Mariusz Iwaniuk

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These are answers submitted by Mariusz Iwaniuk

This integral, in most cases, cannot be expressed in terms of elementary functions,but we can expressed in terms of GAMMA function.

data_v1.mw
 

 

 

 

 

"D1(s,t) :=P- (alpha1-beta*S) +  alpha2 + beta2 *q(t)^();"

proc (s, t) options operator, arrow, function_assign; P+beta*S-alpha1+alpha2+beta2*q(t) end proc

(1)

"(->)"

dem

(2)

NULL

ode1 := diff(q(t), t)+theta*q(t)/(1+N-t) = -D1(s, t)

diff(q(t), t)+theta*q(t)/(1+N-t) = -P-beta*S+alpha1-alpha2-beta2*q(t)

(3)

fn1 := q(t)

q(t)

(4)

ic1 := q(T) = 0

q(T) = 0

(5)

sol1 := simplify(dsolve({ic1, ode1}, fn1))

q(t) = (-S*beta-P+alpha1-alpha2)*(Int(exp(beta2*_z1)*(1+N-_z1)^(-theta), _z1 = T .. t))*exp(-beta2*t)*(1+N-t)^theta

(6)

Int(exp(beta2*_z1)*(1+N-_z1)^(-theta), _z1 = T .. t) = -exp(N*beta2+beta2)*((1+N-t)^(-theta)*(beta2*(1+N-t))^theta*(theta*GAMMA(-theta)+GAMMA(1-theta, beta2*(1+N-t)))+(1+N-T)^(-theta)*(beta2*(1+N-T))^theta*(GAMMA(1-theta)-GAMMA(1-theta, beta2*(1+N-T))))/beta2

Int(exp(beta2*_z1)*(1+N-_z1)^(-theta), _z1 = T .. t) = -exp(N*beta2+beta2)*((1+N-t)^(-theta)*(beta2*(1+N-t))^theta*(theta*GAMMA(-theta)+GAMMA(1-theta, beta2*(1+N-t)))+(1+N-T)^(-theta)*(beta2*(1+N-T))^theta*(GAMMA(1-theta)-GAMMA(1-theta, beta2*(1+N-T))))/beta2

(7)

lprint(Int(exp(beta2*_z1)*(1+N-_z1)^(-theta), _z1 = T .. t) = -exp(N*beta2+beta2)*((1+N-t)^(-theta)*(beta2*(1+N-t))^theta*(theta*GAMMA(-theta)+GAMMA(1-theta, beta2*(1+N-t)))+(1+N-T)^(-theta)*(beta2*(1+N-T))^theta*(GAMMA(1-theta)-GAMMA(1-theta, beta2*(1+N-T))))/beta2)

Int(exp(beta2*_z1)*(1+N-_z1)^(-theta),_z1 = T .. t) = -1/beta2*exp(N*beta2+
beta2)*((1+N-t)^(-theta)*(beta2*(1+N-t))^theta*(theta*GAMMA(-theta)+GAMMA(1-
theta,beta2*(1+N-t)))+(1+N-T)^(-theta)*(beta2*(1+N-T))^theta*(GAMMA(1-theta)-
GAMMA(1-theta,beta2*(1+N-T))))

 

NULL


 

Download data_v1.mw

 


 

 

restart

with(inttrans)

expr := exp(a-sqrt(a^2+b*s))/s

exp(a-(a^2+b*s)^(1/2))/s

(1)

`assuming`([invlaplace(exp(a-(a^2+b*s)^(1/2))/s, s, t)], [b > 0])

(1/2)*(b/Pi)^(1/2)*(int(exp(-a^2*_U1/b+a-(1/4)*b/_U1)/_U1^(3/2), _U1 = 0 .. t))

(2)

