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Consider the Cauchy problem of finding $u=u(x,t)$ such that $$\frac{\partial{u}}{\partial{t}}+u\frac{\partial{u}}{\partial{x}}=0\text{ for }x\in\mathbb{R},t>0\\u(x,0)=u_0(x),\;x\;\epsilon\;\mathbb{R}$$

which choice(s) of the following functions for $u_0$ yield a $C^1$ solution $u(x,\ t)$ for all $x\in\mathbb{R}$ and $t>0.$

  1. $u_0(x)=\frac{1}{1+x^2}$
  2. $u_0(x)=x$
  3. $u_0(x)=1+x^2$
  4. $u_0(x)=1+2x$

Because characteristic of the given PDE is $U=U_0(x-ut)$ from option 2, 4 only the satisfies the given PDE is i am right

1 Answers1

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Using method of characteristics, we set $t=t(r),x=x(r)$, then $u=u(r)$ along characteristic curve. Then $$\frac{du}{dr}=\frac{\partial u}{\partial t}\frac{dt}{dr}+\frac{\partial u}{\partial y}\frac{dy}{dr}$$ Letting $\frac{dt}{dr}=1$, $\frac{dx}{dr}=u$ and thus $\frac{du}{dr}=0$.

$\frac{dt}{dr}=1\implies t=r$. Now, $u$ is constant along characteristic curve, so we have $\frac{dx}{dr}=\frac{dx}{dt}=c\implies x-ct=d$.

$\frac{du}{dr}=\frac{du}{dt}=0\implies u=F(d)=F(x-ut)$

Now, checking initial conditions one by one, only (2) and (4) gives a solution which satisfies the equation.

Jesse P Francis
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  • Can you show me why any one of $1$ and $3$ does not satisfy the given equation? – Kushal Bhuyan Jun 03 '16 at 08:03
  • I am missing something may be, lets see, for $u(x,t)=1+(x-ut)^2$, then $\partial u/\partial x=2x-2ut$ and $\partial u/\partial t=-2xu+2u^2t=u(-2x+2ut)=-u\partial u/\partial x$ so it satisfies the equation. Where is my mistake? – Kushal Bhuyan Jun 03 '16 at 11:39
  • @KushalBhuyan..check boundary condition for ur solution.. –  Jun 05 '16 at 12:40