[Math] Find the solution of the Cauchy problem $u_t-uu_x=0$.

cauchy problemordinary differential equationspartial differential equations

For the Cauchy problem $u_t-uu_x=0, ~x\in \mathbb{R}, t>0$ with $u(x,0)=x,~x\in \mathbb{R} $, which of the following statements is true?

  1. the solution $u$ exists for all $t>0$.
  2. the solution $u$ exists for $t<\frac{1}{2}$ and breaks down at $t=\frac{1}{2}$.
  3. the solution $u$ exists for $t<1$ and breaks down at $t=1$
  4. the solution $u$ exists for $t<2$ and breaks down at $t=2$.

My work

I used the method of characteristics to find out the solution.

Given equation is $u_t-uu_x=0$, then
$$\frac{du}{ds}=\frac{\partial u }{\partial x}\frac{dx}{ds}+\frac{\partial u }{\partial t}\frac{dt}{ds}$$.
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)$

$\dfrac{dx}{ds}=-u$ , letting $x(0)=x_0$ , we have $x=-us+x_0$ ,

Therefore $u(x,t)=f(x_0)=f(x+ut)$. Now applying the initial condition $u(x,0)=f(x)=x$ we get $u(x,t)=x+ut$.

Now the problem is from the solution I am unable to conclude anything from the option given precisely. It seems the option 1 is true but I don't know for sure. How can I solve this? Please help.Thanks.

Best Answer

We will solve this problem using the method of characteristic.

Suppose we could rewrite the PDE as \begin{align} \frac{d}{dt}u(t, X(t)) = u_t +X'(t)u_x = u_t-uu_x=0 \end{align} then it follows \begin{align} u(t, X(t))=\text{ const} \ \ \text{ and } \ \ X'(t) = -u(t, X(t)) =\text{ const}=-u(0, x_0). \end{align} Imposing the condition $X(0) = x_0$ yields \begin{align} X(t) = (1-t)x_0 \end{align} which means \begin{align} u(t, X) = u(t, ct+x_0) = u(0, x_0) = x_0. \end{align} Hence it follows \begin{align} u(t, X) = \frac{X}{1-t}. \end{align}

Hence the solution break down when $t=1$.

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