## Differential Equations – Complex Solutions for ODEs

How do I solve the following IVP problem in Mathematica, so I get real solutions?

$$Q ((t) = b – dfrac {Q (t)} {100-t}; quad Q (0) = 250$$

I tried the following:

$$text { assumptions} = b> 0; text { assumptions} = t> 0;$$

$$f = text {DSolve} left[left{Q'(t)=b-frac{Q(t)}{100-t},Q(0)=250right},Q,tright][[1,1,2]]$$

$$f (t)$$

$$frac {1} {2} (-2 i pi b t-2 bt log (100) -200 b log (t-100) +2 bt log (t-100) +200 i pi b + 200 b log (100) -5 t + 500)$$

Any help would be appreciated. Many Thanks!!!

## approx. classical analysis and odes – Sturm-Liouville equation with finite number of eigenvalues?

Consider the following Storm Liouville (SL) intrinsic value problem $$x in (- infty, 0)$$ $$(py & # 39;) & # 39; – qy = – lambda ^ 2wy,$$ in which $$p (x) = x ^ 2$$, $$w (x) = 1$$, and $$q (x) = (x / 2 + a) ^ 2 + a$$ with parameters $$a> 0$$, It has a regular singularity $$x = 0$$, We basically hope for something like a homogeneous Dirichlet b.c.

It is solved by the substitution $$y (x) = e ^ {x / 2} x ^ {- frac {1} {2} + sqrt {(a + frac {1} {2}) ^ 2- lambda ^ 2}} u (x)$$This leads directly to a standard confluent hypergeometric equation $$xu & # 39; & # 39; (x) + ( gamma-x) u & # 39; (x) – alphau (x) = 0,$$ in which $$alpha = sqrt {(a + frac {1} {2}) ^ 2- lambda ^ 2} -a + frac {1} {2}$$ and $$gamma = 1 + 2 sqrt {(a + frac {1} {2}) ^ 2- lambda ^ 2}$$, There are two independent solutions (1st type & 2nd type).
Let's take a few ubiquitous arguments when solving the eigensystem related to confluent hypergeometric equations. Non-divergence required $$x = 0$$the 2nd kind is dropped. Non-divergence required $$infty$$the 1st kind is reduced to a polynomial if $$– alpha$$ is a non-negative integer and eigenvalue $$lambda ^ 2$$ is reached.

However, seen from this condition for $$alpha$$obviously We have only a finite and finite series of eigenvalues ​​that is different from the infinite eigen spectrum that the SL theory claims.

What is wrong here? Do I miss a few solutions?

## Differential Equations – Solving a Nonlinear System of 2nd Order ODEs with NDSolve

I'm trying to solve a second-order ODE system with five dependent variables.

The equations are:

``````eqn1 = D[z]* C & # 39; & # 39;[z] - u * C & # 39;[z] == a * r1[z]
eqn2 = k[z]* T & # 39; & # 39;[z] - q * T & # 39;[z] == a * (r1[z]* b + r2[z]* c + r3[z]* d)
``````

from where `A B C D` are constants; `k[z], D[z]ri[z]` (from where `i = 1,2,3`) are dependent variables of `T[z]` and `C[z]`, and `C[z]` and `T[z]` depend on each other and where

``````r1[z] = (kc1[z]* keb[z]* (y1[z]* p - (y3[z]* p + y4[z]* p) / keq[z]* P)) / (1 + keb[z]* y1[Z*p+kh2[Z*p+kh2[z*p+kh2[z*p+kh2[z]* y4[z]* p + kst[z]* y3[z]* p) ^ 2
``````

`r2` and `r3` are like `r1[z]`,

I have the following constraints:

``````initConds = {T[0] == 893.15, T & # 39;[L] == 0, C[0] 0.992, C & # 39;[L] == 0};
``````

from where `L = 9`,

I use it to use `NDSolve` as shown below, but the code did not trigger the system, and the error message was also displayed.

``````s = NDSolve[{}{TeqnsinitConds[{}{TeqnsinitConds[{eqnsinitConds}{T[{eqnsinitConds}{T[z]C[z]}, {z, 0, 9, 0.01}]
``````

NDSolve :: ntdv: Unable to resolve to find an explicit formula for the derivatives. Consider using the Method -> {"EquationSimplification" -> "Residual"} option.

Suppose Mathematica can solve my system, how could I write code? `NDSolve` or another method?

