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I need to use NDSolve which in turn uses the solution from another ODE as function in terms of output from another NDSolve.
If I use the exact solution from the first differential equation inside the NDSolve, it's OK. But when I use the same solution in the form of function (which uses InterpolatingFunction) it does not work.
I believe, it's got to do with the structure of NDSolve output. Could anyone please enlighten me on this. Will be of great help!
The code is:
feq = 2 V alpha fip F''[fi] - (V^2 - (V^2 + sigma - 2 fi) (F'[fi])^2 + (F'[fi])^4
Frange[lo_, hi_] :=
Module[{fii, sol},
sol = NDSolve[{(feq == 0 /.fi -> fii), F[0] == 0}, F, {fii, lo, hi}]]
eqpois = fi''[x] == ne[x] - F[fi[x]]/.sol
NDSolve[{eqpois, fi'[0] == 0, fi[0] == 0}, fi, {x,0,1}]
Here in order to find F[phi], I need to solve the 1st diff eq that is feq, which is solved by NDSolve inside the function Frange[lo,hi]. The solution is then used inside the second equation eqpois, which has to be solved using NDSolve again. The problem comes up in the second NDSolve, which does not produce the result. If I use the analytical solution of F[phi] in eqopis, then there is no problem.
Example Problem
I have done a little experiment with this. Let's take an example of coupled ODEs
1st eqn : dg/dx = 2f(g) with initial condition g(0) = 1
The function f(y) is a solution from another ODE, say,
2nd eqn : df/dy = 2y with IC f(0) = 0
The solution of the 2nd ODE is f(y) = y^2 which when put into the the 1st ODE becomes
dg/dx = 2 g^2 and the final solution is g(x) = 1/(1-2x)
The issue:
When I use DSolve, it finds the answer correctly
In[39]:= s = DSolve[{f'[y] == 2 y, f[0] == 0}, f, y]
Out[39]= {{f -> Function[{y}, y^2]}}
In[40]:= ss = DSolve[{g'[x] == 2 (f[g[x]]/.First#s), g[0] == 1}, g, x]
Out[40]= {{g -> Function[{y}, 1/(1 - 2 x)]}}
The problem comes when I use NDSolve
In[41]:= s = NDSolve[{f'[y] == 2 y, f[0] == 0}, f, {y, 1, 5}]
Out[41]= {{f -> InterpolatingFunction[{{1., 5.}}, <>]}}
In[42]:= ss1 = NDSolve[{g'[x] == 2 (Evaluate[f[g[x]]/.First#s1]), g[0] == 1}, g, {x, 1, 2}]
Out[42]= {}
The erros are:
During evaluation of In[41]:= InterpolatingFunction::dmval: Input value {2.01726} lies outside the range of data in the interpolating function. Extrapolation will be used. >>
During evaluation of In[41]:= InterpolatingFunction::dmval: Input value {2.01726} lies outside the range of data in the interpolating function. Extrapolation will be used. >>
During evaluation of In[41]:= InterpolatingFunction::dmval: Input value {2.04914} lies outside the range of data in the interpolating function. Extrapolation will be used. >>
During evaluation of In[41]:= General::stop: Further output of InterpolatingFunction::dmval will be suppressed during this calculation. >>
During evaluation of In[41]:= NDSolve::ndsz: At y == 0.16666654771477857, step size is effectively zero; singularity or stiff system suspected. >>
During evaluation of In[41]:= NDSolve::ndsz: At y == 0.16666654771477857, step size is effectively zero; singularity or stiff system suspected. >>
Any help in this regard will be highly appreciated!
--- Madhurjya
I got your simple example to work with a little mod ..
f0 = First#First#DSolve[{f'[y] == 2 y, f[0] == 0}, f, y]
g0 = g /.
First#First#DSolve[{g'[x] == 2 (f[g[x]] /. f0), g[0] == 1}, g, x]
fn = f /. First#First#NDSolve[{f'[y] == 2 y, f[0] == 0}, f, {y, 0, 10}]
gn = g /.
First#First#
NDSolve[{g'[x] == 2 (fn[g[x]]), g[0] == 1}, g, {x, 0, 9/20}]
GraphicsRow[{
Plot[{g0#x, gn#x}, {x, 0, 9/20},
PlotStyle -> {{Thick, Black}, {Thin, Red, Dashed}}],
Plot[{f#x /. f0, fn#x}, {x, 0, 2},
PlotStyle -> {{Thick, Black}, {Thin, Red, Dashed}}]}]
note we need to ensure the y range in the first NDSolve is sufficient to cover the expected range of g from the second. That is where all those interpolation range errors come from.
