Is this correct that the following Cauchy problem has infinitely many solutions?

\begin{cases}‎ ‎xu_t+u_x=0 \\‎ ‎u(x,0)=\cos x‎ ‎\end{cases}

Using the method of characteristics it is obvious that it has a local solution around the curve $t=0$. But I am puzzled why it should have infinitely many solutions.

It seems that we can construct many solutions by Laplace transform method but I am not confident.


Using the method of characteristics we obtain that $u$ is constant along the parametrised by $s$ curves: $$ (t,x)=\left(\frac{s^2}{2}+c,s\right), $$ and $u$ is of the form $u=f(x^2-2t)$.

This means that the initial data cover only the region: $$ \{(t,x): 0\le t\le x^2/2\}. $$ In other words, the characteristics which start from the $x-$axis never arrive in the region. $$ \{(t,x): t> x^2/2\}. $$ In particular, if $$ f(x)=\left\{ \begin{array}{lll} \cos(x^{1/2}) & \text{if} & x\ge 0, \\ 1+xg(x) & \text{if} & x< 0, \end{array} \right. $$ where $g$ is an arbitrary continuously differentiable function, then $u(t,x)=f(x^2-2t)$ satisfies the given initial value problem.

  • $\begingroup$ @ Yiorgos S. Smyrlis: thanks. I have a similar broblem which says If in the above cauchy problem we replace $\cos x$ by $\sin x $. there is no solution. How to handle it? $\endgroup$ – Finish Apr 8 at 18:59
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    $\begingroup$ @Finish Because, due to the fact that $u=f(x^2-2t)$, the solution satisfies $$ u(t,-x)=u(t,x), \text{for all}\,\, t,x\in\mathbb R.$$ $\endgroup$ – Yiorgos S. Smyrlis Apr 8 at 19:02
  • $\begingroup$ @YiorgosS.Smyrlis Out of curiosity, when using the method of characteristics, how do you know whether to parametrize as $(x(t),t)$ or $x(s),t(s)$? If you try the former on this you get $\frac{dt}{dt} = x$ and hence $t=tx+C$, which causes issues $\endgroup$ – Hushus46 Apr 8 at 19:06
  • $\begingroup$ @Hushus46 You always parametrise with respect to a new parameter, say $s$. Next, it might be possible to replace $s$ by one of the arguments. In this case, the $s$ CANNOT be replaced by $t$, but it can be replaced by $x$. $\endgroup$ – Yiorgos S. Smyrlis Apr 9 at 8:46

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