Cómo resolver un sistema de EDP $u_t+u_x=v, v_t+v_x=-u$

Question

Resolver el siguiente problema de valor inicial:

$$u_t+u_x=v, \\v_t+v_x=-u, \\u(0,x)=u_0(x), \\ v(0,x)=v_0(x).$$

No he aprendido ningún método para resolver un sistema de EDP, así que supongo que hay un "truco". Hasta ahora hemos aprendido a resolver EDP de la forma:

$$u_t+a(t,x)u_x=f(t,x,u)$$

Me vendría muy bien la ayuda.

Gracias.

Answer 1

Dejemos que $\partial_a$ denotan la derivada direccional en el $({1 \over \sqrt{2}},{1 \over \sqrt{2}})$ dirección. Entonces sus ecuaciones son $$\partial_a u = {1 \over \sqrt{2}} v$$ $$\partial_a v = -{1 \over \sqrt{2}} u$$ Combinando se obtiene $$\partial_a^2 u + {1 \over 2} u = 0$$ $$\partial_a^2 v + {1 \over 2} v = 0$$ Esto es fácil de resolver. Ten en cuenta que tus condiciones iniciales combinadas con las ecuaciones también darán expresiones $\partial_a u(0,x)$ y $\partial_a v(0,x)$ que junto con $u(0,x)$ y $v(0,x)$ implicará soluciones únicas.

Answer 2

El presente sistema puede resolverse por el método de las características. En efecto, las dos ecuaciones proporcionan las mismas curvas características:

• $t'(s) = 1$ , dejando que $t(0)=0$ da $t=s$ ;
• $x'(s) = 1$ , dejando que $x(0)=x_0$ da $x=s+x_0$ .

Por lo tanto, las características son líneas rectas con pendiente 1 en el $x$ - $t$ plano. La primera ecuación conduce a $u'(s) = v(s)$ y la segunda ecuación conduce a $v'(s) = -u(s)$ para que $u''(s) + u(s) = 0$ y $v''(s) + v(s) = 0$ . Estas ecuaciones diferenciales de segundo orden pueden resolverse imponiendo las condiciones iniciales $u(0) = u_0(x_0)$ , $u'(0) = v_0(x_0)$ y $v(0) = v_0(x_0)$ , $v'(0) = -u_0(x_0)$ : $$ \begin{aligned} u(s) &= u_0(x_0) \cos s + v_0(x_0) \sin s\\ v(s) &= v_0(x_0) \cos s - u_0(x_0) \sin s \end{aligned} $$ Combinando todas las ecuaciones deducidas del método de las características, tenemos $$ \begin{aligned} u(x,t) &= u_0(x-t) \cos t + v_0(x-t) \sin t\\ v(x,t) &= v_0(x-t) \cos t - u_0(x-t) \sin t \end{aligned} $$ Para este tipo de sistema lineal, también debería ser posible proceder por transformada de Fourier.

Answer 3

Una pista:

$u_{tt}+u_{xt}=v_t$

$u_{xt}+u_{xx}=v_x$

$\therefore v_t+v_x=u_{tt}+2u_{xt}+u_{xx}=-u$

$u_{tt}+2u_{xt}+u_{xx}+u=0$

Dejemos que $\begin{cases}p=x+t\\q=x-t\end{cases}$ ,

Entonces $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial t}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial t}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$

$\dfrac{\partial^2u}{\partial x\partial t}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial q^2}$

$\therefore\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}+2\left(\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial q^2}\right)+\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}+u=0$

$4\dfrac{\partial^2u}{\partial p^2}+u=0$

$u(p,q)=f(q)\sin\dfrac{p}{2}+g(q)\cos\dfrac{p}{2}$

$u(t,x)=f(x-t)\sin\dfrac{x+t}{2}+g(x-t)\cos\dfrac{x+t}{2}$

$u_t(t,x)=-f_t(x-t)\sin\dfrac{x+t}{2}+\dfrac{f(x-t)}{2}\cos\dfrac{x+t}{2}-g_t(x-t)\cos\dfrac{x+t}{2}-\dfrac{g(x-t)}{2}\sin\dfrac{x+t}{2}$

$u_x(t,x)=f_x(x-t)\sin\dfrac{x+t}{2}+\dfrac{f(x-t)}{2}\cos\dfrac{x+t}{2}+g_x(x-t)\cos\dfrac{x+t}{2}-\dfrac{g(x-t)}{2}\sin\dfrac{x+t}{2}$

$\therefore v(t,x)=f_{x-t}(x-t)\sin\dfrac{x+t}{2}+g_{x-t}(x-t)\cos\dfrac{x+t}{2}+f(x-t)\cos\dfrac{x+t}{2}-g(x-t)\sin\dfrac{x+t}{2}$

Por lo tanto, $\begin{cases}u(t,x)=f(x-t)\sin\dfrac{x+t}{2}+g(x-t)\cos\dfrac{x+t}{2}\\v(t,x)=f_{x-t}(x-t)\sin\dfrac{x+t}{2}+g_{x-t}(x-t)\cos\dfrac{x+t}{2}+f(x-t)\cos\dfrac{x+t}{2}-g(x-t)\sin\dfrac{x+t}{2}\end{cases}$