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mehdi jafari

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These are answers submitted by mehdi jafari

see maple help...

mehdi jafari 749
April 07 2014
0 0


``

Examples
 
> 

with(DynamicSystems)

Example 1: Basic usage

> 

sys1 := [diff(x[1](t), t) = x[2](t)^2-4, diff(x[2](t), t) = x[1](t)-1+u(t), y(t) = x[1](t)+x[2](t)]

sys1 := [diff(x[1](t), t) = x[2](t)^2-4, diff(x[2](t), t) = x[1](t)-1+u(t), y(t) = x[1](t)+x[2](t)]

 (1.1)
> 

eq_point1 := EquilibriumPoint(sys1, [u(t)], constraints = [0 < x[1](t)], initialpoint = [u(t) = 0, x[1](t) = 2, x[2](t) = 4])

eq_point1 := [x[1](t) = 1.49999999768708, x[2](t) = -2.00000001053627], [diff(x[1](t), t) = 0.421450963017378e-7, diff(x[2](t), t) = -0.462584648364128e-8], [u(t) = -.500000002312923], [y(t) = -.500000012849197]

 (1.2)
> 

lin_point1 := [op(eq_point1[1]), op(eq_point1[3])]

lin_point1 := [x[1](t) = 1.49999999768708, x[2](t) = -2.00000001053627, u(t) = -.500000002312923]

 (1.3)
> 

lin_model1a := Linearize(sys1, [u(t)], [y(t)], lin_point1)

"lin_model1a:=[[[`State Space`],[continuous],[`1 output(s); 1 input(s); 2 state(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t)]],[statevariable=[x[1](t),x[2](t)]]],[[x[1]([t])=x[1]([t])-1.499999998,x[2]([t])=x[2]([t])+2.000000011]],[[u[1]([t])=u([t])+0.500000002312923]],[[y[1]([t])=y([t])+0.5000000130]]"

 (1.4)
> 

PrintSystem(lin_model1a[1])

"[[[`State Space`],[continuous],[`1 output(s); 1 input(s); 2 state(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t)]],[statevariable=[x[1](t),x[2](t)]],[a=[[[0,-4.000000021],[1,0]]]],[b=[[[0],[1]]]],[c=[[[1,1]]]],[d=[[[0]]]]]"

 (1.5)

The linearization point lin_point does not give exact zero. Compute linear model for a tighter tolerance

> 

lin_model1b := Linearize(sys1, [u(t)], [y(t)], lin_point1, 'equilibriumtolerance' = 1.*10^(-10))

Warning, linpoint is not an equilibrium point

  

"lin_model1b:=[[[`Diff. Equation`],[continuous],[`1 output(s); 1 input(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t)]]],[[x[1]([t])=x[1]([t]),x[2]([t])=x[2]([t])]],[[u[1]([t])=u([t])]],[[y[1]([t])=y([t])]]"

 (1.6)
> 

PrintSystem(lin_model1b[1])

"[[[`Diff. Equation`],[continuous],[`1 output(s); 1 input(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t)]],[de={[[[(x[1])(t)=-8.000000042-4.000000021 x[2](t),],[ (x[2])(t)=x[1](t)-1+u[1](t),],[ y[1](t)=x[1](t)+x[2](t)]]]]]"

 (1.7)

Disable the checkpoint option for the same setting of the tolerance

> 

lin_model1c := Linearize(sys1, [u(t)], [y(t)], lin_point1, 'equilibriumtolerance' = 1.*10^(-10), 'checkpoint' = false)

"lin_model1c:=[[[`State Space`],[continuous],[`1 output(s); 1 input(s); 2 state(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t)]],[statevariable=[x[1](t),x[2](t)]]],[[x[1]([t])=x[1]([t])-1.499999998,x[2]([t])=x[2]([t])+2.]],[[u[1]([t])=u([t])+0.499999998]],[[y[1]([t])=y([t])+0.500000002]]"

 (1.8)

Example 2: Use of user-defined functions

> 

sys2 := {piecewise(x[1](t)<0, x[1](t), x[2](t) + x[1](t)^2) * piecewise(u(t)<0, cos(y(t)), sin(y(t))) = sin(x[1](t)^2) + 5 * y(t) + diff(x[1](t), t, t), y(t) - x[1](t)^2 + u(t)*x[1](t), diff(x[2](t), t) = f(x[1](t), u(t))};
user_function := [
    f,
    [float, float],
    float,
    proc(x, y)
    local d1, d2;
        d1 := cos(x)+x^2;
        d2 := y*d1 + y^2;
        return d1*x+d2*y- exp(d1);
    end proc
    ];

 

user_function := [f, [float, float], float, proc (x, y) local d1, d2; d1 := cos(x)+x^2; d2 := y*d1+y^2; return d1*x+d2*y-exp(d1) end proc]

 (1.9)
> 

eq_point2 := EquilibriumPoint(sys2, [u(t)], functions = [user_function], initialpoint = [x[1](t) = 1, x[2](t) = 1, u(t) = 1])

eq_point2 := [x[1](t) = .853420831346895, x[2](t) = .687832129312234], [diff(x[1](t), `$`(t, 2)) = 0.243748043970982e-8, diff(x[1](t), t) = 0., diff(x[2](t), t) = -0.114152598484907e-8], [u(t) = 1.07055911392119], [y(t) = -.185310333631790]

 (1.10)
> 

lin_point2 := [op(eq_point2[1]), op(eq_point2[3])]

lin_point2 := [x[1](t) = .853420831346895, x[2](t) = .687832129312234, u(t) = 1.07055911392119]

 (1.11)
> 

lin_model2a := Linearize(sys2, [u(t)], [y(t)], lin_point2, functions = [user_function])

"lin_model2a:=[[[`State Space`],[continuous],[`1 output(s); 1 input(s); 3 state(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t)]],[statevariable=[x[1](t),x[2](t),x[3](t)]]],[[x[1]([t])=x[1]([t])-0.8534208313,x[2]([t])=diff(x[1]([t])),x[3]([t])=x[2]([t])-0.6878321293]],[[u[1]([t])=u([t])-1.07055911392119]],[[y[1]([t])=y([t])+0.1853103334]]"

 (1.12)
> 

PrintSystem(lin_model2a[1])

"[[[`State Space`],[continuous],[`1 output(s); 1 input(s); 3 state(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t)]],[statevariable=[x[1](t),x[2](t),x[3](t)]],[a=[[[0,1,0],[-3.88404789416651974,0,-0.184251562819176457],[-0.5192123250,0,0]]]],[b=[[[0],[3.07921630860619411],[6.405315917]]]],[c=[[[0.636282549078812876,0,0]]]],[d=[[[-0.8534208313]]]]]"

