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 December 11th, 2017, 01:24 PM #1 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 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: 7,683 Thanks: 2664 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: 645 Thanks: 408 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: 7,683 Thanks: 2664 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 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{2xy}$ $\displaystyle \frac{dx}{x}=\frac{dy}{y} \rightarrow y=c_{1}x$ $\displaystyle \frac{dx}{x}=\frac{du}{2xy}\rightarrow \frac{dx}{x}=\frac{du}{2c_{1}x^{2}} \rightarrow x^{2}= \frac{u}{c_{1}}+c_{2}$ 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: 7,683 Thanks: 2664 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. Tags characteristics, differentials, manipulating, method Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Meta Differential Equations 2 July 29th, 2014 05:19 AM razzatazz Calculus 1 March 24th, 2013 04:23 AM restin84 Algebra 2 March 30th, 2012 05:06 PM inventorybox Abstract Algebra 1 July 22nd, 2011 11:14 AM marioooo Applied Math 0 June 2nd, 2011 01:55 AM

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