Community RSS Feed https://community.wolfram.com RSS Feed for Wolfram Community showing any discussions in tag Calculus sorted by active NIntegrate::inumr error while integrating interpolated functions https://community.wolfram.com/groups/-/m/t/2372900 Hello guys, i have a problem with double Integral length = 10*10^-6; dz = 5*10^-7; anzahl = length/dz; z = N[Range[0,length,dz]]; f = Range[anzahl+1]; For[i=1,i&lt;(anzahl+2),i++,f[[i]]=RandomReal[{-3*10^-9,3*10^-9}]]; koordinaten = Transpose[{z, f}]; finterpol = Interpolation[koordinaten]; finterpol2[zn_] := Piecewise[{{0,zn&lt;0},{finterpol[zn],0&lt;=zn&lt;=length},{0,zn&gt;length}}] I use this Piecewise function &#034;finterpol2[zn_]&#034; with an interpolated function in it for the following integration. (beta, n1, n2 and k0 are constants ) Ftilde[theta_] := NIntegrate[finterpol2[zn]*Re[Exp[-I*(beta-(n2*k0*Cos[theta]))*zn]],{zn,-Infinity,Infinity}]; After that i want to do a definit integration with Ftilde (\[Phi][d] is defined and outputs a constant): alpha = \[Phi][d]^2*(n2^2-n1^2)^2*(k0^3)/(4*Pi*n1)*NIntegrate[(1/length)*Abs[Ftilde[theta]]^2,{theta,0,Pi}] For the integral which calculates alpha I get an error message: &#034;NIntegrate::inumr: The integrand (\[Piecewise] &lt;&lt;1&gt;&gt;) Re[E^(-I zn (1.09246*10^7-5.83728*10^6 Cos[theta]))] has evaluated to non-numerical values for all sampling points in the region with boundaries {{0.,5.*10^-7}}.&#034; and so on. How can I solve this? Thank you in advance for all the answers. Andrei Costina 2021-09-25T14:25:59Z Temperature evolution differ in two models while it should be the same https://community.wolfram.com/groups/-/m/t/2375242 I&#039;m trying to model the heat teansfer of an object with cross sectional shape of a 25mm by 25mm square. The top and bottom surface are insulated, so only the side dessipates heat convectively. In the Mathematica code I first tried non-dimensionalize the 2D heat equation but with the time variable unchanged, and then I also tried non-dimensionalize the time variable. These two approach gives me different modeling results, can someone help me take a look of my code and give me some insight? Really appreciated! The two governing equations I used are listed here for reference: with dimensional time variable: D[θ[t, ξ, ψ], t] - αmodified*(D[θ[t, ξ, ψ], {ξ, 2}] + D[θ[t, ξ, ψ], {ψ, 2}]) dimensionless time variable: D[θ[tau, ξ, ψ], tau] -(D[θ[tau, ξ, ψ], {ξ, 2}] + D[θ[tau, ξ, ψ], {ψ, 2}]) the dimensionless temperature θ=(T-T∞)/(T0-T∞), dimensionless width=x/L, L is the side length of object, equals 25mm, same for dimensionless length ψ=y/L To make the problem more clear, here&#039;s two image, modeled at position ξ = 0.85, ψ = 0.85, the temperature curve from the model with/without dimensionless time looks significantly different although it&#039;s all plotted for the real time 0 to 20 sec. with dimensional time ![enter image description here] with