UNIVERSIDAD NACIONAL DEL ALTIPLANO FACULTAD DE INGENIERIA ECONOMICA ECONOMICA (FIE) ESCUELA PROFESIONAL DE INGENIERIA ECONOMICA
TRABAJO ENCARGADO
ECONOMIA MATEMATICA MATEMATICA IV
C
PUNO - PERÚ
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3.1.ResolverlossiguientesproblemasdeCálculodeVariaciones:
̇ ∫ + ˙˙ +, ∶ . ∫ + ˙ ̇ + , ∶ ,
a)
b)
en los casos: (i)
0 3 ≥ 2, 1 , , con
a)
Función intermedia Aplicar la ecuación de Euler Euler
Por tanto
, , con x(3) libre. libre.
,˙, ˙ , . + 2˙˙ 2˙ ˙ 2˙ + 2 ˙ 2¨ + 2 + 2˙ . ˙ 2˙ 2¨ + 2 + 2˙ " + 0. "+ 0 +10
Se trata de una ecuación diferencial de segundo orden, lineal, con coeficientes Constantes, homogénea. Homogenizar la ecuación; Raíces características; Raíz imaginaria
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≠ 0
→ → 0 → → 1 0 → → ( .. + .. ) − ℎ 0 41 0 ℎ 1 0 2 1 → → ( + ) → → + + + 0 2 0+20 2 4 →∆ ≠ 0 → ∆ 0 ˙ 0 ˙ 2˙ + 2. ˙ + ⌈2´2´+2⌉2⌉== 0 ⌈22 + + 2 + ⌉== 0
Solución particular
Solución complementaria
Donde:
Solución general
Aplicamos la condición condición inicial
Aplicamos la condición condición de transversalidad es, en este este caso
CASO 3(de lo avanzado en clases);
Calculemos, paso a paso, cada uno de los elementos que necesitamos:
Reemplazando:
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Aplicando
⌈22 + 2 + 22 + 2⌉== 0 4 2 + 16 4 + 840 444 + 2 42484 4+44 2 2 + 2484 4+44 ∶ ˙˙ 2 > 0 → → ,,˙ , + 2˙˙ 22 20 2 4 y
Factor izando
Por lo tanto, la solución general será:
Aplicando la condición condición de segundo orden Condición de Legendre Caso de mínimo.
. Se cumple dicha condición para el
MATRIZ HESSIANO: Tenemos que ver si la función Es convexa en
En los casos:
∫ + ˙ ̇ + , ∶ ,
0 33 ≥ 2, 1 + ˙ ̇
Función Intermedia;
, , con
, , con x (3) libre.
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Aplicar la ecuación de Euler
Por tanto
Integrando tenemos;
˙ 12′ ˙ 1 2′′ 112′ ˙ 0 ′ ∫ ∫ 0 ∫+ ∫ 1 1 + 1 … … … … . 1 3 ≥ 2 3+≥2……… 3+≥2……… 2 3+ 3 + 1 ≥ 2 3+1≥2 ≥ 12 3 →∆ ≠ 0 → ∆ 0 ˙˙ ≤0……… 0, 3 > 2 2 == ≤ 0 2 == ≤ 0 32≤0 ≥ 32 3 > 2
Imponemos la condición inicial y final
De (1) y (2) se tiene:
La condición de transversalidad es, en este caso:
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≥ 32 3 11 2 2 32 12 1 ≤ ≤ 3 ∶ ˙˙ 2 < 0 → → + ˙ ̇ 202 00 2 0
,, pues si fuera x (3) = 2, entonces sería Así que
y no cumpliría que
Por lo tanto, la solución general o la senda optima seria:
Condición de segundo orden Condición de Legendre Caso de máximo.
. Se cumple dicha condición para el
MATRIZ HESSIANO: Tenemos que ver si la función
es cóncava en
Por lo tanto, se cumple que es un máximo.
Se considera el problema:
∫ ,, , + +
Deducir la condición de transversalidad en cada uno de los siguientes casos: (i). La condición final es:
≥ .
