1

Discuss the solution of $u_t+ uu_x= 0$ with the following Cauchy data:

a) $u(x,0)= x ; 0\leq x\leq 1$

b)$u(x,0)=1/2; 0\leq x\leq 1$

Sketch the domains in the $x-t $ plane where the solutions are determined. Does the solution develop a singularity at any time? If yes, discuss the reasons for the break down of the solution.

My trials: For a) by the methos of characteristic I got the solution u(x,t)=x/(1+t). So it has singularity at t=-1, is it possible? And then for b) methos of characteristics is not giving a solution. So if someone explain the solution of this kind problems it will be of great help. Thanks in advance.

EditPiAf
  • 20,898
Germain
  • 2,010
  • What is the initial condition outside the interval $[0, 1]$ in each case? – Tunococ Jul 21 '14 at 08:41
  • there is no other condition..I have written the complete as it has been asked – Germain Jul 21 '14 at 19:37
  • I see. Your answer for a) seems correct. For b), I believe the solution is $u(x, t) = 1/2$ for $0 \le x - \frac 12 t \le 1$. – Tunococ Jul 22 '14 at 07:43
  • can you just write your solution, or take a photo of your solution and upload it here for part b), I can then accept that as an answer. – Germain Jul 22 '14 at 15:52

1 Answers1

2

Suppose the parameters for characteristics are $\eta$ and $\xi$. The characteristics are defined by $$ \frac{\partial t}{\partial \eta} = 1, \frac{\partial x}{\partial \eta} = u, \frac{\partial u}{\partial \eta} = 0. $$ Solving gives $t(\eta, \xi) = \eta + c_1(\xi)$, $u(\eta, \xi) = c_3(\xi)$, $x(\eta, \xi) = c_3(\xi)\eta + c_2(\xi)$.

Pick the parametrization of the initial solution curve: $t(0, \xi) = 0$ and $x(0, \xi) = \xi$. It follows that $c_1 = 0$, $c_2(\xi) = \xi$ and $t = \eta$. The initial condition determines $c_3$:

a) $u(0, \xi) = \xi$, so $c_3(\xi) = \xi$. It follows that $x = \xi(\eta + 1)$. Since the initial condition is only given for $\xi \in [0, 1]$, the domain in which the solution may exist is $(x, t) \in \{(\xi(\eta + 1), \eta) \mid \xi \in [0, 1], \eta \in \mathbb R\}$. Inverting the transformation $(\xi, \eta) \mapsto (x, t)$, we get $\xi = \frac{x}{t+ 1}$, so the singularity exists at $t = -1$. The solution is $u = \xi = \frac{x}{t+1}$.

b) $u(0, \xi) = \frac 12$, so $c_3(\xi) = \frac 12$. It follows that $x = \frac 12\eta + \xi$. The domain in which the solution may exist is $(x, t) \in \{(\frac 12\eta + \xi, \eta) \mid \xi \in [0, 1], \eta \in \mathbb R\}$. Inverting the transformation, we get $\xi = x - \frac 12 t$. The solution is $u = \frac 12$. There is no singularity. The solution is valid throughout the domain $\{ (x, t) \mid \frac t2 \le x \le \frac t2 + 1, t \in \mathbb R \}$.

Tunococ
  • 10,303