Items tagged with derivative derivative Tagged Items Feed

Determine using determinants the range of values of a (if any) such that
f(x,y,z)=4x^2+y^2+2z^2+2axy-4xz+2yz
has a minimum at (0,0,0).

From the theory, I understand that if the matrix corresponding to the coefficients of the function is positive definite, the function has a local min at the point. But, how do I get the range of values of a such that f is a min? Is this equivalent to finding a such that det(A) > 0?

 

2.

Now modify the function to also involve a parameter b: g(x,y,z)=bx^2+2axy+by^2+4xz-2a^2yz+2bz^2. We determine conditions on a and b such that g has a minimum at (0,0,0).
By plotting each determinant (using implicitplot perhaps, we can identify the region in the (a,b) plane where g has a local minimum.

Which region corresponds to a local minimum?

Now determine region(s) in the (a,b) plane where g has a local maximum.

I don't understand this part at all..

I have the function:   f(x,y) = 1/(sqrt(2*Pi)) * e-1/2(x^2+y^2)

I need to take the total derivative of this function in maple, but I don't know the syntax.

I've plotted the graph for this max function. Is there any way I can find the points of discontinuity in general and then use that to compute the derivatives at points where it exists?

Hi Everyone,

I have an expression that contains a second order derivative: EXPR:=ay''(x) + bz'(x) + cf'(x)+... The variable y  obeys an ordinary differential equation, y''(x) = f(y,x). I would like to replace the second order deriavtive in my expression with f(y,x). So far I have tried applyrule([y''(x)=f(y,x)],EXPR), subs(y''(x)=f(y,x),EXPR) and algsubs(y''(x)=f(y,x),EXPR) and nothing seems to work. Any helpful suggestions?

 

Hello,

 

could you help me solve this error ? I don't understand what it means.

 


> eq3:=diff(x(t),t,t)+Gamma*diff(x(t),t)+omega[0]^2*(x(t)-(diff(x(t),t,t)+Gamma*diff(x(t),t)+omega[0]^2*x(t)+omega[0]^2*X[0])/omega[0]^2) = -omega[0]^2*X[0]:
> dsolve(eq3);
Warning, it is required that the numerator of the given ODE depends on the highest derivative. Returning NULL.

 

Thanks.

Hello i want to sort according to u derivatives (k) system.  And finding determining equations system and solving this system. Thank you very much.  

restart

with(PDEtools)

[CanonicalCoordinates, ChangeSymmetry, CharacteristicQ, CharacteristicQInvariants, ConservedCurrentTest, ConservedCurrents, ConsistencyTest, D_Dx, DeterminingPDE, Eta_k, Euler, FromJet, InfinitesimalGenerator, Infinitesimals, IntegratingFactorTest, IntegratingFactors, InvariantEquation, InvariantSolutions, InvariantTransformation, Invariants, Laplace, Library, PDEplot, PolynomialSolutions, ReducedForm, SimilaritySolutions, SimilarityTransformation, Solve, SymmetrySolutions, SymmetryTest, SymmetryTransformation, TWSolutions, ToJet, build, casesplit, charstrip, dchange, dcoeffs, declare, diff_table, difforder, dpolyform, dsubs, mapde, separability, splitstrip, splitsys, undeclare]

(1)

U := diff_table(u(x, y, t))

table( [(  ) = u(x, y, t) ] )

(2)

declare(U[])

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

(3)

pde := diff(U[t]-(3/2)*U[x]-6*U[]^2*U[x]+U[x, x, x], x)+U[y, y] = 0

diff(diff(u(x, y, t), t), x)-(3/2)*(diff(diff(u(x, y, t), x), x))-12*u(x, y, t)*(diff(u(x, y, t), x))^2-6*u(x, y, t)^2*(diff(diff(u(x, y, t), x), x))+diff(diff(diff(diff(u(x, y, t), x), x), x), x)+diff(diff(u(x, y, t), y), y) = 0

(4)

NULL

w := phi(x, y, t, U[])

phi(x, y, t, u(x, y, t))

(5)

w*(-12*U[x]^2-12*U[]*U[x, x])+12*w*U[x]^2+12*U[]*w*U[x, x]+(diff(w, x, x))*(-3/2-6*U[]^2)+diff(diff(w, t), x)+diff(w, y, y)+diff(w, x, x, x, x)-lambda*(diff(U[t]-(3/2)*U[x]-6*U[]^2*U[x]+U[x, x, x], x)+U[y, y])

