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In the $(x,t)$- plane, the characteristic of the initial value problem $$u_t+uu_x=0$$ with $$u(x,0)=x,0\leq x\leq 1$$ are

$1$. parallel straight lines .

$2.$ straight lines which intersects at $(0,-1)$.

$3.$ non- intersecting parabolas.

$4.$ concentric circles with center at origin.

I am learning partial differential equation so don’t have good knowledge of it . According to me characteristic equations are

$$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$$ Now $u=c$ by last fraction. So by first two fractions I have $x-ct=k$, where $c$ and $k$ are constants. Now I don’t known how to use initial condition of $u(x,0)=x$ and what is final answer? I see that $x-ct-k=0$ are straight lines in $(x,t)$-plane. Please help me to reach at final option . Thank you.

EditPiAf
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neelkanth
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  • I am confused by the "characteristic equation" you've stated, what does $\frac{du}0$ represent? also you have stated the domain of $x$ but are there any restrictions on t? normally $t\ge 0$ – Henry Lee Feb 05 '22 at 10:16
  • @HenryLee I have written the question exactly. Yes there may be mistake I had as I am not good in PDE. – neelkanth Feb 05 '22 at 10:20
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    @HenryLee Writing $\frac{du}{0}$ in this context just implies $u$ is constant in some variable. It isn't literal division by zero. – Matthew Cassell Feb 05 '22 at 10:27
  • @neelkanth You found the characteristics $u = C_{2}$ and $x - ut = C_{1}$, hence the solution to the PDE is given implicitly by $$u = f(x-ut)$$ Applying the IC gives $x = u(x,0) = f(x - u(x,0) \cdot 0) = f(x)$ and hence the explicit solution is $$u = x - ut \implies u = \frac{x}{1+t}$$ – Matthew Cassell Feb 05 '22 at 10:28
  • @mattos so what is answer ?? – neelkanth Feb 05 '22 at 10:29
  • @mattos sir question is not about solutions. It’s about characteristics . I am confused about that . – neelkanth Feb 05 '22 at 10:32
  • 'Now I don’t known how to use initial condition of $u(x,0) = x$ and what is final answer?' Seems to me you asked about the solution. Anyway, you have the characteristics. Do you think they look like parabolas? Or concentric circles? – Matthew Cassell Feb 05 '22 at 10:33
  • @mattos seems to be straight lines but how to use initial condition to answer option $1$ or option $2$? – neelkanth Feb 05 '22 at 10:37
  • @mattos final answer according to me means final option to be correct. – neelkanth Feb 05 '22 at 10:40
  • @mattos as I get $x-C_2t=C_1$ but here $C_1$ and $C_2$ are arbitrary constants, so what will be answer from first two options ? – neelkanth Feb 05 '22 at 11:12
  • @neelkanth Please don't tag me in a comment in order to get me to look at your other questions. – Matthew Cassell Feb 10 '22 at 09:38
  • Sorry for that sir . – neelkanth Feb 10 '22 at 11:06

2 Answers2

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The resolution of the initial value problem is discussed in this post. So we end up with the set of curves $$ u=C_1, \qquad x-ut=C_2 $$ where $C_1$, $C_2$ are constants. These curves are straight lines in the $x$-$t$ plane, thus options 3. and 4. are eliminated. Now we implement the boundary condition $u(x,0)=x$ at $t=0$: $$ x=C_1, \qquad x-x\cdot 0 = C_2 , $$ i.e. $C_1=C_2=c$. To see if the curves $x-ct=c$ are parallel, we look at the slope in $x$-$t$ coordinates, whose value equals $$c=u = \frac{x}{1+t}.$$ Are these curves parallel? Lastly you could check for the intersection of two characteristics by solving the system $$ x-a t = a, \qquad x-b t = b $$ with respect to $(x,t)$ for $0\leq a\neq b \leq 1$.

EditPiAf
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I'm just going to expand on my comment. This appears to fit the form of the Inviscid Burgers' equation which is: $$\frac{\partial u }{\partial t}+u\frac{\partial u}{\partial x}=0$$ Using the method of characteristics we say: $$u(x,t)=u(x(s),t(s))$$ then assume: $$\frac d{ds}u(x(s),t(s))=F(u,x(s),t(s))$$ and using chain rule we get: $$\frac d{ds}u(x(s),t(s))=\frac{\partial u}{\partial x}\frac{dx}{ds}+\frac{\partial u}{\partial t}\frac{dt}{ds}$$

I am not not particularly versed in this method but I hope this information helped

Henry Lee
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