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Example 1. Let $a \in \mathbb{R}.$ Find the solution of the initial value problem \begin{align*} \frac{dX}{ds}(s) & = 1 \\ \frac{dY}{ds}(s) & = -X(s) \\ \end{align*} subject to the initial conditions \begin{align*} X(0) & = a \\ Y(0) & = 0 \\ \end{align*} Here \(X(s)\) and \(Y(s)\) are unknown functions of one variable \(s\). We use what is learned in an Ordinary Differential Equations course for find a solution.
Solution. The solution of this system is a pair of functions $X(s) = s+a,$ $Y(s) = (-1/2)s^2 - a s,$ with $s \in \mathbb{R}.$ In \(xy\)-plane this pair of functions provides the parametric equation of the parabola \[ (x,y) = \bigl(X(s), Y(s) \bigr) = \bigl(s+a , (-1/2)s^2 - a s \bigr), \quad s \in \mathbb{R}. \] Solving \(x= s+a\) for \(s\) and substituting in \(y = (-1/2)s^2 - a s\) we obtain the equation for this parabola in \(xy\)-coordinates: $y = (-1/2)(x^2 - a^2).$ Since $a \in \mathbb{R}$ is arbitrary, we in fact have a family of parabolas which are obtained by the vertical shift of the parabola $y=(-1/2)x^2$ up by $(1/2)a^2.$
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Example 2.
Let $a \in \mathbb{R}\setminus\{0\}.$ Find the solution of the initial value problem
\begin{align*}
\frac{dX}{ds}(s) & = X(s) \\
\frac{dY}{ds}(s) & = 2X(s) - Y(s) \\
\end{align*}
subject to the initial conditions
\begin{align*}
X(0) & = a \\
Y(0) & = 0 \\
\end{align*}
Solution.
The solution of this system is a pair of functions $X(s) = a \exp(s),$ $Y(s) = a \exp(s) - a \exp(-s),$ with $s \in \mathbb{R}.$ For a fixed \(a\), this pair gives a parametric equation of one branch of the hyperbola whose equation in \(xy\)-coordinates is $y = x - a^2(1/x)$ where $x\neq 0.$ Since $a \in \mathbb{R}\setminus\{0\}$ is arbitrary, we in fact have a family of branches of hyperbolas, pictured below
The following equation is an example of a first order partial differential equation: \begin{equation*} x \, u_x(x,y) - y \, u_y(x,y) = 0 \end{equation*} In this equation the unknown function is $u(x,y)$ with two independent variables $x$ and $y$. As an additional information we can have some information about the domain of $u$. For this particular equation we can set the requested domain to be the entire $xy$-plane. This equation is called a homogeneous linear partial differentail equation since its left-hand side is a linear function of $u$. One particular solution of this equation is for example the function $u(x,y) = xy$. Interestingly, for an arbitrary differentiable function $f$ of a single variable, the function $u(x,y) = f(xy)$ is also a solution of the preceding partial differential equation.
The following equation is an example of a first order nonhomogeneous linear partial differential equation: \begin{equation*} x \, u_x(x,y) - y \, u_y(x,y) = xy \end{equation*} One particular solution of this equation is for example the function $u(x,y) = x y \ln x$. It is a feature of a nonhomogeneous equation that when we know its particular solution adding it to a solution of the corresponding homogeneous equation we get more solutions of the nonhomogeneous equation: $u(x,y) = x y \ln x + f(xy)$.
The following equation is an example of a nonlinear first order partial differential equation: \begin{equation*} u_x(x,y) u_y(x,y) - u(x,y) = 0. \end{equation*} One solution of this equation is $u(x,y) = (1/4) (1+x+y)^2$. You can verify it. Since in this class we will not study nonlinear partial differential equations we will leave it as a mystery what the other solutions are.
