Here we consider the following first order partial differential equation:
\begin{equation} \label{eq:pde}
A(x,y,u) \, u_x(x,y) + B(x,y,u) \, u_y(x,y) = C(x,y,u)
\end{equation}
subject to the initial condition
\begin{equation} \label{eq:ic}
u(x,0) = f(x), \quad x \in \mathbb R.
\end{equation}
For simplicity we assume that the given functions $A$, $B$ and $C$ are smooth functions defined on ${\mathbb R}^3$ with the values in $\mathbb R$ and that the given function $f$ is a smooth function defined on $\mathbb R$ with values in $\mathbb R$.
A geometric interpretation of \eqref{eq:pde}
Assume that a smooth function $z = u(x,y)$ is a solution of \eqref{eq:pde}. Let $x_0,y_0 \in \mathbb R$ and set $z_0 = u(x_0,y_0).$ Then the point $P_0 = (x_0,y_0,z_0)$ is a point on the graph of the function $z = u(x,y)$. From a multivariable calculus course we know that the vector
\begin{equation} \label{eq:tvx}
\bigl\langle 1, 0, u_x(x_0,y_0) \bigr\rangle
\end{equation}
is tangent to the curve
\[
\bigl\{ \bigl( x, y_0, u_x(x,y_0) \bigr) : x_0 -\epsilon \lt x \lt x_0 + \epsilon \bigr\}
\]
at the point $P_0$. Similarly, the vector
\begin{equation} \label{eq:tvy}
\bigl\langle 0, 1, u_y(x_0,y_0) \bigr\rangle
\end{equation}
is tangent to the curve
\[
\bigl\{ \bigl( x_0, y, u_y(x_0,y) \bigr) : y_0 -\epsilon \lt y \lt y_0 + \epsilon \bigr\}
\]
at the point $P_0$. Thus both vectors \eqref{eq:tvy} and \eqref{eq:tvx} are tangent to the graph of $z = u(x,y)$ at the point $P_0$. Consequently the cross product of \eqref{eq:tvy} and \eqref{eq:tvx}, that is the vector
\begin{equation} \label{eq:nv}
\bigl\langle u_x(x_0,y_0), u_y(x_0,y_0), - 1 \bigr\rangle,
\end{equation}
is orthogonal to the graph of $z = u(x,y)$ at the point $P_0$.
Since we assume that $z= u(x,y)$ is a solution of \eqref{eq:pde}, \eqref{eq:pde} holds for the values $x = x_0$, $y= y_0$ and $z_0 = u(x_0,y_0)$. That is we have
\begin{equation*}
A(x_0,y_0,z_0) \, u_x(x_0,y_0) + B(x_0,y_0,z_0) \, u_y(x_0,y_0) = C(x_0,y_0,z_0)
\end{equation*}
The last identity can be written as a dot product of two vectors:
\[
\Bigl\langle A(x_0,y_0,z_0), B(x_0,y_0,z_0), C(x_0,y_0,z_0) \Bigr\rangle \cdot \bigl\langle u_x(x_0,y_0), u_y(x_0,y_0), - 1 \bigr\rangle = 0.
\]
Hence the vector
\begin{equation} \label{eq:vvf}
\Bigl\langle A(x_0,y_0,z_0), B(x_0,y_0,z_0), C(x_0,y_0,z_0) \Bigr\rangle
\end{equation}
is orthogonal to the vector \eqref{eq:nv}. Since the vector \eqref{eq:nv} is orthogonal to the graph of $z = u(x,y)$ at the point $P_0$, we conclude that the vector \eqref{eq:vvf} is tangent to the graph of $z = u(x,y)$ at the point $P_0$.
The point $P_0 = (x_0,y_0,z_0)$, $z_0 = u(x_0,y_0)$ in the previous item was an arbitrary point on the graph of $z = u(x,y)$. Therefore, the previous item shows that the vector field
\begin{equation} \label{eq:vf}
\Bigl\langle A(x,y,z), B(x,y,z), C(x,y,z) \Bigr\rangle
\end{equation}
which consists of the given functions $A$, $B$, $C$ is tangent to the surface $z= u(x,y)$. Thus, to solve \eqref{eq:pde} subject to \eqref{eq:ic} we have to find a surface which is tangent to the vector field \eqref{eq:vf} and passes through the curve
\begin{equation}\label{eq:ic-curve}
\bigl\{ \bigl(\xi, 0, f(\xi) \bigr) : \xi \in \mathbb R \bigr\}.
