Method of characteristics and shock waves

In summary: Thanks for your help, this is really confusing me :(In summary, this person is having difficulty understanding how to solve PDEs with the method of characteristics. They say that they understand the most important part (the ICs), but when it comes to actually solving problems, they only know the steps. They are lost when it comes to the t and x comparison, and the limits for the s(t) equation. They are trying to find help from someone who knows the method of characteristics, but is having difficulty understanding themselves.
  • #1
Leb
94
0
Hi all!

I just wanted to ask if anyone is finding the usage of method of characteristics difficult ? I sort of feel, that it is a very simple approach to solving PDE's, but I get easily lost, when for instance, we have to keep switching back and forth between variables and such. When it comes to introducing shocks, I get lost even more (I think my problem is, I am not comfortable with all these substitutions and expressions for time and speed etc).

I sort of get the idea behind it ( the most important part are the ICs and you can get any value for u in x-t plane etc). But when it comes to solving actual problems, I can only do basic ones and with "pretend" understanding (i.e. I think I know what I am doing, but in reality I only know the steps).
 
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  • #2
What source are you learning from? I've found that most sources on the method of characteristics are terrible at explaining what's going on.
 
  • #3
My university lecture course and a few notes online (I only find them somewhat useful). I got the intuition from this guy here but when it comes to more complex problems my intuition brakes down. As I said, I don't know why, but it really LOOKS as a very easy thing and that's annoying me greatly. Maybe I just need to practice more to get getter basics...
 
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  • #4
I did my MPhil thesis is shock waves and I have a lot of experience with the method of characteristics, what exactly do you need help with?
 
  • #5
A very simple problem that is probably the key problem for me is the fact that x and t are easily compared and interchanged. I do not understand how one can compare time and distance. For example saying u = 1 for x<-t. Should I somehow think in 3D ?

This is a very simple problem with the t and x comparison.
http://s12.postimg.org/xv0wzdcn1/Chars.jpg
 
  • #6
We need the actual equation to be able to give you some advice.
 
  • #7
Sorry, I have not realized that did not fit.

It's [itex]u_{t}+uu_{x}=0[/itex] and u(x,0) = -1 for x<0 and u(x,0) = 1
 
  • #8
This states that u is constant on the characteristics.
 
  • #9
I think I get that (and I thought that was the whole point of characteristics, that u is the same on a characteristic). What I do not get is how we compare time t and space x...
 
  • #10
You have to compute the characteristics, write dt/ds=1, dx/ds=u. with t(0)=r and x(0)=0.
 
  • #11
Yes, I know how to algebraically get the expressions in the picture of the solutions I have attached. What I do not understand is why are we allowed to compare x and t ?
 
  • #12
You're mapping the characteristics.
 
  • #13
Not sure what you mean there, but I will pretend to understand :)

I'm having trouble with this question:
shock1.jpg


The "shock location is obvious" for me it is only obvious if I draw the picture, maybe I should look at the limits ?

But the real problem to me is understanding how were the limits for s(t) (underlined red) established ?
 
  • #14
Do you know how to compute characteristics? If I have a hyperbolic PDE, I can write down the characteristics as
[tex]\frac{dt}{ds}=1\quad\frac{dx}{dt}=u,\quad\frac{du}{ds}=0[/tex]
Then consider what [itex]u[/itex] is in those separate regions.
 
  • #15
We'll take this one step at a time. Solving PDEs of this worm is handle turning, I'll guide you through the process.
 
  • #16
hunt_mat said:
Do you know how to compute characteristics? If I have a hyperbolic PDE, I can write down the characteristics as
[tex]\frac{dt}{ds}=1\quad\frac{dx}{dt}=u,\quad\frac{du}{ds}=0[/tex]
Then consider what [itex]u[/itex] is in those separate regions.

Solving your given Monge equations:
[itex]t = s + t_{0} ; x = ut + x_{0} ; u=F({\sigma}) [/itex]

Then choose [itex] t_{0} = 0 [/itex] (Could have equally done that to [itex] x_{0} [/itex] I assume because it does not matter where we start, might as well chose 0)

Then [itex]u(x,t) = F(x-us)[/itex] s=t
 
  • #17
No, you could not have chosen x(0)=0, why can't you do that?
 
  • #18
I really do not know, my notes said:
"Note that we may choose t_{0} = 0 without loss of generality since the Monge equations are invariant under the change of variable. In other examples it may be more appropriate to choose x_{0} = 0 instead, using the same argument."

I thought that the constants only matter for which characteristic we are looking at ?

P.S.
I appreciate your help!
 
