- This problem is solved in the Mathematica notebook MoC_Burgers_eq1_v12.nb. I printed this notebook as a pdf file MoC_Burgers_eq1_v12.pdf, so that you can read it without access to Mathematica. A short review is below.
- The corresponding characteristic equations are: \begin{equation*} \begin{array}{l} \dfrac{dX}{ds} = Z \\ \dfrac{dT}{ds} = 1 \\ \dfrac{dZ}{ds} = 0 \end{array} \end{equation*} subject to the initial conditions \begin{equation*} \begin{array}{l} X(0) = \xi \\ T(0) = 0 \\ Z(0) = f(\xi) \end{array} \end{equation*} with $\xi \in \mathbb R$.
- The solution of the above system is \begin{equation*} \begin{array}{l} X(s,\xi) = s f(\xi) + \xi \\ T(s,\xi) = s \\ Z(s,\xi) = f(\xi) \end{array} \end{equation*} with $s \geq 0$ and $\xi \in \mathbb R$.
-
For every $\xi \in \mathbb{R}$ we have a projected characteristic
\[
\bigl\langle s f(\xi) + \xi , s \bigr\rangle, \quad s \geq 0,
\]
which is a line in $xt$-plane. Below is a plot of 100 projected characteristics.
-
The equations
\begin{equation*}
x = s f(\xi) + \xi, \quad t = s \quad z = f(\xi), \quad s \geq 0, \ \xi \in \mathbb R.
\end{equation*}
are parametric equations of a surface in ${\mathbb R}^3$. In other words, the vector function
\[
(\xi, s) \mapsto \bigl\langle s f(\xi) + \xi, s, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}, \ s \geq 0,
\]
is a vector equation of this surface. A plot of this surface is below:
Pay attention to the mesh on the yellow surface. The horizontal lines (the lines parallel to $xt$-plane, there are 50 of them) are the characteristics. The curved lines in the mesh are parallel to $xz$-plane. These lines show the evolution of the initial condition (shown as the green curve in the plot).
-
Looking at the surface in the preceding item we asses that if we limit the range of $t$ to small values of $t$, for example $t \in [0,1]$, the yellow surface will be a graph of a function in two variables $x$ and $t$. How is this reflected in projected characteristics?
The plot above is in the $xt$-plane. In this plot, I draw the line parallel to $x$ with $t=1$. We clearly see that distinct projected characteristics do not intersect below the red line. This is another way to see that the surface above is a graph of a function in $x$ and $t$.
The point is to find the largest value of $t_0 \gt 0$ such that the surface determined by the vector function \[ (\xi, s) \mapsto \bigl\langle s f(\xi) + \xi, s, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}, \ s \geq 0, \] is a graph of a function, when restricted to the infinite rectangle \[ U = \mathbb{R} \times [0,t_0) = \bigl\{ (x,t) \in \mathbb{R} : t \in [0,t_0) \bigr\} \]
- Next we have to find the largest value of $t_0 > 0$ with the following property: the surface determined by the parametric equations in the preceding item, when restricted to the infinite rectangle \[ U = \mathbb{R} \times [0,t_0) = \bigl\{ (x,t) \in \mathbb{R} : t \in [0,t_0) \bigr\} \] represents a graph of a function $z = u(x,t)$ (defined on $U$). The value of $t_0$ will depend on the function $f$.
-
To determine the value of $t_0 > 0$ with the property stated in the preceding item we fix a value of $t \geq 0$ by choosing fixed $\tau \geq 0$ and setting $t = \tau$. Now we consider the curve in the plane parallel to the $xz$-plane with fixed value of $t=\tau$. The intersection of this plane and the surface given by the vector equation
\[
(\xi, s) \mapsto \bigl\langle s f(\xi) + \xi, s, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}, \ s \geq 0,
\]
is the curve
\[
\xi \mapsto \bigl\langle \tau f(\xi) + \xi, \tau, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}.
\]
Below is an animation of the evolution of the green curve, the initial condition, in time. One can see that at a certain time $t_0$ the curve turns from being a graph of a function, to not being a graph of the function.
Place the cursor over the image to start the animation.
Now we want to establish whether this curve is a graph of a function. For example, if $\tau = 0$ we have
\[
\xi \mapsto \bigl\langle \xi, 0, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R},
\]
which is certainly a graph of a function since $f$ is a function. For very small values of $\tau \gt 0$ the graph of $f$ will not change that dramatically and it will still be graph of a function. When will things change dramatically. By considering specific graphs, we realize that the dramatic change will happen when a tangent vector to the curve
\[
\xi \mapsto \bigl\langle \tau f(\xi) + \xi, \tau, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R},
\]
becomes vertical. The tangent vector is
\[
\xi \mapsto \bigl\langle \tau f'(\xi) + 1, \tau, f'(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}.
