Skip to main content

numerical integration - Solving wave equation with singular initial conditions



I am trying to compute the solution to the wave equation using NDSolveValue and singular initial displacement (continuous, but not differentiable in a point). I need to propagate as accurately as possible the singularity of the initial condition through time. However the singularity is "smoothened" during the integration, see below.


enter image description here


The red curve is my initial condition in displacement. The purple one is the output at a different time of the computation. As you can see, the purple curve is perfectly smooth, while it shouldn't.


Here you can see a piece of the script I use to get these solutions :


(*initialisation of parameters and initial conditions*)
Sigma0 = 0.002; g0 = 0.001;  L = 1; ρ = 1; Ei = 1; c =Sqrt[Ei/ρ]; epsi = 10^-7.;
qu[x_] := 
Piecewise[{{(-1 Sigma0/Ei) x, 0 <= x <= L/2}, {(-1 Sigma0/Ei) L/2, 
 L/2 <= x <= L}}];
qv[x_] := 

  Piecewise[{{0, 0 <= x <= L/2}, {(c Sigma0/Ei), L/2 <= x <= L}}];

(*Building of a periodic Solution according to the wave equation 
  in Dirichlet-Neumann Boundary condition*)

u0[x_] :=  qu[x];
v0[x_] :=  qv[x];
uif = NDSolveValue[{D[u[x, t], {t, 2}] - D[u[x, t], {x, 2}] == 0, 
    u[x, 0] == u0[x], (D[u[x, t], t] /. t -> 0) == 
     v0[x], (D[u[x, t], x] /. x -> L) == 0, u[0, t] == 0}, 

   u, {x, 0, L}, {t, -10, 10}];
tc = t /. FindRoot[uif[L, t] - g0, {t, 0.1}];

(*Plot of the solution at a time tc in purple and the initial condition*)
Show[Plot[uif[x, 0], {x, 0, L}, PlotPoints -> 200, PlotStyle -> Red], 
 Plot[uif[x, tc], {x, 0, L}, PlotPoints -> 200, PlotStyle -> Purple], 
 PlotRange -> {{0, L}, All}, AxesLabel -> {"Time", "Displacement"}]

I get those warnings during all the computation :




NDSolveValue::mxsst: Using maximum number of grid points 10000 allowed by the MaxPoints or MinStepSize options for independent variable x.


NDSolveValue::eerr: Warning: scaled local spatial error estimate of 185.53982144076966 at t = 20. in the direction of independent variable x is much greater than the prescribed error tolerance. Grid spacing with 25 points may be too large to achieve the desired accuracy or precision. A singularity may have formed or a smaller grid spacing can be specified using the MaxStepSize or MinPoints method options.


NDSolveValue::ndstf: At t == -2.61883, system appears to be stiff. Methods Automatic, BDF, or StiffnessSwitching may be more appropriate.





Comments

Post a Comment

Popular posts from this blog

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...
Post a Comment
Read more

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...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more