dimensionless time ![enter image description here] : https://community.wolfram.com//c/portal/getImageAttachment?filename=100551.jpg&amp;userId=2339710 : https://community.wolfram.com//c/portal/getImageAttachment?filename=77272.jpg&amp;userId=2339710 you can tell the temperature evolution is very very different, which in theory they shouldn&#039;t Y L 2021-09-28T18:01:02Z OGRe: an object-oriented general relativity package https://community.wolfram.com/groups/-/m/t/2374361 *SUPPLEMENTARY WOLFRAM MATERIALS for the ARTICLE:* &gt; Barak Shoshany (2021). &gt; OGRe: An Object-Oriented General Relativity Package for Mathematica &gt; arXiv: [2109.04193] &gt; [Full article PDF] &amp;[Wolfram Notebook] : https://arxiv.org/abs/2109.04193 : https://arxiv.org/pdf/2109.04193 : https://www.wolframcloud.com/obj/4843389e-0425-418a-a31a-45870ffde600 Barak Shoshany 2021-09-27T19:10:23Z Why does the DSolve not solve the PDE giving the 'Arbitrary functions'? https://community.wolfram.com/groups/-/m/t/2373558 Hello, I have two PDEs (strainDisp11 &amp; strainDisp22) in 2 variables x1 and x2. strainDisp11 is a PDE with the partial differential term in x1 whereas, strainDisp22 is a PDE with the partial differential term in x2 I am trying to solve these two PDEs separately using DSolve (Last two command lines in the attached file), however, the solution is not generated along with the required arbitrary functions C1 which should be f1[x2] and C1 which should be f2[x1] in the respective solutions of the PDEs. Attached is Notebook for your reference. Appreciate your help. &amp;[Wolfram Notebook] : https://www.wolframcloud.com/obj/bed9bc71-dd71-4130-94d4-5adaa362e895 Aslam Kittur 2021-09-25T18:12:19Z Unexpected integration result in Wolfram|Alpha? https://community.wolfram.com/groups/-/m/t/2374008 Hello All, I&#039;m working on a problem which I&#039;m currently stuck on. I would like to calculate the volume of fluid in a horizontal end cap. ![Impression of shape horizontal tank end cap] (here k =1 and r = 1.5) What i do is: Get the describing equation. ![Equation of curve] Then rotate it along the x axis. This gives a circle over the x axis on any given point on the curve. And finally integrate over the y axis. To get the volume. ![Equation of integration] Unfortunately if I put in numbers for k and r and try to compute the volume with y i don&#039;t get the correct results. A full volume with k =1 and r=1.5 should give me: 22.88. What have I done wrong? : https://community.wolfram.com//c/portal/getImageAttachment?filename=MImpressionhorizontaltankendcap.png&amp;userId=2373861 : https://community.wolfram.com//c/portal/getImageAttachment?filename=Curveequation.png&amp;userId=2373861 : https://community.wolfram.com//c/portal/getImageAttachment?filename=Integrationpicture.png&amp;userId=2373861 Wytse Haanstra 2021-09-26T11:49:57Z Empty result when solving two differential equation with NDSolve[ ]? https://community.wolfram.com/groups/-/m/t/2371613 q[x_?NumericQ]:=(830397. - 0.375 x^(87/296))/( 4.51833*10^8 x^(34/37) - 1.02416*10^7 x - 1. x^(383/296)); sol1 = NDSolve[ Rationalize[{y1&#039;[x] == y1[x] (-3 P1[x] - q[x] + G&#039;[x]/G[x]), 3 P1[x] y2[x] + y2&#039;[x] == 0.95 y1[x] q[x], 4 P1[x] y3[x] + (y1[x] G&#039;[x])/G[x] + y3&#039;[x] == 0, 4 P1[x] y4[x] +y4&#039;[x] == 0.05 y1[x] q[x], P1[x] == 2.372463129699238*^-19 Sqrt[y2[x] + y4[x] + y1[x] + y3[x]], G&#039;[x] == (-8.722718060807123*^73 + 1.80559635200202*^-76 G[x]^4 y3[x])/G[x]^2, 3 y1[10^-30] == 10^51 \[Pi]^2, y2[10^-30] == 0, y3[10^-30] == 3.3*10^56 \[Pi]^2, y4[10^-30] == 0, G[10^-30] == 5.468912167778173*^27}], {y1, y2, y3, y4, G}, {x, 10^-30, 10^5}, Method -&gt; &#034;Automatic&#034;, AccuracyGoal -&gt; 100, PrecisionGoal -&gt; 5, WorkingPrecision -&gt; 80] How to solve such coupled numerical differential equation, since it always gives back some empty result instead of some Interpolating Function. John Wick 2021-09-22T16:14:28Z Solver for COVID-19 epidemic model with the Caputo fractional derivatives https://community.wolfram.com/groups/-/m/t/1976589 *MODERATOR NOTE: coronavirus resources &amp; updates:* https://wolfr.am/coronavirus ---------- First version of this code been published on https://mathematica.stackexchange.com/questions/221609/solver-for-covid-19-epidemic-model-with-the-caputo-fractional-derivatives As it is known in biological system with memory it would be suitable to use fractional derivatives to describe evolution of the system. In a current version of Mathematica 12.1 there is no special solver for integrodifferential equations. Here we show solver with using Haar wavelets for dynamic system (3) presented in a paper M.A. Khan, A. Atangana, [Modeling the dynamics of novel coronavirus (2019-nCov) with fractional derivative](https://doi.org/10.1016/j.aej.2020.02.033), Alexandria Eng. J. (2020) [![Figure 1]] with differential operator replaced with the Caputo definition for fractional derivative as follows $$\frac {d f}{dt}\rightarrow \frac {1}{\Gamma (1-q)}\int_0^t{\frac{f&#039;(x)dx}{(t-x)^{q}}}$$ The code below allows us to reproduce Figure 7 from the paper linked above. Let define functions h[x_, k_, m_] := WaveletPsi[HaarWavelet[], m x - k]; h1[x_] := WaveletPhi[HaarWavelet[], x]; Then we can calculate integrals Integrate[h[t, k, m], {t, 0, x}, Assumptions -&gt; {k &gt;= 0, m &gt; 0, x &gt; 0}] Integrate[h1[t], {t, 0, x}, Assumptions -&gt; {x &gt; 0}] Integrate[h[x, k, m]/(t - x)^q, {x, 0, t}, Assumptions -&gt; {t &gt; 0, k &gt;= 0, m &gt; 0, q &lt; 1}] Integrate[h1[x]/(t - x)^q, {x, 0, t}, Assumptions -&gt; {t &gt; 0, q &lt; 1}] With these integrals let define functions p[x_, k_, m_] := Piecewise[{{(1 + k - m*x)/m, k &gt;= 0 &amp;&amp; 1/m + (2*k)/m - 2*x &lt; 0 &amp;&amp; 1/m + k/m - x &gt;= 0 &amp;&amp; m &gt; 0}, {(-k + m*x)/m, k &gt;= 0 &amp;&amp; 1/m + (2*k)/m - 2*x &gt;= 0 &amp;&amp; k/m - x &lt; 0 &amp;&amp; 1/m + k/m - x &gt;= 0 &amp;&amp; m &gt; 0}}, 0] p1[x_] := Piecewise[{{1, x &gt; 1}}, x] pc[t_, k_, m_, q_] := Piecewise[{{-(t^(1 - q)/(-1 + q)), k == 0 &amp;&amp; 1/m - 2*t &gt;= 0 &amp;&amp; m &gt; 0 &amp;&amp; t &gt; 0 &amp;&amp; 1/m - t &gt;= 0}, {-((m^(-1 + q)*(1/(-k + m*t))^(-1 + q))/(-1 + q)), k &gt; 0 &amp;&amp; 1/m + (2*k)/m - 2*t &gt; 0 &amp;&amp; k/m - t &lt; 0 &amp;&amp; m &gt; 0 &amp;&amp; 1/m + k/m - t &gt; 0}, {(-t^q + 2*m*t^(1 + q) - m*t*(-(1/(2*m)) + t)^q)/ (t^q*(-(1/(2*m)) + t)^q*(m*(-1 + q))), k == 0 &amp;&amp; m &gt; 0 &amp;&amp; 1/m - 2*t &lt; 0 &amp;&amp; 1/m - t &gt;= 0}, {(1/(-1 + q))*((2^(-1 + q)*m^(-1 + 2*q)*(-(-(k/m) + t)^q - 2*k*(-(k/m) + t)^q + 2*m*t*(-(k/m) + t)^q + 2*k*(-((1/2 + k)/m) + t)^q - 2*m*t*(-((1/2 + k)/m) + t)^ q))/((1 + 2*k - 2*m*t)*(k - m*t))^q), k &gt; 0 &amp;&amp; 1/m + (2*k)/m - 2*t == 0 &amp;&amp; m &gt; 0 &amp;&amp; 1/m + k/m - t &gt; 0}, {-((1/(-1 + q))*((2^(-1 + q)*m^(-1 + 2*q)* (-2*(-((1/2 + k)/m) + t)^ q*((1 + 2*k - 2*m*t)*(k - m*t))^ q - 2*k*(-((1/2 + k)/m) + t)^q* ((1 + 2*k - 2*m*t)*(k - m*t))^q + 2*m*t*(-((1/2 + k)/m) + t)^q*((1 + 2*k - 2*m*t)* (k - m*t))^q + (-((1 + k)/m) + t)^q* ((1 + 2*k - 2*m*t)*(k - m*t))^q + 2*k*(-((1 + k)/m) + t)^q*((1 + 2*k - 2*m*t)*(k - m*t))^ q - 2*m*t*(-((1 + k)/m) + t)^q* ((1 + 2*k - 2*m*t)*(k - m*t))^ q + (-(k/m) + t)^q* ((1 + 2*k - 2*m*t)*(1 + k - m*t))^q + 2*k*(-(k/m) + t)^q*((1 + 2*k - 2*m*t)*(1 + k - m*t))^q - 2*m*t*(-(k/m) + t)^q*((1 + 2*k - 2*m*t)*(1 + k - m*t))^ q - 2*k*(-((1/2 + k)/m) + t)^q* ((1 + 2*k - 2*m*t)*(1 + k - m*t))^q + 2*m*t*(-((1/2 + k)/m) + t)^q*((1 + 2*k - 2*m*t)* (1 + k - m*t))^ q))/(((1 + 2*k - 2*m*t)*(k - m*t))^q* ((1 + 2*k - 2*m*t)*(1 + k - m*t))^q))), k &gt; 0 &amp;&amp; m &gt; 0 &amp;&amp; 1/m + (2*k)/m - 2*t &lt;= 0 &amp;&amp; 1/m + k/m - t &lt;= 0}, {-((1/(2*m*(-1 + q)))*((2^q*m^(2*q)*t^q*(-(1/m) + t)^q* (-(1/(2*m)) + t)^q - 2^(1 + q)*m^(1 + 2*q)*t^(1 + q)* (-(1/m) + t)^q*(-(1/(2*m)) + t)^q - 2^(1 + q)*m^(2*q)* t^q*(-(1/(2*m)) + t)^(2*q) + 2^(1 + q)*m^(1 + 2*q)* t^(1 + q)*(-(1/(2*m)) + t)^(2*q) + t^q*((-1 + m*t)*(-1 + 2*m*t))^q - 2*m*t^(1 + q)* ((-1 + m*t)*(-1 + 2*m*t))^q + 2*m*t*(-(1/(2*m)) + t)^q* ((-1 + m*t)*(-1 + 2*m*t))^q)/(t^ q*(-(1/(2*m)) + t)^q* ((-1 + m*t)*(-1 + 2*m*t))^q))), k == 0 &amp;&amp; 1/m - 2*t &lt; 0 &amp;&amp; 1/m - t &lt; 0 &amp;&amp; m &gt; 0}, {(1/(-1 + q))*((2^(-1 + q)*m^(-1 + q)*((-m^q)*(-(k/m) + t)^q - 2*k*m^q*(-(k/m) + t)^q + 2*m^(1 + q)*t*(-(k/m) + t)^q + 2*k*m^q*(-((1/2 + k)/m) + t)^q - 2*m^(1 + q)*t* (-((1/2 + k)/m) + t)^ q - ((1 + 2*k - 2*m*t)*(k - m*t))^q* (1/(-1 - 2*k + 2*m*t))^q - 2*k*((1 + 2*k - 2*m*t)*(k - m*t))^q* (1/(-1 - 2*k + 2*m*t))^q + 2*m*t*((1 + 2*k - 2*m*t)*(k - m*t))^q* (1/(-1 - 2*k + 2*m*t))^q))/((1 + 2*k - 2*m*t)*(k - m*t))^ q), 1/m + (2*k)/m - 2*t &lt; 0 &amp;&amp; k &gt; 0 &amp;&amp; m &gt; 0 &amp;&amp; 1/m + k/m - t &gt; 0}}, 0] pc1[t_, q_] := Piecewise[{{-(t^(1 - q)/(-1 + q)), t &lt;= 1}}, -(((-1 + t)^q*t + t^q - t^(1 + q))/((-1 + t)^q*t^q*(-1 + q)))] Now we have all functions to solve a problem with the given parametres Np0 = 8266000; μp (*Natural mortality rate*)= 1/(76.79 365); Πp (*Birth rate*)= μp Np0 ; ηp \ (*Contact rate*)= 0.05; ψ (*Transmissibility multiple*) = 0.02; ηw (*Disease transmission coeﬃcient*)= 0.000001231; θp (*The proportion of asymptomatic \ infection*)= 0.1243; ωp (*Incubation period*)= 0.00047876; ρp (*Incubation period*)= 0.005; τp (*Removal or recovery rate of Ip*)= 0.09871; τap (*Removal or recovery rate of Ap *)= 0.854302; ϱp (*Contribution of the virus to M by Ip*)= 0.000398; ϖp (*Contribution of the virus to M by Ap*) = 0.001; πp(*Removing rate of virus from M*) = 0.01; Let define variables var1 = {Sp1, Ep1, Ip1, Ap1, Rp1, Mp1}; var = {Sp, Ep, Ip, Ap, Rp, Mp}; aco = {aS, aE, aI, aA, aR, aM}; aco1 = {aS1, aE1, aI1, aA1, aR1, aM1}; aco0 = {aS0, aE0, aI0, aA0, aR0, aM0}; The problem can be solved on the unit interval, hence we calculate the collocation points as follows J = 4; M = 2^J; dx = 1/(2*M); A = 0; xl = Table[A + l dx, {l, 0, 2 M}]; xcol = Table[(xl[[l - 1]] + xl[[l]])/2, {l, 2, 2 M + 1}]; We can represent our solution with functions defined above as Sp1[x_, q_] := Sum[aS[i, j] pc[x, i, 2^j, q], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aS1 pc1[x, q]; Sp[x_] := Sum[aS[i, j] p[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aS1 p1[x] + aS0; Ep1[x_, q_] := Sum[aE[i, j] pc[x, i, 2^j, q], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aE1 pc1[x, q]; Ep[x_] := Sum[aE[i, j] p[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aE1 p1[x] + aE0; Ip1[x_, q_] := Sum[aI[i, j] pc[x, i, 2^j, q], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aI1 pc1[x, q]; Ip[x_] := Sum[aI[i, j] p[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aI1 p1[x] + aI0; Ap1[x_, q_] := Sum[aA[i, j] pc[x, i, 2^j, q], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aA1 pc1[x, q]; Ap[x_] := Sum[aA[i, j] p[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aA1 p1[x] + aA0; Rp1[x_, q_] := Sum[aR[i, j] pc[x, i, 2^j, q], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aR1 pc1[x, q]; Rp[x_] := Sum[aR[i, j] p[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aR1 p1[x] + aR0; Mp1[x_, q_] := Sum[aM[i, j] pc[x, i, 2^j, q], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aM1 pc1[x, q]; Mp[x_] := Sum[aM[i, j] p[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + aM1 p1[x] + aM0; All unknown variables should be joined together in one list varM = Join[aco0, aco1, Flatten[Table[{aS[i, j], aE[i, j], aI[i, j], aA[i, j], aR[i, j], aM[i, j]}, {j, 0, J, 1}, {i, 0, 2^j - 1, 1}]]]; Now we are ready to solve the problem. Since we solve system of equations on the unit interval we define scaling function (120 is the length of interval time in days) tn[q_]:= (1/120)^q; eq1[t_, q_] := -tn[q]/Gamma[1 - q] Sp1[t, q] + Πp/ Np0 - μp Sp[t] - ηp Sp[ t] (Ip[t] + ψ Ap[t])/(Sp[t] + Ep[t] + Ip[t] + Ap[t] + Rp[t]) - Np0 ηw Sp[t] Mp[t]; eq2[t_, q_] := -tn[q]/Gamma[1 - q] Ep1[t, q] + ηp Sp[ t] (Ip[t] + ψ Ap[t])/(Sp[t] + Ep[t] + Ip[t] + Ap[t] + Rp[t]) + Np0 ηw Sp[t] Mp[t] - (1 - θp) ωp Ep[ t] - θp ρp Ep[t] - μp Ep[t]; eq3[t_, q_] := -tn[q]/Gamma[1 - q] Ip1[ t, q] + (1 - θp) ωp Ep[t] - (τp + μp) Ip[t]; eq4[t_, q_] := -tn[q]/Gamma[1 - q] Ap1[t, q] + θp ρp Ep[ t] - (τap + μp) Ap[t]; eq5[t_, q_] := -tn[q]/Gamma[1 - q] Rp1[t, q] + τp Ip[ t] + τap Ap[t] - μp Rp[t]; eq6[t_, q_] := -tn[q]/Gamma[1 - q] Mp1[t, q] + ϱp Ip[ t] + ϖp Ap[t] - πp Mp[t]; With these equations we can calculate Figure 6 from the paper above with the next piece of code eq[q_] := Flatten[ParallelTable[{eq1[t, q] == 0, eq2[t, q] == 0, eq3[t, q] == 0, eq4[t, q] == 0, eq5[t, q] == 0, eq6[t, q] == 0}, {t, xcol}]]; Do[icv[i] = {Sp == 8065518/Np0, Ep == 200000/Np0, Ip == 282/Np0, Ap == 200/Np0, Rp == 0, Mp == 50000/Np0}; eqM[i] = Join[eq[i], icv[i]]; solv[i] = FindRoot[eqM[i], Table[{varM[[j]], .1}, {j, Length[varM]}], MaxIterations -&gt; 1000]; lstSv[i] = Table[{x 120 , Np0 Evaluate[Sp[x] /. solv[i]]}, {x, 0, 1, .01}]; lstEv[i] = Table[{x 120, Np0 Evaluate[Ep[x] /. solv[i]]}, {x, 0, 1, .01}]; lstIv[i] = Table[{x 120, Np0 Evaluate[Ip[x] /. solv[i]]}, {x, 0, 1, .01}]; lstAv[i] = Table[{x 120, Np0 Evaluate[Ap[x] /. solv[i]]}, {x, 0, 1, .01}]; lstRv[i] = Table[{x 120, Np0 Evaluate[Rp[x] /. solv[i]]}, {x, 0, 1, .01}]; lstMv[i] = Table[{x 120, Np0 Evaluate[Mp[x] /. solv[i]]}, {x, 0, 1, .01}];, {i, {99/100, 9/10, 8/10, 7/10, 6/10}}];] Visualization of solution: {ListLinePlot[Table[lstSv[i], {i, {99/100, 9/10, 8/10, 7/10, 6/10}}], Frame -&gt; True, FrameLabel -&gt; {&#034;t, days&#034;, &#034;\!$$\*SubscriptBox[\(S$$, $$p$$]\)&#034;}, PlotRange -&gt; All], ListLinePlot[ Table[lstEv[i], {i, {99/100, 9/10, 8/10, 7/10, 6/10}}], Frame -&gt; True, FrameLabel -&gt; {&#034;t, days&#034;, &#034;\!$$\*SubscriptBox[\(E$$, $$p$$]\)&#034;}, PlotRange -&gt; All], ListLinePlot[ Table[lstIv[i], {i, {99/100, 9/10, 8/10, 7/10, 6/10}}], Frame -&gt; True, FrameLabel -&gt; {&#034;t, days&#034;, &#034;\!$$\*SubscriptBox[\(I$$, $$p$$]\)&#034;}, PlotRange -&gt; All], ListLinePlot[ Table[lstAv[i], {i, {99/100, 9/10, 8/10, 7/10, 6/10}}], Frame -&gt; True, FrameLabel -&gt; {&#034;t, days&#034;, &#034;\!$$\*SubscriptBox[\(A$$, $$p$$]\)&#034;}, PlotRange -&gt; All], ListLinePlot[ Table[lstRv[i], {i, {99/100, 9/10, 8/10, 7/10, 6/10}}], Frame -&gt; True, FrameLabel -&gt; {&#034;t, days&#034;, &#034;\!$$\*SubscriptBox[\(R$$, $$p$$]\)&#034;}, PlotRange -&gt; All], ListLinePlot[ Table[lstMv[i], {i, {99/100, 9/10, 8/10, 7/10, 6/10}}], Frame -&gt; True, FrameLabel -&gt; {&#034;t, days&#034;, &#034;M&#034;}, PlotRange -&gt; All, PlotLegends -&gt; Automatic]} [![Figure 2]] : https://i.stack.imgur.com/mwuYC.png : https://i.stack.imgur.com/j05bg.png : https://i.stack.imgur.com/3NjrP.png Alexander Trounev 2020-05-16T17:30:58Z