regard. Thanks in advance

Dear Users!

Hope everyone is fine here. Let me explain my problem first for this consider
diff(Y(xi), xi) = mu*(1-Y(xi)^2)
Then the derivative of a function U=u(Y(xi)) using chain rule (and expression menstiones as red) is given as,
diff(U, xi) = (diff(diff(Y, xi), Y))*U and (diff(diff(Y, xi), Y))*U = mu*(1-Y(xi)^2)*(diff(U, Y))
Similarly the second-order derivaitve of U=u(Y(xi)) using chain rule (and expression menstiones as red) is given as,
((&DifferentialD;)^(2))/(&DifferentialD; xi^(2))U=(&DifferentialD;)/(&DifferentialD; xi)(mu (1-Y^(2)(xi))*(&DifferentialD;)/(&DifferentialD; Y)U)=((&DifferentialD;)/(&DifferentialD; Y)*(&DifferentialD;)/(&DifferentialD; xi)Y)(mu (1-Y^(2)(xi))*(&DifferentialD;)/(&DifferentialD; Y)U)=(&DifferentialD;)/(&DifferentialD; Y)(mu^(2) (1-Y^(2)(xi))^(2)*(&DifferentialD;)/(&DifferentialD; Y)U)=-2 Y(xi) mu^(2) (1-Y^(2)(xi))*(&DifferentialD;)/(&DifferentialD; Y)U+ mu^(2) (1-Y^(2)(xi))^(2)*((&DifferentialD;)^(2))/(&DifferentialD; Y^(2))U;
In the similar way I want to compute the higher-order (like 5th order) derivaitve of U w.r.t. xi using the chain rule  (and expression menstiones as red) explained in above. Kindly help me soolve my problem

I am waiting for positive response.