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I'm a newbie with Mathematica, I know it's a basic answer, but I can't solve the problem on my own.

I have the following equation reflecting insect aggregation at low population densities (taken from page 404 of J.D. Murray's Mathematical Biology: I. An Introduction, Third Edition):

$$\partial_t u = \partial_x (\text{sign}(x) u) + \partial_x (u^2\partial_x u)$$ with initial condition $$u(x,0)= e^{-x^2}$$ and boundary conditions $$u(-7,t)=u(7,t)=0$$

and I want to integrate it until time $t=2$. I obtain the following plot with a program I did with Python, but I have no idea if my solution is correct, so I'd like to double check it with Mathematica.

enter image description here

I tried the following snippet:

sol = NDSolveValue[{
   D[u[x, t], t] == D[Sign[x]*u[x,t],x] + D[u[x, t]^2 D[u[x, t], x], x],
   u[-7, t] == 0, u[7, t] == 0, u[x, 0] == Exp[-x^2]}
  , u, {x, -7, 7}, {t, 0, 2}]

but NDSolve spits out NDSolveValue::ndnum warning and fails. Can someone confirm I wrote the right snippet and show the plot I should obtain?


EDIT:

After a check on my Python implementation, here's what I found at $t=2$:

enter image description here

As pointed out by @xzczd, using a finer mesh can help:

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  • 1
    $\begingroup$ sign is not a built-in function of Mathematica, I think you mean Sign, but it's not differentiable at $x=0$, so we need a bit extra work to define a differentiable sign ourselves. Then, are you sure the equation itself is correct? The solution blows up very fast. $\endgroup$ – xzczd Sep 20 at 12:28
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    $\begingroup$ @xzczd I edited with `Sign. Yes, the PDE it's taken from Murray's book (1 volume), page 404. I just have to choose boundary and initial conditions, so I chose those ones, which are compatible. $\endgroup$ – Vefhug Sep 20 at 12:31
  • $\begingroup$ You could replace Sign with Tanh[1000 x] or 2 ArcTan[100 x]/Pi which are close to a Sign function. The plot then works, but only for very small values of t near zero. By the time t=0.04 the solution is already a mess. $\endgroup$ – flinty Sep 20 at 13:03
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    $\begingroup$ @flinty If the equation is correct, then this is another example that special spatial discretization is needed. See my answer below for more info. $\endgroup$ – xzczd Sep 20 at 13:45
  • $\begingroup$ Cross-posted: scicomp.stackexchange.com/q/35955/5331 $\endgroup$ – xzczd Sep 20 at 14:12
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If the equation is correct, then it's probably another example that we need special treatment for the discretization of conservation law.

As mentioned in the comment above, one easy-to-notice issue of OP's trial is Sign[x] isn't differentiable at x == 0. This seems to be easy to resolve: we just need to define a differentiable approximate sign ourselves:

appro = With[{k = 100}, ArcTan[k #]/Pi + 1/2 &];

sign[x_] = Simplify`PWToUnitStep@PiecewiseExpand[Sign[x], Reals] /. UnitStep -> appro

Nevertheless, it just leads to a solution messing up quickly:

soltest = NDSolveValue[{D[u[x, t], t] == 
    D[sign[x]*u[x, t], x] + D[u[x, t]^2 D[u[x, t], x], x], u[-7, t] == 0, u[7, t] == 0, 
   u[x, 0] == Exp[-x^2]}, u, {x, -7, 7}, {t, 0, 2}]

NDSolveValue::ndsz At t == 0.25352360860722767`, step size is effectively zero; singularity or stiff system suspected.

NDSolveValue::eerr

Does this suggest the equation itself is wrong? Not necessarily, because the PDE involves differential form of consevation law, and we already have several examples showing that serious problem may come up if spatial discretization isn't done properly on such type of PDE:

Conservation of area solving a PDE via finite difference scheme

Instability, Courant Condition and Robustness about solving 2D+1 PDE

How to solve the tsunami model and animate the shallow water wave?

Problems with solving PDEs

So, how to resolve the problem? If you've read the answers above, you'll notice the most effective and general solution seems to be avoiding the symbolic calculation of outermost D before discretization, and I've figured out 3 ways to.

