6.2: Ecuación de calor no homogénea

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Aquí consideramos el problema de valor inicial para$u=u(x,t)$$u\in C^\infty(\mathbb{R}^n\times R_+)$,

\ begin {eqnarray*}
u_t-\ triángulo u&=&f (x, t)\\\ mbox {en}\ x\ in\ mathbb {R} ^n,\ t\ ge0,\\
u (x,0) &=&\ phi (x),
\ end {eqnarray*}

donde$\phi$ y$f$ se dan. Desde

$$\ sombrero ancho {u_t-\ triángulo u} =\ sombrero ancho {f (x, t)}\]

obtenemos un problema de valor inicial para una ecuación diferencial ordinaria:

\ begin {eqnarray*}
\ frac {d\ sombrero ancho {u}} {dt} +|\ xi|^2\ sombrero ancho {u} &=&\ sombrero ancho {f} (\ xi, t)\
\ sombrero ancho {u} (\ xi,0) &=&\ sombrero ancho {\ phi} (\ xi).
\ end {eqnarray*}

La solución viene dada por

$$\ sombrero ancho {u} (\ xi, t) =e^ {-|\ xi|^2 t}\ sombrero ancho {\ phi} (\ xi) +\ int_0^t\ e^ {-|\ xi|^2 (t-\ tau)}\ sombrero ancho {f} (\ xi,\ tau)\ d\ tau.\]

Aplicando la transformada inversa de Fourier y un cálculo como en la prueba del Teorema 5.1, paso (vi), obtenemos}

\ begin {eqnarray*}
u (x, t) &=& (2\ pi) ^ {-n/2}\ int_ {\ mathbb {R} ^n}\ e^ {ix\ cdot\ xi}\ Grande (e^ {-|\ xi|^2t}\ sombrero ancho {\ phi} (\ xi)\\
&&\ +\ int_0^t\ ^ {-|\ xi|^2 (t-\ tau)}\ sombrero ancho {f} (\ xi,\ tau)\ d\ tau\ Grande)\ d\ xi.
\ end {eqnarray*}

A partir del cálculo anterior para el problema homogéneo y cálculo como en la prueba del Teorema 5.1, paso (vi), obtenemos la fórmula

\ begin {eqnarray*}
u (x, t) &=&\ frac {1} {(2\ sqrt {\ pi t}) ^n}\ int_ {\ mathbb {R} ^n}\\ phi (y) e^ {-|y-x|^2/ (4t)}\ dy\
& &+\ int_0^t\ int_ {\ mathbb {R} ^n}\ f (y,\ tau)\ frac {1} {\ izquierda (2\ sqrt {\ pi (t-\ tau)}\ derecha) ^n}\ e^ {-|y-x|^2/ (4 (t-\ tau))}\ dy\ d\ tau.
\ end {eqnarray*}

Esta función$u(x,t)$ es una solución del problema de valor inicial no homogéneo anterior proporcionado

$$\ phi\ en C (\ mathbb {R} ^n),\\\ sup_ {\ mathbb {R} ^n} |\ phi (x) |<\ infty\]

y si

$$f\ en C (\ mathbb {R} ^n\ veces [0,\ infty)),\\ M (\ tau) :=\ sup_ {\ mathbb {R} ^n} |f (y,\ tau) |<\ infty,\ 0\ le\ tau<\ infty.\]

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