-lambda*(diff(diff(u(x, y, t), t), x)-(3/2)*(diff(diff(u(x, y, t), x), x))-12*u(x, y, t)*(diff(u(x, y, t), x))^2-6*u(x, y, t)^2*(diff(diff(u(x, y, t), x), x))+diff(diff(diff(diff(u(x, y, t), x), x), x), x)+diff(diff(u(x, y, t), y), y))+(D[1, 1, 1, 1](phi))(x, y, t, u(x, y, t))+(D[1, 1, 1, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))+((D[1, 1, 1, 4](phi))(x, y, t, u(x, y, t))+(D[1, 1, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(u(x, y, t), x))+(D[1, 1, 4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), x), x))+((D[1, 1, 1, 4](phi))(x, y, t, u(x, y, t))+(D[1, 1, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))+((D[1, 1, 4, 4](phi))(x, y, t, u(x, y, t))+(D[1, 4, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(u(x, y, t), x))+(D[1, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), x), x)))*(diff(u(x, y, t), x))+2*((D[1, 1, 4](phi))(x, y, t, u(x, y, t))+(D[1, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(diff(u(x, y, t), x), x))+(D[1, 4](phi))(x, y, t, u(x, y, t))*(diff(diff(diff(u(x, y, t), x), x), x))+((D[1, 1, 1, 4](phi))(x, y, t, u(x, y, t))+(D[1, 1, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))+((D[1, 1, 4, 4](phi))(x, y, t, u(x, y, t))+(D[1, 4, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(u(x, y, t), x))+(D[1, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), x), x))+((D[1, 1, 4, 4](phi))(x, y, t, u(x, y, t))+(D[1, 4, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))+((D[1, 4, 4, 4](phi))(x, y, t, u(x, y, t))+(D[4, 4, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(u(x, y, t), x))+(D[4, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), x), x)))*(diff(u(x, y, t), x))+2*((D[1, 4, 4](phi))(x, y, t, u(x, y, t))+(D[4, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(diff(u(x, y, t), x), x))+(D[4, 4](phi))(x, y, t, u(x, y, t))*(diff(diff(diff(u(x, y, t), x), x), x)))*(diff(u(x, y, t), x))+3*((D[1, 1, 4](phi))(x, y, t, u(x, y, t))+(D[1, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))+((D[1, 4, 4](phi))(x, y, t, u(x, y, t))+(D[4, 4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(u(x, y, t), x))+(D[4, 4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), x), x)))*(diff(diff(u(x, y, t), x), x))+3*((D[1, 4](phi))(x, y, t, u(x, y, t))+(D[4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(diff(diff(u(x, y, t), x), x), x))+(D[4](phi))(x, y, t, u(x, y, t))*(diff(diff(diff(diff(u(x, y, t), x), x), x), x))+(D[2, 2](phi))(x, y, t, u(x, y, t))+(D[2, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), y))+((D[2, 4](phi))(x, y, t, u(x, y, t))+(D[4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), y)))*(diff(u(x, y, t), y))+(D[4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), y), y))+(D[1, 3](phi))(x, y, t, u(x, y, t))+(D[3, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))+((D[1, 4](phi))(x, y, t, u(x, y, t))+(D[4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(u(x, y, t), t))+(D[4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), t), x))+((D[1, 1](phi))(x, y, t, u(x, y, t))+(D[1, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))+((D[1, 4](phi))(x, y, t, u(x, y, t))+(D[4, 4](phi))(x, y, t, u(x, y, t))*(diff(u(x, y, t), x)))*(diff(u(x, y, t), x))+(D[4](phi))(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), x), x)))*(-3/2-6*u(x, y, t)^2)+12*u(x, y, t)*phi(x, y, t, u(x, y, t))*(diff(diff(u(x, y, t), x), x))+12*phi(x, y, t, u(x, y, t))*(diff(u(x, y, t), x))^2+phi(x, y, t, u(x, y, t))*(-12*(diff(u(x, y, t), x))^2-12*u(x, y, t)*(diff(diff(u(x, y, t), x), x)))

(6)

k := simplify(%)

