My Math Forum Manipulating Differentials: Method of Characteristics

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 December 11th, 2017, 01:24 PM #1 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Manipulating Differentials: Method of Characteristics I've been looking over some papers on the Method of Characteristics but I've found something I'm not sure about. For quasi-linear equations of the form $$a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)$$ I have found two methods. One involves writing $$\frac{\mathrm dx}{a(x,y,u)} = \frac{\mathrm dy}{b(x,y,u)} = \frac{\mathrm du}{c(x,y,u)}$$ and then combining pairs of the three terms to produce two differentials that are equal to zero and thus represent characteristics. Thus, for the equation $$x u_x + y u_y = 2xy \quad u = 2 \, \text{on} \, y=x^2$$ we write $$\frac{\mathrm dx}{x} = \frac{\mathrm dy}{y} = \frac{\mathrm du}{2xy}$$ and then an easy starter for ten is \begin{align*} \frac{\mathrm dx}{x} = \frac{\mathrm dy}{y} \implies \frac{\mathrm dx}{x} - \frac{\mathrm dy}{y} &= 0 \\ y\,\mathrm dx - x\,\mathrm dy &= 0 \implies \mathrm d (\tfrac{y}{x}) = 0 \end{align*} For the second, we must use the $\mathrm du$ term getting, for example \begin{align*} \frac{\mathrm dx}{x} = \frac{\mathrm du}{2xy} \implies \frac{\mathrm dx}{x} - \frac{\mathrm du}{2xy} &= 0 \\ 2xy\,\mathrm dx - x\,\mathrm du &= 0 \end{align*} Now, I need to get $\mathrm d(xy-u)=0$ (in order to match the other method which gives a solution that works in the equation). The only way I can see to get there is to write \begin{align*} 2xy\,\mathrm dx - x\,\mathrm du &= 0 \\ 2y\,\mathrm dx - \mathrm du &= 0 \\ 2x\tfrac{y}{x}\,\mathrm dx - \mathrm du &= 0 \\ \left(2x\tfrac{y}{x}\,\mathrm dx + x^2\mathrm d (\tfrac{y}{x})\right) - \mathrm du &= 0 \\ \mathrm d(x^2\tfrac{y}{x}) - \mathrm du &= 0 \\ \mathrm d(xy - u) &= 0 \end{align*} Is that correct? Does that mean that I can never say that I'm holding variables constant in this process? That is, because $y$ is a variable in the equation it must be treated as one even though its differential doesn't appear in the equation? My solution is $$u(x,y) = xy + 2 - \frac{y^3}{x^3}$$ Last edited by skipjack; December 11th, 2017 at 02:01 PM.
December 11th, 2017, 01:50 PM   #2
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Bedtime here, but this may help.
Attached Images
 quasichar1.jpg (94.9 KB, 12 views) quasichar2.jpg (91.7 KB, 8 views)

 December 11th, 2017, 05:24 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Thanks Studiot, but those two pages didn't really tell me anything new, and nothing about manipulating differentials.
 December 11th, 2017, 07:52 PM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 383 Thanks: 206 Math Focus: Dynamical systems, analytic function theory, numerics I'm not 100% sure what you are asking. However, a differential for $y = f(x)$ is defined to be $dy = f'(x) dx$. This extends in the obvious manner to multiple varaibles via the multi-variate chain rule. As a result, the differential $dy$ should be thought of as a mapping on 2 variables of the form: $dy: \mathbb{R}^2 \to \mathbb{R}$ given by $(x,dx) \to dy$. Thus, even if $x$ or $y$ is fixed in a computation, $dx$ may not and thus a computation involving the differential may appear. Contrast this with the partial derivative from calculus. For instance, set $z= x^3 + x^2y^2$ and consider $\frac{\partial z}{\partial y}$ which is obtained by fixing $x$. Nevertheless, this partial is given by $z_y = 2x^2y$ and involves terms with an $x$ in them. The interpertation is that $z_y$ is a linear operator acting on vectors in $\mathbb{R}^2$. The action of his operator on an arbitrary vector depends on $x$ as it should.
December 12th, 2017, 09:52 AM   #5
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Quote:
 Originally Posted by v8archie ... Method of Characteristics... $$a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)$$ One involves writing $$\frac{\mathrm dx}{a(x,y,u)} = \frac{\mathrm dy}{b(x,y,u)} = \frac{\mathrm du}{c(x,y,u)}$$

