# Runge Kutta method for a system

Posted 5 months ago
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 I want to solve x'=y,y'=6x-y+G,x(0)=1,y(0)=1 with Runge-Kutta Method. Here G lies in [0,1]. I want to get the plot of G versus x. is that possible? May any one help me with that? Here is my code. ClearAll["Global*"]; a = 0; b = 1; x = 1; y = 1; h = 0.1; n = (b - a)/h; Do[t[i] = 0.0 + (i)*h, {i, 0, n}] Do[G[j] = 0.0 + (j)*h, {j, 0, n}] f[t_, G_,x_, y_] = y; g[t_,G_, x_, y_] = 6*x - y+G; Do[S1 = f[t[i], G[i], x[i], y[i]]; P1 = g[t[i], G[i], x[i], y[i]]; S2 = f[t[i] + h, G[i] + h, x[i] + h*S1, y[i] + h*P1]; P2 = g[t[i] + h, G[i] + h, x[i] + h*S1, y[i] + h*P1]; S3 = f[t[i] + h/2, G[i] + h/2, x[i] + h*S2/2, y[i] + h*P2/2]; P3 = g[t[i] + h/2, G[i] + h/2, x[i] + h*S2/2, y[i] + h*P2/2]; S4 = f[t[i] + h, G[i] + h, x[i] + h*S3, y[i] + h*P3]; P4 = g[t[i] + h, G[i] + h, x[i] + h*S3, y[i] + h*P3]; x[i + 1] = x[i] + (S1 + 2 S2 + 2 S3 + S4)*h/6; y[i + 1] = y[i] + (P1 + 2 P2 + 2 P3 + P4)*h/6, {i, 0, k}] Table[{G[j], x[j]}, {j, 0, k}]; Answer
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Posted 5 months ago
 What do you mean by "plot of G versus x"? Isn't G constant?And what about this? eq1 = x'[t] == y[t]; eq2 = y'[t] == 6 x[t] - y[t] + g; DSolve[{eq1, eq2, x == 0, y == 1}, {x, y}, t] Answer
Posted 5 months ago
 Thank you for your reply.G is constant.but let me explain in detail: Let us assume G=0. one gets the value of x(t) versus this G right? Again, let us assume G=0.1.one gets the value of x(t) versus this G right? Again, let us assume G=0.2.one gets the value of x(t) versus this G right? and so on. Now, is it logical to examine the behaviolur x(t) versus these G's? I mean: is it possible get the plot of G versus x(t)? Do you think it is well posed problem?Thanks. Answer
Posted 5 months ago
 Hello Vedat, I am afraid I do not understand what you mean. What do you mean by "one gets the value of x[t] versus G" ? x[t] is a function and not a value! x[ .85 ] for example is a value, and this value certainely depends on G. But for a given t x[ t ] is a linear function of G (look for example at fx[.1, g] ), so you get a straight line. The same is true for y, I attach some code which shows how the functions x[t] and y[t] ( and not a value ) change with G (in my code g) eq1 = x'[t] == y[t]; eq2 = y'[t] == 6 x[t] - y[t] + g; sol = DSolve[{eq1, eq2, x == 0, y == 1}, {x, y}, t]; fx[t_, g_] := Evaluate[(x /. sol[[1, 1]])[]] fy[t_, g_] := Evaluate[(y /. sol[[1, 2]])[]] (* Show x[t] depending on g *) Manipulate[ Plot[fx[tt, g], {tt, 0, 2}, PlotRange -> {0, 6}], {g, 0, 1}] (* Show both x[t] (red) y[t] (blue) depending on g *) Manipulate[ Plot[{fx[t, g], fy[t, g]}, {t, 0, 2}, PlotRange -> {0, 6}, PlotStyle -> {Red, Blue}], {g, 0, 1}] (* Show the vector { x[t], y[t] } depending on g *) Manipulate[ ParametricPlot[{fx[t, g], fy[t, g]}, {t, 0, 2}, PlotRange -> {{0, 2}, {0, 4}}], {g, 0, 1}] Answer
Posted 5 months ago
 Hello Hans, thank you for your invaluable help,again. I had to set up this simple problem. Because actually my goal was to solve the problem in the attached article. I hope you have time for this. The authors solve the system (4.1) on page 11 with the fourth order Runge-Kutta method. Here, except x and y, all other parameters are constant.After calculating x(t) and y(t), we can use (4.5) with the help of (4.2). But for this, we have to find out x(t) and y(t). Please see figure 4 now. g versus Q(s). is something like this possible? that's why I had to impose the simple problem I sent you Am I not right? Attachments: Answer
Posted 5 months ago
 Hello Vedat,I had a look at the paper. Equation 4.1 for me reads (please check it) eq11a = x'[t] == y[t] eq11b = lam (x - lam x^3 + lam^2 x^4) (x'[t])^2 - alf gam (1 + lan x^2) x'[t] - w0^2 x - del x^3 - xi x^5 + gam (1 + lam x^2) ( f Cos[w t] + g cos[ww t]) /. {x'[t] -> y[t], x -> x[t]} Then there are values given like (I omit the beta, it doesn't appear in the equation) val = {alf -> .2, w0 -> I, ww -> 9.842, w -> .5, f -> .05} but inserting these leaves several items open, so a numerical integration ist not possible eq2 /. val What about xi, gam and so on? Answer
