- 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}(s) = Z(s) \\ \dfrac{dT}{ds}(s) = 1 \\ \dfrac{dZ}{ds}(s) = 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 $\xi \in \mathbb R$. In the above equations, \(s\) is the parameter and \(\xi \in \mathbb{R}\) is arbitrary, but fixed.
-
For each $\xi \in \mathbb{R}$ we have the following 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. in the language of multivariable calculus these lines are called contour lines, or level curves. 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 objective here is to find the largest value of $\tau \gt 0$ 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 two variable function defined on the infinite rectangle \[ U = \mathbb{R} \times [0,\tau] = \bigl\{ (x,t) \in \mathbb{R} : t \in [0,\tau] \bigr\}. \]
- Next we will find the largest value of $\tau \gt 0$ with the following property: The surface determined by the parametric equations in the preceding item, when restricted to the infinite rectangle \[ U = \mathbb{R} \times [0,\tau] = \bigl\{ (x,t) \in \mathbb{R} : t \in [0, \tau] \bigr\} \] represents a graph of a function $z = u(x,t)$ (defined on $U$). The value of $\tau$ will depend on the function $f$.
-
This is the strategy that we will use to determine the value of $\tau \gt 0$ with the property stated in the preceding item. First fix a value of $t \geq 0$ by choosing fixed $t_0 \geq 0$ and setting $t = t_0$ (think of \(t_0 = 1\), for example; or $t_0=6/5$, to symbolize to yourself that it is fixed). Next consider the curve in the plane parallel to the $xz$-plane with fixed value of $t=t_0$. The intersection of the plane \(t=t_0\) and the surface is 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
\[
\xi \mapsto \bigl\langle t_0 f(\xi) + \xi, t_0, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}.
\]
For different values of \(t_0\) the curve above has different shape. For \(t_0 = 0\) we have the green curve \(z = e^{-\xi^2}\). 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.
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In this item we figure out for which \(t_0\) the blue curve in the animation is graph of a function. For example, if $t_0 = 0$ 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 $t_0 \gt 0$ the graph of $f$ will not change that dramatically and it will still be graph of a function.
When will the graph change dramatically?
I. By considering specific graphs, we realize that the dramatic change will happen when a tangent vector to the curve is vertical. This is the vector equation of the curve: \[ \xi \mapsto \bigl\langle t_0 f(\xi) + \xi, t_0, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}. \] Its tangent vector is \[ \xi \mapsto \bigl\langle t_0 f'(\xi) + 1, 0, f'(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}. \] This vector is vertical when \[ t_0 f'(\xi) + 1 = 0. \] Thus, the (blue) curve \[ \xi \mapsto \bigl\langle t_0 f(\xi) + \xi, t_0, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}, \] is graph of a function whenever $t_0 \geq 0$ satisfies the condition \[ t_0 f'(\xi) + 1 \geq 0 \quad \text{for all} \quad \xi \in \mathbb{R}. \]
II. Since \(t_0 f'(\xi) + 1 \geq 0\) is equivalent to \(-f'(\xi) \leq \frac{1}{t_0}\), we can restate the preceding sentence as follows: The (blue) curve \[ \xi \mapsto \bigl\langle t_0 f(\xi) + \xi, t_0, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}, \] is graph of a function whenever $t_0 \gt 0$ satisfies the condition \[ -f'(\xi) \leq \frac{1}{t_0} \quad \text{for all} \quad \xi \in \mathbb{R}. \] Since we are working with a function \(f\) that has all the derivatives, we can calculate the maximum value \[ \max \bigl\{ - f(\xi) \mkern 2mu : \mkern 2mu \xi \in \mathbb{R} \bigr\} = \mu. \] Now we can restate the condition on \(t_0 \gt 0\) for which the (blue) curve is graph of a function simply as \[ \mu \leq \frac{1}{t_0} \quad \text{or, equivalently} \quad t_0 \leq \frac{1}{\mu}. \]
III. In conclusion, the largest value of $t_0 \gt 0$ (we called this value \(\tau\)) for which the (blue) curve \[ \xi \mapsto \bigl\langle t_0 f(\xi) + \xi, t_0, f(\xi) \bigr\rangle, \quad \xi \in \mathbb{R}, \] is graph of a function is given by \[ \tau = \frac{1}{\mu}. \]
IV. It remains to explain how we find \[ \mu = \max \bigl\{ -f'(\xi) : \xi \in \mathbb{R} \bigr\} \] This problem is 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 \[ -f''(\xi) = 0. \] If this equation has more than one solution, one has to be careful to determine which solution will lead to the maximum. That can be done by looking at the third derivative, or simply looking at the graph of the function \(-f'(\xi)\). Say that \(x_0\) is the correct solution of \(-f''(\xi) = 0\). Then we have \[ \mu = -f'(x_0). \]
- Specific Example. For the Mathematica illustrations presented on this page, 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 $x$ since the function has negative values for \(x \lt 0\)) 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, \[ \tau = \sqrt{\frac{e}{2}} \approx 1.16582 \]
-
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
A visual inspection of the surface above indicates that the plotted surface is graph of a function. We do not have a formula for that function, but it can be studied numerically.
- Above we found the maximal infinite rectangle on which we have the solution of the Burgers' equation. The next items deal with finding the maximal domain (not necessarily rectangular) of the solution of Burgers' equation with the initial condition $f(x) = \exp\bigl(-x^{2}\bigr)$.
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Let us look again at 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).
Using the method explained at the Wikipedia epage Envelope_(mathematics)
we calculate the parametric vector equation of the envelope to be
\[
\left\langle \frac{1}{2 \xi} + \xi, \frac{1}{2 \xi} e^{-\xi^2} \right\rangle, \quad \xi \gt 0.
\]
That this equation is correct is indicated in the above picture.
-
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
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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.