## Differential equations – DSolve for ODEs with terms such as y[t]^ 2

Mathematica does not give me a solution for the ODEs like this:

``````eqns = {t x & # 39;
sol = DSolve[eqns, {x, y}, t]
``````

But if I change it

``````eqns = {t x & # 39;
sol = DSolve[eqns, {x, y}, t]
``````

in which I just change `y`

``` That's really strange. Does anyone know how to solve this? ```
``` ```
``` Author AdminPosted on December 10, 2018Categories ArticlesTags 2, differential, DSolve, equations, ODEs, terms, YT ```
``` Differential Equations – Draw Gradient and Newton Directions for the Parametric System of Nonlinear ODEs I have a bucket (system) of chemical kinetic models, a nonlinear dynamical system, given by: Kf> = 0 and kr> = 0 are the parameters. The initial conditions are A (0) = B (0) = 1 and C (0) = 0. I generated data according to y1 = C (0,5) + noise and y2 = C (2) + noise, where the Noise is normally distributed with mu = 0 and sigma = 0.1 using kf = 0.1 and kr = 2. odes = {A & # 39; B & # 39; C1 & # 39; C1[0] == 0}; odesData = odes /. {kf -> 0,1, kr -> 2}; Solution = NDSolve[odesData, {A, B, C1}, {t, 0, 5}][[1]]; seedRandom[10] y1 = C1[0.5] + RandomVariate[NormalDistribution[0, 0.1]]/. solution seedRandom[30] y2 = C1[2] + RandomVariate[NormalDistribution[0, 0.1]]/. solution Data = {{0.5, y1}, {2, y2}}; I have to visualize the data with an input / output image, a parameter space image and a data space image. For the parameter space and data space images, I also have to plot the gradient and newton directions for several randomly selected parameter values. With ParametricNDSolve I can get an input / output image. kfmax = 1; krmax = 5; number = 100; kfrange = range[0, kfmax, kfmax/numsteps]; krrange = range[0, krmax, krmax/numsteps]; soln = ParametricNDSolve[odes, {A, B, C1}, {t, 0, 5}, {kf, kr}]; eqns = Rate[Table[C1[kf, kr] feweqns = flattening[eqns][[ ;; ;; 1000]]; eqnsplot = plot[feweqns, {t, 0, 2.5}, PlotRange -> All] bestfitplot = plot[model[kf, kr] AxesLabel -> {"t", "C Using ParametricNDSolveValue and FindFit, I can find the best fitting parameters for the model (line is with fit parameters and points are generated data). model = ParametricNDSolveValue[odes, C1, {t, 0, 5}, {kf, kr}] fit = FindFit[data, {model[kf, kr] C1 & # 39; bestfitplot = plot[model[kf, kr] validptplot = ListPlot[{{0.5, y1}, {2, y2}}]; (* the data you are trying to match *) show[bestfitplot, validptplot] I can also visualize the parameter space using the log of the cost function: list1 = eqns /. t -> 0.5; list2 = eqns /. t -> 2; Cost = (list1 - y1) ^ 2 + (list2 - y2) ^ 2; Costs // MatrixForm; (* kf-th row and kr-th column *) parspaceplot = ListContourPlot[Log[cost], PlotLegends -> Automatic, DataRange -> {{0, krmax}, {0, kfmax}}, FrameLabel -> {"kr", "kf"}, Contours -> 50]; bestfitptplot = ListPlot[{{kr, kf}} /. fit]; parspace = show[parspaceplot, bestfitptplot] as well as the data space image max = 2; dy1 = 0.2; dy2 = dy1; modely1 = MapThread[model[#1, #2][0.5] &, Table[{i, j}, {i, 0, max, dy1}, {j, 0, max, dy2}][[#]][Transpose]]& /@ Offer[max/dy1]; modely2 = MapThread[model[#1, #2][2] &, Table[{i, j}, {i, 0, max, dy1}, {j, 0, max, dy2}][[#]][Transpose]]& /@ Offer[max/dy1]; plot1 = ListPlot[{Modely1[{Modely1[{modely1[{modely1[[#]]modely2[[#]]}[Transpose] & /@ Offer[max/dy1], Joined -> True, PlotStyle -> Blue, AxesLabel -> {"y1", "y2"}]; plot2 = ListPlot[{Modely1[{Modely1[{modely1[{modely1[Transpose][[#]]modely2 [Transpose][[#]]}[Transpose] & /@ Offer[max/dy1], Joined -> True, PlotStyle -> Red]; plot3 = ListPlot[{{{y1, y2}}, {{model[kf, kr][0.5] /. fit, model[kf, kr][2] /. fit}}}, PlotStyle -> {{Black, dot, point size[0.025]}, {Green}}]; show[plot1, plot2, plot3] However, if I try to compute the directions of the gradient (j # 39) and Newton ((j # j) .x ==-degrees for x), where j === jacobian, j & # 39; j === Fischer information matrix, r === residuals and j & # 39; denotes the transposition of j, I do not understand what I think I should get. That is, the slope direction is not perpendicular to the contour lines. seedRandom[] Randmax = 10; Randoms = table[{RandomReal[{0, krmax}], RandomReal[{0, kfmax}]}, {i, randmax}]; r = {y1, y2} - (model[kf, kr][#] & / @ {0.5, 2}); j = - (degree[model[kf, kr][#]{kf, kr}]{0.5, 2} /0.1; degree = j [Transpose].r; Fish = j [Transpose].j /. Random[[#, 1]], kf -> Random[[#, 2]]} & /@ Offer[randmax]; grads = grad /. Random[[#, 1]], kf -> Random[[#, 2]]} & /@Offer[randmax]; Newts = LinearSolve[Fish[fish[Fisch[fish[[#]]-grads[[#]]]& /@ Offer[randmax]; gradarrows = graphics[{Black arrow Random[{BlackArrow[{Randoms[{BlackArrow[{randoms[[#]]Random[[#]]+ Normalize[grads[grads[grads[grads[[#]]]/ 2.5}]& /@ Offer[randmax]}]; newtarrows = graphics[{Blue arrow Random[{BlueArrow[{Randoms[{BlueArrow[{randoms[[#]]Random[[#]]+ Normalize[newts[newts[Molche[newts[[#]]]/ 10}]& /@Offer[randmax]}]; show[parspace, gradarrows, newtarrows] Things still do not look good, even if I only have costs for log[cost], This is a homework assignment for a degree in Predictive Modeling. A Jupyter notebook is provided for the class, and I know how to numerically calculate the Jacobian (derive sensitivity equations and solve the 9 equations simultaneously with odeint), but I've invested enough in that code to get it to the end to see. I could do the same with Mathematics NDSolve with the 9 equations, but it seemed like I should be able to get the Jacobi with the parameter equation generated by ParametricNDSolve (similar to the parameter sensitivity section in the ParametricNDSolve reference page). Any suggestions on how I can get the Gradient and Newton directions? (Both math and coding recommendations are welcome). P.S. This is my first post and for the first time I could not solve things with what I could find online and in this forum, which was very helpful! Author AdminPosted on December 5, 2018Categories ArticlesTags differential, Directions, draw, equations, gradient, Newton, nonlinear, ODEs, Parametric, system Classical Analysis and Odes – Is it possible to construct a trigonometric series that converges to \$ (0,1) \$ while diverging in \$ (2,3) \$? By a trigonometric series I mean $$f (x) = sum_ {n = 1} ^ infty a_n e ^ {i b_n x},$$ from where $$a_n$$, $$b_n$$ can be any complex number. As a related question it is possible to do $$f$$ so defined differentiable in $$(0.1)$$ but not differentiable in $$(2,3)$$? Author AdminPosted on November 28, 2018Categories ArticlesTags 01, 23, analysis, classical, construct, converges, diverging, ODEs, series, trigonometric Plotting – How to draw phase porters and the ODES solutions Dear Ladies and Gentlemen, I can draw phase porters for the system of non-linear oodes but I did not know where the result came from show the parametric solutions with phase do not portage well I wrote the code as follows: Sol[{N0_, I0_}?NumericQ] : = First @ NDSolve[{N1' r