Posted 5 months ago
 Hello again Hans; Please see the attached files for my parameters.we can use them to get for example figure 4 c.&By the way, in your eq11b lam^2 x^4 should be replaced with lam^2 x^5Wolfram Notebook Attachments: Answer
Posted 5 months ago
 Hi Vedat,here is a code which gives you some values of Qs and Qc, which of course depend on g. Please have a look about typos and if the results are reasonable.I still don't know what you finally want to do, x[t] is a function and not a value, but I admit I didn't read the paper, only the part around equation ( 4.1 ) eq11a = x'[t] == y[t]; eq11b = y'[t] == lam (x - lam x^3 + lam^2 x^5) (x'[t])^2 - alf gam (1 + lam x^2) x'[t] - w0^2 x - del x^3 - xi x^5 + gam (1 + lam x^2) (f Cos[w t] + g Cos[ww t]) /. {x'[t] -> y[t], x -> x[t]}; val = {alf -> .2, w0 -> I, ww -> 9.842, w -> .5, f -> .05, lam -> 0, beta -> 1, m0 -> 1.5, gam -> 1/m0, del -> beta gam - lam w0^2, xi -> beta gam lam +( lam w0)^2}; eq11b2 = eq11b //. val; sol[gg_?NumericQ, t1_?NumericQ] := Module[{}, eq11b3 = eq11b2 /. g -> gg; NDSolve[{eq11a, eq11b3, x == 1, y == 1}, {x, y}, {t, 0, t1}]] nT = 10; lsg = sol[.3, nT]; fx = x /. lsg[[1, 1]] fy = y /. lsg[[1, 2]] qQs = NIntegrate[fx[t] (Sin[w t] /. val), {t, 0, nT}] 2/nT qQc = NIntegrate[fx[t] (Cos[w t] /. val), {t, 0, nT}] 2/nT Answer
Posted 5 months ago
 Now first I should calculate Qnum given in Eq.(4.5) and then get the plot of g versus Qnum for this m0 given in Fig.4 c color in green interval in [0,200] for g. Answer
Posted 5 months ago
 Hmmmm ?Ok, add to the foregoing code this (but there seems to difficulties during the integration, I didn't have a closer look at this) It doesn't look similar to fig. 4c. qQnum[g_] := Module[{}, nT = 10; lsg = sol[g, nT]; fx = x /. lsg[[1, 1]]; fy = y /. lsg[[1, 2]]; qQs = NIntegrate[fx[t] (Sin[w t] /. val), {t, 0, nT}] 2/nT; qQs = NIntegrate[fx[t] (Sin[w t] /. val), {t, 0, nT}] 2/nT; qQc = NIntegrate[fx[t] (Cos[w t] /. val), {t, 0, nT}] 2/nT; qnum = Sqrt[qQs^2 + qQc^2]] and tt = Table[{qQnum[g], g}, {g, 0, 200}]; ListLinePlot[tt] ListLinePlot[Reverse /@ tt] Answer
Posted 5 months ago
 I think Table[{qQnum[g], g}, {g, 0, 200}] should be replaced with Table[{g,qQnum[g]}, {g, 0, 200}] Right? Answer
Posted 5 months ago
 Hmmmm ?Ok, add to the foregoing code this (but there seems to difficulties during the integration, I didn't have a closer look at this) It doesn't look similar to fig. 4c. qQnum[g_] := Module[{}, nT = 10; lsg = sol[g, nT]; fx = x /. lsg[[1, 1]]; fy = y /. lsg[[1, 2]]; qQs = NIntegrate[fx[t] (Sin[w t] /. val), {t, 0, nT}] 2/nT; qQs = NIntegrate[fx[t] (Sin[w t] /. val), {t, 0, nT}] 2/nT; qQc = NIntegrate[fx[t] (Cos[w t] /. val), {t, 0, nT}] 2/nT; qnum = Sqrt[qQs^2 + qQc^2]] and tt = Table[{qQnum[g], g}, {g, 0, 200}]; ListLinePlot[tt] ListLinePlot[Reverse /@ tt] Answer
Posted 5 months ago
 Reverse@tt interchanges q and qQnum. So you get both plots, Answer
Posted 5 months ago
 Hello Hans, According to the paper, n=1,2,3,...,T=2Pi/w and Ts=nT. I have added them to the code. But I can not run it.I need your help. Attachments: Answer
Posted 5 months ago
 The code works fine for me tt = Table[{g, qQnum[g]}, {g, 0, Ts}] (* {{0, 0.303553}, {1, 0.299951}, {2, 0.296613}, {3, 0.293514}, {4, 0.290635}, {5, 0.287959}, {6, 0.285475}, {7, 0.283173}, {8, 0.281044}, {9, 0.279083}, {10, 0.277284}, {11, 0.275644}, {12, 0.27416}} *) ListLinePlot[tt]  Answer
Posted 5 months ago
 Helo hans, did you add these n = 1; w = 0.5; T = (2N[Pi])/w; Ts = nT; Answer
Posted 5 months ago
 I can not get the same result Could you please as a single code again? Answer
Posted 5 months ago
 I just copied and pasted all of the code you provided above in a new notebook and ran it against a fresh kernel. Generating the tt table takes ~45 sec. to finish on my laptop running "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" Answer
Posted 5 months ago