\end{equation}
The reason why I use $\xi$ instead of $x$ will be clear in the next section.
The related system of ordinary differential equations
From the previous section we learned that to solve \eqref{eq:pde} we need to find a surface which is tangent to the vector field \eqref{eq:vf} and passes through the curve \eqref{eq:ic-curve}.
Now recall that studying systems of ordinary differential equations we learned how to find a curve \[
\Bigl\{ \bigl(X(s), Y(s), Z(s) \bigr) : s \geq 0 \Bigr\}
\]
which is tangent to the vector field \eqref{eq:vf} and starts at the given point
\[
\bigl(X(0), Y(0), Z(0) \bigr).
\]
Such curve is called an integral curve of the vector field \eqref{eq:vf}. It is obtained by solving the system of ordinary differential equations
\begin{equation} \label{eq:sys}
\begin{array}{l}
\dfrac{dX}{ds} = A(X,Y,Z) \\ \dfrac{dY}{ds} = B(X,Y,Z) \\ \dfrac{dZ}{ds} = C(X,Y,Z)
\end{array}
\end{equation}
subject to the initial conditions
\begin{equation}\label{eq:sys-ic}
\begin{array}{l}
X(0) = \xi \\ Y(0) = 0 \\ Z(0) = f(\xi)
\end{array}
\end{equation}
These initial conditions are chosen so that the integral curves that we find start at the curve \eqref{eq:ic-curve}.
Assume that we can solve the system of ODEs \eqref{eq:sys} subject to \eqref{eq:sys-ic}. The solution is a triple of functions $(X, Y, Z)$ which depend on two variables $s$ and $\xi$:
\begin{equation*}
\begin{split}
x &= X(s,\xi) \\ y & = Y(s,\xi) \\ z &= Z(s,\xi) \\
\end{split}
\end{equation*}
For a fixed $\xi$ there above equations represent a curve in $xyz$-space. These curves are called characteristics.
The union of all the characteristics is the surface which passes through the curve in \eqref{eq:ic-curve} and is tangent to the vector field \eqref{eq:vf}. This surface is the graph of the solution that we are seeking.
There is one more step to find the formula for the function $z= u(x,y)$. We need to solve the system
\begin{equation*}
\begin{split}
x &= X(s,\xi) \\ y & = Y(s,\xi)
\end{split}
\end{equation*}
for $s$ and $\xi$ in terms of $x$ and $y$. Solving this system will lead to two functions
\begin{equation*}
\begin{split}
s &= S(x,y) \\ \xi & = \Xi(x,y).
\end{split}
\end{equation*}
Now the desired function $u(x,y)$ is
\[
u(x,y) = Z\bigl( S(x,y), \Xi(x,y) \bigr).
\]
Mathematica notebooks
This
Mathematica notebook contains the examples below worked out in Mathematica. This notebook has been written in Mathematica 8. It should run well in later versions as well.
Here is a pdf print-out of this notebook so that you can view it when Mathematica is not available.
Example 1
Consider the following first order PDE
\begin{equation*}
y \, u_x + 3 \, u_y = -u
\quad \text{in} \quad \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = \cos(x), \quad \text{for} \quad x \in \mathbb R.
\end{equation*}
The corresponding system of ODEs is
\begin{equation*}
\begin{array}{l}
\frac{dX}{ds} = Y \\ \frac{dY}{ds} = 3 \\ \frac{dZ}{ds} = -Z
\end{array}
\end{equation*}
subject to the initial conditions
\begin{equation*}
\begin{array}{l}
X(0) = \xi \\ Y(0) = 0 \\ Z(0) = \cos(\xi)
\end{array}
\end{equation*}
The solution of the above system is
\begin{equation*}
\begin{array}{l}
X(s,\xi) = \frac{3}{2}s^2 +\xi \\ Y(s,\xi) = 3s \\ Z(s,\xi) = e^{-s} \cos(\xi)
\end{array}
\end{equation*}
Now solve the system
\begin{equation*}
\begin{array}{l}
x = \frac{3}{2}s^2 +\xi \\ y = 3s
\end{array}
\end{equation*}
for $s$ and $\xi$ to get
\begin{equation*}
\begin{array}{l}
s = \frac{1}{3} y \\ \xi = x - \frac{1}{6} y^2.