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∫ ,, , + ,, , ,, , + ′′ + ≥ == ≤0…… ≤0…… 0, ∗ > ==∗ 0 ==∗ 0 ∗ == ∗ ∗0 == + 0 ==∗ 0 ∗ + ∗ == 0 0
(ii). No hay condición final, pero Replanteando el problema
Por lo cual se tiene;
Después
es libre, incógnita a determinar.
i.
La condición de transversalidad correspondiente a la condición fina
ii.
Las condiciones de transversalidad correspondientes a
Y
Pero
Por lo tanto, nos queda:
==
libres libres serán:
sera:
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En cada uno de los siguientes casos, calcular los extremales, estudiar las condiciones de Legendre y suficiente:
Sujeto a:
Solución:
:
, ∫ ( ̇ + + ̇ ) ,
12 21 ;; 12 0,0,1.1. ̇ + + ̇ 2 0̇ ̇ ̇ 2̈ 2̈ 0 + : 22̇ ̇ ̇ 2̈ 2 2̇ ̇ 0
La función intermedia es:
Ecuación de Euler para
:
Formando la ecuación de Euler para
se tiene que:
Integrando ambos miembros se tiene:
Ecuación de Euler para
Formando la ecuación de Euler para
se tiene que:
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La solución complementaria y general es:
: + − 1 +1+1 1 22+2 2 +2 2+2 +1 1 , 0 1 + − 0 − 2 + 1 1 1 1 − 1 + 1
Aplicando condición condición inicial y final para
Formando un sistema de ecuaciones para obtener los valores de A y B:
Restando quedaría que:
Tenemos que la extremal para
es:
Aplicando condición condición inicial y final para
Desarrollando se tiene los valores de
Tenemos que la extremal para
es:
:
y
:
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Por lo tanto, viendo que y son positivos se dice que las funciones obtenidas cumplen la condición de legendre para un mínimo y convexidad: Condición suficiente: Matriz Hessiana de f que contenga la segundas segundas derivadas con respecto a
0 0 0 0 , ̇ 000 200 020 002
, , ̇, ̇
Y dado que esta matriz es semidefinida positiva, f es convexa; por ende, las extremales obtenidas constituyen el mínimo global de problema.
Sujeto a:
Solución:
: , ∫ + + ̇ ̇ , − 2112 ;;121 1
La función intermedia es:
Ecuación de Euler para
:
Formando la ecuación de Euler para
2 + ̇ + 13 ̇ 22 ̇ ̇ 2̈ 2 2̈ se tiene que:
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2 2̇ ̈ ̇ ̈ ̈ ̇ ̈ ̇ ̈ ̇̈ : , ∫ + ̇ + ̇ , 0 0; 00, 2 1 ; 2
Formando la ecuación de Euler para
A partir de la ecuación
Reemplazando
Sujeto a:
y
se tiene que:
se obtiene que:
en
obtenemos:
Solución: La función intermedia es:
Ecuación de Euler para
2 2 ̇ ̇ 2 4 2̇ :
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2 2̇ ̇ ̇ 2̈ 2 2̈ ̈ + 0 ̈ + 2 0 ̈ + 0 2 ̈ ̇ 2̇ ̈ 2̈ ̈ ̈ + 0 2 0
Formando la ecuación de Euler para
De las ecuaciones de Euler de
Despejando
Sustituyendo
y
se tiene que:
resulta el siguiente sistema de ecuaciones diferenciales: diferenciales:
en la primera ecuación y derivando dos veces con respecto a , , se tiene que:
en
, , queda:
Obtenemos una ecuación diferencial lineal, con coeficientes constantes, homogénea, de cuarto orden. La ecuación característica asociada es:
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Diferenciando con respecto al tiempo tenemos:
̇ 1.̈ 2.554..1.+2.545−.−. 0.0.6440. 6 4 + 0. 6 40. 6 4 0.640.40.64 2 ̈ 0.4. 0.4:−. +1.60.64+1.60.64 0 ++0 2. + −. +0.32+0.321 0 0.40.4+1.60 [ ̇]== 0 0.62. +0.62−. 1.020.32+1.020.32 0.04, 0.04, 0,0, 0.61
Entonces:
Aplicando condición condición inicial y final para
Para
y
aplicaremos condición de transversalidad:
Resolviendo el sistema se tiene:
Por lo tanto, la extremal para
y
es:
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Sujeto a:
10 0 0, ∝
En donde es la constante de gravitación universal, α y β son constantes. Resolver el problema: (i) (ii)
Por el método de sustitución. Por el método de LaGrange.