 (1.13)

Example of the statevariable, inputvariable, and outputvariable options

> 

lin_model2b := Linearize(sys2, [u(t)], [y(t)], lin_point2, functions = [user_function], statevariable = XX, inputvariable = UU, outputvariable = YY)

"lin_model2b:=[[[`State Space`],[continuous],[`1 output(s); 1 input(s); 3 state(s)`],[inputvariable=[UU[1](t)]],[outputvariable=[YY[1](t)]],[statevariable=[XX[1](t),XX[2](t),XX[3](t)]]],[[XX[1]([t])=x[1]([t])-0.8534208313,XX[2]([t])=diff(x[1]([t])),XX[3]([t])=x[2]([t])-0.6878321293]],[[UU[1]([t])=u([t])-1.07055911392119]],[[YY[1]([t])=y([t])+0.1853103334]]"

 (1.14)
> 

PrintSystem(lin_model2b[1])

"[[[`State Space`],[continuous],[`1 output(s); 1 input(s); 3 state(s)`],[inputvariable=[UU[1](t)]],[outputvariable=[YY[1](t)]],[statevariable=[XX[1](t),XX[2](t),XX[3](t)]],[a=[[[0,1,0],[-3.88404789416651974,0,-0.184251562819176457],[-0.5192123250,0,0]]]],[b=[[[0],[3.07921630860619411],[6.405315917]]]],[c=[[[0.636282549078812876,0,0]]]],[d=[[[-0.8534208313]]]]]"

 (1.15)

Example 3: Inverted pendulum on a moving cart

Variables

`#mi("&theta;")`(t)

counter-clockwise angular displacement of the pendulum from the upright position

`#mi("&phi;")`(t)

angular velocity of the pendulum, `#mi("&phi;")`(t) = diff(`#mi("&theta;")`(t), t)

x(t)

position of the cart

y(t)

velocity of the cart, y(t) = diff(x(t), t)

u(t)

horizontal force applied to the cart
 

Parameters

L

half-length of pendulum

m

mass of the pendulum

M

mass of the cart

g

gravitational constant (9.8 m/s^2)
 

> 

sys3 := [diff(x(t), t) = y(t), diff(theta(t), t) = phi(t), diff(y(t), t) = -(-3*cos(theta(t))*sin(theta(t))*g-2*u(t)+2*m*L*sin(theta(t))*phi(t)^2)/(-3*cos(theta(t))*m+2*M+2*m), diff(phi(t), t) = -(3*(-sin(theta(t))*g*M-sin(theta(t))*g*m-m*u(t)+m^2*L*sin(theta(t))*phi(t)^2))/((-3*cos(theta(t))*m+2*M+2*m)*m*L)]

sys3 := [diff(x(t), t) = y(t), diff(theta(t), t) = phi(t), diff(y(t), t) = -(-3*cos(theta(t))*sin(theta(t))*g-2*u(t)+2*m*L*sin(theta(t))*phi(t)^2)/(-3*cos(theta(t))*m+2*M+2*m), diff(phi(t), t) = -3*(-sin(theta(t))*g*M-sin(theta(t))*g*m-m*u(t)+m^2*L*sin(theta(t))*phi(t)^2)/((-3*cos(theta(t))*m+2*M+2*m)*m*L)]

 (1.16)

Linearization point is given by:

> 

lin_point3 := [phi(t) = 0, x(t) = 0, y(t) = 0, theta(t) = 0, u(t) = 0]

lin_point3 := [phi(t) = 0, x(t) = 0, y(t) = 0, theta(t) = 0, u(t) = 0]

 (1.17)
> 

lin_model3 := Linearize(sys3, [u(t)], [phi(t), x(t), y(t), theta(t)], lin_point3)

"lin_model3:=[[[`State Space`],[continuous],[`4 output(s); 1 input(s); 4 state(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t),y[2](t),y[3](t),y[4](t)]],[statevariable=[x[1](t),x[2](t),x[3](t),x[4](t)]]],[[x[1]([t])=phi([t]),x[2]([t])=theta([t]),x[3]([t])=x([t]),x[4]([t])=y([t])]],[[u[1]([t])=u([t])]],[[y[1]([t])=phi([t]),y[2]([t])=theta([t]),y[3]([t])=x([t]),y[4]([t])=y([t])]]"

 (1.18)

The state-space object given by lin_model3[1] can be used to construct a stabilizing controller using linear control theory.

> 

PrintSystem(lin_model3[1])

"[[[`State Space`],[continuous],[`4 output(s); 1 input(s); 4 state(s)`],[inputvariable=[u[1](t)]],[outputvariable=[y[1](t),y[2](t),y[3](t),y[4](t)]],[statevariable=[x[1](t),x[2](t),x[3](t),x[4](t)]],[a=[[[0,-(3. (-g M-g m))/((-m+2. M) m L),0,0],[1,0,0,0],[0,0,0,1],[0,(3. g)/(-m+2. M),0,0]]]],[b=[[[3./((-m+2. M) L)],[0],[0],[2./(-m+2. M)]]]],[c=[[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]]],[d=[[[0],[0],[0],[0]]]]]"

 (1.19)
> 

 

``


Download linearize.mw

not answer, just as an idea...

mehdi jafari 749
April 07 2014
0 4


> 

restart:
> 

sys:={diff(Y(x, t), x$2) = exp(-2*x*b)*(A(x, t)-Y(x, t)), diff(A(x, t), t) = exp(-2*x*b)*(Y(x, t)-A(x, t))};

{diff(A(x, t), t) = exp(-2*x*b)*(Y(x, t)-A(x, t)), diff(diff(Y(x, t), x), x) = exp(-2*x*b)*(A(x, t)-Y(x, t))}

 (1)
> 

ans:=pdsolve(sys);

