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I am trying to solve the first order problem, namely:

$$ \begin{cases} u_{t}+uu_{x} = 0, \hspace{0.5cm} x>0, t>0 \\ u(0,t)=t, \hspace{0.5cm} t>0 \\ u(x,0)=x^2, \hspace{0.5cm} x>0 \end{cases} $$

First I proceed as always: I introduce $U(t):=u(X(t),t)$ and write down the characteristic system: $$ \begin{cases} \frac{dX}{dt} = U \\ \frac{dU}{dt} = 0 \hspace{0.5cm} \iff U(t)=const=U_{0} \end{cases} $$ thus $X(t)=U_{0}t + C$ (characteristics are straight lines), but now I have a problem. In the presence of only IC I would take $X(t)=\phi(\xi)t + \xi$ (where the $\phi$ is IC and $\xi$ is the intersection point of each of the characteristic with the $Ox$) and probably would obtain some shock there (which I would deal with the help of the R-H condition). However, in my case I have to take account of the BC as well. Do you know how should I proceed? It would be very helpful if someone could include the characteristic plane as well. Thanks in advance for any help!

Mat Dyl
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2 Answers2

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Compute the characteristics through the points $(0,\tau)$, $\tau>0$. On these characteristics, the value of $U$ is constant and equal to $\tau$, the value on $(0,\tau)$.

Julián Aguirre
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The solution to the characteristic system can be written as $x-ut=f(u)$, where $f$ is a function determined by the initial and the boundary conditions. For the initial condition $u(0,t)=t$, we have $$ 0-t^2=f(t) \implies x-ut=-u^2. \tag{1} $$ Solving $(1)$ for $u$, we find two solutions, but only one of them satisfies the initial condition: $$ u(x,t)=\frac{t+\sqrt{t^2-4x}}{2}\qquad(t^2\geq 4x). \tag{2} $$ In a similar fashion, the boundary condition $u(x,0)=x^2$ implies $$ x-0=f(x^2) \implies x-ut=\sqrt{u}. \tag{3} $$ The solution to $(3)$ is$^{(*)}$ $$ u(x,t)=\frac{1+2tx-\sqrt{1+4tx}}{2t^2}. \tag{4} $$ Notice that, for fixed $x>0$, the solution $(4)$ is defined for all $t>0$, whereas the solution $(2)$ is defined only for $t>\sqrt{4x}$. This suggests the formation of a shock. The curve $x=\xi(t)$ along which it develops can be determined with the help of the Rankine-Hugoniot formula$^{(\dagger)}$, but this is beyond the scope of this answer.

$^{(*)}$ See Burgers equation with initial data $u(x,0) = x^2$ or Solving the problem : $z z_x + z_y = 0, \quad z(x,0) = x^2$.

$^{(\dagger)}$ Walter A. Strauss, Partial Differential Equations: An Introduction (Second Edition), John Wiley & Sons, Section 14.1.

Gonçalo
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