\end{array}
\end{equation*}
Finally, substitute the formulas for $s$ and $xi$ into $Z(s,\xi) = (\cos \xi)e^{-s}$ to get
\[
u(x,y) = e^{-y/3} \cos\!\left(x - \frac{1}{6} y^2\!\right)
\]
Clearly the domain of $u$ is the entire plane ${\mathbb R}^2$.
Example 2
Consider the following first order PDE
\begin{equation*}
2\, u_x + u_y = u^2 \quad \text{in} \quad U \subseteq \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = \cos(x), \quad x \in \mathbb R.
\end{equation*}
Notice that the domain $U$ above is not specified. As a part of your solution you should determine the largest "rectangular" box $U$ (whose boundary is the $x$-axis) on which the problem has a solution.
The corresponding system of ODEs is
\begin{equation*}
\begin{array}{l}
\frac{dX}{ds} = 2 \\ \frac{dY}{ds} = 1 \\ \frac{dZ}{ds} = Z^2
\end{array}
\end{equation*}
subject to the initial conditions
\begin{equation*}
\begin{array}{l}
X(0) = \xi \\ Y(0) = 0 \\ Z(0) = \cos(\xi)
\end{array}
\end{equation*}
The solution of the above system is
\begin{equation*}
\begin{array}{l}
X(s,\xi) = 2s +\xi \\ Y(s,\xi) = s \\ Z(s,\xi) = \frac{\cos \xi}{1- s \cos \xi }
\end{array}
\end{equation*}
Now solve the system
\begin{equation*}
\begin{array}{l}
x = 2s +\xi \\ y = s
\end{array}
\end{equation*}
for $s$ and $\xi$ to get
\begin{equation*}
\begin{array}{l}
s = y \\ \xi = x - 2 y.
\end{array}
\end{equation*}
Finally, substitute the formulas for $s$ and $\xi$ into $Z(s,\xi) = \frac{\cos \xi}{1- s \cos \xi }$ to get
\[
u(x,y) = \frac{\cos(x-2y)}{1- y \cos(x-2y)}
\]
The domain of the function $u$ is complicated. However, since for $y \in [0,1)$ we have $y \cos(x-2y) \lt 1$, the function $u$ is defined on the infinite rectangle $\mathbb R \times [0,1)$. Below is its graph on a part of this rectangle.
The above graph shows that the function $u$ becomes large near the points $(2-2\pi,1)$ and $(2,1)$. In fact the function $u(x,1)$ is not defined at $x = 2 + 2k\pi$ with $k \in \mathbb Z$.
Often a partial differential equation is a model for a physical process. In this setting it is common to think of the variable $y$ as time and denote it by $t$. The initial condition $u(x,0) = \cos x$ describes the state of the process at time $t=0$ and the solution $u(x,t)$ describes the state of the process at any time $t \geq 0$. Therefore it is of interest to show the movie of the graph of $u(x,t)$ as $t$ changes
Place the cursor over the image to start the animation.
Or, it might be useful to see all the graphs in one picture
Example 3
Consider the following first order partial differential equation
\begin{equation*}
u_t + u\, u_x = 0 \quad \text{in} \quad U \subseteq \bigl\{(x,t) \in \mathbb R^2 : t \gt 0 \bigr\}
\end{equation*}
subject to the initial condition
\begin{equation*}
u(x,0) = f(x), \quad x \in \mathbb R.
\end{equation*}
You should consider the following three specific functions $f(x) = \arctan x$, $f(x) = -\arctan x$ and $f(x) = \exp(-x^2)$.
Notice that the domain $U$ above is not specified. As a part of your solution, for each specific $f$ you should determine the largest "rectangular" box $U$ (whose boundary is the $x$-axis) on which the problem has a solution.
Solved in the Mathematica file. Some details are below.
The corresponding system of ODEs is
\begin{equation*}
\begin{array}{l}
\frac{dX}{ds} = Z \\ \frac{dT}{ds} = 1 \\ \frac{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 $\xi \in \mathbb R$.
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 ${\mathbb R}^3$. Next we have to find the largest $t_0 > 0$ such that this surface restricted to the rectangle $\bigl\{(x,t) : x \in \mathbb R, \ 0 \leq t \leq t_0 \bigr\}$ represents a graph of a function $z = u(x,t)$ defined on this rectangle. This will depend on the function $f$. So, we will consider some simple functions $f$s next.