I am very new to Mathematica, and I am trying to solve the following problem.
I have a cubic equation of the form Z = aZ^3 + bZ^2 + a + b. The first thing I want to do is to get a function that solves this analytically for Z and chooses the minimal positive root for that, as a function of a and b.
I thought that in order to get the root I could use:
Z = Solve[z == az^3 + bz^2 + a + b, z];
It seems like I am not quite getting the roots, as I would expect using the general cubic equation solution formula.
I want to integrate the minimal positive root of Z over a and b (again, preferably analytically) from 0 to 1 for a and for a to 1 for b.
I tried
Y = Integrate[Z, {a, 0, 1}, {b, a, 1}];
and that does not seem to give any formula or numerical value, but just returns an integral. (Notice I am not even sure how to pick the minimal positive root, but I am playing around with Mathematica to try to figure it out.)
Any ideas on how to do this?
Spaces between a or b and z are important. You can get the roots by:
sol = z /. Solve[z == a z^3 + b z^2 + a + b, z]
However, are you sure this expression has a solution as you expect? For a=0.5 and b=0.5, the only real root is negative.
sol /. {a->0.5, b->0.5}
{-2.26953,0.634765-0.691601 I,0.634765+0.691601 I}
sol = z /. Solve[z == a z^3 + b z^2 + a + b, z];
zz[a0_ /; NumericQ[a0], b0_ /; NumericQ[b0]] :=
Min[Select[ sol /. {a -> a0, b -> b0} ,
Element[#, Reals] && # > 0 & ]]
This returns -infinty when there are no solutions. As sirintinga noted your example integration limits are not valid..
RegionPlot[NumericQ[zz[a, b] ] , {a, -1, .5}, {b, -.5, 1}]
but you can numerically integrate if you have a valid region..
NIntegrate[zz[a, b], {a, -.5, -.2}, {b, .8, .9}] ->> 0.0370076
Edit ---
there is a bug above Select in Reals is throwin away real solutions with an infinitesimal complex part.. fix as:..
zz[a0_ /; NumericQ[a0], b0_ /; NumericQ[b0]] :=
Min[Select[ Chop[ sol /. {a -> a0, b -> b0} ],
Element[#, Reals] && # > 0 & ]]
Edit2, a cleaner approach if you dont find Chop satisfyting..
zz[a0_ /; NumericQ[a0], b0_ /; NumericQ[b0]] :=
Module[{z, a, b},
Min[z /. Solve[
Reduce[(z > 0 && z == a z^3 + b z^2 + a + b /.
{ a -> a0, b -> b0}), {z}, Reals]]]]
RegionPlot[NumericQ[zz[a, b] ] , {a, -2, 2}, {b, -2, 2}]
NIntegrate[zz[a, b], {a, 0, .5}, {b, 0, .5 - a}] -> 0.0491321
I had a math problem I solved like this:
In[1]:= Solve[2x(a-x)==0, x]
Out[1]= {{x->0}, {x->a}}
In[2]:= Integrate[2x(a-x), {x,0,a}]
Out[2]= (a^3)/3
In[3]:= Solve[(a^3)/3==a, a]
Out[3]= {{a->0}, {a->-Sqrt[3]}, {a->Sqrt[3]}}
My question is if I could rewrite this to compute it in one step, rather than having to manually input the result from the previous line. I could easily replace the integral used in step three with the Integrate command from step two. But what I can't figure out is how I would use the result from step 1 as the limits of integration in the integral.
You could combine step 1 and 2 by doing something like
Integrate[2 x (a - x), {x, ##}] & ## (x /. Solve[2 x (a - x) == 0, x]);
If you agree to delegate the choice of the (positive oriented) domain to Integrate, by means of using Clip or Boole:
In[77]:= Solve[
Integrate[
Clip[2 x (a - x), {0, Infinity}], {x, -Infinity, Infinity}] == a, a]
Out[77]= {{a -> 0}, {a -> Sqrt[3]}}
or
In[81]:= Solve[
Integrate[
2 x (a - x) Boole[2 x (a - x) > 0], {x, -Infinity, Infinity}] ==
a, a]
Out[81]= {{a -> 0}, {a -> Sqrt[3]}}
The reason only non-negative roots are found, is that Integrate will integrate from the smallest root to the largest root, i.e. from {x,0,a} for positive a and {x,a,0} for negative a.
