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I want to solve the standard 1-dimensional wave equation $y_{xx}=y_{tt}$ using NDSolve (for $y(x,t)$) with the following conditions:

cond1 = Piecewise[{{1 - Abs[x - 1], Abs[x - 1] < 1}, {0, Abs[x - 1] > 1}}]
cond2 = Piecewise[{{1, 3 < x < 4}}, 0]

Where $y_{t}(x,0)=\mathrm{cond2}$ and $y(x,0)=\mathrm{cond1}$. I used the following code: 

WaveEquation = D[y[x, t], {x, 2}] - D[y[x, t], {t, 2}] == 0;
cond1 = Piecewise[{{1 - Abs[x - 1], Abs[x - 1] < 1}, {0, Abs[x - 1] > 1}}];
cond2 = Piecewise[{{1, 3 < x < 4}}, 0];
sol1 = NDSolve[{WaveEquation, y[x, 0] == cond1, 
Derivative[0, 1][y][x, 0] == cond2}, {y[x, t]}, {x, 0, 10}, {t, 0,
 10}];

I would as well like to find the profiles when $t=0,1,2$. However, when I try to run the code for $\mathrm{sol1}$, I get the following error:

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

What would I be doing wrong in this case? I'm not sure on how to figure it out. The program has been running for a while, so I'm sure it crashed. As well, what would be a simple way to plot the time profiles for the solution to the PDE with the specified initial conditions? Thanks!

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  • $\begingroup$ You want a numeric solution, or just need to solve it no matter it's numeric or analytic? $\endgroup$
    – xzczd ♦
    Commented Oct 12, 2016 at 8:02
  • $\begingroup$ Just a numerical solution, I would in particular like to plot the profiles of $u(x,t)$ for times $t=0,1,2,4$. I think the analytic solution can be obtained by superposition and D'Alembert's formula. $\endgroup$
    – arcbloom
    Commented Oct 12, 2016 at 8:09

3 Answers 3

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I've waited for this question for a long time :)

Fully NDSolve-based Numerical Solution

There actually exist 2 issues here:

  1. NDSolve can't handle unsmooth i.c. very well by default.

  2. NDSolve can't add proper artificial b.c. for the initial value problem (Cauchy problem) for the 1-dimensional wave equation.

The first issue is easy to solve: just make the spatial grid dense enough and fix its size to avoid NDSolve trying using too much points to handle those roughness. I guess finite element method in and after v10 can handle this issue in a even better way, but since I'm still in v9, I'd like not to explore this more.

What's really… big is the second issue. It's easy to solve the initial value problem for the 1-D wave equation analytically, we just need to use DSolve in and after v10.3 or d´Alembert's formula or do a Fourier transform, but when solving it numerically, we need to add proper artificial b.c., which NDSolve doesn't know how to. (NDSolve does add artificial b.c. when b.c. isn't enough, but as far as I know, it seldom works well, actually it's even unclear that what artificial b.c. is added, see this post for more information. )

Adding proper artificial b.c. for wave equation can be troublesome when the equation becomes more complicated (2D, 3D, nonlinear, etc.), but luckily, what you want to solve is just a simple 1D wave equation, then the corresponding artificial b.c. (usually called absorbing boundary condition) is quite simple:

{lb, rb} = {-10, 10};
(* absorbing boundary condition *)
abc = D[y[x, t], x] + direction D[y[x, t], t] == 0 /. 
  {{x -> lb, direction -> -1}, {x -> rb, direction -> 1}}

mol[n_Integer, o_: "Pseudospectral"] := {"MethodOfLines", 
    "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> n, 
        "MinPoints" -> n, "DifferenceOrder" -> o}}

WaveEquation = D[y[x, t], {x, 2}] - D[y[x, t], {t, 2}] == 0;
cond1 = Piecewise[{{1 - Abs[x - 1], Abs[x - 1] < 1}, {0, Abs[x - 1] > 1}}];
cond2 = Piecewise[{{1, 3 < x < 4}}, 0];
ic = {y[x, 0] == cond1, Derivative[0, 1][y][x, 0] == cond2};

nsol = NDSolveValue[{WaveEquation, ic, abc}, y, {x, lb, rb}, {t, 0, 10}, 
  Method -> mol[600, 4](* fix the spatial grid and make it dense enough *)]

Animate[Plot[nsol[x, t], {x, lb, rb}, PlotRange -> {0, 2}], {t, 0, 10}]