To be continued.
Problems
Solve the following first order PDE
\begin{equation*}
y\, u_x + u_y = 0 \quad \text{in} \quad \mathbb R^2
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = f(x) \quad \text{for} \quad x \gt 0.
\end{equation*}
You can try $f(x) = x$, $f(x) = x^2$, $f(x) = \cos x$ as some specific examples.
Solve the following first order PDE
\begin{equation*}
u_x - x u_y = 0 \quad \text{in} \quad \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = f(x) \quad \text{for} \quad x \gt 0.
\end{equation*}
You can try $f(x) = x$, $f(x) = x^2$, $f(x) = \cos x$ as some specific examples.
Solve the following first order PDE
\begin{equation*}
y\, u_x - x u_y = 0 \quad \text{in} \quad \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = f(x) \quad \text{for} \quad x \gt 0.
\end{equation*}
You can try $f(x) = \sin x$, $f(x) = \cos x$, $f(x) = \exp(-x^2)$ as some specific examples.
Solve the following first order PDE
\begin{equation*}
u_x + x u_y = u \quad \text{in} \quad \mathbb R^2
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(0,y) = g(y) \quad \text{for} \quad x \in \mathbb R.
\end{equation*}
You can try $g(y) = y$, $g(y) = \cos y$ as some specific examples.
Solve the following first order PDE
\begin{equation*}
y\, u_x - x u_y = u \quad \text{in} \quad \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = f(x) \quad \text{for} \quad x \geq 0.
\end{equation*}
You can try $f(x) = \sin x$, $f(x) = \cos x$ as some specific examples.
Solve the following first order PDE
\begin{equation*}
x u_x + y\, u_y = u \quad \text{in} \quad \mathbb R^2
\end{equation*}
subject to the condition
\begin{equation*}
u\bigl(\cos \theta, \sin \theta\bigr) = 1 \quad \text{for} \quad 0 \leq \theta \lt 2 \pi.
\end{equation*}
Solve the following first order PDE
\begin{equation*}
y u_x + u_y = x \quad \text{in} \quad \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = x^2 \quad \text{for} \quad x \in \mathbb R.
\end{equation*}
Solve the following first order PDE
\begin{equation*}
x^2 u_x + u_y = 0 \quad \text{in} \quad U \subseteq \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = f(x) \quad \text{for} \quad x \in \mathbb R.
\end{equation*}
Notice that the domain $U$ is not given in the problem.
For an arbitrary differentiable function $f$ find the solution $u(x,y)$ of the above problem and determine its maximum domain $U\subseteq \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}$.
Under which condition on $f$ the solution $u(x,y)$ will be defined on $\bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}$.
Consider the following first order PDE
\begin{equation*}
u_x - u\, u_y = 0 \quad \text{in} \quad \boxed{U \subseteq \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}}
\end{equation*}
subject to the boundary condition
\begin{equation*}
u (x,0) = x \quad \text{for} \quad \boxed{x \in \mathbb R}.
\end{equation*}
Does there exist an open set $U \subseteq \bigl\{(x,y) \in \mathbb R^2 : y \gt 0 \bigr\}$ whose boundary is the $x$-axis such that the above problem has a solution on $U$?
Choose smaller specific sets in the boxed formulas in such a way that the corresponding modified problem has a solution. Prove your claim by providing a solution of the modified problem.
Solve the following first order PDE
\begin{equation*}
(1+x^2) u_x + 2 x y \, u_y = 0 \quad \text{in} \quad {\mathbb R}^2
\end{equation*}
subject to the condition
\begin{equation*}
u(0,y) = g(y) \quad \text{for all} \quad x \in \mathbb R.
\end{equation*}
Solve the following first order PDE
\begin{equation*}
(1+x^2) u_x - 2 x y \, u_y = 0 \quad \text{in} \quad {\mathbb R}^2
\end{equation*}
subject to the condition
\begin{equation*}
u(0,y) = g(y) \quad \text{for all} \quad x \in \mathbb R.
\end{equation*}
Solve the following first order PDE
\begin{equation*}
x \, u_x + y \, u_y = 2 u \ln u \quad \text{in} \quad {\mathbb R}^2
\end{equation*}
subject to the condition
\begin{equation*}
u(x, 1) = e^{x^2-1} \quad \text{for all} \quad x \in \mathbb R.
\end{equation*}