Say I have a list of Rules
rules = {a -> b, c -> d};
which I use throughout a notebook. Then, at one point, it makes sense to want the rules to apply before any other evaluations take place in an expression. Normally if you want something like this you would use
In[2]:= With[{a=b,c=d}, expr[a,b,c,d]]
Out[2]= expr[b, b, d, d]
How can I take rules and insert it into the first argument of With?
Edit
BothSome solutions fail do all that I was looking for - but I should have emphasised this point a little more. See the bold part above.
For example, let's look at
rules = {a -> {1, 2}, c -> 1};
If I use these vaules in With, I get
In[10]:= With[{a={1,2},c=1}, Head/#{a,c}]
Out[10]= {List,Integer}
Some versions of WithRules yield
In[11]:= WithRules[rules, Head/#{a,c}]
Out[11]= {Symbol, Symbol}
(Actually, I didn't notice that Andrew's answer had the Attribute HoldRest - so it works just like I wanted.)
You want to use Hold to build up your With statement. Here is one way; there may be a simpler:
In[1]:= SetAttributes[WithRules, HoldRest]
In[2]:= WithRules[rules_, expr_] :=
With ## Append[Apply[Set, Hold#rules, {2}], Unevaluated[expr]]
Test it out:
In[3]:= f[args___] := Print[{args}]
In[4]:= rules = {a -> b, c -> d};
In[5]:= WithRules[rules, f[a, c]]
During evaluation of In[5]:= {b,d}
(I used Print so that any bug involving me accidentally evaluating expr too early would be made obvious.)
I have been using the following form of WithRules for a long time. Compared to the one posted by Andrew Moylan, it binds sequentially so that you can say e.g. WithRules[{a->b+1, b->2},expr] and get a expanded to 3:
SetAttributes[WithRules, HoldRest]
WithRules[rules_, expr_] := ReleaseHold#Module[{notSet}, Quiet[
With[{args = Reverse[rules /. Rule[a_, b_] -> notSet[a, b]]},
Fold[With[{#2}, #1] &, Hold#expr, args]] /. notSet -> Set,
With::lvw]]
This was also posted as an answer to an unrelated question, and as noted there, it has been discussed (at least) a couple of times on usenet:
A version of With that binds variables sequentially
Add syntax highlighting to own command
HTH
EDIT: Added a ReleaseHold, Hold pair to keep expr unevaluated until the rules have been applied.
One problem with Andrew's solution is that it maps the problem back to With, and that does not accept subscripted variables. So the following generates messages.
WithRules[{Subscript[x, 1] -> 2, Subscript[x, 2] -> 3},
Power[Subscript[x, 1], Subscript[x, 2]]]
Given that With performs syntactic replacement on its body, we can set WithRules alternatively as follows:
ClearAll[WithRules]; SetAttributes[WithRules, HoldRest];
WithRules[r : {(_Rule | _RuleDelayed) ..}, body_] :=
ReleaseHold[Hold[body] /. r]
Then
In[113]:= WithRules[{Subscript[x, 1] -> 2,
Subscript[x, 2] -> 3}, Subscript[x, 1]^Subscript[x, 2]]
Out[113]= 8
Edit: Addressing valid concerns raised by Leonid, the following version would be safe:
ClearAll[WithRules3]; SetAttributes[WithRules3, HoldRest];
WithRules3[r : {(_Rule | _RuleDelayed) ..}, body_] :=
Developer`ReplaceAllUnheld[Unevaluated[body], r]
Then
In[194]:= WithRules3[{Subscript[x, 1] -> 2, Subscript[x, 2] -> 3},
Subscript[x, 1]^Subscript[x, 2]]
Out[194]= 8
In[195]:= WithRules3[{x -> y}, f[y_] :> Function[x, x + y]]
Out[195]= f[y_] :> Function[x, x + y]
Edit 2: Even WithRules3 is not completely equivalent to Andrew's version:
In[206]:= WithRules3[{z -> 2}, f[y_] :> Function[x, x + y + z]]
Out[206]= f[y_] :> Function[x, x + y + z]
In[207]:= WithRules[{z -> 2}, f[y_] :> Function[x, x + y + z]]
Out[207]= f[y$_] :> Function[x$, x$ + y$ + 2]
I am doing a brute force search for "gradient extremals" on the following example function
fv[{x_, y_}] = ((y - (x/4)^2)^2 + 1/(4 (1 + (x - 1)^2)))/2;
This involves finding the following zeros
gecond = With[{g = D[fv[{x, y}], {{x, y}}], h = D[fv[{x, y}], {{x, y}, 2}]},
g.RotationMatrix[Pi/2].h.g == 0]
Which Reduce happily does for me:
geyvals = y /. Cases[List#ToRules#Reduce[gecond, {x, y}], {y -> _}];
geyvals is the three roots of a cubic polynomial, but the expression is a bit large to put here.