{A(x, t) = -(9/4)*_C3*exp(_c[2]*t)*((8/3)*(((2/3)*I)*b*((exp(x*b))^2*b^2+1)*_c[2]^(5/2)+((2/3)*I)*_c[2]^(3/2)*b^3+(I*_c[2]^(7/2)*b-2*b^4*_c[2]^2-(3/2)*_c[2]^3*b^2+(1/6)*_c[2]^4)*(exp(x*b))^2-(2/3)*(b^2-(1/4)*_c[2])*(b^2+_c[2])*_c[2])*_C2*exp(2*x*b)*hypergeom([(1/2)*(-I*_c[2]^(1/2)+4*b)/b, (1/2)*(-I*_c[2]^(1/2)+4*b)/b], [(2*b-I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))-_C2*((I*_c[2]^(9/2)*b-(32/9)*b^4*_c[2]^3-(22/9)*_c[2]^4*b^2+(1/9)*_c[2]^5)*(exp(x*b))^2+I*_c[2]^(7/2)*b-(32/9)*b^4*_c[2]^2-(22/9)*_c[2]^3*b^2+(1/9)*_c[2]^4)*(exp(2*x*b))^2*hypergeom([(1/2)*(-I*_c[2]^(1/2)+6*b)/b, (1/2)*(-I*_c[2]^(1/2)+6*b)/b], [(-I*_c[2]^(1/2)+3*b)/b], -_c[2]*exp(2*x*b))-(8/3)*(exp(I*_c[2]^(1/2)*x))^2*_C1*(((2/3)*I)*b*((exp(x*b))^2*b^2+1)*_c[2]^(5/2)+((2/3)*I)*_c[2]^(3/2)*b^3+(I*_c[2]^(7/2)*b+2*b^4*_c[2]^2+(3/2)*_c[2]^3*b^2-(1/6)*_c[2]^4)*(exp(x*b))^2+(2/3)*(b^2-(1/4)*_c[2])*(b^2+_c[2])*_c[2])*exp(2*x*b)*hypergeom([(1/2)*(I*_c[2]^(1/2)+4*b)/b, (1/2)*(I*_c[2]^(1/2)+4*b)/b], [(2*b+I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))+(exp(I*_c[2]^(1/2)*x))^2*((I*_c[2]^(9/2)*b+(32/9)*b^4*_c[2]^3+(22/9)*_c[2]^4*b^2-(1/9)*_c[2]^5)*(exp(x*b))^2+I*_c[2]^(7/2)*b-(1/9)*_c[2]^4+(22/9)*_c[2]^3*b^2+(32/9)*b^4*_c[2]^2)*_C1*(exp(2*x*b))^2*hypergeom([(1/2)*(I*_c[2]^(1/2)+6*b)/b, (1/2)*(I*_c[2]^(1/2)+6*b)/b], [(I*_c[2]^(1/2)+3*b)/b], -_c[2]*exp(2*x*b))-(16/9)*_C2*(-(1/4)*(_c[2]*(exp(x*b))^2+1)*(b^2+_c[2])*exp(-2*x*b)+I*_c[2]^(5/2)*(exp(x*b))^2*b+I*_c[2]^(3/2)*(exp(x*b))^2*b^3-(b^2+_c[2])*_c[2]*(-1/4+(b^2-(1/4)*_c[2])*(exp(x*b))^2))*hypergeom([-(1/2)*(I*_c[2]^(1/2)-2*b)/b, -(1/2)*(I*_c[2]^(1/2)-2*b)/b], [(b-I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))+(16/9)*(exp(I*_c[2]^(1/2)*x))^2*_C1*((1/4)*(_c[2]*(exp(x*b))^2+1)*(b^2+_c[2])*exp(-2*x*b)+I*_c[2]^(5/2)*(exp(x*b))^2*b+I*_c[2]^(3/2)*(exp(x*b))^2*b^3+(b^2+_c[2])*_c[2]*(-1/4+(b^2-(1/4)*_c[2])*(exp(x*b))^2))*hypergeom([(1/2)*(2*b+I*_c[2]^(1/2))/b, (1/2)*(2*b+I*_c[2]^(1/2))/b], [(b+I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b)))/(exp(I*_c[2]^(1/2)*x)*(b+I*_c[2]^(1/2))*(I*_c[2]^(1/2)-b)*exp(-2*x*b)), Y(x, t) = _C3*exp(_c[2]*t)*_C1*hypergeom([(1/2)*(2*b+I*_c[2]^(1/2))/b, (1/2)*(2*b+I*_c[2]^(1/2))/b], [(b+I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))*_c[2]*exp(I*_c[2]^(1/2)*x)*(exp(x*b))^2+_C3*exp(_c[2]*t)*_C2*hypergeom([-(1/2)*(I*_c[2]^(1/2)-2*b)/b, -(1/2)*(I*_c[2]^(1/2)-2*b)/b], [-(I*_c[2]^(1/2)-b)/b], -_c[2]*exp(2*x*b))*_c[2]*(exp(x*b))^2/exp(I*_c[2]^(1/2)*x)+_C3*exp(_c[2]*t)*_C1*hypergeom([(1/2)*(2*b+I*_c[2]^(1/2))/b, (1/2)*(2*b+I*_c[2]^(1/2))/b], [(b+I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))*exp(I*_c[2]^(1/2)*x)+_C3*exp(_c[2]*t)*_C2*hypergeom([-(1/2)*(I*_c[2]^(1/2)-2*b)/b, -(1/2)*(I*_c[2]^(1/2)-2*b)/b], [-(I*_c[2]^(1/2)-b)/b], -_c[2]*exp(2*x*b))/exp(I*_c[2]^(1/2)*x)}

 (2)
> 

PDEtools[build](op(ans));