  • #19
The characteristic is defined by the initial value of the x in this case. Use int inition conditions u(0,x)=..., what does that tell you for the initial conditions for the characteristic equations?
 
  • #20
hunt_mat said:
The characteristic is defined by the initial value of the x in this case. Use int inition conditions u(0,x)=..., what does that tell you for the initial conditions for the characteristic equations?

Not really sure how to answer that, but whatever u(t=0,x) is equal to, will stay the same along the characteristic (note, that I am answering using the definition, not understanding). Is that what you are asking about ?
 
  • #21
when s=0, this will be the starting point of your characteristic curve, that is when t=0, so this should imply that t=... at s=0 and also we assign a value to x at x=0 (the solution calls this [itex]\sigma[/itex]).
 
  • #22
Thank you for your help. However, I think I am just to daft to understand these. I am not sure what I do not understand. I bet it has to do with understanding the surfaces. The Monge equations, as far as I understand, are about the tangent vector ? Sort of like streamlines in fluid mechanics ?

I'm learning bit by bit how to solve it mechanically (i.e. without thinking), I wonder, though, how far this will take me...
 
  • #23
To see which variable has the zero initial condition, look at the initial condition for the PDE u(0,x). The characteristics should come from that initial curve, so t=0 when s=0, because s=0 is when you are at the initial condition. However x is not-zero there so set it to some (as yet) unknown value (the solution calls it [itex]\sigma[/itex]), so you have to solve the equations:
[tex]\frac{dt}{ds}=1\quad t(0)=0,\quad\frac{dx}{ds}=u\quad x(0)=\sigma ,\quad\frac{du}{dx}=0,\quad u(0,\sigma )=...[/tex]
Where ... are the two cases mentioned in the problem. Let's examine the case for ...=0, what does this make the equation solution?
 
  • #24
I assume you are talking about the most recent problem, so

[itex]u(0,\sigma) = \sigma[/itex] and that will let us solve for x i.e. [itex]x = \sigma t + \sigma [/itex] for the given limits (with now [itex] 0 \leq \sigma < 1[/itex] ). I then rearrange for [itex] \sigma = x / 1+t [/itex] now look at limits [itex]0 \leq x/1+t < 1[/itex] and then consider the two cases when t > -1 and t < -1 (the latter just inverts the order of inequalities) (But I think we assume t > 0, because in the solution it does not consider this case). Then, since [itex]\sigma = 1 [/itex] is not included in the limits, there has to be a fan of chars. The form of u will be [itex] u = \alpha [/itex]. Hence the "fan" will be [itex]x = \alpha t + 1 (\text{here 1 is the sigma which was not included in the limits})[/itex]. Now the way I think one should get alpha (apart from drawing and looking for the jump) is to look at the case when t = 0 (not sure if this is general, but the discont. in this case occurs at t = 0 ). In this case alpha ranges from 0 (since that's the u which is after sigma > 1 ) to 1 (since at t=0 we have u=1 for the top expression of u). So we get the expression for u of the fan of chars. Now it also makes sense, why the shock location is at 1.

I now think that the reason for the limits on s(t) i.e. [itex]1<s(t)<1+t[/itex] is because we will always insert the shock, where the fan is (but not on the end points, because I think that has something to do with mass conservation (i.e. equal areas rule?))?

Does that sound OK ?P.S.

I think I found out when setting x_{0} to zero works. It's when considering boundary condition problem instead of an IVP...
 
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  • #25
I was hoping you would just solve the problem. If [itex]u_{0}(0,x)=0[/itex], then [itex]t(s)=s[/itex] and [itex]du/ds=0[/itex] which then implies [itex]u(\sigma ,s)=0[/itex], then [itex]dx/ds=0[/itex], which then implies [itex]x=\sigma[/itex]. So we have solved the problem when [itex]u_{0}(0,x)=0[/itex]. The next case to tackle is when [itex]u_{0}(0,x)=x[/itex]. How do you think we should proceed?
 
  • #26
Following the same idea, would it not just be [itex]x = \sigma t + \sigma[/itex] ? (And I expanded the solution in my previous post). Sorry, I thought I solved the problem in the previous post.
 
  • #27
It's good to get these things done, step by step rather than ploughing headlong into things. So what is sigma in this case?
 
  • #28
OK [itex]\sigma = x / (1+t)[/itex]
 
  • #29
In terms of u?
 
  • #30
I assume you are asking for this: [itex] u(0,\sigma) = \sigma [/itex] ?

Otherwise, maybe this [itex]u(\sigma,t) = F(\sigma) = u(x,t) = F(x/(1+t))[/itex] ?
 