- 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*} \Bigl\langle 1, 0, u_x(x_0,y_0) \Bigr\rangle \end{equation*} is tangent to the curve \[ \bigl\{ \bigl( x, y_0, u(x,y_0) \bigr) : x_0 -\epsilon \lt x \lt x_0 + \epsilon \bigr\} \quad (\text{where} \quad \epsilon \gt 0 \quad \text{is small}) \] at the point $P_0$. Similarly, the vector \begin{equation*} \Bigl\langle 0, 1, u_y(x_0,y_0) \Bigr\rangle \end{equation*} is tangent to the curve \[ \bigl\{ \bigl( x_0, y, u(x_0,y) \bigr) : y_0 -\epsilon \lt y \lt y_0 + \epsilon \bigr\} \quad (\text{where} \quad \epsilon \gt 0 \quad \text{is small}) \] at the point $P_0$.
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Thus both vectors
\begin{equation*}
\Bigl\langle 1, 0, u_x(x_0,y_0) \Bigr\rangle \quad \text{and} \quad \Bigl\langle 0, 1, u_y(x_0,y_0) \Bigr\rangle
\end{equation*}
are tangent to the graph of the function $u$ at the point $P_0$.
The reasoning from the previous item applies to an arbitrary point $P = \bigl(x,y,u(x,y)\bigr)$ on the graph of the function $u.$ Therefore the vecors \begin{equation} \label{eq:tvxy} \Bigl\langle 1, 0, u_x(x,y) \Bigr\rangle \quad \text{and} \quad \Bigl\langle 0, 1, u_y(x,y) \Bigr\rangle \end{equation} are tangent to the graph of $u$ at the point $P$.
Two vectors in \eqref{eq:tvxy} are linearly independent. (This follows from the definition of linear independence. The only way to get the zero vector $\langle 0,0,0 \rangle$ as a linear combination of the vectors in \eqref{eq:tvxy} is with both coefficients equal to $0.$) Therefore, the tangent plane to the graph of $u$ at the point $P$ is spanned by the vectors in \eqref{eq:tvxy}. That is, a vector is tangent to the graph of the function $u$ at the point $P$ if and only if it is a linear combination of the vectors in \eqref{eq:tvxy}. - Since we assume that the function $u$ is a solution of \eqref{eq:pde} we have that \eqref{eq:pde} holds for points $(x,y)$ in the domain of $u$. That is we have \begin{equation} \label{eq:pdee} A\bigl(x,y,u(x,y)\bigr) \, u_x(x,y) + B\bigl(x,y,u(x,y)\bigr) \, u_y(x,y) = C\bigl(x,y,u(x,y)\bigr). \end{equation}
- The following step might look a little bit like a Deus ex machina action. But it is like that sometimes in math. Consider the vector field \begin{equation} \label{eq:vf} (x,y,z) \mapsto \Bigl\langle A(x,y,z), B(x,y,z), C(x,y,z) \Bigr\rangle. \end{equation} Next consider the vectors in this field that have tails at the points on the graph of the function $u,$ that is the vectors \begin{equation} \label{eq:vfu} \Bigl\langle A\bigl(x,y,u(x,y)\bigr), B\bigl(x,y,u(x,y)\bigr), C\bigl(x,y,u(x,y)\bigr) \Bigr\rangle. \end{equation} Next we use \eqref{eq:pdee} to establish a connection between the vector in \eqref{eq:vfu} and the vectors in \eqref{eq:tvxy}. But, for the sake of brevity of the notation we set $P = \bigl(x,y,u(x,y)\bigr)$ and write \eqref{eq:pdee} as \begin{equation*} A(P) \, u_x(x,y) + B(P) \, u_y(x,y) = C(P) \end{equation*} and \eqref{eq:vfu} as \begin{equation*} \Bigl\langle A(P), B(P), C(P) \Bigr\rangle. \end{equation*} Now we calculate \begin{align*} \Bigl\langle A(P), B(P), C(P) \Bigr\rangle & = \Bigl\langle A(P), B(P), A(P) \, u_x(x,y) + B(P) \, u_y(x,y) \Bigr\rangle \\ & = \Bigl\langle A(P), 0, A(P) \, u_x(x,y) \Bigr\rangle + \Bigl\langle 0, B(P), B(P)\, u_y(x,y) \Bigr\rangle \\ & = A(P) \Bigl\langle 1, 0, u_x(x,y) \Bigr\rangle + B(P) \Bigl\langle 0, 1, u_y(x,y) \Bigr\rangle. \end{align*} Notice that for the first equality in the preceding calculation we used \eqref{eq:pdee} and for the second and the third we used the vector algebra. To celebrate what we calculated we write: \begin{equation} \label{eq:vfut} \Bigl\langle A(P), B(P), C(P) \Bigr\rangle = A(P) \bigl\langle 1, 0, u_x(x,y) \bigr\rangle + B(P) \bigl\langle 0, 1, u_y(x,y) \bigr\rangle. \end{equation} Equality \eqref{eq:vfut} states that the vector $\Bigl\langle A(P), B(P), C(P) \Bigr\rangle$ is a linear combination of the vectors in \eqref{eq:tvxy}. Therefore, the vector $\Bigl\langle A(P), B(P), C(P) \Bigr\rangle$ is tangent to the graph of the function $u.$
- The conclusion from the preceding item is the following: 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.
