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.

  • $\begingroup$ @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

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:


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.