# Show that the initial value problem PDE has no solution

$$u_t+u_x=0,$$ $$u=x$$ on $$x^2+t^2=1$$ Show that the initial value problem has no solution.

From my understanding of the method of characteristics:$$\frac{dx}{ds}=1 \\\frac{dt}{ds}=1$$ so $$x=s+c_1$$ and $$t=s+c_2$$. I'm not sure how to use:$$\frac{du}{ds}=0$$ From past questions, I would integrate this as it would equal some constant and find $$u(s,r)$$ using the initial condition. However, I would have $$du=0$$?

• You will have constant in respect to $s$ – Holo Sep 27 '18 at 21:09
• Ok, I see that now, because by letting $u=c$, where $c$ is some constant, then $\frac{du}{ds}=0$. So, does this imply that $x=c$, given the initial condition $u=x$? – 383930283423 Sep 27 '18 at 21:18
• I posted full answer – Holo Sep 27 '18 at 21:40

The point is, your function is a constant in respect to $s$, but not in respect to $r$, set $s=t+x, r=t-x$, if you do this you will get again $\frac{du}{ds}=0$, but integrating this you will get $u(s,r)=c(r)\implies u(x,t)=c(t-x)$ for some arbitrary function $c$.

Now we look on the condition: $\forall x,t\quad x^2+t^2=1\implies u(x,t)=c(t-x)=x$, we want to show that no such function exists, to do so we just need to find $x_0,t_0$ and $x_1,t_1$ such that $x_1\ne x_0$ and $t_0-x_0=t_1-x_1$.

Take $x_0=y_0=\frac{1}{\sqrt2}$, $x_1=y_1=-\frac{1}{\sqrt2}$, from the first tuple we get $c(0)=\frac{1}{\sqrt2}$ and from the second we get $c(0)=-\frac{1}{\sqrt2}$, thus $c$ is ill defined