A(x, t) = -(9/4)*_C3*exp(_c[2]*t)*((8/3)*(((2/3)*I)*b*((exp(x*b))^2*b^2+1)*_c[2]^(5/2)+((2/3)*I)*_c[2]^(3/2)*b^3+(I*_c[2]^(7/2)*b-2*b^4*_c[2]^2-(3/2)*_c[2]^3*b^2+(1/6)*_c[2]^4)*(exp(x*b))^2-(2/3)*(b^2-(1/4)*_c[2])*(b^2+_c[2])*_c[2])*_C2*exp(2*x*b)*hypergeom([(1/2)*(-I*_c[2]^(1/2)+4*b)/b, (1/2)*(-I*_c[2]^(1/2)+4*b)/b], [(2*b-I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))-_C2*((I*_c[2]^(9/2)*b-(32/9)*b^4*_c[2]^3-(22/9)*_c[2]^4*b^2+(1/9)*_c[2]^5)*(exp(x*b))^2+I*_c[2]^(7/2)*b-(32/9)*b^4*_c[2]^2-(22/9)*_c[2]^3*b^2+(1/9)*_c[2]^4)*(exp(2*x*b))^2*hypergeom([(1/2)*(-I*_c[2]^(1/2)+6*b)/b, (1/2)*(-I*_c[2]^(1/2)+6*b)/b], [(-I*_c[2]^(1/2)+3*b)/b], -_c[2]*exp(2*x*b))-(8/3)*(exp(I*_c[2]^(1/2)*x))^2*_C1*(((2/3)*I)*b*((exp(x*b))^2*b^2+1)*_c[2]^(5/2)+((2/3)*I)*_c[2]^(3/2)*b^3+(I*_c[2]^(7/2)*b+2*b^4*_c[2]^2+(3/2)*_c[2]^3*b^2-(1/6)*_c[2]^4)*(exp(x*b))^2+(2/3)*(b^2-(1/4)*_c[2])*(b^2+_c[2])*_c[2])*exp(2*x*b)*hypergeom([(1/2)*(I*_c[2]^(1/2)+4*b)/b, (1/2)*(I*_c[2]^(1/2)+4*b)/b], [(2*b+I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))+(exp(I*_c[2]^(1/2)*x))^2*((I*_c[2]^(9/2)*b+(32/9)*b^4*_c[2]^3+(22/9)*_c[2]^4*b^2-(1/9)*_c[2]^5)*(exp(x*b))^2+I*_c[2]^(7/2)*b-(1/9)*_c[2]^4+(22/9)*_c[2]^3*b^2+(32/9)*b^4*_c[2]^2)*_C1*(exp(2*x*b))^2*hypergeom([(1/2)*(I*_c[2]^(1/2)+6*b)/b, (1/2)*(I*_c[2]^(1/2)+6*b)/b], [(I*_c[2]^(1/2)+3*b)/b], -_c[2]*exp(2*x*b))-(16/9)*_C2*(-(1/4)*(_c[2]*(exp(x*b))^2+1)*(b^2+_c[2])*exp(-2*x*b)+I*_c[2]^(5/2)*(exp(x*b))^2*b+I*_c[2]^(3/2)*(exp(x*b))^2*b^3-(b^2+_c[2])*_c[2]*(-1/4+(b^2-(1/4)*_c[2])*(exp(x*b))^2))*hypergeom([-(1/2)*(I*_c[2]^(1/2)-2*b)/b, -(1/2)*(I*_c[2]^(1/2)-2*b)/b], [(b-I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b))+(16/9)*(exp(I*_c[2]^(1/2)*x))^2*_C1*((1/4)*(_c[2]*(exp(x*b))^2+1)*(b^2+_c[2])*exp(-2*x*b)+I*_c[2]^(5/2)*(exp(x*b))^2*b+I*_c[2]^(3/2)*(exp(x*b))^2*b^3+(b^2+_c[2])*_c[2]*(-1/4+(b^2-(1/4)*_c[2])*(exp(x*b))^2))*hypergeom([(1/2)*(2*b+I*_c[2]^(1/2))/b, (1/2)*(2*b+I*_c[2]^(1/2))/b], [(b+I*_c[2]^(1/2))/b], -_c[2]*exp(2*x*b)))/(exp(I*_c[2]^(1/2)*x)*(b+I*_c[2]^(1/2))*(I*_c[2]^(1/2)-b)*exp(-2*x*b))

 (3)
> 

# up to here,the answer witouht inital conditons and boundaries #
> 

 

> 

 

> 

 

> 

SYS:={isolate(op(1,sys),Y(x,t)),isolate(op(2,sys),A(x,t))};

{A(x, t) = (diff(diff(Y(x, t), x), x))/exp(-2*x*b)+Y(x, t), Y(x, t) = (diff(A(x, t), t))/exp(-2*x*b)+A(x, t)}

 (4)
> 

ibc:={A(x, 0) = 0, Y(0, t) = 0.1, D[1](Y)(0, t) =0};

{A(x, 0) = 0, Y(0, t) = .1, (D[1](Y))(0, t) = 0}

 (5)
> 

 

> 

Y(x,t):=(rhs(op(2,SYS)));

(diff(A(x, t), t))/exp(-1.0*x)+A(x, t)

 (6)
> 

PDE:=diff(Y(x, t), x$2) = exp(-2*x*b)*(A(x, t)-Y(x, t));pdsolve(PDE);PDEtools[build](op(%));

(diff(diff(diff(A(x, t), t), x), x))/exp(-1.0*x)+2.0*(diff(diff(A(x, t), t), x))/exp(-1.0*x)+1.00*(diff(A(x, t), t))/exp(-1.0*x)+diff(diff(A(x, t), x), x) = -(diff(A(x, t), t))

  

PDESolStruc(A(x, t) = _F1(x)*_F2(t), [{diff(_F2(t), t) = _c[2]*_F2(t), diff(diff(_F1(x), x), x) = (-2.0*(diff(_F1(x), x))*exp(1.0*x)-1.00*_F1(x)*exp(1.0*x)-1.*_F1(x))*_c[2]/(1+exp(1.0*x)*_c[2])}])

  

A(x, t) = _F1(x)*_F2(t)

 (7)
> 

# this PDE actually is implification of system of PDEs, i think if u can convert your boundary like this, u can solve numerically your system, but actually i do not know how #
> 

 

> 

 

> 

 


Download pdsolve.mw

use evalr...

mehdi jafari 749
April 07 2014
1 5

 

restart; with(Tolerances); ALL := [`$`(1 .. 3)]; solution_k := simplify(solve(map(proc (i) options operator, arrow; 1/k[i] = (cos(Phi)*(x[i]-Tx)+sin(Phi)*(y[i]-Ty))/(-sin(Phi)*(x[i]-Tx)+cos(Phi)*(y[i]-Ty)) end proc, ALL), [Tx, Ty, Phi]), size); assign(solution_k); k := map(proc (i) options operator, arrow; tan(phi[i]) end proc, ALL); deg2rad := Tolerances:-`*`(Pi, Tolerances:-`^`(180, Tolerances:-`-`(1))); phi0[1] := Tolerances:-`*`(Pi, Tolerances:-`^`(3, Tolerances:-`-`(1))); phi0[2] := Pi; phi0[3] := Tolerances:-`-`(Tolerances:-`*`(Pi, Tolerances:-`^`(3, Tolerances:-`-`(1)))); alpha := Tolerances:-`*`(Pi, Tolerances:-`^`(10, Tolerances:-`-`(1))); L := 1; phi := phi0; x := map(proc (i) options operator, arrow; cos(phi0[i]+alpha)*L end proc, ALL); y := map(proc (i) options operator, arrow; sin(phi0[i]+alpha)*L end proc, ALL); angleError := Tolerances:-`&+-`(0, Tolerances:-`^`(10, -4)); locError := Tolerances:-`&+-`(0, Tolerances:-`^`(10, -4)); phi := map(proc (i) options operator, arrow; phi[i]+angleError end proc, ALL); x := map(proc (i) options operator, arrow; x[i]+locError end proc, ALL); y := map(proc (i) options operator, arrow; y[i]+locError end proc, ALL); evalr(Phi)