- 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 we 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}(s) = A\bigl(X(s),Y(s),Z(s)\bigr) \\ \dfrac{dY}{ds}(s) = B\bigl(X(s),Y(s),Z(s)\bigr) \\ \dfrac{dZ}{ds}(s) = C\bigl(X(s),Y(s),Z(s)\bigr) \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} The ordinary differential equations in \eqref{eq:sys} are called the characteristic equations for partial differential equation \eqref{eq:pde}. 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 depends 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 $\xi$ with $s$ being the independent variable the preceding three equations represent a curve in $xyz$-space. These curves are called the characteristics of the partial differential equation \eqref{eq:pde} subject to the initial condition \eqref{eq:ic}. Now we can vary $\xi.$ By varying $\xi$ we get a set of curves in $xyz$-space that all start at points on the curve \eqref{eq:ic-curve}. The union of all the these curves 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. However, we still do not have the algebraic formula for the solution since the surface is given by its parametric equations depending on $s$ and $\xi.$
- Another set of curves that plays important role for the partial differential eqation \eqref{eq:pde} is the set of projected characteristics. The projected characteristics are the curves in $xy$-plane. The significance of the projected characteristics is as follows: From the equations of the characteristics, it is often "easy" to deduce how the solution $u(x,y)$ behaves along the projected characteristics.
- 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 \[ u(x,y) = Z\bigl( S(x,y), \Xi(x,y) \bigr). \]
- This Mathematica notebook contains the examples below worked out in Mathematica. 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.
- The characteristic equations are: \begin{equation*} \begin{array}{l} \dfrac{dX}{ds}(s) = 1 \\ \dfrac{dY}{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 \\ Y(0) = 0 \\ Z(0) =f(\xi) \end{array} \end{equation*}
- The solution of the above system is \begin{equation*} \begin{array}{l} X(s,\xi) = s +\xi \\ Y(s,\xi) = -s \\ Z(s,\xi) = f(\xi) \end{array} \end{equation*}
- Now solve the system \begin{equation*} \begin{array}{l} x = s +\xi \\ y = -s \end{array} \end{equation*} for $s$ and $\xi$ to get \begin{equation*} \begin{array}{l} s = - y \\ \xi = x + y. \end{array} \end{equation*}
- Finally, substitute the formulas for $s$ and $xi$ into $Z(s,\xi) = f(\xi)$ to get \[ u(x,y) = f(x+y) \] Since the domain of \(f\) is \(\mathbb{R}\), the domain of $u$ is the entire plane ${\mathbb R}^2$.
- The characteristic equations are: \begin{equation*} \begin{array}{l} \dfrac{dX}{ds}(s) = Y(s) \\ \dfrac{dY}{ds}(s) = 3 \\ \dfrac{dZ}{ds}(s) = -Z(s) \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 ${\mathbb R}^2$.
- The characteristic equations are \begin{equation*} \begin{array}{l} \dfrac{dX}{ds}(s) = 2 \\ \dfrac{dY}{ds}(s) = 1 \\ \dfrac{dZ}{ds}(s) = Z^2(s) \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) = \dfrac{\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 $Z(s,\xi) = \dfrac{\cos \xi}{1- s \cos \xi }$ to get \[ u(x,y) = \frac{\cos(x-2y)}{1- y \cos(x-2y)} \]
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The domain of the function $u(x,y)$ is complicated. Notice that for $y \in [0,1)$ we have $y \cos(x-2y) \lt 1$. Therefore, the function $u(x,y)$ is defined on the infinite rectangle
\[
\mathbb{R} \times [0,1) = \bigl\{(x,y) \in \mathbb{R} : x \in \mathbb{R}, y \in [0,1) \bigr\}.