Now to my question: For different values of x, different numbers of these roots are real, and I would like to pick out the values of x where the solutions branch in order to piece together the gradient extremals along the valley floor (of fv). In the present case, since the polynomial is only cubic, I could probably do it by hand -- but I am looking for a simple way of having Mathematica do it for me?
Edit: To clarify: The gradient extremals stuff is just background -- and a simple way to set up a hard problem. I am not so interested in the specific solution to this problem as in a general hand-off way of spotting the branch points for polynomial roots. Have added an answer below with a working approach.
Edit 2: Since it seems that the actual problem is much more fun than root branching: rcollyer suggests using ContourPlot directly on gecond to get the gradient extremals. To make this complete we need to separate valleys and ridges, which is done by looking at the eigenvalue of the Hessian perpendicular to the gradient. Putting a check for "valleynes" in as a RegionFunction we are left with only the valley line:
valleycond = With[{
g = D[fv[{x, y}], {{x, y}}],
h = D[fv[{x, y}], {{x, y}, 2}]},
g.RotationMatrix[Pi/2].h.RotationMatrix[-Pi/2].g >= 0];
gbuf["gevalley"]=ContourPlot[gecond // Evaluate, {x, -2, 4}, {y, -.5, 1.2},
RegionFunction -> Function[{x, y}, Evaluate#valleycond],
PlotPoints -> 41];
Which gives just the valley floor line. Including some contours and the saddle point:
fvSaddlept = {x, y} /. First#Solve[Thread[D[fv[{x, y}], {{x, y}}] == {0, 0}]]
gbuf["contours"] = ContourPlot[fv[{x, y}],
{x, -2, 4}, {y, -.7, 1.5}, PlotRange -> {0, 1/2},
Contours -> fv#fvSaddlept (Range[6]/3 - .01),
PlotPoints -> 41, AspectRatio -> Automatic, ContourShading -> None];
gbuf["saddle"] = Graphics[{Red, Point[fvSaddlept]}];
Show[gbuf /# {"contours", "saddle", "gevalley"}]
We end up with a plot like this:
Not sure if this (belatedly) helps, but it seems you are interested in discriminant points, that is, where both polynomial and derivative (wrt y) vanish. You can solve this system for {x,y} and throw away complex solutions as below.
fv[{x_, y_}] = ((y - (x/4)^2)^2 + 1/(4 (1 + (x - 1)^2)))/2;
gecond = With[{g = D[fv[{x, y}], {{x, y}}],
h = D[fv[{x, y}], {{x, y}, 2}]}, g.RotationMatrix[Pi/2].h.g]
In[14]:= Cases[{x, y} /.
NSolve[{gecond, D[gecond, y]} == 0, {x, y}], {_Real, _Real}]
Out[14]= {{-0.0158768, -15.2464}, {1.05635, -0.963629}, {1.,
0.0625}, {1., 0.0625}}
If you only want to plot the result then use StreamPlot[] on the gradients:
grad = D[fv[{x, y}], {{x, y}}];
StreamPlot[grad, {x, -5, 5}, {y, -5, 5},
RegionFunction -> Function[{x, y}, fv[{x, y}] < 1],
StreamScale -> 1]
You may have to fiddle around with the plot's precision, StreamStyle, and the RegionFunction to get it perfect. Especially useful would be using the solution for the valley floor to seed StreamPoints programmatically.
Updated: see below.
I'd approach this first by visualizing the imaginary parts of the roots:
This tells you three things immediately: 1) the first root is always real, 2) the second two are the conjugate pairs, and 3) there is a small region near zero in which all three are real. Additionally, note that the exclusions only got rid of the singular point at x=0, and we can see why when we zoom in:
We can then use the EvalutionMonitor to generate the list of roots directly:
Map[Module[{f, fcn = #1},
f[x_] := Im[fcn];
Reap[Plot[f[x], {x, 0, 1.5},
Exclusions -> {True, f[x] == 1, f[x] == -1},
EvaluationMonitor :> Sow[{x, f[x]}][[2, 1]] //
SortBy[#, First] &];]
]&, geyvals]
(Note, the Part specification is a little odd, Reap returns a List of what is sown as the second item in a List, so this results in a nested list. Also, Plot doesn't sample the points in a straightforward manner, so SortBy is needed.) There may be a more elegant route to determine where the last two roots become complex, but since their imaginary parts are piecewise continuous, it just seemed easier to brute force it.
