Triangular Profile

We modified the initialize.f subroutine to include a triangular profile as an option for initial conditions.

We know that at later stages of the evolution the profile assumes the form of a time dependent propagating right triangle, as shown in the assignment sheet. So we proceed to derivate an analytic weak solution for this profile, as an asymptotic solution to the evolution of the Gaussian profile. We will compare this exact (somewhat guessed) solution with the Roe solver approximation. We can set the following initial conditions for the triangular profile:

$${ q(x_o,0) } = \left\{ \begin{array}{rl} 0 & if x < -0.5\\ 2x+1 & if -0.5 < x < 0\\ 0 & if x > 0 \end{array}\right.$$

We have to solve for the characteristics of the system, in the domain $-0.5 < x < 0$, So we proceed as follows:

$$q(x,0)=\frac{\partial x}{\partial \eta} = 2x_o+1 \hspace{.5c... ...rtial t}{\partial \eta} = 1 \rightarrow \frac{dx}{dt} = 2x_o+1$$ (36)

which has a solution

$$x = (2x_o+1)t+x_o$$ (37)

We solve now for $x_o$ (a constant of integration),

$$x_o = \frac{x+t}{2t+1}$$ (38)

substitutingi we get,

$$q(x,0) = 2x_o+1 \rightarrow q(x,t) = \frac{2x+1}{2t+1}$$ (39)

This characteristic solution is analytic in our domain of interest, but this does not hold for the long term evolution. The solutions does not produce a shock, and behaves well in the long term evolution. For $x < -0.5$ the solution will be zero, but for $x > -0.5$ a rarefaction is produced, which necessarily evolves in time.

Now we use the fact that the solution must be area preserving and does not depend on Burgers equation, except for the speed of the shock,

$$Area = \frac{1}{2}(x+\frac{1}{2})(q-0)$$ (40)

We state the problem as following,

$$\frac{dArea}{dt} = \frac{1}{2} \left(\frac{d(x+ \frac{1}{2})}{t}q + (x- \frac{1}{2})\frac{dq}{dt}\right) =$$ (41)

$$\frac{1}{2} \left(\frac{d(x)}{t}q + (x- \frac{1}{2})\frac{dq}{dt}\right)= 0$$ (42)

So we must solve the equation,

$$\frac{1}{(x+ \frac{1}{2})}\frac{dx}{dt} = - \frac{1}{q}\frac{dq}{dt}$$ (43)

and we know the speed of the shock from Burger equation,

$$v_s = \frac{1}{2}(q_L+q_R) = \frac{dx}{dt} \rightarrow \frac{1}{q}\frac{dq}{dt} = -\frac{q_L+q_R}{2(x+0.5)}$$ (44)

what is even better is that the value of $q_R$ at the shock front is always 0, so we can simplify further;

$$\frac{1}{q}\frac{dq}{dt} = -\frac{q}{2(x+0.5)}$$ (45)

Relating this to the area of the triangle we get,

$$\frac{1}{2}q(x+\frac{1}{2}) = Area_o = \frac{1}{4} \rightarrow (x+\frac{1}{2}) = \frac{1}{2q}$$ (46)

$$\frac{dq}{dt} = -q^3\rightarrow q(t) = \frac{1}{\sqrt{2(t+\lambda)}}$$ (47)

where $\lambda$ is the integration constant, $\lambda=0.5$ in our case.

The key property here is the conservation of the area under the curve during evolution. The value of $q_C$ and the original value of the area are used to compute the size of the triangle at every time step. Subroutine exact_triangle.f³ implements this exact solution and the mpeg can be see on the website.

Figure 7: Triangle profile evolution, showing the exact and Roe approximation. We see both agree.