Solución: (i)
Método de sustitución.
Como a partir de la restricción se tiene que:
1>̇ ̇ ∝ ̇ ̇ 1 1 1+ + 22 ∫ √
Hay que resolver, tras sustituir en el objetivo:
Sujeto a:
0 0, ∝ 1+1 +̇
En este caso la función intermedia será:
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Es decir:
Por lo tanto:
̇ 1 22 1 √ 1 ∫ √ 1 ⇒ √ 1 √ 1⇔ 1 ⇔ 1⇔1 1 2 1 (√ 1 1 √ +√ +√ 1∝) 1 ∝) + 2
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De la restricción de tiene que:
Es decir:
1>̇ ̇ ̇ 22√ 1+1 +̇ 1 (√ 1√ 1 √ + + √ √ 1) 1 ) + 2 1 (√ 1√ 1 √ + + √ √ 1) 1 ) + 2 1
Esta última ecuación la obtuvimos en el apartado (i) y sabemos que se llega a que:
250
√12 , , , , ∫ + ̇ + ̇
Calcular la mínima mínima distancia entre los puntos A (4,3,0) y B (2.3, . (Indicación: la distancia entre dos puntos A ( y y B ( se determina por la fórmula:
En donde
+∅,,+,
) ) en la esfera en en la superficie
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Tenemos que:
2; ̇ 1+1 +̇ ̇ + ̇ 2 1+1+̇ ̇ + ̇ 0
Para la función : Teniendo en cuenta que:
Tenemos que:
2; ̇ 1+1+̇ ̇ + ̇ 22 1+1+̇ ̇ + ̇ 0
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,̇,, ̇, − ̇ + √ ̇ 0 ∫ √ − 2√ ; ̇ 0 −
1) utilizando el multiplicador, se define la función:
En donde es constante.
Para la función , , se tiene que cumplir la ecuación ecuación de Euler:
Además, se tiene que cumplir cumplir la restricción:
Como
La ecuación de Euler es:
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1 ̇, 210 √ 00
Suponiendo que x>0, dicha matriz es semidesnuda positiva, ya que condición suficiente de mínimo global. 2) definimos la siguiente función auxiliar:
Para la que se verifica:
∫ ⁄
̇ 0⁄0 , ∫−
Por lo tanto, el problema inicial se puede expresar como:
> 0. Por lo tanto, se vírica la
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Imponiendo la restricción;
Se tiene que:
Por tanto;
∫ 2 2 1 ∗ 1 2 ∗ 1 0≤≤
a) Definimos función intermedia:
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0
Siendo
4 04 0
Por lo tanto, el problema no tiene solución óptima. b) Definimos función intermedia:
Ecuación de Euler:
20 ̇ ̇ ̇, 2̈
Para
̇ + ̇ 4̇ 4
:
̇,
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Condición inicial y final: Para
:
Reemplazando:
0 00+0 1 11 1
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Condición de Legendre
Por lo tanto;
̇, ̇ 22 >0>0 ̇, ̇
La función es estrictamente convexa y es un caso de minimización.
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1: Mediante la ecuación de Euler, halle la senda óptima en los siguientes casos:
EJERCICIO1a).
⌊⌋ ∫( )
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Por lo tanto, la Senda Optima será:
11 22+911 22 1 +9
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Aplicando la condición condición inicial
Aplicando la condición condición final
000 1 2 56 1 + 1 +12
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Integrando:
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̇, 3̇8̈
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√ 22
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Por lo tanto, la función no posee extremal.