-(3/2)*(D[1, 1](phi))(x, y, t, u(x, y, t))+(D[1, 3](phi))(x, y, t, u(x, y, t))+(D[2, 2](phi))(x, y, t, u(x, y, t))+(D[1, 1, 1, 1](phi))(x, y, t, u(x, y, t))+4*(D[1, 1, 1, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)+6*(D[1, 1, 4](phi))(x, y, t, u(x, y, t))*(D[1, 1](u))(x, y, t)+4*(D[1, 4](phi))(x, y, t, u(x, y, t))*(D[1, 1, 1](u))(x, y, t)+(D[4](phi))(x, y, t, u(x, y, t))*(D[1, 1, 1, 1](u))(x, y, t)+2*(D[2, 4](phi))(x, y, t, u(x, y, t))*(D[2](u))(x, y, t)+(D[4](phi))(x, y, t, u(x, y, t))*(D[2, 2](u))(x, y, t)+(D[3, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)+(D[4](phi))(x, y, t, u(x, y, t))*(D[1, 3](u))(x, y, t)-3*(D[1, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)-(3/2)*(D[4](phi))(x, y, t, u(x, y, t))*(D[1, 1](u))(x, y, t)-lambda*(D[1, 3](u))(x, y, t)+(3/2)*lambda*(D[1, 1](u))(x, y, t)-lambda*(D[1, 1, 1, 1](u))(x, y, t)-lambda*(D[2, 2](u))(x, y, t)+6*(D[1, 1, 4, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)^2+4*(D[1, 4, 4, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)^3+(D[4, 4, 4, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)^4+3*(D[4, 4](phi))(x, y, t, u(x, y, t))*(D[1, 1](u))(x, y, t)^2+(D[4, 4](phi))(x, y, t, u(x, y, t))*(D[2](u))(x, y, t)^2+(D[3](u))(x, y, t)*(D[1, 4](phi))(x, y, t, u(x, y, t))-(3/2)*(D[4, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)^2-6*(D[1, 1](phi))(x, y, t, u(x, y, t))*u(x, y, t)^2+12*lambda*u(x, y, t)*(D[1](u))(x, y, t)^2+6*lambda*u(x, y, t)^2*(D[1, 1](u))(x, y, t)+12*(D[1](u))(x, y, t)*(D[1, 4, 4](phi))(x, y, t, u(x, y, t))*(D[1, 1](u))(x, y, t)+6*(D[1](u))(x, y, t)^2*(D[4, 4, 4](phi))(x, y, t, u(x, y, t))*(D[1, 1](u))(x, y, t)+4*(D[1](u))(x, y, t)*(D[4, 4](phi))(x, y, t, u(x, y, t))*(D[1, 1, 1](u))(x, y, t)+(D[3](u))(x, y, t)*(D[4, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)-12*(D[1, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)*u(x, y, t)^2-6*(D[4, 4](phi))(x, y, t, u(x, y, t))*(D[1](u))(x, y, t)^2*u(x, y, t)^2-6*(D[4](phi))(x, y, t, u(x, y, t))*(D[1, 1](u))(x, y, t)*u(x, y, t)^2

(7)

frontend(coeff, [k, U[x]^2]);

0

(8)

frontend(coeff, [k, U[x]*U[x, x]])

Error, invalid input: coeff received O*O, which is not valid for its 2nd argument, x

 

NULL


Download det.eq..mw

 

with(plots):

a:=polarplot(3-3*cos(theta),theta=0..2*Pi):

c:=plot((3*sqrt(2)+3)/2 + ((-3*sqrt(2))/(-3*sqrt(2)-6))*(x+((3*sqrt(2)+3)/2)),x=-10..10):

display(a,c,view=[-10..10,-10..10]);

 

a:= is the polar plot of the cardiod (3-3cos(theta)


In order to plot the tangent line to the cardiod in theta= 3Pi/4, I find the point (x,y) in rectangular coord x=(3-3cos(theta)cos(theta) and y=(3-3cos(theta)sin(theta); then I find the derivative of dx/dy=

[(3-3*cos(theta)*cos(theta)+sin(theta)(3sin(theta)]/[-(3-3cos(theta)sin(theta)+cos(theta)(3sin(theta)]and from here I get the slope.So I can plot c:= tangent line to the cardiod in 3Pi/4.

How can I avoid having to convert everyting to rectangular coords, and plot the tangent line in polars?

