- #1
Hyperreality
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- 0
I think I've solved this problem, but just need someone to check my answers.
[tex]u_{x}u+u_{t}=2[/tex]
With initial data
[tex]u(x,0)=f(x)[/tex]
Use the method of charactersitics [tex]u(x,t)=u(\xi,\tau)[/tex], we get
[tex]\tau_{t}=t[/tex], [tex]x_{u}=u[/tex] and [tex]u_{\tau}=2[/tex].
So, by choice of our initial condition,
[tex]\tau=t[/tex] and [tex]u=2t + f(\xi)[/tex].
Since [tex]\xi=\xi(x,t)[/tex]. We have
[tex]u(x,t)=u(\xi(x,t),t)[/tex]
At [tex]t=0[/tex] we have [tex]u(x,0)=u(\xi(x,0),0)[/tex], therefore
[tex]x=\xi(x,0)[/tex]
Since we do not know what [tex]f(x)[/tex] is, our solution for [tex]u(x,t)[/tex] is just
[tex]u(x,t)=2t+f(\xi(x,t))[/tex]
with initial condition
[tex]\xi(x,0)=x[/tex]
Is this right??
[tex]u_{x}u+u_{t}=2[/tex]
With initial data
[tex]u(x,0)=f(x)[/tex]
Use the method of charactersitics [tex]u(x,t)=u(\xi,\tau)[/tex], we get
[tex]\tau_{t}=t[/tex], [tex]x_{u}=u[/tex] and [tex]u_{\tau}=2[/tex].
So, by choice of our initial condition,
[tex]\tau=t[/tex] and [tex]u=2t + f(\xi)[/tex].
Since [tex]\xi=\xi(x,t)[/tex]. We have
[tex]u(x,t)=u(\xi(x,t),t)[/tex]
At [tex]t=0[/tex] we have [tex]u(x,0)=u(\xi(x,0),0)[/tex], therefore
[tex]x=\xi(x,0)[/tex]
Since we do not know what [tex]f(x)[/tex] is, our solution for [tex]u(x,t)[/tex] is just
[tex]u(x,t)=2t+f(\xi(x,t))[/tex]
with initial condition
[tex]\xi(x,0)=x[/tex]
Is this right??