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Exercise :

Given the PDE IVP : $$\begin{cases} zz_x + z_y = \quad0 \\ z(x,0) \; \; \; =-3x \end{cases}$$ a) Find the solution of it. b) Determine the lines of the $(x,y)$ plane on which the solution of the IVP is constant. c) Are there shock waves for $y \geq 0$ ?

Attempt :

a)We form the characteristics problem : $$\frac{\mathrm{d}x}{z}=\frac{\mathrm{d}y}{1}=\frac{\mathrm{d}z}{0}$$

From the second pair, we yield :

$$\frac{\mathrm{d}y}{1}=\frac{\mathrm{d}z}{0} \implies z_1 = z$$

From the first pair, we yield :

$$\frac{\mathrm{d}x}{z}=\frac{\mathrm{d}y}{1} \implies z_2 = x-yz$$

Then, the general solution will be given as an expression of a differentiable function $F$, such that :

$$z_1 = F(z_2) \Rightarrow z = F(x-yz)$$

Now, taking into account the initial value, we have :

$$z(x,0) = -3x \implies F(x) = -3x$$

Letting $x := x-yz$, we yield $F(x-yz) = -3(x-yz)$ and thus the solution of the IVP is : $$z = -3(x-yz) \implies z(x,y) = \frac{3x}{3y-1}$$

Note : The above calculations are correct, verified with Wolfram Alpha.

b) One can easily see that for $x=0$, regardless $y$, the solution is constant as $z(0,y) = 0$. How would I proceed though to determine the lines along which the solution is constant ?

c) The solution $z(x,y)$ becomes undetermined for $3y-1=0\Rightarrow y=1/3$ and thus at this point there develops a shock wave, which means that there is a shock wave for $y \geq 0$. Is this approach correct, or am I missing something regarding shock waves ?

Summing up my questions : (1) How should I proceed with determining the lines which I am being asked in part (b) - (2) Is my approach to part (c) correct ?

I would really appreciate a thorough explanation as this is an exams problem that I am preparing for.

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  • $\begingroup$ @MariuszIwaniuk It's called a typo. That's where it is. $\endgroup$ – Rebellos Jun 1 '18 at 9:28
  • $\begingroup$ @Harry49 I have solved the IVP as clearly shown in my attempt, thus this post is not helpful. Please take the time to read my elaboration and figure out what I am asking. $\endgroup$ – Rebellos Jun 1 '18 at 9:33
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This is the inviscid Burgers' equation. The solution obtained by the method of characteristics looks fine (cf. similar post). To determine the lines along which $z$ is constant, we set $x = ay+x_0$. Thus, $$\frac{\text d z}{\text d y} = az_x + z_y = (a-z)\, z_x \, .$$ Therefore, $z = z|_{y=0} = -3x_0$ is constant along the lines which satisfy $a = z$, i.e., the lines with equation $x = -3x_0\, y + x_0$, or $y = -\frac{1}{3x_0} x + \frac{1}{3}$ for $x_0$ in $\Bbb R^\star$. These characteristic curves are displayed in the picture below:

curves

One can observe that they intersect at $y=1/3$, where a shock wave occurs. The breaking time $y_b$ is indeed $$ y_b = \frac{-1}{\min z_x(x,0)} = \frac{1}{3} \, . $$ The argument given in OP (nonzero denominator) is fine too.

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  • $\begingroup$ Thanks a lot for your detailed answer ! The method of finding the constant lines is standard ? $\endgroup$ – Rebellos Jun 1 '18 at 10:49
  • $\begingroup$ @Rebellos It is standard for conservation laws $u_t+f(u)_x = 0$, where the quantity $u$ is conserved along characteristic curves. But basically, this is nothing else as the method of characteristics. $\endgroup$ – Harry49 Jun 1 '18 at 11:42

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