 Dear Hans, Now it is ok. I fixed it. As a last thing, was my question logical? I mean the plot of G versus qQnum is logical? Right? Answer
Posted 5 months ago
 Hello Vedat,I am afraid I cannot comment on that. I didn't read the whole paper you attached some posts above. What do you mean by "Is it logical to plot quantity1 against quantity2" ? Of course it is. The question is, what do you want to show.I still don't know what you essentially want to do. Reproduce some figures of the paper? Doing some new things? Please excuse me, but I have no idea what you wanted to do. Rohit says, the code (which version of code exactly) works on his system, and he gets a reasonable plot of qQnum against g. If it this what you wanted I dare say, everything is ok. Or isn't it?@Rohit: Did you not encounter error messages during the numerical integration? I got some errors concerning convergence of the integrals. Answer
Posted 5 months ago
 Hi Hans,There were no error messages. I believe you are running V8? Perhaps fixed in a later version. If you have a Wolfram Cloud account (free), you can see the results here. Answer
Posted 5 months ago
 @ RohitHello Rohit, thank you for your answer. " I believe you are running V8? " It is even worse, I am very old fashioned and only have V 7. So this is my situation: NDSolve gave solutions for x[t] and y[t], but NIntegrate gave some errors, and so it is very likely that my qQnum is wrong. And I couldn't reproduce any of the pics in the paper, what seems to be what Vedat wants to do. Answer
Posted 5 months ago
 Hello Hans, Possibly there is some mistyping errors. So, thank you for your help. Vedat. Take care. Answer
Posted 5 months ago
 Hello Hans, Thank you for your prompt replies so far. I was just not familiar with this kind of systems. I mean G is not constant in the beginning. That is why I was trouble with that. In fact my aim was to reproduce some figures to understand the paper well. May be the main trouble is missing initial conditions for the considered systems of odes. For this reason I can not be sure of the results exactly. For example, why can not I still obtain one of the curves given in Fig4.c? I hope I can tell what I mention. This is my case. Answer
Posted 5 months ago
 Dear Hans and Rohit, it should be qnum=Sqrt[qQs^2 + qQc^2]/f instead qnum = Sqrt[qQs^2 + qQc^2] to the paper. Thanks. Answer
Posted 5 months ago
 Hello Hans, I hope you are fine and can speed it up my attached code. For large K's and the small values of T/M, my code runs slowly. Please do not hesitate to ask me anything. Attachments: Answer
Posted 5 months ago
 Hello Vedat,a pdf is not really better than a foto. No chance to run or edit it. And I am afraid I really can't speed up your code. For me it is too complicated in fact. Some general remarks: you should avoid variables starting with a capital letter. Such a variable could be an internal symbol. You don't get an error, but results are not what you expect. So: no captial letters. You should avoid variables with super- and/or subscripts, or both. As far as I remember it occurred sometimes that they were not handled in the way you want to.You should analyze your sums. Are there constants common to some sums and could they be calculated once and outside the sum? I noticed there are power-sums: rewrite them according to Horner's scheme.And are you sure your Do[ For [....constructs is really what you want?Greetings, Hans Answer
Posted 5 months ago
 Dear Hans, Thank you for reply. I will remove capital letters.I have not used Horner's scheme for power-sums before.could you please give a simple example? Moreover I am sure my Do[ For [.... constructs what I really want. Answer
Posted 5 months ago
 Horner seems to be about 10% faster (at least on my machine). And you should change all variables with super and/or subscripts or both. cc = RandomReal[{1, 10}, 6] p1 = Sum[cc[[i]] x^(i - 1), {i, 1, Length[cc]}] p2 = HornerForm[p1] Timing[ Do[ e1 = p1 /. x -> .245, {i, 100000}]; e1] Out= {0.858, 5.16398} Timing[ Do[ e2 = p2 /. x -> .245, {i, 100000}]; e2] Out= {0.749, 5.16398} ` Answer