\]
Below is its graph on a part of this rectangle.
The above graph shows that the function $u(x,y)$ 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 $k \in \mathbb Z$.
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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
- 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.
- My thinking about Burgers' Equation is on the website Burgers' Equation.
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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 \in \mathbb{R}.
\end{equation*}
You can try $f(x) = x$, $f(x) = x^2$, $f(x) = \cos x$ as some specific examples.
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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 $f(x) = x^2$, $f(x) = (\sin x)^2$, $f(x) = \cos x$ as some specific examples.
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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 $f(x) = x^2$, $f(x) = (\sin x)^2$, 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 \theta\in [0, 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 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*}
Apply the Method of Characteristics to give yourself enough tools to answer the questions below.
- Does there exist an open and convex 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 defined on $U$?
- Can you modify the boxed formula in the statement of the problem in such a way that your explorations in part a. lead to the solution of such a modified problem? State explicitly the problem and its solution.
-
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 y \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 y \in \mathbb R.
\end{equation*}
-
Solve the following first order PDE
\begin{equation*}
x \, u_x(x,y) + y \, u_y(x,y) = 2 u(x,y) \ln \bigl( u(x,y) \bigr) \quad \text{in} \quad (x,y) \in {\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*}
-
Let \(c \gt 0\) be arbitrary. Let
\[
f:[0,+\infty) \to \mathbb{R} \quad \text{and} \quad h:[0,+\infty) \to \mathbb{R}
\]
be given differentiable functions such that
\[
f(0) = h(0) \quad \text{and} \quad c f'(0) = - h'(0)
\]
Consider the following quasi-linear first-order PDE
\begin{equation*}
c \mkern 2mu u_x(x,y) + u_y(x,y) = 0 \quad \text{on} \quad \bigl\{ (x,y) \in {\mathbb R}^2 : x \geq 0, \ y \geq 0 \bigr\},
\end{equation*}
subject to the condition
\begin{align*}
u(x,0) & = f(x) \quad \text{for} \quad x \geq 0, \\
u(0,y) & = h(y) \quad \text{for} \quad y \geq 0.
\end{align*}
- Solve the given PDE. Notice that the solution will be a piecewise defined function with the domain \[ \bigl\{ (x,y) \in {\mathbb R}^2 : x \geq 0, \ y \geq 0 \bigr\}. \]
- Prove that the partial derivatives $u_x(x,y)$ ang $u_y(x,y)$ of the solution $u(x,y)$ provided in the preceding item are defined on the open quadrant \[ \bigl\{ (x,y) \in {\mathbb R}^2 : x \gt 0, \ y \gt 0 \bigr\} \] and satify the given PDE on this open quadrant.
- Illustrate the solution in Mathematica with the functions \[ f(x) = h(x) = \cos x \quad x \in [0,+\infty). \] and \[ f(x) = - \sin x, \qquad h(y) = c \sin y \quad x,y \in [0,+\infty). \] For $c \gt 0$, consider several different values, for example $c \in \{1/2, 1, 2\}.$
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Consider the following initial value problem
\[
x \frac{\partial u}{\partial x}(x,y) - y \, \frac{\partial u}{\partial y}(x,y) = u(x,y), \qquad u(x,1) = \sin x, \ \ x \in \mathbb{R}.
\]
- Use the method of characteristics to solve the given initial value problem.
- What is the largest domain in which the solution of this problem is defined?
- Give an explicit formula for the solution $u(x,y)$.
- Think of the variable \(y\) as being time. Explain how the solution \(u(x,y)\) evolves in time. That is, consider \(u(x,y)\) as a function of \(x\) and explain how this function changes with increasing \(y \gt 1\) and decreasing time \(y \lt 1\). Please pay special attention to the maximums, minimums, and zero values of the function \(u(x,y)\) at a specific time \(y\) with \(y\) a large positive number and \(y\) a small positive number.