Skip to main content

differential equations - DSolve gives complex function although the solution is a real one


I have a problem with the DSolve[] command in mathematica 8. Solving the the following 4th order differential equation spits out a complex solution although it should be a real one. The equation is:


y''''[x] + a y[x] == 0

Solving this equation by hand yields a solution with only real parts. All constants and boundary conditions are also real numbers.


The solution I get by hand is:


y1[x_] = (C[5] E^(Power[a, (4)^-1]/Power[2, (2)^-1] x) + 
C[6] E^(-(Power[a, (4)^-1]/Power[2, (2)^-1]) x)) Cos[
Power[a, (4)^-1]/Power[2, (2)^-1]
x] + (C[7] E^(Power[a, (4)^-1]/Power[2, (2)^-1] x) +

C[8] E^(-(Power[a, (4)^-1]/Power[2, (2)^-1]) x)) Sin[
Power[a, (4)^-1]/Power[2, (2)^-1] x];

Now I have to solve for the constants C[5]...C[8]. This arises a similar issue. I use the Solve[] command with the boundary conditions


Solve[{y1''[-c] == ic0, y1''[c] == ic0 , y1'''[-c] == ic1 , 
y1'''[c] == - ic1 }, {C[5], C[6], C[7], C[8]} ];

The constants C[5]...C[8] are now real if using //Simplify and complex if using //FullSimplify.


Any idea what the reasons are? The notebook with my calculations can be downloaded under: http://dl.dropbox.com/u/4920002/DGL_4th_Order_with_own_solution.nb


In further work I have to use DSolve[] and I would like to understand the issue here.




Answer



Here is the reply I got from the Mathematica support team:



Thank you for your prompt reply.


Please find attached a small notebook which demonstrates using FullSimplify directly on the output of your first DSolve expression to obtain a symbolic real-valued formula.


The formula itself does contain imaginary values, but these zero each other out as seen when the formula is evaluated numerically. Unfortunately, the corresponding "hidden zero" in the imaginary component cannot be eliminated by either of two typical techniques:


(1) increasing the internal precision available to the N function when evaluating the formula numerically, or


(2) using FullSimplify which is sometimes successful in removing hidden zeros from symbolic formulas.


The recommended approach in cases like this is to use the Chop function, which will eliminate near-zero numerical values in any given expression.


For more information, please see the Possible Issues subsection of the $MaxExtraPrecision page.



For details on the Chop function and how its behavior can be customized, please see its corresponding Documentation Center page.


Please let me know if you have any questions. Thank you.


{sol} = DSolve[{y''''[x] + a y[x] == 0, y''[-c] == ic0, y''[c] == ic0,
y'''[-c] == ic1, y'''[c] == -ic1}, y[x], x];

val = FullSimplify[sol, a > 0 && {ic0, ic1, c, x} \[Element] Reals]

N[val]
% // Chop


Block[{$MaxExtraPrecision = 10000}, N[val]]
% // Chop

Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....