NULL

(1/2)*(b/Pi)^(1/2)*(int(exp(-a^2*_U1/b+a-(1/4)*b/_U1)/_U1^(3/2), _U1 = 0 .. t)) = (1/2)*erfc((-2*a*t+b)/(2*sqrt(b*t)))+(1/2)*exp(2*a)*erfc((2*a*t+b)/(2*sqrt(b*t)))

(1/2)*(b/Pi)^(1/2)*(int(exp(-a^2*_U1/b+a-(1/4)*b/_U1)/_U1^(3/2), _U1 = 0 .. t)) = (1/2)*erfc((1/2)*(-2*a*t+b)/(b*t)^(1/2))+(1/2)*exp(2*a)*erfc((1/2)*(2*a*t+b)/(b*t)^(1/2))

(3)

NULL

a := 2; b := 5

2

 

5

(4)

evalf[20](invlaplace(exp(a-sqrt(a^2+b*s))/s, s, 1.0))

.49675487848107438114

(5)

evalf[20](eval((1/2)*erfc((-2*a*t+b)/(2*sqrt(b*t)))+(1/2)*exp(2*a)*erfc((2*a*t+b)/(2*sqrt(b*t))), t = 1.0))

.49675487848107438112

(6)

NULL


 

Download invLaplace.mw

 


Do you have any reason to think there is a closed form?

Most integrals don't have one.

Maybe the best you can do is numerical methods.

See atthached file.

ode.mw

 

See teory of First Order Differential Equations. Only one initial value problem can be not two.

See attached file:

integral.mw

 

See attached file:

PDE_by_Elziki_Transform.mw

For first question: 

 

f := x -> 36*x^6 + 2665*x^4 + 240*x - 675 + 4534*x^2 - 5836*x^3 - 516*x^5;

minimize(f(x), x = 0 .. 4, location);
#-675, {[{x = 0}, -675]}

evalf(maximize(f(x), x = 0 .. 4, location));
#703.9550742, {[{x = 3.800387934}, 703.9550742]}

 

Try:

ode := diff(U(z), z $ 4) + c^2*diff(U(z), z $ 2) + k*c*diff(U(z), z $ 2) - (3*U(z)^2 + a)*diff(U(z), z $ 2) = 0;
Order := 5;dsolve(ode, U(z), type = 'series');


#U(z) = U(0) + D(U)(0)*z + 1/2*(D@@2)(U)(0)*z^2 + 1/6*(D@@3)(U)(0)*z^3 + (U(0)^2*(D@@2)(U)(0)/8 - c^2*(D@@2)(U)(0)/24 - k*c*(D@@2)(U)(0)/24 + (D@@2)(U)(0)*a/24)*z^4 + O(z^5)

With initial conditions 

Order := 5;dsolve([ode, U(A) = A1, D(U)(A) = B1, (D@@2)(U)(A) = C1], U(z), type = 'series');

#U(z) = A1 + B1*(z - A) + 1/2*C1*(z - A)^2 + 1/6*(D@@3)(U)(A)*(z - A)^3 + (1/8*A1^2*C1 - 1/24*c^2*C1 - 1/24*k*c*C1 + 1/24*C1*a)*(z - A)^4 + O((z - A)^5)

 

As a workround using fourier transform:

(inttrans:-invfourier(int((inttrans:-fourier(sin(p*r), p, s) assuming (0 <= r))*sin(q*r)/(p*q), r = 0 .. infinity), s, p) assuming (q < p));

#-Pi*Dirac(p + q)/(2*p*q)

Try:

simplify(pdetest(sol, sys));

gives:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Maybe this helps, see attached file:

General_formula_.mw

 

simplify(diff(int(JacobiSN(x, k)^2, x), x));

#JacobiSN(x, k)^2

See attached file:

EQ_v3.mw

Only solution,not  phase portrait.See atached file.

Solution.mw

One way is:

[seq(rhs(op(1, rootsq0[[n]])), n = 1 .. numelems([rootsq0]))];

 

Maybe like this:



Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/integrals.mw .
 

Download integrals.mw

 

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