Manipulating Differentials: Method of Characteristics

v8archie

Math Team
Dec 2013
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Colombia
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 $$u(x,y) = xy + 2 - \frac{y^3}{x^3}$$
 
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v8archie

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Dec 2013
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Thanks Studiot, but those two pages didn't really tell me anything new, and nothing about manipulating differentials.
 

SDK

Sep 2016
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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.
 
Mar 2015
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... 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:

v8archie

Math Team
Dec 2013
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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.
 
Mar 2015
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\(\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{2xy}\)

\(\displaystyle \frac{dx}{x}=\frac{dy}{y} \rightarrow y=c_{1}x\)

\(\displaystyle \frac{dx}{x}=\frac{du}{2xy}\rightarrow \frac{dx}{x}=\frac{du}{2c_{1}x^{2}} \rightarrow x^{2}= \frac{u}{c_{1}}+c_{2}\)

but \(\displaystyle c_{1}=\frac{y}{x}\), so that

\(\displaystyle u=xy+c\frac{y}{x}\)


Ref: [youtube]LpHqrlrU5pM[/youtube]
 
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v8archie

Math Team
Dec 2013
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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.
 
Aug 2011
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\(\displaystyle u_x-6u_y=y\)
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.
\(\displaystyle u(x,y)=-\frac{y^2}{12}+F(y+6x) \)
 
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v8archie

Math Team
Dec 2013
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Colombia
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.