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We have the following equation with initial condition \begin{align*} u_y + uu_x &= 0\\ u(x,0)&=\frac{1}{1+x^2}\quad x\in\mathbb{R} \end{align*} let's say $g(x)=\dfrac{1}{1+x^2}$ and corresponding to the manifold $\Gamma$ in the $xyz$ space given by \begin{equation} x = s,\quad y = 0,\quad z = g (s) \end{equation} The characteristic differential equations \begin{equation*} \frac{dx}{dt}=z\quad \frac{dy}{dt}=1\quad \frac{dz}{dt}=0 \end{equation*} Integrating each of the expressions we have \begin{align*} \int dx&=\int z dt\\ x&=zt \end{align*} also \begin{align*} \int dy&=\int 1 dt\\ y&=t \end{align*} and \begin{align*} \int dz&=\int 0 dt\\ z&=0 \end{align*} combined with the initial condition for $t=0$ lead to the parametric representation \begin{equation*} x=s+zt,\quad y=t, \quad z=g(s) \end{equation*} now as \begin{equation*} s=x-zt\quad \text{and}\quad y=t \quad \text{and}\quad z=g(s) \end{equation*} For the solution $z=u(x,y)$ then yields the implicit equation \begin{equation*} u=g(x-uy)=\frac{1}{1+(x-uy)^2} \end{equation*} now the study is going to be done of Weak (or Integral) Solutions as the characteristic is given by \begin{equation*} x=\frac{y}{1+\xi^2}+\xi \end{equation*} it is necessary that this characteristic presents shock to do the study according to the study made by the book (artial Differential Equations in Action From Modelling to Theory) page 199 to find $y_s$ (breaking time) and the location $x_s$ the following is done where $q(u)=u^2 / 2$ then $q'(u)=u$ and $q''(u)=1$ \begin{equation*} z(\xi)=-q''(g(\xi))g'(\xi)=-1\cdot \frac{-2x}{\left(1+x^2\right)^2}=\frac{2x}{\left(1+x^2\right)^2} \end{equation*} and the maximum point of $z(\xi)$ is $\left(\sqrt{\frac{1}{3}},\:\frac{3\sqrt{3}}{8}\right)$ then $\xi_M=\sqrt{\frac{1}{3}}$ and $z(\xi_M)=\frac{3\sqrt{3}}{8}$ then \begin{equation*} y_s=\frac{1}{z(\xi_M)}=\frac{1}{\frac{3\sqrt{3}}{8}}=\frac{8}{3\sqrt{3}} \end{equation*} and \begin{equation*} x_s=q'(g(\xi_M))y+\xi_M=\frac{1}{1 + (\sqrt{1/3})^2}\left(\frac{8}{3\sqrt{3}}\right)+\sqrt{\frac{1}{3}}=\frac{2}{\sqrt{3}}+\sqrt{\frac{1}{3}} \end{equation*}

therefore $\displaystyle (x_s,y_s)=(\frac{2}{\sqrt{3}}+\sqrt{\frac{1}{3}},\frac{8}{3\sqrt{3}})$

1)My question is already having all these data, how can I do the study of Rankine-Hugoniot condition?

Your answers would be of great help. Thank you very much. I remain attentive to your answer.

EditPiAf
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ducarita
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1 Answers1

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The characteristic curves satisfy $$ x= x_0 + g(x_0) y, \qquad u = g(x_0) $$ where $g(x) = (1+x^2)^{-1}$ is the initial data, i.e. $ u = g(x-uy) $ in implicit form. The shock time is given by $$ y_s = \frac{-1}{\inf_{x_0} g'(x_0) } = \frac{8}{3\sqrt{3}} \simeq 1.54 $$ for $x_0 = 1/\sqrt{3}$. Therefore, the results in OP look correct, except the potential mistake in the expression of $x_s$. The Rankine-Hugoniot condition for the inviscid Burgers equation is still valid in the usual form $$ x_s'(y) = \tfrac12 \left( u_\ell + u_r\right),$$ $$\text{with}\qquad x_s(y_s) = \frac{1}{\sqrt 3} + \frac{8}{3\sqrt{3}}\, g\left(\frac{1}{\sqrt 3}\right) = \sqrt 3 \simeq 1.73 $$ where $u_\ell$, $u_r$ represent the value of $u = g(x-uy)$ on the left and on the right of the discontinuity $(x_s(y), y)$ with $y \geq y_s$. See this post where a simpler case is tackled analytically.

EditPiAf
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