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The picture above is the question I have been asked to do.

For part ii) a) what I have done is that I have said that $g'(v) = v^{\frac{1}{2}}=u$ therefore it does satisfy the scalar conservation laws and it does satisfy Burgers' equation.

Now for part ii) b) I have an answer but I don't think I am correct.

What I have said is that

$$v_0(x) = \begin{cases} u_l^2, & x<0\\[2ex] u_r^2, & x>0 \end{cases}$$

This would give me that $$\dot\sigma = \frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}$$

which comes from the Rankine Hugionot condition. This means, along with initial condition that $\sigma(0)=0$, that the shock comes at

$$\sigma = \frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}t$$

which gives me weak solution

$$v(x,t) = \begin{cases} u_l^2, & x<\frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}t\\[2ex] u_R^2, & x>\frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}t \end{cases}$$

Which means that $v\neq u^2$

What I am looking for here is for someone to tell me whether my solution is correct or not. And if I am wrong, what have I do wrong and how do I rectify it.

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Your solution is correct. Written differently:

(a) If $u$ is smooth, positive and $v=u^2$, then \begin{aligned} v_t &= 2 u u_t \\ &= -2 u^2 u_x \\ &= - 2v ({v}^{1/2})_x \\ &= -v^{1/2} v_x \\ &= -\left(\tfrac{2}{3}v^{3/2}\right)_x \, . \end{aligned} Therefore, $v$ satisfies the conservation law $v_t + g(v)_x = 0$ with $g(v) = \tfrac{2}{3}v^{3/2}$.

(b) We have to solve the Cauchy problem for $v_t + g(v)_x = 0$ with initial data $v(x,0) = v_0(x)$ (Riemann problem), where $$ v_0(x) = \left\lbrace \begin{aligned} &v_l &&\text{if } x<0 \\ &v_r &&\text{if } x>0 \end{aligned} \right. \qquad \text{and}\qquad v_l>v_r. $$ The Rankine-Hugoniot condition gives the shock speed \begin{aligned} s &= \frac{2}{3} \frac{(v_r)^{3/2} - (v_l)^{3/2}}{v_r-v_l}\\ &= \frac{2}{3} \frac{(u_r)^{3} - (u_l)^{3}}{(u_r)^2-(u_l)^2} \\ &= \frac{2}{3}\left( u_l+u_r - \frac{u_r u_l}{u_r+u_l} \right) , \end{aligned} so that $$ v(x,t) = \left\lbrace \begin{aligned} &v_l = (u_l)^2 &&\text{if } x<s t \\ &v_r = (u_r)^2 &&\text{if } x>s t \end{aligned}\right. $$ and $v\neq u^2$. Note that $u$ is not smooth here, which invalidates the derivation in (a).

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  • $\begingroup$ Ah. Nice. That's good to hear. I had made a big mistake but edited it out before you saw lol. Thanks $\endgroup$
    – Gragbow
    May 21 '18 at 14:31

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