`&plusmn;`(.314, 0.401e-3)

  

NominalValue(%);

(1/2)*arctan(((1/5000+sin((13/30)*Pi)+sin((1/10)*Pi))*tan(1/10000+(1/3)*Pi)-tan(1/10000)*(-1/5000+sin((7/30)*Pi)+sin((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)))+tan(1/10000+(1/3)*Pi)*(1/5000+sin((1/10)*Pi)-sin((7/30)*Pi)-tan(1/10000+(1/3)*Pi)*(-1/5000-cos((7/30)*Pi)+cos((13/30)*Pi))-tan(1/10000)*(-1/5000-cos((1/10)*Pi)-cos((13/30)*Pi))))/((-1/5000+cos((13/30)*Pi)+cos((1/10)*Pi))*tan(-1/10000+(1/3)*Pi)+tan(1/10000)*(-1/5000-cos((7/30)*Pi)+cos((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000-sin((1/10)*Pi)+sin((7/30)*Pi)))+tan(-1/10000+(1/3)*Pi)*(-1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)-tan(-1/10000+(1/3)*Pi)*(1/5000-sin((7/30)*Pi)-sin((13/30)*Pi))-tan(1/10000)*(1/5000+sin((13/30)*Pi)+sin((1/10)*Pi)))))+(1/2)*arctan(((-1/5000+sin((13/30)*Pi)+sin((1/10)*Pi))*tan(-1/10000+(1/3)*Pi)+tan(1/10000)*(-1/5000+sin((7/30)*Pi)+sin((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)))+tan(-1/10000+(1/3)*Pi)*(-1/5000+sin((1/10)*Pi)-sin((7/30)*Pi)-tan(-1/10000+(1/3)*Pi)*(1/5000-cos((7/30)*Pi)+cos((13/30)*Pi))+tan(1/10000)*(-1/5000-cos((1/10)*Pi)-cos((13/30)*Pi))))/((1/5000+cos((13/30)*Pi)+cos((1/10)*Pi))*tan(1/10000+(1/3)*Pi)-tan(1/10000)*(-1/5000-cos((7/30)*Pi)+cos((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000-sin((1/10)*Pi)+sin((7/30)*Pi)))+tan(1/10000+(1/3)*Pi)*(1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)-tan(1/10000+(1/3)*Pi)*(-1/5000-sin((7/30)*Pi)-sin((13/30)*Pi))+tan(1/10000)*(1/5000+sin((13/30)*Pi)+sin((1/10)*Pi)))))

.3141594479

  

ToleranceValue(`%%%`);

(1/2)*arctan(((-1/5000+sin((13/30)*Pi)+sin((1/10)*Pi))*tan(-1/10000+(1/3)*Pi)+tan(1/10000)*(-1/5000+sin((7/30)*Pi)+sin((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)))+tan(-1/10000+(1/3)*Pi)*(-1/5000+sin((1/10)*Pi)-sin((7/30)*Pi)-tan(-1/10000+(1/3)*Pi)*(1/5000-cos((7/30)*Pi)+cos((13/30)*Pi))+tan(1/10000)*(-1/5000-cos((1/10)*Pi)-cos((13/30)*Pi))))/((1/5000+cos((13/30)*Pi)+cos((1/10)*Pi))*tan(1/10000+(1/3)*Pi)-tan(1/10000)*(-1/5000-cos((7/30)*Pi)+cos((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000-sin((1/10)*Pi)+sin((7/30)*Pi)))+tan(1/10000+(1/3)*Pi)*(1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)-tan(1/10000+(1/3)*Pi)*(-1/5000-sin((7/30)*Pi)-sin((13/30)*Pi))+tan(1/10000)*(1/5000+sin((13/30)*Pi)+sin((1/10)*Pi)))))-(1/2)*arctan(((1/5000+sin((13/30)*Pi)+sin((1/10)*Pi))*tan(1/10000+(1/3)*Pi)-tan(1/10000)*(-1/5000+sin((7/30)*Pi)+sin((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)))+tan(1/10000+(1/3)*Pi)*(1/5000+sin((1/10)*Pi)-sin((7/30)*Pi)-tan(1/10000+(1/3)*Pi)*(-1/5000-cos((7/30)*Pi)+cos((13/30)*Pi))-tan(1/10000)*(-1/5000-cos((1/10)*Pi)-cos((13/30)*Pi))))/((-1/5000+cos((13/30)*Pi)+cos((1/10)*Pi))*tan(-1/10000+(1/3)*Pi)+tan(1/10000)*(-1/5000-cos((7/30)*Pi)+cos((13/30)*Pi)-tan(1/10000+(1/3)*Pi)*(1/5000-sin((1/10)*Pi)+sin((7/30)*Pi)))+tan(-1/10000+(1/3)*Pi)*(-1/5000+cos((1/10)*Pi)+cos((7/30)*Pi)-tan(-1/10000+(1/3)*Pi)*(1/5000-sin((7/30)*Pi)-sin((13/30)*Pi))-tan(1/10000)*(1/5000+sin((13/30)*Pi)+sin((1/10)*Pi)))))

-0.4006133e-3

  

``


Download evalr.mw

with sum and index notation...

mehdi jafari 749
April 07 2014
0 2


> 

restart:
> 

B[t+1]:=B[t]+X[t,t+1]-D[t,t+1];

B[t]+X[t, t+1]-D[t, t+1]

 (1)
> 

T[t+1]:=(m+1)/(m*sum(X[t-i,t+1-i],i=1..m-1));

(m+1)/(m*(sum(X[t-i, t+1-i], i = 1 .. m-1)))

 (2)
> 

 


Download sum.mw

i think u just want assign s members to ...

mehdi jafari 749
April 07 2014
0 0

i think u just want assign s members to constants , u can do it like this :

for i to nops(s) do
C||i:=rhs(op(i,s));od;



> 

restart:
> 

Eq1 := diff(psi(y), y, y, y, y)-M*(diff(psi(y), y, y))-Gr*b*y = 0;
> 

 

