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Characteristic Eqn of PDE?

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Old   April 16, 2007, 16:43
Default Characteristic Eqn of PDE?
  #1
Bob
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Hello folks, I've come accross a paper that talks about a 2nd order quasi-linear PDE and discusses the "characteristic equation" of the pde. Could someone please explain what this characteristic equation represents.

The eqn I'm looking at is A*u_tt + 2B*u_xt = C*u_xx + u_yy

the authors state that the characteristic equation is C*t^2 - A*x^2 - (AC + B^2)*y^2 + 2B*x*t = 0

I have no idea where this is coming from. I'm hoping that this isn't a trivial problem and that I'm not wasting anyone's time asking stupid questions. I have looked through all my textbooks (of which I have quite a few) and can't find any reference that has been helpful. I'm pretty sure that this is related to the method of characteristics but haven't managed to verify this yet.

If anyone could point what's happening here, or refer me to a relevant text, then I'd be extremely grateful. B

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Old   April 17, 2007, 06:50
Default Re: Characteristic Eqn of PDE?
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Trinity
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this would be found in some math/numerical methods textbook which talks about PDE's. you are right, this IS related to the method of characteristics. Try Engineering Mathematics by Kreyzig or Anderson,Tannehill, Pletcher.
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Old   April 17, 2007, 07:29
Default Re: Characteristic Eqn of PDE?
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Bob
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Thanks for the references. Unfortunately I've already looked through both of these books and they've not been very useful (

I'm familiar with the method of characteristics for 2nd order equations involving x & y. The vast majority of the literature seems to focus on 1st order equations although I've found a few pieces that refer to 2nd order quasi-linear equations with 2 variables (x & y).

The equation I'm dealing with involves 3 variables (x, y and t) and I'm having difficulty determining the characteristics. I think that the problem is caused by the temporal terms and in determining the equation linking x, y and t.
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Old   April 17, 2007, 08:18
Default Re: Characteristic Eqn of PDE?
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Trinity
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oohh, now i see. well, as far as i remember, it being difficult to analyze an eqn in three variables, mostly only t and x are chosen. i think the nature of the eqn does not change by introducing y, since both x and y are spatial variables. this is for a general fluids problem, where there is flow symmetry in x and y, for example a problem where both x and y derivatives are equally important. So the method of characteristics can be used to get a 'feel' for the problem. Not sure whether you would be able to determine actual characteristics though.

However, if we are looking at a steady flow problem (no t), with one direction being more important than the other, for example boundary layers, then x and y are not equivalent. but once again, we are down to 2 variables, so we can use the method of characteristics to analyze the problem.
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Old   April 17, 2007, 11:05
Default Re: Characteristic Eqn of PDE?
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Tom
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You should be able to find the derivation in either volume 2 of Courant & Hilbert or the book on pdes by Garabedian. It's basically an equation for the eigenvalues of a matrix; i.e. the directions along which the partial differential equation becomes degenerate.
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Old   April 18, 2007, 05:33
Default Re: Characteristic Eqn of PDE?
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Tom
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"oohh, now i see. well, as far as i remember, it being difficult to analyze an eqn in three variables, mostly only t and x are chosen. i think the nature of the eqn does not change by introducing y, since both x and y are spatial variables."

Not quite - a counter example is (quasi)-steady compressible flow in 2D. Here you only have 2 variables (x,y) both of which are spatial. However the problem is elliptic for subsonic flow and hyperbolic for supersonic flows. This means that the two spatial directions are very different in the supersonic case; i.e. one of them is time-like! So spatial directions are not really equivalent when you change the order of a pde.
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Old   April 19, 2007, 08:03
Default Re: Characteristic Eqn of PDE?
  #7
Bob
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Thanks Tom that's very helpful.

I'm still a little confused, I've looked through the textbooks suggested and sadly they've not proven very helpful. If I take my original equation:

A*u_tt + 2B*u_xt = C*u_xx + u_yy (1)

and set the characteristic equation given by the original authors as some variable "eta":

eta=C*t^2 - A*x^2 - (AC + B^2)*y^2 + 2B*x*t (2)

Then does this mean that the general solution to (1) should be given by u(x,y,t) = function(eta) ?

When I tried this I found that the function(eta) did not satisfy the pde (1). I understand the theory of characteristics well enough in first order cases but feel I might be missing something here.
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Old   April 19, 2007, 11:50
Default A similar question
  #8
Frank
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Hello Forum Users, I'm sorry to piggy-back this query onto Bob's post but I was wondering if one of you kind people could perhaps help me with a similar problem.

I am trying to solve a similar equation, albeit one that has only a first derivative w.r.t time:

A*p_xt = B*p_xx + p_yy

and the 'characteristic equation' is given as B*k_x*k_x + k_y*k_y - A*k_x*k_t = 0

k is a variable that represents the 'characteristic plane'. If someone could please advise me as to what these terms mean, and in particular where the characteristic equation comes from, I'd be eternally grateful.

Once again I appologise for piggybacking a post but I feel the two queries are closely related.

F

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Old   April 19, 2007, 14:56
Default Re: Characteristic Eqn of PDE?
  #9
Tom
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As an example consider the simpler 2nd order problem (there's more algebra to do in the full problem which would be a pain to type in)

A.u_tt + 2B.u_xt - C.u_xx = 0. (1)

Now consider the curve t=t(s), x=x(s) on which

u_t(x(s),t(s)) = f(s) & u_x(x,y) = g(s)

where f and g are known functions. Differentiate these with respect to s to obtain ('=d/ds)

f' = t'.u_tt + x'.u_xx & g' = t'u_xt + x'.u_xx (2)

(2) along with (1) are 3 equations in (3) unknowns u_xx, u_tt & u_xt evaluated on x=x(s), t=t(s). This system will fail to be invertible when

A.(x')^2 - 2B.(x't') - C.(t')^2 = 0 (3)

Which is the definition of the characteristic directions - this is why the Cauchy problem requires that data is not prescribed on a characteristic!

Note that in this example dividing (3) through by (t')^2 gives

A.p^2 -2B.p -C = 0, where p=dx/dt. (4)

For higher order it's more messy, in your case the characteristics are surfaces and not lines. Most advanced books on pdes should have a discussion of this (Linear and Nonlinear Waves by G.B. Whitham has a short discussion of this).

The theory of characteristics is different for 2nd order and 1st order equations (although the basic idea is the same; i.e. degenaracy of the problem along a characteristic direction). Basically the roots to (4) can be distinct and real (hyperbolic), complex conjugate (elliptic) or a single real (repeated) root (parabolic).
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