Additionally, a method that doesn't require one to transform the equation is found, but this only works in or before v11.2.

FiniteElement Based Solution

Thanks to the new-in-v12 nonlinear FiniteElement method, it's possible to resolve the problem completely inside NDSolve with the help of Inactive:

With[{u = u[x, t]}, 
 neweq = D[u, t] == 
   Inactive[Div][{{Sign[x] u/D[u, x] + u^2}}. Inactive[Grad][u, {x}], {x}]]

{bc, ic} = {{u[-7, t] == 0, u[7, t] == 0}, u[x, 0] == Exp[-x^2]}

solFEM = NDSolveValue[{neweq, ic, bc}, u, {t, 0, 2}, {x, -7, 7}, 
  Method -> {MethodOfLines, 
    SpatialDiscretization -> {FiniteElement, MeshOptions -> MaxCellMeasure -> 0.1}}]

p1 = Plot[solFEM[x, 2], {x, -7, 7}, PlotRange -> All]

enter image description here

Several warnings will pop up, but don't worry.

Tested on v12.0.0, v12.1.1.

Semi-NDSolve Based Solution

You may be suspicious of the result above because it's different from your first one. Also, you may find NDSolveValue fail for certain setting of MaxCellMeasure (say MaxCellMeasure -> 0.01), which seems to make the result more suspicious, so let's double check it with another method i.e. a self-implementation of method of lines, as I've done in answers linked above.

I'll use pdetoode for the discretization in $x$ direction.

domain = {L, R} = {-7, 7}; tend = 2;
With[{u = u[x, t], mid = mid[x, t]}, eq = {D[u, t] == D[mid, x],
   mid == Sign[x] u + u^2 D[u, x]};
 {bc, ic} = {u == 0 /. {{x -> L}, {x -> R}}, u == Exp[-x^2] /. t -> 0};]

points = 100;
grid = Array[# &, points, domain];
difforder = 2;
(* Definition of pdetoode isn't included in this post,
   please find it in the link above. *)
ptoofunc = pdetoode[{u, mid}[x, t], t, grid, difforder];
del = #[[2 ;; -2]] &;

Block[{mid}, Evaluate@ptoofunc@eq[[2, 1]] = ptoofunc@eq[[2, -1]]; 
  ode = ptoofunc@eq[[1]] // del];
odeic = ptoofunc[ic] // del;
odebc = ptoofunc[bc];

sollst = NDSolveValue[{ode, odebc, odeic}, u /@ grid, {t, 0, tend}];
sol = rebuild[sollst, grid, 2]

p2 = Plot[sol[x, tend], {x, L, R}, PlotRange -> All, PlotStyle -> {Dashed, Red}]

enter image description here

Tested on v9.0.1, v12.0.0, v12.1.1.

TensorProductGrid Based Solution

It's a bit surprising that the following method even work in v9, because pdord is just equivalent to failure in my memory:

{L, R} = {-7, 7}; tend = 2;

With[{u = u[x, t], mid = mid[x, t]},
 eq = {D[u, t] == D[mid, x], mid == Sign[x] u + u^(2) D[u, x]};
 {bc, ic} = {u == 0 /. {{x -> L}, {x -> R}}, u == Exp[-x^2] /. t -> 0};]

icadditional = mid[x, 0] == eq[[2, 2]] /. u -> Function[{x, t}, Evaluate@ic[[2]]]

solTPG = NDSolveValue[{eq, ic, bc, icadditional}, {u, mid}, {t, 0, tend}, {x, L, R}, 
  Method -> {MethodOfLines, 
    SpatialDiscretization -> {TensorProductGrid, MaxPoints -> 500}}]

p3 = Plot[solTPG[[1]][x, 2], {x, L, R}, PlotRange -> All, PlotStyle -> {Orange, Thin}]

enter image description here

Again, you'll see several warnings, simply ignore them.