> 

bcs1:=psi(0)=0,(D@@2)(psi)(0)=0,psi(h)=-F/2,D(psi)(h)=A;
> 

 

> 

res1:=(dsolve(Eq1));
> 

 

> 

 

diff(diff(diff(diff(psi(y), y), y), y), y)-M*(diff(diff(psi(y), y), y))-Gr*b*y = 0

  

psi(0) = 0, ((D@@2)(psi))(0) = 0, psi(h) = -(1/2)*F, (D(psi))(h) = A

  

psi(y) = -((1/6)*Gr*b*y^3-exp(M^(1/2)*y)*_C2-exp(-M^(1/2)*y)*_C1)/M+_C3*y+_C4

 (1)
> 

res2:=(dsolve({Eq1,bcs1},psi(y)));

psi(y) = -((1/6)*Gr*b*y^3-(1/6)*exp(M^(1/2)*y)*(2*Gr*b*h^3+6*A*M*h+3*F*M)/(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h))+(1/6)*exp(-M^(1/2)*y)*(2*Gr*b*h^3+6*A*M*h+3*F*M)/(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h)))/M-(1/6)*(-Gr*exp(M^(1/2)*h)*M^(1/2)*b*h^3-Gr*exp(-M^(1/2)*h)*M^(1/2)*b*h^3+3*Gr*exp(M^(1/2)*h)*b*h^2-3*Gr*exp(-M^(1/2)*h)*b*h^2+3*F*M^(3/2)*exp(M^(1/2)*h)+3*F*M^(3/2)*exp(-M^(1/2)*h)+6*A*M*exp(M^(1/2)*h)-6*A*M*exp(-M^(1/2)*h))*y/(M*(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h)))

 (2)
> 

 

> 

match(rhs(res2)=rhs(res1),y,s);

true

 (3)
> 

s;

{_C1 = -(1/6)*(2*Gr*b*h^3+6*A*M*h+3*F*M)/(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h)), _C2 = (1/6)*(2*Gr*b*h^3+6*A*M*h+3*F*M)/(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h)), _C3 = -(1/6)*(-Gr*exp(M^(1/2)*h)*M^(1/2)*b*h^3-Gr*exp(-M^(1/2)*h)*M^(1/2)*b*h^3+3*Gr*exp(M^(1/2)*h)*b*h^2-3*Gr*exp(-M^(1/2)*h)*b*h^2+3*F*M^(3/2)*exp(M^(1/2)*h)+3*F*M^(3/2)*exp(-M^(1/2)*h)+6*A*M*exp(M^(1/2)*h)-6*A*M*exp(-M^(1/2)*h))/(M*(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h))), _C4 = 0}

 (4)
> 

for i to nops(s) do
C||i:=rhs(op(i,s));od;

-(1/6)*(2*Gr*b*h^3+6*A*M*h+3*F*M)/(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h))

  

(1/6)*(2*Gr*b*h^3+6*A*M*h+3*F*M)/(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h))

  

-(1/6)*(-Gr*exp(M^(1/2)*h)*M^(1/2)*b*h^3-Gr*exp(-M^(1/2)*h)*M^(1/2)*b*h^3+3*Gr*exp(M^(1/2)*h)*b*h^2-3*Gr*exp(-M^(1/2)*h)*b*h^2+3*F*M^(3/2)*exp(M^(1/2)*h)+3*F*M^(3/2)*exp(-M^(1/2)*h)+6*A*M*exp(M^(1/2)*h)-6*A*M*exp(-M^(1/2)*h))/(M*(exp(M^(1/2)*h)*M^(1/2)*h+exp(-M^(1/2)*h)*M^(1/2)*h-exp(M^(1/2)*h)+exp(-M^(1/2)*h)))

  

0

 (5)
> 

 



Download maple_(3).mw

 

the picture is not seen...

mehdi jafari 749
April 06 2014
1 1

it is almost correct, i just made i also local ,


> 

restart;
> 

 

> 

with(plots);
> 

 

> 

g := proc (F, A, B, X0, g::evaln) local f, a, b, E, x0,i, erreur, N, n, x1, ps, pp, pic; f := F; a := A; b := B; x0 := X0; E :=evalf(abs(f(x0)));
> 

 

> 

erreur := 0.1e-3;
> 

 

> 

N := 100;
> 

 

> 

n := 0;
> 

 

> 

for i to N do x1 := evalf(x0-f(x0)/(D(f))(x0));
> 

 

> 

E := abs(f(x1));
> 

 

> 

n := n+1; ps[n] := plot([[x0, 0.], [x0, f(x0)], [x1, 0.]], color = green, thickness = 2);
> 

 

> 

x0 := x1 end do; g := evalf(x1); pp := plot(f(x), x = a .. b, color = red); pic := seq(ps[i], i = 1 .. n); display(pp, pic) end proc;
> 

f := proc (x) options operator, arrow; sin(x)-cos(2*x)+.2 end proc;
> 

a := 0; b := 6;
> 

 

[animate, animate3d, animatecurve, arrow, changecoords, complexplot, complexplot3d, conformal, conformal3d, contourplot, contourplot3d, coordplot, coordplot3d, densityplot, display, dualaxisplot, fieldplot, fieldplot3d, gradplot, gradplot3d, implicitplot, implicitplot3d, inequal, interactive, interactiveparams, intersectplot, listcontplot, listcontplot3d, listdensityplot, listplot, listplot3d, loglogplot, logplot, matrixplot, multiple, odeplot, pareto, plotcompare, pointplot, pointplot3d, polarplot, polygonplot, polygonplot3d, polyhedra_supported, polyhedraplot, rootlocus, semilogplot, setcolors, setoptions, setoptions3d, spacecurve, sparsematrixplot, surfdata, textplot, textplot3d, tubeplot]

  

proc (F, A, B, X0, g::evaln) local f, a, b, E, x0, i, erreur, N, n, x1, ps, pp, pic; f := F; a := A; b := B; x0 := X0; E := evalf(abs(f(x0))); erreur := 0.1e-3; N := 100; n := 0; for i to N do x1 := evalf(x0-f(x0)/(D(f))(x0)); E := abs(f(x1)); n := n+1; ps[n] := plot([[x0, 0.], [x0, f(x0)], [x1, 0.]], color = green, thickness = 2); x0 := x1 end do; g := evalf(x1); pp := plot(f(x), x = a .. b, color = red); pic := seq(ps[i], i = 1 .. n); plots:-display(pp, pic) end proc

  

proc (x) options operator, arrow; sin(x)-cos(2*x)+.2 end proc

  

0

  

6

 (1)
> 

 


Download correct.mw

maplesoft help page...

mehdi jafari 749
April 05 2014
2 0

please read this article or i copy some of it here :
http://www.maplesoft.com/support/faqs/detail.aspx?sid=87675

Changing the default set of fonts Maple uses in the Standard interface is a two step process.