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Solución general;
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0 →→∆∆≠00 [ ]== 0 22== 0 2+ 2 +2 == 0 2 2 2 + + = 0 == 0 +2 2 ==0 0 2 2 √ 22 √ ∫ 2 + ˙2 0 0, √ √2
Aplicando la condición condición de transversalidad
Reemplazando
Por lo tanto, la senda optima seria
Calculado T
Replanteando
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SOLUCIONES DE LOS EJERCICIOS PROPUESTOS DEL LI BRO DE BONIFAZ OPTIMIZACION DINAMICA
Hallar las sendas óptimas
∫32 ẏ 3 1 0 1 2
A:
Solución: Función Hamiltoniana
32 31 → 0 430 34 4<0 →
Aplicando condición de primer orden:
Aplicando condición de segundo orden:
Ecuación de movimiento de estado
ẏ 31ẏ ̇ 3 ̇ ̇ 3
Ecuación de movimiento de coestado
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Reemplazando en
ẏ
Reemplazando en
Integrando
Aplicando condición inicial
=∆Δ 0 2 → ∆ 0 →∆ ≠ 0 Δ 0 0 2 0 32 0 6 ⟹ 36 336 4 918 4 3 918 4 1ẏ ẏ 2754 4 1 ẏ 27544 4 ẏ 274 252 ẏ 274 252 27 8 252 0 1
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−+ 1 36 1
4 0 .ẏ41 0 1 1 4 41 → 0 44 40 41 1 <0 →
:
Función Hamiltoniana
Condición de primer orden:
Condición de segundo orden:
Ecuación de movimiento de estado:
ẏ 41 ẏ
Ecuación de movimiento de coestado:
̇
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̇ 41 1 0
Reemplazando u:
1 0 ̇ 4 4 1 41 ̇ 4 1 1 0 ̇ 40
Resolviendo Raíces características
Reemplazando en
Reemplazando
Resolviendo
40 4 − − 41 − 4 ẏ4 ẏ4 1 4 ẏ4
[ ∫ ∫ ] ] −
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1 0 1 1 1 1 11 1
Aplicando condición Inicial
Aplicando condición Final
Las sendas óptimas son:
1 (−)−) − :
Planteamiento del Hamiltoniano:
.
..: 0 1 10
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Reemplazando en :
Reemplazando en :
Condición Inicial:
Condición final:
0 10 1 − − 2 2− 2− − ∫− − − − . ∫ 2 −. − −
2 2− − 4 − 0 4 1 4 1 −
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:
1 2 . ..:
0 1 1
Planteamiento del Hamiltoniano:
12 0 → 1< 1 < 0 → á á óó → ′
Ecuación de Movimiento de Estado:
Ecuación de Movimiento de Coestado:
→ 0 0 0 10 01 [] 11 11 00 0 11 11 00
Sistema de ecuaciones:
Solución particular:
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1 1 11 110 2 √ 2 √ 2 √ .. −√ .. [ ] √ .. −√ .. Relacionar constantes: Para
√ 2
Para
√ 2
Solución General:
{10 01 √ 2 2 11 11}[} [ ] 00 [√ 21 1 √ 21 1] [ ] 00 (√ 21) 2 1) 0 (√ 21) 2 1) {10 01 √ √ 2 2 11 11}[} [ ] 00 [√ 12 1 √ 21 1] [ ] 00 (1√ (1√ 2) 2) 0 (1√ (1√ 2) 2) √ √ .... −√ .. √ .. [ ] (√ 21) 2 1) (1√ (1 √ 2) 2)
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Entonces:
1∆ 0 1 0 1 (√ 21) 2 1)√ (1√ (1 √ 2) 2)−√ 0 9.93 0.1 0 0.00997 0.99003
Entonces la trayectoria dinámica de la variable de estado es:
0.00997√ .. 0.99003 −√ .. :
:ẏ : ẏ 1 ; 0 5; 11 11 ; ;∈ ∈ 0,1 SOLUCION Hamiltoniana
H11 H ⟶ u es lineal neal H
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Reemplazando:
̇ 1̇ 1 ̇ ̇ 1 − 1 ̇ ̇ ̇ ̇ 0 0 5 − 15 4
Ecuación de movimiento de la variable de coestado
Reemplazando:
Aplicando condición inicial
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Los óptimos son:
4− 1
ut1
101
∞ 1
: Sujeto a
̇ 1 0 0∈0,1
Solución La función Hamiltoniana del problema es la siguiente:
1
H = (1-y)u (1-y)u + λ
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̇ 1 ̇ 1 10 1 − 11 1 − 1 0
Obteniendo las raíces características
Solución complementaria
Solución particular
Solución general
Aplicando condición inicial
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Solución complementaria
Solución particular
Solución general
11 1 1 l→∝im 10
Aplicando condición de transversalidad
∴
La variable coestado coestado no converge a cero.