Solution is a surface: $\displaystyle u=u(x,y)$, or $\displaystyle F(x,y,u)=u(x,y)-u=0$
An increment of displacement on the surface has to satisfy dF=0, from which $\displaystyle N=(u_{x},u_{y},-1)$ is normal to surface.
You are given $\displaystyle A\cdot N=0$ so that $\displaystyle A=(a,b,c)$ is tangent to surface. A solution curve on the surface has a tangent parallel to A, so that the equation of a solution (characterestic) curve is
$\displaystyle \frac{dx}{a}=\frac{dy}{b}=\frac{dz}{c}$

Now you can start thinking about the problem. For example, if a,b,c are constants, what is the equation of a solution (characteristic) curve? A straight line on the solution surface.

Ref: https://en.wikipedia.org/wiki/Method_of_characteristics

EDIT: Basically you have reduced the problem to asking, given the tangent to a curve, what is the curve? Hmm, a nice thread topic.

Last edited by zylo; December 12th, 2017 at 10:01 AM.

 December 12th, 2017, 06:25 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Thanks for the replies, but they all seem to be telling me what I already know - the basics of the theory. I'm really looking for some insight on manipulating the differentials. This is the source for the method: Method of Characteristics (University of Alberta). What the file doesn't give in any detail is any method, strategy or techniques for manipulating the equalities. I personally prefer the (apparently) more standard method of creating a system of PDEs, but I can see that they have differing strengths and weaknesses. Each is better for certain types of problem.
 December 13th, 2017, 07:47 AM #7 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 $\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{2xy}$ $\displaystyle \frac{dx}{x}=\frac{dy}{y} \rightarrow y=c_{1}x$ $\displaystyle \frac{dx}{x}=\frac{du}{2xy}\rightarrow \frac{dx}{x}=\frac{du}{2c_{1}x^{2}} \rightarrow x^{2}= \frac{u}{c_{1}}+c_{2}$ but $\displaystyle c_{1}=\frac{y}{x}$, so that $\displaystyle u=xy+c\frac{y}{x}$ Ref: Thanks from v8archie
December 13th, 2017, 06:20 PM   #8
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Quote:
 Originally Posted by studiot Bedtime here, but this may help.
Looking at the first question on Studiot's pages.

We have $$u_x - 6u_y=y$$
This has characteristic curves $6x+y=c$ of which two of the boundary conditions are examples. This seems awfully strange.

I assume that, in the first, $y = -6x + 2$ having \begin{align*}\frac{\mathrm du}{\mathrm dx} = u_x + \frac{\mathrm dy}{\mathrm dx}u_y &= y \\ u' &= -6x+2 \\ u &= -3x^2 + 2x + c \end{align*}
Means that there is no solution with $u=e^x$ on the given characteristic.

The same would be true for the third \begin{align*}\frac{\mathrm du}u' &= -6x \\ u &= -3x^2 + c \end{align*}
Where $u = -4x$ is not possible.

The calculation for the second seems extremely messy.

 December 13th, 2017, 11:40 PM #9 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 $\displaystyle u_x-6u_y=y$ The characteristics system of ODEs is : $\displaystyle \quad\frac{dx}{1}=\frac{dy}{(-6)}=\frac{du}{y}$ A first family of characteristic curves comes from : $\displaystyle \quad\frac{dx}{1}=\frac{dy}{(-6)} \quad\to\quad y+6x=c_1$ A second family of characteristic curves comes from : $\displaystyle \quad\frac{dy}{(-6)}=\frac{du}{y} \quad\to\quad u+\frac{y^2}{12}=c_2$ The general solution expressed in the form of implicit equation is : $\displaystyle \quad\Phi\left((y+6x)\:,\: (u+\frac{y^2}{12}) \right)=0$ $\displaystyle \Phi$ is any differentiable function of two variables. Or equivalently in explicit form : $\displaystyle u+\frac{y^2}{12}=F(y+6x)$ $\displaystyle F$( ) is any differentiable function. $\displaystyle u(x,y)=-\frac{y^2}{12}+F(y+6x)$ Last edited by skipjack; December 14th, 2017 at 03:15 AM.
 December 14th, 2017, 04:23 AM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Why is everybody so keen to tell me what I already know? Yes, $$\displaystyle u(x,y)=-\frac{y^2}{12}+F(y+6x)$$ is the general solution, but that doesn't address the point that I was making that the boundary conditions appear to be given along characteristic curves and are not solutions of the ODE along those curves.

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