Re: boundary conditions

*To*: mathgroup at smc.vnet.net*Subject*: [mg121601] Re: boundary conditions*From*: Oliver Ruebenkoenig <ruebenko at wolfram.com>*Date*: Thu, 22 Sep 2011 07:23:46 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201109210935.FAA13289@smc.vnet.net>

The message means following: In the PDE you have m, but not a derivative of m. Now, the order of the boundary condition must be less than the lowest order of m appearing in the PDE. The same applies for initial conditions. Just specifying Neumann type boundary conditions leave the system in a floating state. While to following code returns the interpolation functions, the boundary and initial conditions are more certainly not what you would like to model: Clear[v] alpham[v_] = (0.1 (-40 - v))/(Exp[(-40 - v)/10] - 1); betam[v_] = 4*Exp[(-65 - v)/18]; alphah[v_] = 0.07*Exp[(-65 - v)/20]; betah[v_] = 1/(Exp[(-35 - v)/10] + 1); alphan[v_] = 0.01 (-55 - v)/(Exp[(-55 - v)/10] - 1); betan[v_] = 0.125 Exp[(-65 - v)/80]; eqv = Derivative[1, 0][v][t, x] == (0.0238/(2*35400)) Derivative[0, 2][v][t, x] - 120 (v[t, x] - 50)*h[t, x]*m[t, x]^(3) - 36 (v[t, x] + 77)*n[t, x]^(4) - 0.3 (v[t, x] + 54.387) + 200; eqm = Derivative[1, 0][m][t, x] == alpham[v[t, x]]*(1 - m[t, x]) - betam[v[t, x]]*m[t, x]; eqh = Derivative[1, 0][h][t, x] == alphah[v[t, x]]*(1 - h[t, x]) - betah[v[t, x]]*h[t, x]; eqn = Derivative[1, 0][n][t, x] == alphan[v[t, x]]*(1 - n[t, x]) - betan[v[t, x]]*n[t, x]; iniconds = {v[0, x] == 0, m[0, x] == 0, h[0, x] == 0, n[0, x] == 0}; bconds = {Derivative[0, 1][v][t, 0] == 0, m[t, 0] == 0, h[t, 0] == 0, n[t, 0] == 0, Derivative[0, 1][v][t, 5] == 0, m[t, 5] == 0, h[t, 5] == 0, n[t, 5] == 0}; NDSolve[Join[{eqv, eqm, eqh, eqn}, iniconds, bconds], {v[t, x], m[t, x], h[t, x], n[t, x]}, {t, 0, 5}, {x, 0, 5}, Method -> {"MethodOfLines", "SpatialDiscretization" -> {"TensorProductGrid", MinPoints -> 1000}}, PrecisionGoal -> 2] I agree, that NDSolve can produce quite cryptic message - but I find this one a particularly clear one.... Oliver On Wed, 21 Sep 2011, DrMajorBob wrote: > I didn't solve the problem, but I did make a few changes that may be > useful: > > Clear[v] > alpham[v_] = (0.1 (-40 - v))/(Exp[(-40 - v)/10] - 1); > betam[v_] = 4*Exp[(-65 - v)/18]; > alphah[v_] = 0.07*Exp[(-65 - v)/20]; > betah[v_] = 1/(Exp[(-35 - v)/10] + 1); > alphan[v_] = 0.01 (-55 - v)/(Exp[(-55 - v)/10] - 1); > betan[v_] = 0.125 Exp[(-65 - v)/80]; > eqv = Derivative[1, 0][v][t, > x] == (0.0238/(2*35400)) Derivative[0, 2][v][t, x] - > 120 (v[t, x] - 50)*h[t, x]*m[t, x]^(3) - > 36 (v[t, x] + 77)*n[t, x]^(4) - 0.3 (v[t, x] + 54.387) + 200; > eqm = Derivative[1, 0][m][t, x] == > alpham[v[t, x]]*(1 - m[t, x]) - betam[v[t, x]]*m[t, x]; > eqh = Derivative[1, 0][h][t, x] == > alphah[v[t, x]]*(1 - h[t, x]) - betah[v[t, x]]*h[t, x]; > eqn = Derivative[1, 0][n][t, x] == > alphan[v[t, x]]*(1 - n[t, x]) - betan[v[t, x]]*n[t, x]; > iniconds = {Derivative[1, 0][v][0, x] == 0, > Derivative[1, 0][m][0, x] == Derivative[1, 0][h][0, x] == 0, > Derivative[1, 0][n][0, x] == 0}; > bconds = {Derivative[0, 1][v][t, 0] == 0, > Derivative[0, 1][m][t, 0] == 0, Derivative[0, 1][h][t, 0] == 0, > Derivative[0, 1][n][t, 0] == 0, Derivative[0, 1][v][t, 5] == 0, > Derivative[0, 1][m][t, 5] == 0, Derivative[0, 1][h][t, 5] == 0, > Derivative[0, 1][n][t, 5] == 0}; > > NDSolve[Join[{eqv, eqm, eqh, eqn}, iniconds, bconds], {v[t, x], > m[t, x], h[t, x], n[t, x]}, {t, 0, 5}, {x, 0, 5}, > Method -> {"MethodOfLines", > "SpatialDiscretization" -> {"TensorProductGrid", > MinPoints -> 1000}}, PrecisionGoal -> 2] > > SetDelayed wasn't needed and, in defining alpham and alphan, it makes no > difference what the value is at a single point -- v==-40 or -55. "Which" > is also NOT differentiable, so your definitions could cause problems > (potentially). If you meant v <= -40 or something like that, try using > UnitStep or HeavisideTheta, rather than Which. > > Reversing the order of the variables in NDSolve -- {t, 0, 5}, {x, 0, 5} > rather than the other way around -- caused the error message to mention m > rather than something unrecognizable, like TemporaryVariable$3339. The > complaint is about (m^(0,1))[t,0], which DOES appear in your boundary > conditions, but it seems to match the condition that WORKS (not the one > that doesn't) in the ONLY help example for the error message. Obviously, > more examples are needed. > > Beyond that, sorry, I don't understand the problem well enough to help. > > Bobby > > On Tue, 20 Sep 2011 01:56:26 -0500, tahereh tekieh <golesang80 at gmail.com> > wrote: > >> I had attached the nb file but anyway i write it here as well. >> >> alpham[v_] := Which [v == -40, 1, True, (0.1 (-40 - v))/(Exp[(-40 - >> v)/10] >> - 1) ]; >> >> betam[v_] := 4*Exp[(-65 - v)/18]; >> >> alphah[v_] := 0.07*Exp[(-65 - v)/20]; >> >> betah[v_] := 1/(Exp[(-35 - v)/10] + 1); >> >> alphan[v_] := Which [v == -55, 1, True, 0.01 (-55 - v)/(Exp[(-55 - >> v)/10] >> - 1)] ; >> >> betan[v_] := 0.125 Exp[(-65 - v)/80]; >> >> eqv = Derivative[1, 0][v][t, x] == (0.0238/(2*35400)) Derivative[0, >> 2][v][t, >> x] - 120 (v[t, x] - 50)*h[t, x]*m[t, x]^(3) - >> 36 (v[t, x] + 77)*n[t, x]^(4) - 0.3 (v[t, x] + 54.387) + 200; >> >> eqm = Derivative[1, 0][m][t, x] == alpham[v[t, x]]*(1 - m[t, x]) - >> betam[v[t, x]]* m[t, x]; >> >> eqh = Derivative[1, 0][h][t, x] == alphah[v[t, x]]*(1 - h[t, x]) - >> betah[v[t, x]]* h[t, x]; >> >> eqn = Derivative[1, 0][n][t, x] == alphan[v[t, x]]*(1 - n[t, x]) - >> betan[v[t, x]]* n[t, x]; >> >> iniconds = {Derivative[1, 0][v][0, x] == Derivative[1, 0][m][0, x] == >> Derivative[1, 0][h][0, x] == Derivative[1, 0][n][0, x] == 0}; >> >> bconds = {Derivative[0, 1][v][t, 0] == Derivative[0, 1][m][t, 0] >> ==Derivative[0, 1][h][t, 0] == Derivative[0, 1][n][t, 0] == 0, >> Derivative[0, 1][v][t, 5] == Derivative[0, 1][m][t, 5] ==Derivative[0, >> 1][h][t, 5] == Derivative[0, 1][n][t, 5] == 0}; >> >> here's the last statement which the error occures: >> >> solution =NDSolve[Join[{eqv, eqm, eqh, eqn}, iniconds, bconds], {v[t, >> x],m[t, x], h[t, x], n[t, x]}, {x, 0, 5}, {t, 0, 5}, >> Method -> {"MethodOfLines", "SpatialDiscretization" -> >> {"TensorProductGrid",MinPoints -> 1000}}, PrecisionGoal -> 2]; >> >> and this is the error: >> NDSolve::bdord: Boundary condition (TemporaryVariable$2704^(1,0))[0,t] >> should have derivatives of order lower than the differential order of the >> partial differential equation. >> >> >> On Mon, Sep 19, 2011 at 3:56 PM, DrMajorBob <btreat1 at austin.rr.com> >> wrote: >> >>> Show us the statements in which the error occurs. >>> >>> We're not mind-readers. >>> >>> Bobby >>> >>> >>> On Mon, 19 Sep 2011 03:32:26 -0500, tahereh tekieh >>> <golesang80 at gmail.com> >>> wrote: >>> >>> Dear Bobby, >>>> Thanks for the clue, I have attached my notebook file, the problem is >>>> with >>>> this legal expression I'll receive an bdord error. I have no idea how >>>> to >>>> impose my boundary condition the way it is. I am looking to hearing >>>> from >>>> you. >>>> regards >>>> >>>> On Sun, Sep 18, 2011 at 10:29 PM, DrMajorBob <btreat1 at austin.rr.com> >>>> wrote: >>>> >>>> Neither one is a legal expression, so I suppose there is no >>>> difference. >>>>> >>>>> Try instead: >>>>> >>>>> Derivative[0, 1][v][t, 0] >>>>> >>>>> Bobby >>>>> >>>>> On Sun, 18 Sep 2011 07:19:50 -0500, goli <golesang80 at gmail.com> wrote: >>>>> >>>>> what is the difference between these boundary conditions? do they >>>>> >>>>>> represent the same thing? >>>>>> Derivative[0,1]v[t,0]==0 >>>>>> v(0,1)[t,0]==0 >>>>>> >>>>>> >>>>>> >>>>> -- >>>>> DrMajorBob at yahoo.com >>>>> >>>>> >>> >>> -- >>> DrMajorBob at yahoo.com >>> > > >

**References**:**Re: boundary conditions***From:*DrMajorBob <btreat1@austin.rr.com>