N1 I1 & # 39; m + N1 I1[0] == I0}, {N1, I1}, {t, 0, 365}]; P1 = Parametric Plot[ assess[{N1[{N1[{N1[{N1 30}, PlotRange -> All, AspectRatio -> Full, PlotRange -> Full, Frame -> True, MaxRecursion -> 8] from where r = 0.431201; [Beta] = 2.99 * 10 ^ -6; [Eta] = 0.2; [Sigma] = 0.7; [Rho] = 0.003; m = 0.427; [Delta] = 0.57; [Mu] = 0.82; and I needed StreamPlot as follows: f[N1_, I1_] = r N1 (1 - [Beta] N1) - [Eta] N1 I1; G[N1_, I1_] = [Sigma] + ([Rho] N1 I1) / ( m + N1) - [Delta] I1 - [Mu] N1 I1; G[{N1_, I1_}] = {f[N1, I1]G[N1, I1]}; then StreamPlot[{F[{F[{f[{f[N1, I1]G[N1, I1]}, {N1, 0.30}, {I1, 0, 30}, StreamStyle -> Blue, AspectRatio -> Automatic, Frame -> True, Axes -> False, AxesLabel -> {"N1", "I1"}]show[StreamPlot[{F[StreamPlot[{F[StreamPlot[{f[StreamPlot[{f[N1, I1]G[N1, I1]}, {N1, 0.30}, {I1, 0, 30}, StreamPoints -> Fine, StreamStyle -> Blue, AspectRatio -> 1/2, Frame -> True, AxesLabel -> {"N1", "I1"}, StreamPoints -> Fine, PlotRange -> All], P1] the final result was Can someone help me to improve my result Author AdminPosted on November 26, 2018Categories ArticlesTags draw, ODEs, Phase, Plotting, porters, Solutions mathematical modeling – finding the nullclines of a two connected ODE’s so I’ve given the following problem: $$dn_1/dt=N(t)n_1-n_1$$ $$dn_2/dt=N(t)n_2-4n_2$$ $$N(t)=25-6n_1-3n_2$$ so I’ve found the fixed points of the 2 equations and got: [(0,0),(0,7),(4,0)] now I want to find the nullclines of them: so for $$dn_1/dt=0$$ we get $$n_1(24-6n_1-3n_2)$$ which means: $$n_1=0$$ for all $$n_2$$ $$n_1=4-0.5n_2$$ so for $$dn_2/dt=0$$ we get $$n_2(21-6n_1-3n_2)$$ which means: $$n_2=0$$ for all $$n_1$$ $$n_2=7-2n_2$$ plotting the following using matlab in order to find the phase plane and basin of attraction gives peculiar results which I can’t find to be able to understand. (fig attached below) I thought the the 3 nullcline need to be intersecting both the fixed points which is more intuitive to me but instead I get the following. the Matlab code generating this plot is as follows: clear, close all; clc %Const N0=25; G1=1;G2=1; a1=6;a2=3; k1=1;k2=4; %Params t0 = 0; % starting time dt = 0.5; % step size tEnd = 50; % end time timeVec=t0:dt:tEnd; NumSteps=floor(tEnd./dt); %Function Handlers N_t=@(n1,n2) N0-a1.*n1-a2.*n2; n1_dot=@(n1,n2) G1.*N_t(n1,n2).*n1-k1.*n1; n2_dot=@(n1,n2) G2.*N_t(n1,n2).*n2-k2.*n2; %%finding the fixed point's %%%%%%%%% system=@(n1,n2) [n1_dot(n1,n2);n2_dot(n1,n2)]; s = solve(system); %%%%%%%%% %TODO RECHECK!!!! %%nulclines %%%%%%%%% %n1_dot=0 %case 1: n1 = 0 %case 2: n1 != 0 %n2_dot=0 %case 1: n2 = 0 %case 2: n2 != 0 %%%%%%%%% ptsMesh=-2:0.2:10; figure [X,Y]=meshgrid(ptsMesh); U=n1_dot(X,Y); V=n2_dot(X,Y); normalizedFactor=sqrt(U.^2+V.^2); quiver(X,Y,U./normalizedFactor,V./normalizedFactor,0.8,'b') hold on plot(s.n1,s.n2,'k*','lineWidth',8) %plot the trivial nullclines on the axis [x,y]=size(ptsMesh); nullclineN1_zero=[0,min(ptsMesh);0,max(ptsMesh)]; nullclineN2_zero=[min(ptsMesh),0;max(ptsMesh),0]; plot(nullclineN2_zero(:,1),nullclineN2_zero(:,2),'g','lineWidth',2); plot(nullclineN1_zero(:,1),nullclineN1_zero(:,2),'r','lineWidth',2); t1=@(x2) 4-(0.5).*x2; res1=t1(ptsMesh./2); plot(res1,ptsMesh,'c','lineWidth',2); t2=@(x1) 7-(2).*x1; res2=t2(ptsMesh./2); plot(ptsMesh,res2,'k','lineWidth',2); will appreciate some help here thanks Author AdminPosted on November 14, 2018Categories ArticlesTags connected, finding, mathematical, modeling, nullclines, ODEs ```