  • #31
Really it's [itex]u=\sigma[/itex] from whence you can obtain the solution. So the only other case is for x=1, What are the characteristic equations for this case (as I set them up)?
 
  • #32
hunt_mat said:
Really it's [itex]u=\sigma[/itex] from whence you can obtain the solution. So the only other case is for x=1, What are the characteristic equations for this case (as I set them up)?

I assume you mean the [itex]\sigma = 1[/itex] which will give the characteristic [itex]x = \alpha t + 1[/itex], where [itex]\alpha = u[/itex] and in range from 0 to 1. [itex]\alpha = x-1/t[/itex] so [itex]u = x-1/t[/itex]
 
  • #33
The function [itex]u_{0}(0,x)[/itex] isn't actually defined for x=1 oddly enough and I would expect it to be. The solution seems to indicate that it's 0, so let's go with that. the equations are then:
[tex]\frac{dt}{ds}=1\quad t(0)=0,\quad\frac{dx}{ds}=u\quad x(0)=1,\quad\frac{du}{ds}=0\quad u(0)=\alpha[/tex], these equations can be solved to yield the equation you said, the equation for x will give you the ranges for the validity of your solution.
 
  • #34
So where was I wrong ? (I have not really seen x being set to something before (except when playing around with limits)) Did you mean it like sigma (I take it these can be interchanged for t=0) ?
 
  • #35
You weren't wrong, I just pointed out the usual process that someone would do when solving these characteristics problems.
 
<h2>What is the method of characteristics?</h2><p>The method of characteristics is a mathematical technique used to solve partial differential equations. It involves finding curves, known as characteristics, along which the solution to the equation is constant. By solving for these characteristics, the solution to the equation can be determined.</p><h2>How is the method of characteristics used to solve shock wave problems?</h2><p>The method of characteristics is particularly useful in solving problems involving shock waves. This is because shock waves are characterized by sudden changes in the solution, which can be represented by discontinuities in the characteristics. By using the method of characteristics, these discontinuities can be accurately calculated, allowing for the solution to the shock wave problem to be determined.</p><h2>What are the limitations of the method of characteristics?</h2><p>While the method of characteristics is a powerful tool for solving partial differential equations, it does have some limitations. One limitation is that it can only be applied to certain types of equations, such as hyperbolic equations. Additionally, the method can become computationally expensive for complex problems with many characteristics.</p><h2>How is the method of characteristics related to shock wave physics?</h2><p>The method of characteristics is closely related to shock wave physics, as it is a mathematical technique commonly used to model and understand shock waves. By using the method of characteristics, scientists and engineers can predict the behavior of shock waves and design systems to mitigate their effects.</p><h2>What are some real-world applications of the method of characteristics and shock waves?</h2><p>The method of characteristics and shock waves have a wide range of real-world applications. Some examples include studying the behavior of supersonic aircraft, designing shock absorbers for vehicles, and modeling the flow of fluids in pipes and channels. Additionally, these techniques are used in fields such as astrophysics, meteorology, and geology to better understand natural phenomena. </p>

What is the method of characteristics?

The method of characteristics is a mathematical technique used to solve partial differential equations. It involves finding curves, known as characteristics, along which the solution to the equation is constant. By solving for these characteristics, the solution to the equation can be determined.

How is the method of characteristics used to solve shock wave problems?

The method of characteristics is particularly useful in solving problems involving shock waves. This is because shock waves are characterized by sudden changes in the solution, which can be represented by discontinuities in the characteristics. By using the method of characteristics, these discontinuities can be accurately calculated, allowing for the solution to the shock wave problem to be determined.

What are the limitations of the method of characteristics?

While the method of characteristics is a powerful tool for solving partial differential equations, it does have some limitations. One limitation is that it can only be applied to certain types of equations, such as hyperbolic equations. Additionally, the method can become computationally expensive for complex problems with many characteristics.

How is the method of characteristics related to shock wave physics?

The method of characteristics is closely related to shock wave physics, as it is a mathematical technique commonly used to model and understand shock waves. By using the method of characteristics, scientists and engineers can predict the behavior of shock waves and design systems to mitigate their effects.

What are some real-world applications of the method of characteristics and shock waves?

The method of characteristics and shock waves have a wide range of real-world applications. Some examples include studying the behavior of supersonic aircraft, designing shock absorbers for vehicles, and modeling the flow of fluids in pipes and channels. Additionally, these techniques are used in fields such as astrophysics, meteorology, and geology to better understand natural phenomena.

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