Skip to main content

differential equations - Initial condition trouble with NDSolve for a 2nd order PDE


In general, when solving a 2nd order PDE (such as the wave equation below) for u(x,t),x∈(−∞,∞),t∈(0,∞)

it should be sufficient to provide initial conditions u(x,0) and ∂tu(x,0). However when I try solve this numerically in Mathematica


NDSolve[{D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}], 
u[x, 0] == E^(-x^2), Derivative[0, 1][u][x, 0] == 1}, u, {x, -7, 7}, {t, 0, 4}]

I get the error


NDSolve::bcart: Warning: an insufficient number of boundary conditions 
have been specified for the direction of independent variable x.

Artificial boundary effects may be present in the solution.

The only thing I can think of is that when solving numerically I have introduced fictitious new boundaries to the problem by simply specifying a finite range {x,-7,7}. If this is the problem, how do I get around it? (NB the real PDE I'm trying to solve is more complicated and does not describe localized waves).


Also, oddly, if I take out the second I.C.


Derivative[0, 1][u][x, 0] == 1

It doesn't complain but it seems to invent its own boundary conditions and gives the wrong solution.


Edit: the actual PDE I'm solving is of the form −∂2tϕ+√2Mr(32r∂tϕ+2∂t∂rϕ)+1r2∂r[r(r−2M)∂tϕ]−m2ϕ=0,


with the initial conditions ϕ(r>>1,t)=10−t1/3 and \partial_r\phi(r>>1,t) = 0$. Any tips regarding the best way to get around the issues above in this case would be amazing.




Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

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}]