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Consider the

Problem. Let $\gamma$ be a parameterized curve in $\mathbb{R}^{2}$ by $\gamma: I \to \Omega$, where $I$ is an interval of $\mathbb{R}$ and $\Omega$ an open in $\mathbb{R}^{2}$. Let $a,b,c: \Omega \to \mathbb{R}$ be given functions. Determine a function $\varphi(x,y)$ solution of the equation $$a(x,y)\frac{\partial \varphi}{\partial x} + b(x,y)\frac{\partial \varphi}{\partial y} = c(x,y) \tag{Eq 1}$$ where $\varphi(\gamma(t)) = \varphi_{0}(t)$ with $\varphi_{0}: I \to \mathbb{R}$ is a given function.

Proof. (just a loosely idea) Fixed a point $\gamma_{0} = \gamma_{0}(s_{0}) = \gamma_{0}(x_{0},y_{0})$ of $\gamma$, consider the curve $\Gamma(t) = (x(t),y(t))$ passing through $\gamma_{0}$, that is, $\Gamma(0) = \gamma_{0}$. Define $z(t) = \varphi(x(t),y(t))$ where $\varphi$ is a solution of Eq 1. If $\Gamma$ is differentiable, by the Chain Rule, $$\frac{dz}{dt} = \langle \Gamma'(t), \nabla\varphi(\Gamma(t)) \rangle = \frac{dx}{dt}\frac{d\varphi}{dx} + \frac{dy}{dt}\frac{d\varphi}{dy}.$$ Therefore, if $\Gamma$ satisfies the system of ODE $$\begin{cases} \frac{dx}{dt} = a(x,y),&x(0) = x_{0}\\ \frac{dy}{dt} = b(x,y),&y(0) = y_{0}, \end{cases}\tag{Sy 1}$$ we can the solution $\varphi$ solving $$\frac{dz}{dt} = c(x,y),\quad z(0)=\varphi(s_{0}).$$ If we repeat the previous argument for all points $\gamma(s)$, $s \in I$, we obtain a family of curves on which the solution $\varphi$ can be determined.

The solutions of the Sy 1 define a change of variables, that is, a function $$f: \mathbb{R}^{2} \to \mathbb{R}^{2}$$ $$(t,s) \mapsto (x,y)$$ and the Inverse Function Theorem ensures a solution if $(a,b)$ is transversal to curve $\gamma$ where $(a,b)$ represents $(x,y) \mapsto (a(x,y),b(x,y))$.



My objective is to apply the Method of Characteristics (the Problem) to solve the one-dimensional linear Wave Equation that is given by

$$\frac{\partial^{2} u}{\partial^{2} x} - c_{0}^{2}\frac{\partial^{2} u}{\partial^{2} y} = 0$$

and we write $$\frac{\partial^{2} u}{\partial^{2} x} - c_{0}^{2}\frac{\partial^{2} u}{\partial^{2} y} = \left(\frac{\partial }{\partial x} + c_{0}\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x} - c_{0}\frac{\partial }{\partial y}\right)u = 0.$$ Let $$\left(\frac{\partial }{\partial x} - c_{0}\frac{\partial }{\partial y}\right)u = v(x,y) = v.$$ Then it's enough to solve

$$\underbrace{\frac{\partial u}{\partial x} - c_{0}\frac{\partial u}{\partial y} = v}_{(1)}\quad\text{and}\quad\underbrace{\frac{\partial v}{\partial x} + c_{0}\frac{\partial v}{\partial y} = 0}_{(2)}$$

I broke the Wave Equation in (1) and (2) because the Problem is only for order $1$, since the partial derivatives are of order $1$. Then I think that solving the two equations, I'll find a solution for Wave Equation. This is the way that I found to use the Problem.

Now, I want to use the Problem for solve (2) and so, use the solution for (2) to solve (1).

I couldn't apply the above problem to get a solution to the equations (1) and (2). Hans Lundmark gave me a good reference, but in it, the author does the construction with some different details and my professor wants me to use the problem, exactly as it is, to get the solution. I wish someone could help me with this.

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    $\begingroup$ PDE by Pinchover Rubinstein? If that doesn't help, then perhaps I don't understand your question. $\endgroup$ – BCLC Sep 5 '18 at 3:04
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    $\begingroup$ I wrote the problem wrong. Corrected due to Hans observation. I'll read what you suggested! Thank you! $\endgroup$ – Lucas Corrêa Sep 5 '18 at 3:36
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There is a mistake in the equation that you wrote :

$$\frac{\partial^{2} u}{\partial^{2} x} + c\frac{\partial^{2} u}{\partial^{2} y} = \left(\frac{\partial }{\partial x} + c\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x} - c\frac{\partial }{\partial y}\right)u = 0.$$ The correct equation is : $$\frac{\partial^{2} u}{\partial^{2} x} - c^2\frac{\partial^{2} u}{\partial^{2} y} = \left(\frac{\partial }{\partial x} + c\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x} - c\frac{\partial }{\partial y}\right)u = 0.$$ And moreover it is true only if $c=$constant.

