Plese help me to find the solution of te following equation.

For values of $x$ in the interval $[-2,3]$ and $t>0$ we consider the one way wave equation $$u_t+u_x=0$$ with initial data \begin{align*} u(0,x)= \left\{\begin{array}{ll}1-|x|\;\; \text{if}\; 0\le |x|\le 1\\0\;\;\;\text{otherwise} \end{array} \right.\end{align*} and boundary data $u(t,-2)=0.$

I tried to use the method of characteristics in the following way:

$\frac{dx}{ds}=1$ and $\frac{dt}{ds}=1$

but i didn't manage to find the solution.

Please help me to do so. Thanks

  • $\begingroup$ What do you mean, "didn't manage to find the solution"? Can you solve those ODEs? $\endgroup$ – Calvin Khor Feb 2 at 21:37

I guess that a graphical answer is more in the spirit of this exercise. By using the method of characteristics, we find the set of curves $x = t + x_0$ along which $u$ is constant. These curves are represented in the $x$-$t$ plane below:


The boundary is represented by the thick black line. All curves starting on the boundary carry the value $u=0$, besides the curves in the red area defined by $-1\leq x-t \leq 1$ which carry the value $1 - |x-t|$. Therefore the solution for $x \in [-2,3]$, $t>0$ reads $$ u(x,t) = \left\lbrace \begin{aligned} &1 - |x-t| && \text{if}\quad {-1}+t\leq x \leq 1+t,\quad (0\leq t\leq 4) \\ &0 && \text{otherwise} \end{aligned}\right. $$


The correct method of characteristics, follows as :

$$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}x}{1} = \frac{\mathrm{d}u}{0}$$

Now, the first characteristic curve is given as :

$$\mathrm{d}t = \mathrm{d}x \implies u_1 = x-t$$

Note that the solution for the second one, is trivial, as :

$$\frac{\mathrm{d}u}{0} = \text{(something)} \implies u_2 = u$$

Now, the general solution shall be given as a $C^1$ function $F$, such that :

$$u_2=F(u_1) \Leftrightarrow u = F(x-t) \equiv u(x,t)$$

That's straighforward and doable since $u_2 = u \equiv u(x,t)$.

By applying the initial conditions given, are you now able to form the solution for the given IVP ?

  • $\begingroup$ The solution that i find is $u(x,t)=\begin{align*} u(0,x)= \left\{\begin{array}{ll}1-|x-t|\;\; \text{if}\; 0\le |x-|\le 1\\0\;\;\;\text{otherwise} \end{array} \right.\end{align*}$ but in my answer i didn't used the fact that $x\in[-2,3]$ nor the boundary condition $u(t,-2)=0$ that is way i am comfused. Can you please show me how shoud i use those conditions? thanks. $\endgroup$ – user640960 Feb 2 at 21:42
  • $\begingroup$ You would have to check whether the solution you have found actually satisfies the extra boundary condition. If not, your system would have no solution. In this case it does, see my answer. $\endgroup$ – Ivo Terek Feb 2 at 21:52
  • $\begingroup$ First, it should be "if $|x-1| \leq 1$. Secondly, substitute your other given conditions and see if it satisfies that, namingly if $u(t,-2) = 0$ indeed implies the given conditions and vice versa. $\endgroup$ – Rebellos Feb 2 at 21:55

If $\nabla u =(u_t,u_x)$, the equation says $u_t+u_x=0$ becomes $\nabla u \cdot (1,1) = 0$. In other words, the directional derivative of $u$ in the direction of $(1,1)$ is zero, meaning that $u$ is constant along this line. Solving $$\frac{{\rm d}x}{{\rm d}t} = \frac{1}{1}=1 \implies x = t+c,$$or equivalently, $t = x-c$. We can find the value of $u(x-c, x)$ by moving along the characteristic curve, choosing values for $x$ and doing $$u(t,x) = u(x-c, x) = u(0,c) = \begin{cases} 1-|c|, \mbox{ if } |c| \leq 1, \\ 0, \mbox{ else}.\end{cases}$$Solving for $c$, we get $c = x-t$, so $$u(t,x) = \begin{cases} 1-|x-t|, \mbox{ if } |x-t| \leq 1, \\ 0, \mbox{ else}.\end{cases}$$Now, in particular you have $$u(t,-2) =\begin{cases} 1-|t+2|, \mbox{ if } |t+2| \leq 1, \\ 0, \mbox{ else}.\end{cases}$$But $|t+2|\leq 1$ is equivalent to $-3 \leq t \leq -1$, which never happens since you're assuming $t>0$. Then we conclude that $u(t,-2) = 0$ as wanted.

  • $\begingroup$ I think this is literally the same answer as mine but expanding it on the case question that was left for the OP to conclude, since he showed no considerable effort. $\endgroup$ – Rebellos Feb 2 at 21:54
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    $\begingroup$ I do not follow why you have to check that $u(t,-2)=0$. It should have been already baked into the initial data, and in particular since the characteristic lines coming from the line parallel to the $t$-axis $(t,-2)$ never cross those coming from the $x$-axis, you can prescribe literally anything on the line $(t,-2)$, and the method of characteristics will give you a solution (suitably interpreted, OP's problem doesn't have a differentiable solution anyway) $\endgroup$ – Calvin Khor Feb 2 at 22:03

The initial data and boundary data for the PDE together form initial data for the characteristic equations, prescribed on an "L" shaped curve, infinite upwards that we shall call $L$. Since the equation is $\binom{1}{1} \cdot \nabla u = 0$ and $\binom{1}{1}$ is never tangent to $L$, this means that $L$ is non-characteristic, so the method of characteristics will work. Let's parameterise $L$ as $$ L: (-\infty,3] \to \mathbb R^2, $$ $$ L(z) = \begin{cases} \binom{0}{z} & z\in[-2,3] \\ \binom{-(z+2)}{-2} & z\le -2\end{cases}$$ So combining the "initial data" and the "boundary data" we obtain the equivalent data for $z\in(-\infty,3]$, $$ u(L(z)) = b(z) := \begin{cases} 1-|z| & |z|\le 1 \\ 0 & |z|>1\end{cases}$$ The characteristic equations are

$$ \partial_s \binom{t}{x} = \binom{1}{1} ,\quad \binom{t(0,z)}{x(0,z)} = L(z),$$ $$ \partial_s y = 0, \quad y(0,z) = b(z)$$ and the solution is given by $$ u(t(s,z),x(s,z) ) = y(s,z)$$ The setup is done and the rest is mechanical work.


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