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Ecuaciones Diferenciales en Derivadas Parciales
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  • Métodos Matemáticos
  • Institution
  • Universidad Politécnica De Madrid

Incluye los métodos de resolución de las siguientes ecuaciones: -Ecuaciones diferenciales en derivadas parciales de primer orden. -Ecuación de Ondas. -Ecuación de Calor. -Ecuación de Laplace. También incluye la teoría acerca de las series de Fourier para resolver las ecuaciones.

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Document information

  • February 17, 2023
  • 23
  • 2022/2023
  • Class notes
  • María higuera
  • All classes

Subjects

  • ecuación del calor
  • ecuación de ondas
  • ecuaciones diferenciales
  • ecuación de laplace
  • series de fourier

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  • Universidad Politécnica de Madrid
  • Ingeniería Aeroespacial
  • Métodos Matemáticos
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ECUACIONES DIFERENCIALES ENDERIVADAS PARCIALES DE
PRIMERORDEN



CONSERVADON
LEYES DE AD



·
u(x.t) Magnitud(masa, energia etc.


·
p(x, t, u,u(...) Flujo


f(x, t, u) Fuentes
·




Let be conservacise forma integral
u(x.t)dx 8(x2 0/15
1)"20x
·
on =

-




20
Lex be conservacios forma diferencial (i.e., AX):
Gv(X,t) g
·
X2 x1
*
en + +
=




2/




ONDAS ESTACIONARIAS



La ecuacio's diferencial en derivadas parciales mas simple para una funcion de dos variables
es 0
=




t


Para resolverla integramos a ambos lados de la ecacion:
2u(s,y)ds u(t,t) u(0,4
= -
0




por esola solucion dela forma:u(t,x) f(x),
as conde
f(x) w(0.4)
= =




El lnico requerimiento es
que f(x) sea sufficientements regular.


La solucion representa una anda estacionaria (no depende del tiempol.



ONDAS VIAJERAS. ECUACION DETRANSPORTE



Considers la siguiente ecuacion:
Gu cayto +




Aesta ecuacion se la conoce can nombred ecuacion de transporte.

Es necesario especificar la solucion en untiempo inicial, dando lugar al problema devalores iniciales. ults, x) f(x)
=
&tCR

xU
u(t,y)
x(t)"du-2u+Oudoet
11x
x =




d* (t
=(t)
a
=
=
+
1 >
Curva caracteristic
LX (t k
+




considere unobservador describiendo una trajectoria (x (t), 4). Como varia what) segan la perspectiva del observator?

du du(x(t),t)
uxdy ut
=
=




dt dt

,dy c
=




ydy u++..
=




Por tanto las soluciones de la ecuacion de transporte son las mismas
que las del sistema de EDOs.


Ut (uX 0
+
=
dx 0
=




dt
⑤



u(t,y) f(X) =



du=o
Eiemple.

xU
+chy=
Gu 0;u(0,X) f(X
=




2t




u(t,y)
x(t)du-2u+Oudoet
11x
x = LX (t k
+




d* (t caracteristic
=(t)
u
a
=
=
1
+ >
Curva
f(x) w(0,X)
=




x(t) (t 11
= +

1(0) 5 = +
k7 5 =




↑ ↑




u(t,X(t)) k2u(0) f(s)
f(s) ke
=
=
= =




+
it
X(t,s) (t =
s
+
+
5 1 = -
ct

Yu(t, x) f(x (t)
=

ult,x) f(x ct.)
=




u(t,s) f(s);u(x,t)
f(x ct)
=

=
-




ECUACIONES CASI-LINEALES:METODO DE LAS CARACTERISTICAS


a(x,y,u)ux p(X,y,u)ay ((X,y,u)(I) +
=




15) (a,p,c).(ux,uy, 1)
+
= 0
Gualquier curva superficie S tiene vector tangente a (a,p,c)


has curvas caracteristicas satisfacen el signiente sistema:



EX a(x,y,u);d b(x,y,u);du can see
=
= -




La elacion (1) tiene condiciones iniciales:u/f(s), g(s)) h(s),
=
se



Entonces el sistema tiene condiciones iniciales:



1(0) 10(s,0)
=
=

f(s),y(0) y0(s.0) =g(s), =
w(0) 40(s,0) h(s)
=
=

, Ejemplo:

S.u u(X,y)
=




5:H(x,y,u) u(x,y)
=
-
u 0
=




dy ((X,y,u) u(t,s) 0
=

=




dy a(X,y,n) x(t,5) 0
=
=




dy=b(x,y,u) y(t,s) 0 =




y(0) X(s);y(0) y(s);u() u(s)
= = =




S:(X,y,u)(t,s)



METODO PARA RESOLVER LAS ECUACIONES CASILINEALES
DE LAGRANGE




Ejemploi



u,isent
1,ux x*7 yu
1
+
=
10(s) 8
=




tolst=s
no(s) sen(s)




u1o."1.
s

e
=




·
La solucion aparece de form a
(0-426
/4 (00*7c8R
7




( (, (1,1,u) x
-
=
11i
=
=
=




by X
=
implicita, como interseccion de dos
Dt


(* (2u*
du u
dly: /0u (0 4(0,4,u) = Superficies.In
=
su +
u
=
=
=
= ·




bt ↑
&




solucion general:FCP IFFFuncion = arbritarial sF( = =0 f(4242) ve*
= =




u(0,y) sen(y)= 5(4e(2) senye*;5(4)= seny
+
=




ne*k senle*(2):u(,1) eksenlie*2)
= =

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