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Question: Variables with different number of arguments in a system of partial differential equations

Question:Variables with different number of arguments in a system of partial differential equations

Delali Accolley 20 Maple 2022
pde differential-equations incomplete-question
February 04 2023
0

Hi, I would appreciate some help in solving the following system of partial differential equations. The particularity of the system is that only the individual variables (i.e. a, c, and l) depend on two arguments (t: current time period and v: date of birth). The other variables are either rates or aggregates. The aggregate variables (i.e. K, L, and Y) depend only on time, t, because they are the integrals of the individual variables over v. Thanks for your help.

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NULL
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NULL
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restart; with(PDEtools)
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alpha := .3306341; delta := solve(1+0.4877489e-1 = exp(delta_a)); T := 1
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l0 := 42.375*(33.16667/(52*(24-9.95)*7)); L0 := l0*T
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x0 := 0
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IOR := .2093723865
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KOR := IOR/delta
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r0 := alpha/KOR-delta; rho := r0; fsolve({K0 = Y0*KOR, w0 = (1-alpha)*Y0/L0, alpha*exp(x0)*K0^(alpha-1)*L0^(1-alpha) = r0+delta}); assign(%)

0.2758153402e-1

  

{K0 = 2.510740299, Y0 = .5710794198, w0 = 1.390997088}

 (1)
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C0 := -K0*delta+Y0; c0 := C0/T

.4515111588

  

.4515111588

 (2)
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sigma := w0*(1-l0)/c0

2.234133040

 (3)
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eq1 := diff(c(t, v), t) = (r(t)-rho)*c(t, v)
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eq2 := sigma*c(t, v) = w(t)*(1-l(t, v))
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eq3 := r(t)*a(t, v)+w(t)*l(t, v) = c(t, v)+diff(a(t, v), t)
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eq4 := Y(t) = exp(x(t))*K(t)^alpha*L(t)^(1-alpha)
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eq5 := r(t)-delta = alpha*Y(t)/K(t)
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eq6 := w(t) = (1-alpha)*Y(t)/L(t)
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eq7 := K(t) = int(a(t, v), v = t-T .. t)
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eq8 := L(t) = int(l(t, v), v = t-T .. t)
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eq9 := x(t) = piecewise(t = 0, 0, t > 0, 0.1e-1)
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eq := {eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9}
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SV := {K(0) = K0, L(0) = L0, Y(0) = Y0, a(t, 0) = 0, c(0, 0) = c0, l(0, 0) = l0, r(0) = r0, w(0) = w0}
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pdsolve(eq, SV, numeric, 'time' = t, 'range' = 0 .. T)

Error, (in pdsolve/numeric/process_PDEs) variable(s) {v} are in the PDE system but are not dependent or independent variables

  
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sol := pdsolve(eq, SV, 'time' = t, 'range' = 0 .. T)

(4)
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NULL

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