Author Topic: characteristic vs. integral lines  (Read 3489 times)

Thomas Nutz

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characteristic vs. integral lines
« on: September 21, 2012, 05:57:04 PM »
Heythere,
I am confused by the terms characteristic lines and integral lines. The book introduces characteristic lines as the curves along which a function is constant. Now in the notes integral lines are curves to which the vector field is tangential, i.e. in the case of the gradient vector field the lines along which the function changes most (in abs. value).

So I thought these two should be orthogonal in the case of $au_t+bu_x=0$.

The characteristic lines of u are given by $x=\frac{b}{a}y+c$, but this is also the solution to the ODE
$$\frac{dt}{a}=\frac{dx}{b}$$, which is given for the integral lines in the notes (First order PDEs)...

So are integral lines the same as characteristic lines?

Thomas Nutz

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Re: characteristic vs. integral lines
« Reply #1 on: September 21, 2012, 05:59:31 PM »
havin some trouble there... $$x(t)=\frac{b}{a}t+c$$

Victor Ivrii

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Re: characteristic vs. integral lines
« Reply #2 on: September 21, 2012, 06:36:18 PM »
Heythere,
I am confused by the terms characteristic lines and integral lines. The book introduces characteristic lines as the curves along which a function is constant. Now in the notes integral lines are curves to which the vector field is tangential, i.e. in the case of the gradient vector field the lines along which the function changes most (in abs. value).

So I thought these two should be orthogonal in the case of $au_t+bu_x=0$.
So are integral lines the same as characteristic lines?

If we consider 1-st order PDEs in the form
\begin{equation}
a_0\partial_t u + a_1\partial_x u + a_2\partial_y u=0
\label{eq-1}
\end{equation}
then characteristics of the equation (\ref{eq-1}) are integral lines of the vector field $(a_0,a_1,a_2)$ i.e. curves
\begin{equation}
\frac{dt}{a_0}=\frac{dx}{a_1}=\frac{dy}{a_2}.
\label{eq-2}
\end{equation}
There could be just two variables $(t,x)$ or more ... and coefficients are not necessary constant. Yes, for (\ref{eq-1}) characteristics are lines along which $u$ is constant.

But we preserve the same definition of characteristics as integral lines for more general equation f.e.
\begin{equation}
a_0\partial_t u + a_1\partial_x u + a_2\partial_y u =f (t,x,y,u)
\label{eq-3}
\end{equation}
and here $u$ is no more constant along characteristics but solves ODE
\begin{equation}
\frac{dt}{a_0}=\frac{dx}{a_1}=\frac{dy}{a_2}=\frac{du}{f}.
\label{eq-4}
\end{equation}
Further, notion of characteristics generalizes to higher order equations. Definition curves along which solution is constant goes to the garbage bin almost immediately.


Miranda Jarvis

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Re: characteristic vs. integral lines
« Reply #3 on: September 22, 2012, 12:53:38 AM »
I feel like I understand what is being said here, but I am confused by the second set of lecture notes,  specifically equation (7) which seems to be doing the same thing as equation (4) in the reply to this post except with a + sign instead of a equal. Am I misunderstanding and these are actually different scenarios, or is one a misprint?

Victor Ivrii

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Re: characteristic vs. integral lines
« Reply #4 on: September 22, 2012, 02:20:38 AM »
I feel like I understand what is being said here, but I am confused by the second set of lecture notes,  specifically equation (7) which seems to be doing the same thing as equation (4) in the reply to this post except with a + sign instead of a equal. Am I misunderstanding and these are actually different scenarios, or is one a misprint?

Miranda, thanks!  There was a misprint (mistype) in Lecture 2 (equation (7)), should be =.
Now it has been corrected.