NDSolve still spits out eerri and eerr warning, but it's not a big deal in this case:

enter image description here

Fourier-transform-based Analytical Solution

As shown by rewi, DSolve can solve the problem after v10.3. Here I just want to show how to solve it with Fourier transform (the ft function is from here):

teqn = ft[{WaveEquation, ic}, x, w] /. HoldPattern@FourierTransform[a_, __] :> a

tsol = y[x, t] /. First@DSolve[teqn, y[x, t], t]

asol[x_, t_] = InverseFourierTransform[tsol, w, x]
(* 1/4 ((2 + t - x) Sign[2 + t - x] + (4 + t - x) Sign[
     4 + t - x] + (3 + t - x) Sign[-3 - t + x] + 
   2 (1 + t - x) Sign[-1 - t + x] + (-t + x) Sign[-t + x] - (-4 + t + x) Sign[-4 + t + 
      x] + (-3 + t + x) Sign[-3 + t + x] + (-2 + t + x) Sign[-2 + t + x] - 
   2 (-1 + t + x) Sign[-1 + t + x] + (t + x) Sign[t + x]) *)

Compared to the solution given by DSolve, this solution doesn't involve Integrate so it's more suitable for numeric evaluation.

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  • 1
    $\begingroup$ I see, I ended up doing brute force method by just putting it straight into NDSolve without specifying an auxiliary boundary condition. I ended up getting a similar movie to yours, except that mine has much more noise in comparison (the one you produced is slick). I'm still unsure on what does the "absorbing boundary condition" represent though, am I missing something simple? Thanks! $\endgroup$
    – arcbloom
    Commented Oct 12, 2016 at 9:32
  • 2
    $\begingroup$ @Arcbloom No, it's not simple at all, I think. In short, it allows the wave passing through the boundary without being reflected, in 1D case, we're actually using a transport equation as the boundary. Well, currently I don't find a very good reference (I myself learned the basic of this issue when reading Chapter 6 of this free book), you may search "absorbing boundary condition wave equation" to find more. $\endgroup$
    – xzczd ♦
    Commented Oct 12, 2016 at 9:50
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Here is a FEM-based solution:

cond1 = Piecewise[{{1 - Abs[x - 1], Abs[x - 1] < 1}, {0, 
     Abs[x - 1] > 1}}];
cond2 = Piecewise[{{1, 3 < x < 4}}, 0];
WaveEquation = 
  D[y[x, t], {t, 2}] - 
    D[y[x, t], {x, 2}] == -NeumannValue[Derivative[0, 1][y][x, t], 
     x == 0 || x == 10];

The absorbing boundary is modeled with a NeumannValue on the derivative of t. This is best understood by remembering that a wave equation can be transformed into a system of two first order equations. The NeumannValue on the time derivative then becomes a NeumannValue on the auxiliary variable.

nsol = NDSolveValue[{WaveEquation, y[x, 0] == cond1, 
    Derivative[0, 1][y][x, 0] == cond2}, y, {x, 0, 10}, {t, 0, 10}, 
   Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"FiniteElement", 
       "MeshOptions" -> {"MaxCellMeasure" -> 0.05}}}];

Animate[Plot[nsol[x, t], {x, 0, 10}, PlotRange -> {0, 2}], {t, 0, 10}]

animation

Update: For more information on how to model with the wave equation you can have a look at the (Finite Element Method) Acoustics in the Time Domain tutorial. This tutorial was added in V12.0 but the content presented should work in previous versions. The wave equation is one of the models presented there and many details about the equation and it's boundary conditions are presented. Hope this is useful. Search for Acoustics in the help system and the tutorial will come up.

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This can be done with DSolve or DSolveValue.

WaveEquation = D[y[x, t], {x, 2}] - D[y[x, t], {t, 2}] == 0;
cond1 = Piecewise[{{1 - Abs[x - 1], Abs[x - 1] < 1}, {0, Abs[x - 1] > 1}}];
cond2 = Piecewise[{{1, 3 < x < 4}}, 0];
sol1 = DSolveValue[{WaveEquation, y[x, 0] == cond1, 
   Derivative[0, 1][y][x, 0] == cond2}, y, {x, 0, 10}, {t, 0, 10}]

enter image description here

Table[Plot[sol1[x, t], {x, 0, 10}, PlotRange -> All], {t, 0, 2}]

enter image description here

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    $\begingroup$ Thanks, I haven't thought about that! If we were to have a more complicated PDE though, would the only method be invoking NSolve? (it seems that DSolveValue for a rectangle in $\mathbb{R}^{2}$ works, but not for DSolve on $\mathbb{R}^{2}$) $\endgroup$
    – arcbloom
    Commented Oct 12, 2016 at 9:24
  • $\begingroup$ @Arcbloom DSolve and DSolveValue are very powerful procedures. Many PDE can be solved. Let's see what still gives us the future. $\endgroup$
    – user36273
    Commented Oct 12, 2016 at 9:37

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