 

 

Hi All,

I have a problem with regard to partial differential equations. I am using Lagrangian dynamics for a problem. First i have a function First i defined a function with two speeds of angles (first derivatives):

ODE := 5*(diff(theta1(t), t))+diff(theta2(t), t). This gives:

Now this gives an output. Lagrange (just a simple example now) demands that i now derive the obtained function with regard to the first derivative of theta1. In this case, the answer i want is 5. Now, if i give the command: 

diff(ODE, diff(theta1(t),t)), maple says go home. Does anybody know how to solve this? I have been searching for a solution all afternoon.

 

Thnx in advance!

Hi everyone,

I have a question regarding the derivation of tensors/matrices.
Let's assume for simplicity, that I have a vector (6x1) s and a matrix A (6x6)defining Transpose(s)*Inverse(A)*s. From this function I want to calculate the derivative w.r.t. s. My approach would be

restartwith(Physics):
with(LinearAlgebra):
Define(s,A)

Diff(
Transpose(s)*Inverse(A)*s, s)

As a result I get

though I'd rather expect something like Inverse(A)*s + Transpose(s)*Inverse(A)

Now as I'm pretty new to Maple, I can imagine that my approach is wrong, but I don't know any better and can't seem to get any information out of the help documents.

Thanks in advance for any of your suggestions!

Hi

I need you to help me in writting procedure with input "r" ( order of derivative) and some coefficients seq(alpha[i],i=1..N).  My code work very well, need only put all the element in procedure with output The Taylor series obtained in last line of my code and the order of error.  I want the procedure return the coefficients beta used in the series and the order of Error  and the coefficients beta[i]

May thinks.

Second_Question.mw

 

I will not modify my previous question: only I add this remark. I tried to write these lines of procedure.
Add_display.mw

I need only to add( if the procedure is true) these lines in my procedure.  These lines gives the degree of error computed in the code, but when I put these lines in the codes, there is an error. Thanks for your help.

Degree := degree(Error,stepsize);
    if (showorder) then:
       print(cat(`This stencil is of order `,Degree));
    fi:
    if (showerror) then:
       print(cat(`This leading order term in the error is `,Error));
    fi:
    convert(D[r$n](f)(vars) = stencil,diff);

 

 

 

Dear All, I need your help to plot the numerical solution. many thanks.

The variable t in [0,T], x in [0,1], b in [0,2].

Difference finie for waves equation is :

pde:=diff(u(x, y,t), t$2) = c^2*(diff(u(x, y,t),x$2)+diff(u(x,y,t),y$2));

i: according to x, j according to y, and k according to t.

u[i,j,k+1]=2*u[i,j,k]-u[i,j,k-1]+(c*dt/dx)^2*(u[i-1,j,k]-2*u[i,j,k]+u[i+1,j,k])+ (c*dt/dy)^2*(u[i,j-1,k]-2*u[i,j,k]+u[i,j+1,k])

 

Boundary condition: u(t=0)=1, diff(u(x,y,t),t=0)=0, and the normal derivative on the boundary of Omega =0.

How can solve this problem and plot the numerical solution.

 

 

 

I'm simplifying this Reynolds Equation starting from here:

Reynolds:=Diff(p(x)*h(x)^3/(12*mu)*Diff(p(x),x),x)-u(x)/2*Diff(p(x)*h(x),x)+Diff(p(x)*h(x)^3/(12*mu)*Diff(p(x),z),z)=Diff(p(x)*h(x),t):

 

Then I apply the dchange command:

dchange({p(x)=P(X)*Pa,x=Lx*X,h(x)=H(X)*h2},Reynolds,{P,h,X,u,H});

 

The problem is that it expands the new derivatives and I need the equation in its compact form.

How do I tel Maple only to make the substitution but not to expand the result?

I know that it expands by default, can I modify that?

 

 

I'm writing a simple Maple program to test the Generalized Finite Element Method: main_screened_Poisso.mw

When trying to define the Neumann boundary conditions, I have to define a directional derivative dudn=dudx*n. However, I can't seem to define a unit vector normal to Gamma, which is defined by a LineSegments objects.

Other than that, the row reduction is very slow, even though I'm using floating point arithmatic and not exact arithmatic, I believe.

How can I solve these problems? Thanks in advance!

 

I want  to verify the following function expression

is indeed an antiderivative of the function expression

where  A>0 with B^2-4*A*C<0.

I have tried the command

>diff((2),x);simplify(%);

where (2) is the label of the antiderivative expression.  But the result is very awkarward and lengthy. How could I verify the antiderivative expression is indeed an antiderivative of the other function by using Maple 17? 

 

1 2 3 4 5 6 7 Last Page 1 of 22