Tested on v9.0.1, 11.3.0, v12.0.0, v12.1.1.

fix Based Solution (Only Works Before v11.3)

Luckily, my fix turns out to be effective on the problem. If you're in or before v11.2, then this is probably the simplest solution (but as you can see, it's not quite economical, 2000 grid points are used to obtain a good enough result):

appro = With[{k = 100}, ArcTan[k #]/Pi + 1/2 &];

sign[x_] = Simplify`PWToUnitStep@PiecewiseExpand[Sign[x], Reals] /. UnitStep -> appro

solpreV112 = 
 fix[tend, 4]@
  NDSolveValue[{D[u[x, t], t] == D[sign[x] u[x, t], x] + D[u[x, t]^2 D[u[x, t], x], x], 
    u[-7, t] == 0, u[7, t] == 0, u[x, 0] == Exp[-x^2]}, u, {x, -7, 7}, {t, 0, 2}, 
   Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"TensorProductGrid", "MaxPoints" -> 2000, 
       "MinPoints" -> 2000, "DifferenceOrder" -> 4}}]

Plot[solpreV112[x, tend], {x, -7, 7}, PlotRange -> All]

enter image description here

Tested on v9.0.1.

The 4 solutions agree well. Your first result in Python is wrong.

Remark

If you want to check the $m=\frac{1}{2}$ case mentioned in p404 of the book, remember to add a Re to the code to avoid tiny imaginary number generated by numeric error. To be more specific, you need to use

With[{u = u[x, t]}, 
 neweq = D[u, t] == 
   Inactive[Div][{{Sign[x] u/D[u, x] + Re[u^(1/2)]}}. Inactive[Grad][u, {x}], {x}]]

in the FiniteElement based approach,

With[{u = u[x, t], mid = mid[x, t]}, eq = {D[u, t] == D[mid, x],
    mid == Sign[x] u + Re[u^(1/2)] D[u, x]};]

in the semi-NDSolve based and TensorProductGrid based approach, and

Plot[solpreV112[x, tend] // Re, {x, -7, 7}, PlotRange -> All]

in the fix based approach. (Yeah in fix approach you just need to add Re into Plot. )

BTW the obtained result seems to be consistent with the one in the book:

enter image description here

enter image description here

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  • $\begingroup$ Thanks so much for your detailed answer. I indeed used a finite element approach (with degree one) in my Python code, so there must be something wrong in my implementation, or should I expect that a FEM approach fails because the one of Mathematica has something special? @xzczd $\endgroup$ – Vefhug Sep 20 at 16:54
  • $\begingroup$ I just edited my post with what I obtain with FEM (with degree 1 elements), this seems right now. How do you think one can remove oscillations? $\endgroup$ – Vefhug Sep 20 at 22:53
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    $\begingroup$ @Vefhug Using a denser mesh should help. (Given that you're using degree 1 element, the mesh may need to be quite dense. ) $\endgroup$ – xzczd Sep 21 at 1:55
  • $\begingroup$ @xzcdz thanks, that's right. Since oscillations appear also in Mathematica, it should be a problem of these method itself, due to the non-differentiability of $\text{sgn(x)}$. Btw, why do you state that my result Python in "wrong". It seems that same of Mathematica, actually. $\endgroup$ – Vefhug Sep 21 at 7:04
  • $\begingroup$ @Vefhug That line is written before your EDIT. $\endgroup$ – xzczd Sep 21 at 7:07
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If only the steady state is desired, it can be obtained easily by

sa = Values[DSolve[1 + u[x] D[u[x], x] == 0, u[x], x] /. C[1] -> c][[2, 1]]

and c determined from conservation of the the integral over u.

scint = Integrate[sa, {x, 0, c}];
int = Integrate[Exp[-x^2], {x, 0, Infinity}];
sc = Solve[scint == int, c] // Flatten
{c -> (3^(2/3) Pi^(1/3))/(2 2^(2/3))}

Plot[Re[sa /. sc], {x, 0, 7}, AxesLabel -> {x, u}, 
    ImageSize -> Large, LabelStyle -> {15, Black, Bold}]

enter image description here

If desired, the time-dependent solution can be obtained by a do-it-yourself method of lines applied to

{D[u[x, t], t] == D[u[x,t],x] + D[u[x, t]^2 D[u[x, t], x], x], 
u[7, t] == 0, Integrate[u[x,t], {x, 0, 7}] == Sqrt[Pi]/2, u[x, 0] == Exp[-x^2]}

over the domain {0, 7}.

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