Skip to main content

differential equations - DSolve not finding solution I expected


Try to solve the following ODE via DSolve


$$ \left\{\begin{aligned} y'(x)+2 y(x) e^x-y(x)^2 &= e^{2 x}+e^x \\ y'(0) &=1 \end{aligned}\right. $$


The expected solution is $y(x)=e^x$, but Mathematica produces the message:



DSolve::bvfail: For some branches of the general solution, the given boundary conditions lead to an empty solution




Note: $y(x)=e^x$ is obviously a special solution to the original ODE and satisfies the initial/boundary value condition which can be easily verified by substituting it into the ODE.


Additionally, it is easy to deduct that: $y'(0)=1\Leftrightarrow y(0)=1$ for this ODE;


How to handle it?



Answer



The problem - Genericity of solutions


The problem we encounter here is that DSolve can return only a generic solution however that general solution cannot satisfy such an initial condition as y'[0] == 1. The issue is related to an arbitrary choice of constants of integration i.e. such constants that are specific to certain types of a differential equations DSolve tries to solve during the integration process.


DSolve[{y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x, y'[0] == 1}, y[x], x]


DSolve::bvnul: For some branches of the general solution, the given boundary conditions 

lead to an empty solution. >>

{}

Solving the ODE without the initial condition we will see that the general solution excludes exceptional ones:


y[x_, c_] = y[x] /. First @ DSolve[{ y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x},
y[x], x] /. C[1] -> c


 E^x + 1/(c - x)


Namely there is no constant c reducing the general solution to E^x however putting e.g. y'[0] == 2 we will find a solution:


DSolve[{y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x, y'[0] == 2}, y[x], x]// First


 {y[x] -> (-1 - E^x + E^x x)/(-1 + x)}

This is quite similar issue to one we can find here: Solving a differential equation with initial conditions.


The problem cannot be fixed changing the variables in the ODE to :


y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x /. 

{y[x] -> z[x] + E^x, y'[x] -> z'[x] + E^x} // Simplify


z[x]^2 == z'[x]

DSolve[{z'[x] == z[x]^2, z'[0] == 0}, z[x], x]


DSolve::bvnul: For some branches of the general solution, the given boundary conditions 
lead to an empty solution. >>


{}

Remedy


Introducing an initial condition in a general symbolic way we can find also the special solution:


f[x_, c_] = z[x] /. DSolve[{z'[x] == z[x]^2, z'[0] == c}, z[x], x]


 { -(Sqrt[c]/(-1 + Sqrt[c] x)), -(Sqrt[c]/(1 + Sqrt[c] x))}


f[x, c] /. Solve[D[f[x, c] == 0], c]


{ 0, 0}

I.e. we get the special solution y[x] == E^x.


we can see that the same trick can work also in the original equation.


Alternatively the general solution can be reparametrized in a different way:


E^x + c/(1 - x)


yields also the special solution although this excludes e.g. E^x - 1/x


Comments

Popular posts from this blog

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 - 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 - 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.