If $c=c(x,y)$ some terms with $\frac{\partial c}{\partial x}$ and $\frac{\partial c}{\partial y}$ would appear.

In the elementary case of $c=$constant one can separate $$\frac{\partial }{\partial x} + c\frac{\partial }{\partial y}=0$$ and $$\frac{\partial }{\partial x} - c\frac{\partial }{\partial y}=0$$ The method of characteristics leads to the solutions, respectively : $$u_1(x,y)=f(y+cx)$$ and $$u_2(x,y)=g(y-cx)$$ where $f$ and $g$ are independent arbitrary functions.

So, the general solution of $\frac{\partial^{2} u}{\partial^{2} x} - c^2\frac{\partial^{2} u}{\partial^{2} y} =0$ is : $$u(x,y)=f(y+cx)+g(y-cx)$$

If the equation is $\frac{\partial^{2} u}{\partial^{2} x} + c\frac{\partial^{2} u}{\partial^{2} y} =0$, the general solution is : $$u(x,y)=f(y+i\sqrt{c}\:x)+g(y-i\sqrt{c}\:x)$$

ADDITION after the discussion in the comments section :

Solving $\left(\frac{\partial }{\partial x} + c_{0}\frac{\partial }{\partial y}\right)u=0$ $$\begin{cases} \frac{dx}{dt}=1\:,\quad x(0)=x_0 \quad\text{leads to}\quad x=t+x_0 \\ \frac{dy}{dt}=c_0\:,\quad y(0)=y_0 \quad\text{leads to}\quad y=c_0t+y_0 \\ \frac{du}{dt}=0\:,\quad u(0)=u_0 \quad\text{leads to}\quad u=u_0 \end{cases}$$ A first family of characteristic curves comes from $x=t+x_0$ and $y=c_0t+y_0$ which are the parametric equation of $y=c_0(x-x_0)+y_0$ $$y-c_0x=c_1\quad\text{with}\quad c_1=y_0-c_0x_0.$$ A second family of characteristic curves comes from $\quad u=u_0=$constant$=c_2$

The general solution of the PDE $\left(\frac{\partial }{\partial x} + c_{0}\frac{\partial }{\partial y}\right)u=0$ expressed on the form of implicit equation is $$F(c_1,c_2)=F(y-c_0x\:,\: u)=0$$ where $F$ is an arbitrary function of two variables.

Or, equivalently on explicit form : $$u=g(y-c_0x)$$ where $g$ is an arbitrary function.

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  • $\begingroup$ Yeah, the signal was wrong! So, my doubt is in the application of the method of characteristic. I was unable to apply it as described in the problem. If you could help me with one of the two equations would be great. $\endgroup$ – Lucas Corrêa Sep 13 '18 at 7:04
  • $\begingroup$ I cannot understand your question. From what equations are you taking about ? $\endgroup$ – JJacquelin Sep 13 '18 at 7:45
  • $\begingroup$ $\frac{\partial}{\partial x} + c\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial x} - c\frac{\partial}{\partial y}$. Besides, should not I do $\left(\frac{\partial}{\partial x} - c\frac{\partial}{\partial y}\right)u(x,y) = v(x,y)$ before? $\endgroup$ – Lucas Corrêa Sep 13 '18 at 7:50
  • $\begingroup$ Sorry, I cannot understand exactly what your specific problem is. From what PDE those two equations are coming from? What is $c$, a function or a constant ? $\endgroup$ – JJacquelin Sep 13 '18 at 7:59
  • $\begingroup$ $c$ is a constant. These two equations came when I "broke" the wave equation in two parts. So I'm trying to use the system Sy 1, applying the method of characteristics, to find the solution of the wave equation from these two PDEs. So I called the first part of $ v $ to find a solution to equation (2). Then, use the solution of equation (2) to solve equation (1). $\endgroup$ – Lucas Corrêa Sep 13 '18 at 8:08
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I'm not sure I understand exactly what your specific problem is, but for a general description of the method of characteristics for the one-dimensional wave equation $$u_t+c(u)u_x=0$$ (which includes the linear case with $c(u)=c_0$ constant), I can recommend Chapter 2 of Whitham's Linear and Nonlinear Waves.

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    $\begingroup$ Yeah! I thought the general case follows the same idea. Because I only tried this case, I just wrote it. It was my mistake, it was really confusing. I will correct. Thanks for the book! $\endgroup$ – Lucas Corrêa Sep 5 '18 at 3:32

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