Message Boards

WOLFRAM COMMUNITY

3245 Views

8 Replies

3 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Improved Eulers Method Problem

Athanasios Paraskevopoulos

Athanasios Paraskevopoulos, Hellenic Open University

Posted 1 year ago

It is given the following problem: $$X'=Z+(Y-\alpha)X$$ $$Y'=1-\beta Y-X^2$$ $$Z'=-X-\gamma Z$$ with initial conditions $(X(0),Y(0),Z(0)=(1,2,3)$. Where X: interest rate Υ:investment demand Z: price index $\alpha$: savings, $\beta$: cost per investment, $\gamma$: the absolute value of the elasticity of demand \[Alpha]=0.9; \[Beta]=0.2; \[Gamma]=1.2; f[x_,y_,z_]:=z+(y-\[Alpha])x g[x_,y_,z_]:=1-\[Beta]y-x^2 h[x_,y_,z_]:=-x-\[Gamma]z I know how to define a system with two functions so I could use these Method A: accuracy of order h S[a_, b_, h_, N_] := (u[0] = a; u[1] = a + hb; Do[u[n + 1] = 2 u[n] - u[n - 1] + hhf[nh, u[n], (u[n] - u[n - 1])/h], {n, 1, N}]) Method B: accuracy of order h^2 Q[a_, b_, h_, N_] := (u[0] = a; v[0] = b; Do[{u[n + 1] = u[n] + h F[u[n] + (h/2)F[u[n], v[n]], v[n] + (h/2) G[u[n], v[ n]]], \ v[n + 1] = v[n] + h* G[u[n] + (h/2)F[u[n], v[n]], v[n] + (h/2)G[u[n], v[n]]]}, {n, 0, N}])

It is given the following problem: $$X'=Z+(Y-\alpha)X$$ $$Y'=1-\beta Y-X^2$$ $$Z'=-X-\gamma Z$$ with initial conditions $(X(0),Y(0),Z(0)=(1,2,3)$.

Where

X: interest rate

Υ:investment demand

Z: price index

$\alpha$: savings, $\beta$: cost per investment, $\gamma$: the absolute value of the elasticity of demand

\[Alpha]=0.9;
\[Beta]=0.2;
\[Gamma]=1.2;
f[x_,y_,z_]:=z+(y-\[Alpha])*x
g[x_,y_,z_]:=1-\[Beta]*y-x^2
h[x_,y_,z_]:=-x-\[Gamma]*z

I know how to define a system with two functions so I could use these

Method A: accuracy of order h

S[a_, b_, h_, N_] := (u[0] = a; u[1] = a + h*b; 
Do[u[n + 1] = 
2 u[n] - u[n - 1] + h*h*f[n*h, u[n], (u[n] - u[n - 1])/h], {n, 1, 
N}])

Method B: accuracy of order h^2

    Q[a_, b_, h_, N_] := (u[0] = a; v[0] = b; 
    Do[{u[n + 1] = 
    u[n] + h*
    F[u[n] + (h/2)*F[u[n], v[n]], 
    v[n] + (h/2)*
    G[u[n], v[
    n]]],                                                     \

    v[n + 1] = 
    v[n] + h*
    G[u[n] + (h/2)*F[u[n], v[n]], v[n] + (h/2)*G[u[n], v[n]]]}, {n,
    0, N}])

POSTED BY: Athanasios Paraskevopoulos

8 Replies

Sort By:

Athanasios Paraskevopoulos

Athanasios Paraskevopoulos, Hellenic Open University

Posted 1 year ago

Another Way I have thought is the following

POSTED BY: Athanasios Paraskevopoulos

Athanasios Paraskevopoulos

Athanasios Paraskevopoulos, Hellenic Open University

Posted 1 year ago

That works but it is not accurate as your solution

POSTED BY: Athanasios Paraskevopoulos

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 1 year ago

Athanasios, sorry - obviously I misunderstood your question! As far as I could understand on the quick the "improved Euler method" I simply would do it like so: \[Alpha] = 0.9; \[Beta] = 0.2; \[Gamma] = 1.2; f[{x_, y_, z_}] := {z + (y - \[Alpha])x, 1 - \[Beta]y - x^2, -x - \[Gamma]*z} h = 0.1; impEuler[v_List] := v + h/2 (f[v] + f[v + h f[v]]) data = NestList[impEuler, {1, 2, 3}, 1000]; Graphics3D[{Arrowheads[{.03, #} & /@ Range[0, 1, .05]], Arrow[data]}, BoxRatios -> Automatic, Axes -> True, AxesLabel -> {"X", "Y", "Z"}] I do hope that this now is more helpful! At least my "solution" above can be used as reference. Best regards -- Henrik

Athanasios,

sorry - obviously I misunderstood your question!

