Consider the P.D.E. $u_x + u_y = 1$ subject to the initial condition $u(x, y) = h(x, y)$ for $(x, y) ∈ Γ$ where $Γ$ is a given smooth curve and $h : Γ → \mathbb{R}$ is a given smooth function.

a. Find a smooth initial curve $Γ$ passing through the origin and a smooth function $h : Γ →\mathbb{R}$ such that a solution to the problem exists in a neighborhood of every point of $Γ$ except the origin. Verify this non-existence at the origin for your example.

b. Now find an initial curve $Γ$ and function $h : Γ → \mathbb{R}$ such that the problem has infinitely many solutions. Describe explicitly this infinite family of solutions.

I know that this is a 1st order PDE so I have to use method of characterization.

So far what I got is $\displaystyle \frac{\partial X}{\partial \tau }=1,\frac{\partial Y}{\partial \tau }=1,\frac{\partial Z}{\partial \tau }=1$ and $X(s,0)=x_0(s),Y(s,0)=y_0(s),Z(s,0)=h(x_0(s),y_0(s))$.

So, $X(s,\tau)=\tau+x_0(s),Y(s,\tau)=\tau+y_0(s),Z(s,\tau)=\tau+h(x_0(s),y_0(s))$

That is what my teacher taught me. I don't know how to proceed now toward part a. I really want to learn what is going on there and how to handle this kind of problems. Any help is appreciated.

  • $\begingroup$ I was going to post an answer but then a doubt popped up in my mind: have you ever seen the characteristic ODE? (I think the answer is 'no', but I want to be sure) $\endgroup$ – Uskebasi Apr 10 '17 at 16:19
  • $\begingroup$ No. What is that? May be I know it but not by that name. $\endgroup$ – Extremal Apr 10 '17 at 16:25
  • $\begingroup$ Other tiny question: does the term noncharacteristic boundary condition say something to you? $\endgroup$ – Uskebasi Apr 10 '17 at 16:31
  • $\begingroup$ I have no idea about the things you are mentioning. Sorry about that. $\endgroup$ – Extremal Apr 10 '17 at 16:46
  • $\begingroup$ I think it will be difficult for you to have intuition on "where to look for a curve that solves point a)" without the concept of noncharacteristic boundary condition (or, at least, it would be difficult for me). The first ten pages of this should explain you what's happening and give you intuition on how should a curve satisfying a) look like. $\endgroup$ – Uskebasi Apr 10 '17 at 17:23


  • Change coordinates $x=s+t,y=s-t$ to new coordinates $(s,t)$. Then the PDE becomes $u_s=1$.

  • Next change variable $u(s,t)=s+v(s,t)$ and $h(s,t)=s+k(s,t)$. Then the PDE becomes $v_s=0$. In other words, solutions $v(s,t)$ can only depend on $t$.

  • Exercise (a): Choose $\Gamma$ to be the graph of $s\mapsto \exp(-1/s^2)\sin(1/s)$ and $k(s,t)=s$. [The main idea is that $\Gamma$ then is of class $C^{\infty}$ and intersects the $s$-axis infinitely many times in all neighborhoods of the origin. Hence the $s$-independent $v$-solution is forces to depend on $s$. Not possible.]

  • Exercise (b): Choose $\Gamma$ to be the $s$-axis with $k(s,t)=k_0$ a constant. The full solution is $v(t)$ such that $v(0)=k_0$.


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