- #1
evinda
Gold Member
MHB
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Hello! (Wave)
I want to solve the non-homogeneous equation $u_t+uu_x=0$ with the initial condition $u(x,0)=x$. Also I want to draw some of the characteristic curves.
I have tried the following so far:
The characteristic curves for $u_t+uu_x=0$ are the curves that are given by the solutions of the ode $\frac{dx}{dt}=u(x,t)$.
We have that $\frac{d}{dt}[u(x(t),t)]=u_t+uu_x=0$ and so $u(x(t),t)=c$.
We consider the line that passes through the points $(x_0,0)$ and $(x,t)$.
The slope of the line is
$\frac{x-x_0}{t-0}=\frac{dx}{dt}=u(x,t)=u(x_0,0)=x_0$, thus $x-x_0=tx_0 \Rightarrow x_0=\frac{x}{t+1}$.
Then we would get that $u(x,t)=\frac{x}{t+1}$. But this cannot be right, since $u$ should be only a function of $x$.
So have I done something wrong? (Thinking)
I want to solve the non-homogeneous equation $u_t+uu_x=0$ with the initial condition $u(x,0)=x$. Also I want to draw some of the characteristic curves.
I have tried the following so far:
The characteristic curves for $u_t+uu_x=0$ are the curves that are given by the solutions of the ode $\frac{dx}{dt}=u(x,t)$.
We have that $\frac{d}{dt}[u(x(t),t)]=u_t+uu_x=0$ and so $u(x(t),t)=c$.
We consider the line that passes through the points $(x_0,0)$ and $(x,t)$.
The slope of the line is
$\frac{x-x_0}{t-0}=\frac{dx}{dt}=u(x,t)=u(x_0,0)=x_0$, thus $x-x_0=tx_0 \Rightarrow x_0=\frac{x}{t+1}$.
Then we would get that $u(x,t)=\frac{x}{t+1}$. But this cannot be right, since $u$ should be only a function of $x$.
So have I done something wrong? (Thinking)
