mskalsi

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These are questions asked by mskalsi

Dear all

I have an operator given below in the image. I want expand this operator upto order 5, when there are three independent variables x, y, t(i=1,2,3). The expansion can also be obtained manually, but that is always prone mistake, is the expansion posible using any routine of Differential Geometry package?

Suppose I have an expression like

Is it possible to reverse the divergence expression using homotopy operator to obtain Psi and Phi?

The above divergence expression is possible since Euler operator of the left side is zero.

I am trying to solve system linear partial differential equations using command "pdsolve". I am surprised to see that the solution given by this command is not satisfying the system, instead, an additional constraint is obtained for an arbitrary function, is there something about "pdsolve" I am missing? 


 

with(PDEtools):

DepVars := [f(x, y, t, u)]

[f(x, y, t, u)]

(1)

Sys := {diff(f(x, y, t, u), u, t)-(diff(f(x, y, t, u), x, y)) = 0, diff(f(x, y, t, u), u, u) = 0, diff(f(x, y, t, u), u, y) = 0, diff(f(x, y, t, u), x, u) = 0, diff(f(x, y, t, u), x, x) = 0, diff(f(x, y, t, u), y, y, y) = 0}

{diff(diff(f(x, y, t, u), t), u)-(diff(diff(f(x, y, t, u), x), y)) = 0, diff(diff(diff(f(x, y, t, u), y), y), y) = 0, diff(diff(f(x, y, t, u), u), u) = 0, diff(diff(f(x, y, t, u), u), x) = 0, diff(diff(f(x, y, t, u), u), y) = 0, diff(diff(f(x, y, t, u), x), x) = 0}

(2)

pdsolve(Sys)

{f(x, y, t, u) = (_F3(t)*y+_F4(t))*x+(_F3(t)+_C1)*u+(1/2)*_F7(t)*y^2+_F8(t)*y+_F9(t)}

(3)

f := proc (x, y, t, u) options operator, arrow; (_F3(t)*y+_F4(t))*x+(_F3(t)+_C1)*u+(1/2)*_F7(t)*y^2+_F8(t)*y+_F9(t) end proc

proc (x, y, t, u) options operator, arrow; (_F3(t)*y+_F4(t))*x+(_F3(t)+_C1)*u+(1/2)*_F7(t)*y^2+_F8(t)*y+_F9(t) end proc

(4)

Sys

{0 = 0, diff(_F3(t), t)-_F3(t) = 0}

(5)

``


 

Download pdsolve_command.mw

Dear All

I have a trignometric function and I plotted it in 2D. It is visible from the graph that the function has Maxima and Minima. My question is, can I located all values
 

0.12981e-1+0.80285e-1*cos(.9519256799*x)+0.41370e-1*cos(1.903851360*x)+0.35690e-1*cos(2.855777040*x)+0.147e-3*cos(3.807702720*x)

0.12981e-1+0.80285e-1*cos(.9519256799*x)+0.41370e-1*cos(1.903851360*x)+0.35690e-1*cos(2.855777040*x)+0.147e-3*cos(3.807702720*x)

(1)

plot(diff(0.12981e-1+0.80285e-1*cos(.9519256799*x)+0.41370e-1*cos(1.903851360*x)+0.35690e-1*cos(2.855777040*x)+0.147e-3*cos(3.807702720*x), x), x = -6.2 .. 6.2)

 

``


 

Download Max_Min_from_Graph.mw

of "x" corresponding to these Max. and Min. ?

Dear All

I am trying to use differential operator two times in sucession but I am not getting desired differentiation. Please see content below:


 

with(PDEtools):

DepVars := [u(x, t), v(x, t), phi(x, t), psi(x, t)]; 1; declare(u(x, t), v(x, t), phi(x, t), psi(x, t))

[u(x, t), v(x, t), phi(x, t), psi(x, t)]

 

u(x, t)*`will now be displayed as`*u

 

v(x, t)*`will now be displayed as`*v

 

phi(x, t)*`will now be displayed as`*phi

 

psi(x, t)*`will now be displayed as`*psi

(1)

_local(I):

I := phi(x, t)*(diff(v(x, t), x)+b*(diff(u(x, t), x, x)))+psi(x, t)*(diff(u(x, t), x, x, x)+d*(diff(v(x, t), x, x)))

phi(x, t)*(diff(v(x, t), x)+b*(diff(diff(u(x, t), x), x)))+psi(x, t)*(diff(diff(diff(u(x, t), x), x), x)+d*(diff(diff(v(x, t), x), x)))

(2)

ToJet(I, DepVars)

phi*(b*u[x, x]+v[x])+psi*(d*v[x, x]+u[x, x, x])

(3)

T[1] := proc (f) options operator, arrow; diff(f, u[x]) end proc; 1; T[2] := proc (f) options operator, arrow; diff(f, u[x, x]) end proc; 1; T[3] := proc (f) options operator, arrow; diff(f, u[x, x, x]) end proc; 1; U[1] := proc (f) options operator, arrow; diff(f, v[x]) end proc; 1; U[2] := proc (f) options operator, arrow; diff(f, v[x, x]) end proc

proc (f) options operator, arrow; diff(f, u[x]) end proc

 

proc (f) options operator, arrow; diff(f, u[x, x]) end proc

 

proc (f) options operator, arrow; diff(f, u[x, x, x]) end proc

 

proc (f) options operator, arrow; diff(f, v[x]) end proc

 

proc (f) options operator, arrow; diff(f, v[x, x]) end proc

(4)

d := proc (f) options operator, arrow; diff(f, x)+u[x]*(diff(f, u))+u[x, x]*(diff(f, u[x]))+u[x, x, x]*(diff(f, u[x, x])) end proc

proc (f) options operator, arrow; diff(f, x)+u[x]*(diff(f, u))+u[x, x]*(diff(f, u[x]))+u[x, x, x]*(diff(f, u[x, x])) end proc

(5)

T[2](I)

0

(6)

Why this is giving zero result ?

If I change Ito jet notation then I get result non zero result like:

T[2](phi*(b*u[x, x]+v[x])+psi*(d*v[x, x]+u[x, x, x]))

phi*b

(7)

d(T[2](phi*(b*u[x, x]+v[x])+psi*(d*v[x, x]+u[x, x, x])))

0

(8)

I was expecting this to be "phi[x]*b, "but instead I am getting zero result.``


 

Download operator_for_differentiation.mw

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