Dear all, I was wondering if the following approach is good. I find it easy, maybe I miss a point:
Assume that the following non linear PDE is uniquely solvable for example in the classe $\mathcal{C}^{1,2}( [0,T]\times[0,a]$:
[math]\partial_tu(t,x)+F(x,u(t,x),\partial_xu(t,x),\partial_x^2u(t,x))=0,~~(t,x)\in(0,T)\times(0,a)[/math] and boundary conditions (not very important) : $u(0,.)=g,~~u(.,a)=u(.,0)=\phi$
The main concern of my question is how to formulate a result of solvability when the coefficient $F$ has a time dependence namely: $F(t,x,u(t,x),\partial_xu(t,x),\partial_x^2u(t,x))$ without going into sequences with schemes and a kind of point fixed argument; namely define:
$\forall s\in[0,T],~~s\mapsto u(s,t,x)$ the unique solution of [math]\partial_tu(s,t,x)+F(s,x,u(s,t,x),\partial_xu(s,t,x),\partial_x^2u(t,x))=0,~~(t,x)\in(0,T)\times(0,a)[/math] and boundary conditions : $u(0,.)=g,~~u(.,a)=u(.,0)=\phi(s)$, forall $s\in[0,T]$. It seems that we have a kind of symetrie property: $\forall (s,t)\in[0,T]^2,~~x\in[0,a],~~u(s,t,x)=u(t,s,x)$ by uniqueness.
Hence if we define $(s,x)\mapsto 1/2u(s,s,x)$, it seems that we have the required solution? Or maybe should I use a fixed-point argument?
Thank you in advance.
Assume that the following non linear PDE is uniquely solvable for example in the classe $\mathcal{C}^{1,2}( [0,T]\times[0,a]$:
[math]\partial_tu(t,x)+F(x,u(t,x),\partial_xu(t,x),\partial_x^2u(t,x))=0,~~(t,x)\in(0,T)\times(0,a)[/math] and boundary conditions (not very important) : $u(0,.)=g,~~u(.,a)=u(.,0)=\phi$
The main concern of my question is how to formulate a result of solvability when the coefficient $F$ has a time dependence namely: $F(t,x,u(t,x),\partial_xu(t,x),\partial_x^2u(t,x))$ without going into sequences with schemes and a kind of point fixed argument; namely define:
$\forall s\in[0,T],~~s\mapsto u(s,t,x)$ the unique solution of [math]\partial_tu(s,t,x)+F(s,x,u(s,t,x),\partial_xu(s,t,x),\partial_x^2u(t,x))=0,~~(t,x)\in(0,T)\times(0,a)[/math] and boundary conditions : $u(0,.)=g,~~u(.,a)=u(.,0)=\phi(s)$, forall $s\in[0,T]$. It seems that we have a kind of symetrie property: $\forall (s,t)\in[0,T]^2,~~x\in[0,a],~~u(s,t,x)=u(t,s,x)$ by uniqueness.
Hence if we define $(s,x)\mapsto 1/2u(s,s,x)$, it seems that we have the required solution? Or maybe should I use a fixed-point argument?
Thank you in advance.