Edit: Since you've mentioned that you want an automatic method for generating where some of the roots become complex, I've been exploring what happens when you substitute in y -> p + I q. Now this assumes that x is real, but you've already done that in your solution. Specifically, I do the following
In[1] := poly = g.RotationMatrix[Pi/2].h.g /. {y -> p + I q} // ComplexExpand;
In[2] := {pr,pi} = poly /. Complex[a_, b_] :> a + z b & // CoefficientList[#, z] & //
Simplify[#, {x, p, q} \[Element] Reals]&;
where the second step allows me to isolate the real and imaginary parts of the equation and simplify them independent of each other. Doing this same thing with the generic 2D polynomial, f + d x + a x^2 + e y + 2 c x y + b y^2, but making both x and y complex; I noted that Im[poly] = Im[x] D[poly, Im[x]] + Im[y] D[poly,[y]], and this may hold for your equation, also. By making x real, the imaginary part of poly becomes q times some function of x, p, and q. So, setting q=0 always gives Im[poly] == 0. But, that does not tell us anything new. However, if we
In[3] := qvals = Cases[List#ToRules#RReduce[ pi == 0 && q != 0, {x,p,q}],
{q -> a_}:> a];
we get several formulas for q involving x and p. For some values of x and p, those formulas may be imaginary, and we can use Reduce to determine where Re[qvals] == 0. In other words, we want the "imaginary" part of y to be real and this can be accomplished by allowing q to be zero or purely imaginary. Plotting the region where Re[q]==0 and overlaying the gradient extremal lines via
With[{rngs = Sequence[{x,-2,2},{y,-10,10}]},
Show#{
RegionPlot[Evaluate[Thread[Re[qvals]==0]/.p-> y], rngs],
ContourPlot[g.RotationMatrix[Pi/2].h.g==0,rngs
ContourStyle -> {Darker#Red,Dashed}]}]
gives
which confirms the regions in the first two plots showing the 3 real roots.
Ended up trying myself since the goal really was to do it 'hands off'. I'll leave the question open for a good while to see if anybody finds a better way.
The code below uses bisection to bracket the points where CountRoots changes value. This works for my case (spotting the singularity at x=0 is pure luck):
In[214]:= findRootBranches[Function[x, Evaluate#geyvals[[1, 1]]], {-5, 5}]
Out[214]= {{{-5., -0.0158768}, 1}, {{-0.0158768, -5.96046*10^-9}, 3}, {{0., 0.}, 2}, {{5.96046*10^-9, 1.05635}, 3}, {{1.05635, 5.}, 1}}
Implementation:
Options[findRootBranches] = {
AccuracyGoal -> $MachinePrecision/2,
"SamplePoints" -> 100};
findRootBranches::usage =
"findRootBranches[f,{x0,x1}]: Find the the points in [x0,x1] \
where the number of real roots of a polynomial changes.
Returns list of {<interval>,<root count>} pairs.
f: Real -> Polynomial as pure function, e.g f=Function[x,#^2-x&]." ;
findRootBranches[f_, {xa_, xb_}, OptionsPattern[]] := Module[
{bisect, y, rootCount, acc = 10^-OptionValue[AccuracyGoal]},
rootCount[x_] := {x, CountRoots[f[x][y], y]};
(* Define a ecursive bisector w/ automatic subdivision *)
bisect[{{x1_, n1_}, {x2_, n2_}} /; Abs[x1 - x2] > acc] :=
Module[{x3, n3},
{x3, n3} = rootCount[(x1 + x2)/2];
Which[
n1 == n3, bisect[{{x3, n3}, {x2, n2}}],
n2 == n3, bisect[{{x1, n1}, {x3, n3}}],
True, {bisect[{{x1, n1}, {x3, n3}}],
bisect[{{x3, n3}, {x2, n2}}]}]];
(* Find initial brackets and bisect *)
Module[{xn, samplepoints, brackets},
samplepoints = N#With[{sp = OptionValue["SamplePoints"]},
If[NumberQ[sp], xa + (xb - xa) Range[0, sp]/sp, Union[{xa, xb}, sp]]];
(* Start by counting roots at initial sample points *)
xn = rootCount /# samplepoints;
(* Then, identify and refine the brackets *)
brackets = Flatten[bisect /#
Cases[Partition[xn, 2, 1], {{_, a_}, {_, b_}} /; a != b]];
(* Reinclude the endpoints and partition into same-rootcount segments: *)
With[{allpts = Join[{First#xn},
Flatten[brackets /. bisect -> List, 2], {Last#xn}]},
{#1, Last[#2]} & ### Transpose /# Partition[allpts, 2]
]]]