To change the default font for a type of input or output:

1. Select Format -> Styles

2. Select the style you wish to change and click the "Modify" button.

3. Change the font type, size, colour, etc. as desired and click "OK"

The default input/output modes for Maple are:

2D Math input: "C 2D Math"

1D Math input: "C Maple Input"

2D Math output: "C 2D Output"

The above steps will apply the change for the current worksheet only. If you were to restart Maple, the old default values will be restored. You must make a user defined style set if you want this change permanently. Here is how you can save the changes made above:

1. Select Format -> Manage Style Sets -> New Style Set

2. Select "Select All". Save the file.

3. Select the "User-defined Style Set" radio button. Browse to the style set file you created in the previous step and select it. Click "OK".

This makes Maple use the style set you've defined in future sessions.

(Note that you will need to keep that style set worksheet in its current location. If you move it, the font will go back to its default)

To return to the defaults, go back to:

Format -> Manage Style Sets

and select “Default Maple Style Set”.

good luck!

oh, you are so respective and cute...

mehdi jafari 749
April 05 2014
0 2

one of many way is this :


> 

restart:n:=5;# number of arguments#

5

 (1)
> 

f:=Sum((a[i])*x^i,i=0..n);

Sum(a[i]*x^i, i = 0 .. 5)

 (2)
> 

 

> 

# giving values to coefficients#
> 

a[1]:=1.3:a[2]:=5.73:a[3]:=7:a[4]:=9:a[5]:=11:
> 

value(eval(f));

a[0]+1.3*x+5.73*x^2+7*x^3+9*x^4+11*x^5

 (3)
> 

#or u can do like this
> 

restart:
> 

f:=a*x^2+b*x+c:
> 

params:={a=1,b=2,c=3}:
> 

eval(f,params);

x^2+2*x+3

 (4)
> 

 

for editing your post, in the down of your question,there is more,click on more and edit it . hope bests for you,since you decerve it . good luck

Download value.mw

use a program which adds virtual printer...

mehdi jafari 749
April 05 2014
2 0

u can use a program which adds virtual printer so that u can print any thing u want to pdf files. for example adobe acrobat pro.
also u can use these pdfs.

M15QRC_win.pdf

maple-ref.pdf

maple_18.pdf

use hint...

mehdi jafari 749
April 05 2014
0 2

use hint for sepration of varibales, you should also write Pi with capital P .


``

> 

restart:
> 

PDE:=diff(u(x,t),t)=diff(u(x,t),x$2);

diff(u(x, t), t) = diff(diff(u(x, t), x), x)

 (1)
> 

BCs:={u(0,t)=0, u(Pi/4,t)=0 , u(x,0)=x*(Pi/4-x)};

{u(0, t) = 0, u(x, 0) = x*((1/4)*Pi-x), u((1/4)*Pi, t) = 0}

 (2)
> 

 

> 

ans:=pdsolve(PDE, HINT = h(x)*g(t));

PDESolStruc(u(x, t) = h(x)*g(t), [{diff(diff(h(x), x), x) = _c[1]*h(x), diff(g(t), t) = _c[1]*g(t)}])

 (3)
> 

PDEtools:-build(ans);

u(x, t) = exp(_c[1]^(1/2)*x)*_C1*exp(_c[1]*t)*_C2+_C1*exp(_c[1]*t)*_C3/exp(_c[1]^(1/2)*x)

 (4)
> 

NULL
> 

pds:=pdsolve(PDE,BCs,numeric):
> 

p1 := pds:-plot(t = 0):p2 := pds:-plot(t = 1/10):p3 := pds:-plot(t = 1/2):p4 := pds:-plot(t = 1):``NULLp5 := pds:-plot(t = 2):plots[display]({p1, p2, p3, p4, p5});

 
> 

 

> 

PDEtools:-Laplace(PDE,u(x,t));
> 

``

NULL

``


 actually i do not know why Laplace does not solve this.

Download pde_(2).mw

 

 

two procedures...

mehdi jafari 749
April 04 2014
0 0

 

> 

restart:
> 

Max:=proc() local m,i;
m:=-infinity;
for i in (args) do
if i>m then
m:=i
end if;od;m;
end proc:
> 

triangle:=proc(a,b,c) local A,S;
A:=evalf({(args)} minus {Max(args)});
S:=sqrt(op(1,A)^2+op(2,A)^2);
if Max(args)<S then;
print("the triangle is acute");
elif Max(args)=S then;
print("the triangle is right");
else;
print("the triangle is obtuse");
end if;
end proc:
> 

triangle(3,5,10);triangle(3,5,4);

 

  

(1)
> 

 

> 

 

> 

 

> 

 

 

 two_procedures.mw

the problem solved...

mehdi jafari 749
April 04 2014
2 9


> 

restart:
with(Optimization):
with(LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= DiagonalMatrix(P20):
P3:= DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Matrix(%id = 18446744073932440990)

Matrix(%id = 18446744073932442670)

Matrix(%id = 18446744073936503742)

Matrix(%id = 18446744073936503982)

Vector[row](%id = 18446744073936504342)

Vector[row](%id = 18446744073936504462)

Vector[row](%id = 18446744073936504582)

Vector[row](%id = 18446744073936504702)

Vector[row](%id = 18446744073936506878)

 (1)
> 

J:=X.E.Transpose(X)+U.E.Transpose(U);

a[1]^2+(1/2)*a[2]^2+(1/2)*a[3]^2+(1/2)*a[4]^2+(1/2)*a[5]^2+b[1]^2+(1/2)*b[2]^2+(1/2)*b[3]^2+(1/2)*b[4]^2+(1/2)*b[5]^2

 (2)
> 

indets(J); indets(C)

{a[1], a[2], a[3], a[4], a[5], b[1], b[2], b[3], b[4], b[5]}

{a[1], a[2], a[3], a[4], a[5], b[1], b[2], b[3], b[4], b[5]}

 (3)
> 

Minimize(J,{seq(C[i]=0,i=1..nops(C))});

