The function $u(x,y)$ satisfies $u_y + u_x = 0$ in $x > 0$, $y > 0$ together with the initial condition $u(x, 0) = \sin(x)$, $x > 0$ and the boundary condition $u(0, y) = \sin(y)$, $y > 0$. Determine the values of $u$ in the whole quarter plane $x > 0$, $y > 0$.

My problem is that, although I think I know how to use the initial condition, I don't know how to use the boundary condition?

The characteristic equations are

$$\frac{dy}{dt} = 1, \frac{dx}{dt} = 1, \frac{du}{dt} = 0$$

Writing the initial condition parametrically (parameterising $t = 0$ as the initial condition), we get

$$\Gamma = \Gamma(s) = (x_0(s)), y_0(s), u_0(s)) = (s, 0, \sin(s))$$

The curve $\Gamma$ is called the initial curve.

$\therefore \frac{\partial y(t, s)}{\partial t} = 1, \frac{\partial x(t, s)}{\partial t} = 1, \frac{\partial u(t, s)}{\partial t} = 0$

with initial conditions

$$y = t + C_1(s)$$ $$\therefore y(0, s) = C_1(s) = 0$$

$$x = t + C_2(s)$$ $$\therefore x(0, s) = C_2(s) = s$$

$$u(0, s) = C_3(s) = \sin(s)$$

So our solution is given by

$$x(t, s) = t + s, y(t, s) = t, u(t, s) = \sin(s)$$

I'm not sure entirely sure how to do this, and my textbooks don't discuss how to use method of characteristics to solve problems that have initial conditions and boundary values. I would be very grateful for help.


1 Answer 1


The solutions to the PDE $u_x + u_y = 0$ are functions that are constant in the $(1,1)$ direction.

Recall how you use the initial data $u_0$ to define the solution: for any point $(x,y)$ you follow the characteristic line back to the initial condition at $y=0$ (so the point $(x-y,0)$), and since it was constant along the characteristic, the answer is $u(x,y) = u_0(x-y)$. Since the initial data is defined on $[0,\infty)$, can do this for all points $x\ge y$ i.e. points below or on the line $y=x$.

Now have a look at this graph of the boundary condition, initial condition and characteristic lines- enter image description here

How would you now define the values of the solution above the $y=x$ line?

(Full solution.) It seems that you don't have any issue using just the "initial condition", i.e. the points $\color{red}{\{(x,0): x\ge 0\}}$. You can also use the "boundary data" i.e. the points $\color{blue}{\{(0,y): y\ge 0\}}$ to define the full "initial curve"

$$ \Gamma = \color{blue}{\{ (0,y, u_{\text{bdry}}(y)) : y\ge 0 \}} \cup \color{red}{\{ (x,0, u_{\text{init}}(x)) : x\ge 0\}} $$ You can parameterise this using $s\in\mathbb R$, $$ \Gamma(s) = \begin{cases} (s,0, \color{red}{u_{\text{init}}(s)}) & s\in [0,\infty) \\ (0,-s, \color{blue}{u_{\text{bdry}}(-s)}) & s\in (-\infty,0] \end{cases}$$ with a compatability condition $\color{red}{u_{\text{init}}(0)} = \color{blue}{u_{\text{bdry}}(0)}$.

Now, we implement the method of characteristics. To find the characteristic curves $(x,y) = (x(t),y(t))$, we solve $$ \frac{d}{dt} u(x(t),y(t)) = 0$$ Of course, $ \frac{d}{dt} u(x(t),y(t)) = u_x(x(t)) x'(t) + u_y(y(t)) y'(t) $ so we just need $x'(t) = y'(t) = 1$. Thus all characteristic curves are given by $$ (x(t),y(t)) = (x_0,y_0) + (t,t) , \quad t\geq t_0 $$ where $t_0$ is the time we leave the set $\Gamma$. Using the same parameter $s$ as before, we determine $t_0$ using $x_0=x_0(s),y_0=y_0(s)$, \begin{align} s\ge 0 &\implies x_0 = s,y_0 = 0, t_0 = 0 \\ s<0 &\implies x_0 = 0, y_0 = -s, t_0 = -s\end{align}

and the full solution is $$u(s,t) = u(x(s,t),y(s,t)) = \begin{cases} \color{red}{u_{\text{init}}(s)} & s\ge 0, t\ge 0 \\ \color{blue}{u_{\text{bdry}}(-s)} & s < 0, t>-s \end{cases}$$

We have defined the solution on the set $\{ (s,t) : s \in \mathbb R, t \ge \max(-s,0) \} $, and they are related to $x,y$ by $$x = s+t , y=t $$ So that $s = x-y$, and it should be clear that this defines the region $x\geq 0$ and $y\ge 0$. In particular $\color{red}{x>y>0}$ iff $\color{red}{s>0}$, and $\color{blue}{y>x>0}$ iff $\color{blue}{s<0}$. Therefore, the answer in the original coordinates is $$ u(x,y) = \begin{cases} \color{red}{u_{\text{init}}(x-y)} & 0\le y\le x \\ \color{blue}{u_{\text{bdry}}(y-x)} & 0 \le x<y\end{cases}$$

(Final remark.) There are many different ways to write this down. However, the graph above captures the main idea that has to be expressed by all proofs of this fact. Regardless of whichever proof you learn, my opinion is that you should make sure that your understanding matches the graph.

  • $\begingroup$ I would say that the values of the solution above the $y = x$ line first intersect the $y$-axis, so I would use the condition $u(0, y) = \sin(y)$? I guess, in this case, these would be classified as the "initial condition" for those lines? Not sure if I'm on the right track here. $\endgroup$
    – user423167
    Aug 1, 2018 at 16:52
  • $\begingroup$ @handler'shandle Yes, exactly. The "boundary condition" is nothing more than an initial condition specified along a different curve. $\endgroup$ Aug 1, 2018 at 16:53
  • 1
    $\begingroup$ @handler'shandle I have added a full solution. $\endgroup$ Aug 1, 2018 at 20:19
  • 1
    $\begingroup$ @handler'shandle yes, this is the effect of the term $f$. $\endgroup$ Aug 2, 2018 at 7:15
  • 1
    $\begingroup$ @handler'shandle you're welcome! $\endgroup$ Aug 2, 2018 at 7:17

You must log in to answer this question.