Entonces para que converge tiene que cumplir el siguiente caso:
Aplicando límites
≅ 0 lim lim→∝ 11 →∝
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,,,, , > 0 Solución Función Hamiltoniana:
→ 0 20 2 2>0 → ̇ → ̇ ̇
Ecuación de movimiento de estado
Integrando
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Remplazando
Remplazando u en x
0 2 2
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1. Aplicar el principio del máximo a los siguientes problemas:
∫ ̇
a) max J =
Sujeto a: =u x(1)=3
Planteamos el hamiltoniano:
uλu
H=x+tu-
*Aplicación del principio del máximo
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∗t = 1 para t ε [1,2] ∗t = t 2 para t ε [1,2] ∫ , en donde α > 0, ̇
b) max J = s.a : =1 =1
x(0) =
El hamiltoniano es: H = -x -
λu
A continuación se aplica el principio del máximo: * Ecuación de movimiento de coestado:
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C) max J = s.a
̇
∫
= u-x
con x(0) =5
El hamiltoniano es : H=
λu
-x)
* A continuación aplicamos el principio del máximo: *planteamos la ecuación de movimiento de coestado: -
= ̇
con λ1 = 0
Resolvemos la ecuación diferencial:
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Reemplazando (1) en (2) tenemos la ecuación diferencial:
̇ x = αt − αt –α– α imponiendo la condición inicial se obtiene que A = 5 α por lo que queda ∗t = 5 α − αt –α– α ∫2 43 = ̇ 1≤u≤1
Cuya solucion es: x(t) = A
finalmente que:
d) max J = s.a
x(0) = 1 , -
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= 0 …………………………………………… …………………………………………… 1 ≤ 0, ≤ 0 …………………………………………… …………………………………………………………. …………….. 2 1 ≤ u ≤ 1 ………………………………………………… ……………………………………………………………3 …………3 – – u= 0 …………………………………………. ………………………………………….. 4 –
-4 u +7-3t +
-
-
(u 1)= 0 ,
(-1
A partir de la condición (4) consideramos el caso U 1 = 0,
=0
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* planteamos la ecuación de movimientos de coestado
= ̇ = 0 , con λ1 = x1 -1 será: ∗ = , con ∗(1) = ∗(1) -1 -
* minimizamos el hamiltoniano
= 0
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∫ ̇ = x0=2 0 ≤ u ≤ 3 max J =
s.a
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∗
(t) = t + 2
min J = h [x(T)] s.a
̇
= = u
1 ≤ u ≤ 1, en donde h es una función estrictamente
x(0) = convexa,diferenciable, no negativa, con h(0) = 0.
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∗ – ∗T ≥ 0, entonces por las propiedades de la función h, se verificará que: ℎ[] < 0 ∗ = 1 ∗ = 1 ≥0 =
Si
T
-
Pero entonces debería ser este razonamiento:
,que se contradice con el punto de partida de
.
Por tanto,hemos obtenido que si
,entonces
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-
= ̇ = -
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Resolviendo el sistema, se obtiene que:
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Formular el programa que de a la persona la tasa de consumo adecuada enel resto
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0 =
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S(x) = 1> 0
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