\]
This vector is vertical when
\[
\tau f'(\xi) + 1 = 0.
\]
Thus, the curve
\[
\xi \mapsto \bigl\langle \tau f(\xi) + \xi, \tau, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R},
\]
is graph of a function for all $\tau \geq 0$ such that
\[
\tau f'(\xi) + 1 \gt 0 \quad \text{for all} \quad \xi \in \mathbb{R}.
\]
That is, the curve
\[
\xi \mapsto \bigl\langle \tau f(\xi) + \xi, \tau, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R},
\]
is graph of a function for all $\tau \gt 0$ such that
\[
-f'(\xi) \lt \frac{1}{\tau} \quad \text{for all} \quad \xi \in \mathbb{R}.
\]
Thus, the largest value of $\tau \gt 0$ such that the curve
\[
\xi \mapsto \bigl\langle \tau f(\xi) + \xi, \tau, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R},
\]
is graph of a function is obtained by calculating
\[
\max \bigl\{ -f'(\xi) : \xi \in \mathbb{R} \bigr\} = \mu.
\]
Provided that $\mu \gt 0$, we have
\[
t_0 = \frac{1}{\mu}.
\]
- But, how do we find \[ \max \bigl\{ -f'(\xi) : \xi \in \mathbb{R} \bigr\} = \mu. \] That is what was discussed in Math 124, the first course in Calculus. We use the derivative of the function which is involved to find the critical points, at some of which the maximum will occur. In this case the derivative is $-f''(\xi)$, and the critical points are found by solving \[ -f''(\xi) = 0. \] At one of the solutions of the preceding equation, we will find the maximum which we called $\mu$.
- For the Mathematica illustrations used in this Example I used the function $f(x) = \exp\bigl(x^{-2}\bigr)$. For this function, we have \[ - f'(x) = 2 x \exp\bigl(x^{-2}\bigr). \] To find the maximum of this function (which will occur for a positive $x$) we find its derivative \[ - f''(x) = 2 \exp\bigl(x^{-2}\bigr) - 4 x^2 \exp\bigl(x^{-2}\bigr) = 2 \bigl(1 - 2 x^2 \bigr) \exp\bigl(x^{-2}\bigr). \] Thus, the critical point giving the maximum is $x = 1/\sqrt{2}$. Hence, what I called $\mu$ earlier is \[ \mu = 2 \frac{1}{\sqrt{2}} \exp\bigl(-1/2\bigr) = \sqrt{\frac{2}{e}}. \] Consequently, \[ t_0 = \sqrt{\frac{e}{2}} \approx 1.16582 \]
-
Finally, we present a graph of the solution of Burgers' equation with the initial condition given as $f(x) = \exp\bigl(x^{-2}\bigr)$. It is the function defined on the infinite rectangle
\[
U = \mathbb{R} \times \left[0, \sqrt{\frac{e}{2}} \right) = \left\{ (x,t) \in \mathbb{R} : t \in \left[0, \sqrt{\frac{e}{2}} \right) \right\}
\]
whose graph is below
- The next items deal with finding the maximal domain of the solution of Burgers' equation with the initial condition $f(x) = \exp\bigl(x^{-2}\bigr)$.
-
Let us look again at the projected characteristics.
Looking at the above plot it is clear that the projected characteristics intersect each other in a specific region of the $xt$-plane.
-
The boundary of the specific region of the $xt$-plane in which the projected characteristics intersect is called the envelope of the family of projected characteristics.
More about the envelopes and how to calculate them you can find at the Wikipedia page Envelope_(mathematics).
In this particular case, the parametric vector equation of the envelope is
\[
\left\langle \frac{1}{2 \xi} + \xi, \frac{1}{2 \xi} e^{-\xi^2} \right\rangle, \quad \xi \gt 0.
\]
-
Now, we are wondering what is happening above the envelope on the surface that we calculated above. We have two interesting curves that I colored navy blue and maroon.
There is no closed form expression for the maroon curve. The navy blue curve has the following vector equation in $xtz$-space
\[
\left\langle \frac{1}{2 \xi} + \xi, \frac{1}{2 \xi} e^{-\xi^2} , e^{-\xi^2} \right\rangle, \quad \xi \gt 0.
\]
-
Finally, to get a graph of the solution of Burgers' equation on its maximal domain we need to remove the region bounded by the envelope. Fortunately and amazingly, Mathematica has a Plot3D Option to do exactly that: RegionFunction
-
Another ViewPoint, to make sure that the maroon curve is directly above the navy envelope.
- I made all these pictures in a little sloppy, but I hope interesting Mathematica notebook MoC_Burgers_eq2_v12.nb.