As far as I could understand on the quick the "improved Euler method" I simply would do it like so:

\[Alpha] = 0.9;
\[Beta] = 0.2;
\[Gamma] = 1.2;

f[{x_, y_, z_}] := {z + (y - \[Alpha])*x, 1 - \[Beta]*y - x^2, -x - \[Gamma]*z}

h = 0.1;
impEuler[v_List] := v + h/2 (f[v] + f[v + h f[v]])

data = NestList[impEuler, {1, 2, 3}, 1000];
Graphics3D[{Arrowheads[{.03, #} & /@ Range[0, 1, .05]], Arrow[data]}, 
 BoxRatios -> Automatic, Axes -> True, AxesLabel -> {"X", "Y", "Z"}]

enter image description here

I do hope that this now is more helpful! At least my "solution" above can be used as reference.

Best regards -- Henrik

POSTED BY: Henrik Schachner

Athanasios Paraskevopoulos

Athanasios Paraskevopoulos, Hellenic Open University

Posted 1 year ago

Thank you that was very useful for sure. At my master with have an assignment and we study the Dynamic analysis and control of a new hyperchaotic finance system. For this reason we use Improved Euler's Method

POSTED BY: Athanasios Paraskevopoulos

Cornel B.

Posted 1 year ago

Hello, impEuler[v_List] := v + h/2 (f[v] + f[v + h f[v]]) - Henrik Schachner, Radiation Therapy Center, Weilheim, Germany So, initially we have a function with 3 parameters: x, y, and z (f[{x , y , z _}]), and then we call the function only with one single parameter: v (f[v]). Could you explain what is the meaning of v List, and how f[v] is related to f[{x , y , z }]? Formula for improved Euler method:

POSTED BY: Cornel B.

Rohit Namjoshi

Posted 1 year ago

So, initially we have a function with 3 parameters: x, y, and z (f[{x , y , z _}]) No, it is a function that takes a single parameter, which is a `List` of 3 elements that are locally bound to the symbols `x`, `y`, and `z`. f[{x_, y_, z_}] If it took three parameters it would be f[x_, y_, z_] This question is unrelated to this thread. You should have asked a new question.

POSTED BY: Rohit Namjoshi

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 1 year ago

Athanasios, you did not say so, but I am assuming that you want the above system of DGL get solved: \[Alpha] = 0.9; \[Beta] = 0.2; \[Gamma] = 1.2; eqs = {x'[t] == z[t] + (y[t] - \[Alpha])x[t], y'[t] == 1 - \[Beta]y[t] - x[t]^2, z'[t] == -x[t] - \[Gamma]*z[t]}; {fx, fy, fz} = NDSolveValue[{eqs, x[0] == 1, y[0] == 2, z[0] == 3}, {x, y, z}, {t, 0, 500}]; Plot[{fx[t], fy[t], fz[t]}, {t, 0, 50}, ImageSize -> Large, GridLines -> Automatic] ParametricPlot3D[{fx[t], fy[t], fz[t]}, {t, 0, 500}, ImageSize -> Large, AxesLabel -> {"X", "Y", "Z"}] I did not know that Euler was dealing with interest rates, etc., but interesting though! Does that help? Regards -- Henrik

Athanasios,

you did not say so, but I am assuming that you want the above system of DGL get solved:

\[Alpha] = 0.9;
\[Beta] = 0.2;
\[Gamma] = 1.2;

eqs = {x'[t] == z[t] + (y[t] - \[Alpha])*x[t],
   y'[t] == 1 - \[Beta]*y[t] - x[t]^2,
   z'[t] == -x[t] - \[Gamma]*z[t]};

{fx, fy, fz} = NDSolveValue[{eqs, x[0] == 1, y[0] == 2, z[0] == 3}, {x, y, z}, {t, 0, 500}];
Plot[{fx[t], fy[t], fz[t]}, {t, 0, 50}, ImageSize -> Large, GridLines -> Automatic]
ParametricPlot3D[{fx[t], fy[t], fz[t]}, {t, 0, 500}, ImageSize -> Large, AxesLabel -> {"X", "Y", "Z"}]

enter image description here