[1.52587355672056, [a[1] = HFloat(1.1444051675404217), a[2] = HFloat(0.07357433784470904), a[3] = HFloat(0.019083716254587775), a[4] = HFloat(-0.23114059939913092), a[5] = HFloat(-0.11990652561630644), b[1] = HFloat(-0.3814683891801405), b[2] = HFloat(0.0), b[3] = HFloat(-8.695843524419988e-16), b[4] = HFloat(-0.23114059939913095), b[5] = HFloat(-0.1199065256163064)]]

 (4)
> 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):
> 

 

> 

 

> 

 

> 

 

> 

 


Download solved.mws

correct your matrix bounds...

mehdi jafari 749
April 04 2014
1 2

at first,i should request to upload the worksheet insetead of copying commands with this format, just copy and paste it in your worksheet to see it is not directly executable. when you load LinearAlgebra packgar,there is no need everytime to load part of package , if you need to load part of the package, just to it like this : 

with(LinearAlgebra,DiagonalMatrix):

your matrix bounds are not correct so you can not multiply them correctly ! please change your matrix bounds,

 

> 

restart:with(Optimization):with(LinearAlgebra,DiagonalMatrix):
> 

L := 1: r := 2: tau := 1:
> 

interface(rtablesize = 2*r+1):
> 

 Z:= Matrix(2*r+1, 2*r+1,[tau,seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)],scan= columns,datatype = float[8]);

Z := Matrix(5, 5, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (1, 5) = 0., (2, 1) = 0., (2, 2) = 0., (2, 3) = 0., (2, 4) = 0., (2, 5) = 0., (3, 1) = 0., (3, 2) = 0., (3, 3) = 0., (3, 4) = 0., (3, 5) = 0., (4, 1) = 0., (4, 2) = 0., (4, 3) = 0., (4, 4) = 0., (4, 5) = 0., (5, 1) = 0., (5, 2) = 0., (5, 3) = 0., (5, 4) = 0., (5, 5) = 0.})

 (1)
> 

Dtau00 := `<,>`(1); Dtau01 := Vector[row](r); Dtau02 := Vector[row](r); Dtau10 := Vector(r); Dtau20 := Vector(r);
> 

 

Dtau00 := Vector(1, {(1) = 1})

  

Dtau01 := Vector[row](2, {(1) = 0, (2) = 0})

  

Dtau02 := Vector[row](2, {(1) = 0, (2) = 0})

  

Dtau10 := Vector(2, {(1) = 0, (2) = 0})

  

Dtau20 := Vector(2, {(1) = 0, (2) = 0})

 (2)
> 

Dtau1:= DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]): Dtau2:= DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):  Dtau3:= -Dtau2: Dtau4:= copy(Dtau1): Dtau:= < < Dtau00 | Dtau01 | Dtau02 >>, < Dtau10 | Dtau1 | Dtau2 >, < Dtau20 | Dtau3 | Dtau4 > ;

Dtau := Matrix(1, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0}), Matrix(2, 5, {(1, 1) = 0, (1, 2) = 1., (1, 3) = 0, (1, 4) = 0., (1, 5) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 1., (2, 4) = 0, (2, 5) = 0.}), Matrix(2, 5, {(1, 1) = 0, (1, 2) = -0., (1, 3) = -0., (1, 4) = 1., (1, 5) = 0, (2, 1) = 0, (2, 2) = -0., (2, 3) = -0., (2, 4) = 0, (2, 5) = 1.})

 (3)
> 

P00:= < L/2 >: P01:= Vector[row](r): P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):P10:= Vector(r): P20:= Vector(r, i-> evalf(L/2/i/Pi)): P1:= Matrix(r,r): P2:= DiagonalMatrix(P20): P3:= DiagonalMatrix(-P20): P4:= Matrix(r,r): P:= < < P00 | P01 | P02 >> ,< P10 | P1 | P2 >, < P20 | P3 | P4 > ;

P := Matrix(1, 5, {(1, 1) = 1/2, (1, 2) = 0, (1, 3) = 0, (1, 4) = -.318309886100000, (1, 5) = -.159154943000000}), Matrix(2, 5, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = .1591549430, (1, 5) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0.7957747152e-1}), Matrix(2, 5, {(1, 1) = .1591549430, (1, 2) = -.159154943000000, (1, 3) = 0., (1, 4) = 0, (1, 5) = 0, (2, 1) = 0.7957747152e-1, (2, 2) = 0., (2, 3) = -0.795774715200000e-1, (2, 4) = 0, (2, 5) = 0})

 (4)
> 

interface(rtablesize = 2*r+1); J := Vector([L, `$`((1/2)*L, 2*r)]); E := DiagonalMatrix(J); X := Vector[row](2*r+1, symbol = a); U := Vector[row](2*r+1, symbol = b); X0 := Vector[row](2*r+1, [1]); G := Vector[row](2*r+1, [1]); C := simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

5

  

J := Vector(5, {(1) = 1, (2) = 1/2, (3) = 1/2, (4) = 1/2, (5) = 1/2})

  

E := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1/2, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1/2, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1/2, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 1/2})

  

X := Vector[row](5, {(1) = a[1], (2) = a[2], (3) = a[3], (4) = a[4], (5) = a[5]})

  

U := Vector[row](5, {(1) = b[1], (2) = b[2], (3) = b[3], (4) = b[4], (5) = b[5]})

  

X0 := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

  

G := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

 (5)
> 

J := X.E.Transpose(X)+U.E.Transpose(U);
> 

Minimize(J, {C = 0});
> 

 


Download problem.mw

actually the only way is to go back from...

mehdi jafari 749
April 02 2014
0 0

you can remove differentiation independent parameter by using declare and prime, like this :

with(PDEtools,declare):
declare(A(t), prime = t);

but actually it is not what you want, what you want is this :

> 

restart:
> 

 

> 

 

> 

d:=diff(A(t), t);
> 

 

diff(A(t), t)

 (1)
> 

A(t)=((Int(d,t)));

A(t) = Int(diff(A(t), t), t)

 (2)
> 

``

maybe it is so obvious ,but this is the only direct way to relate a function to its differentiation !!! and u can not do it in any other way i think ! good luck !

Download diff.mw

refer to these links, hops it helps you....

mehdi jafari 749
April 02 2014
1 2

Hi sir,please refer to these links, hope it helps you . good luck Sir.

http://www.mapleprimes.com/questions/150659-How-To-Plot-With-The-Results-Of-Desolve-Command

http://www.mapleprimes.com/questions/201243-Solution-Of-System-Of-Nonlinear-ODEs#answer204818

good luck!

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