Method of characteristics and shock waves

[Leb]

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).

 

[Ben Niehoff]

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.

 

[Leb]

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...

 

[hunt_mat]

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?

 

[Leb]

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

 

[hunt_mat]

We need the actual equation to be able to give you some advice.

 

[Leb]

Sorry, I have not realized that did not fit.

It's u_{t}+uu_{x}=0 and u(x,0) = -1 for x<0 and u(x,0) = 1

 

[hunt_mat]

This states that u is constant on the characteristics.

 

[Leb]

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...

 

[hunt_mat]

You have to compute the characteristics, write dt/ds=1, dx/ds=u. with t(0)=r and x(0)=0.

 

[Leb]

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 ?

 

[hunt_mat]

You're mapping the characteristics.

 

[Leb]

Not sure what you mean there, but I will pretend to understand :)

I'm having trouble with this question:

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 ?

 

[hunt_mat]

Do you know how to compute characteristics? If I have a hyperbolic PDE, I can write down the characteristics as
\frac{dt}{ds}=1\quad\frac{dx}{dt}=u,\quad\frac{du}{ds}=0
Then consider what u is in those separate regions.

 

[hunt_mat]

We'll take this one step at a time. Solving PDEs of this worm is handle turning, I'll guide you through the process.

 

[Leb]

Do you know how to compute characteristics? If I have a hyperbolic PDE, I can write down the characteristics as
\frac{dt}{ds}=1\quad\frac{dx}{dt}=u,\quad\frac{du}{ds}=0
Then consider what u is in those separate regions.

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

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

Then u(x,t) = F(x-us) s=t

 

[hunt_mat]

No, you could not have chosen x(0)=0, why can't you do that?

 

[Leb]

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!

 

[hunt_mat]

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?

 

[Leb]

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 ?

 

[hunt_mat]

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 \sigma).

 

[Leb]

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...

 

[hunt_mat]

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 \sigma), so you have to solve the equations:
\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 )=...
Where ... are the two cases mentioned in the problem. Let's examine the case for ...=0, what does this make the equation solution?

 

[Leb]

I assume you are talking about the most recent problem, so

u(0,\sigma) = \sigma and that will let us solve for x i.e. x = \sigma t + \sigma for the given limits (with now 0 \leq \sigma &lt; 1 ). I then rearrange for \sigma = x / 1+t now look at limits 0 \leq x/1+t &lt; 1 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 \sigma = 1 is not included in the limits, there has to be a fan of chars. The form of u will be u = \alpha. Hence the "fan" will be x = \alpha t + 1 (\text{here 1 is the sigma which was not included in the limits}). 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. 1&lt;s(t)&lt;1+t 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...

 

[hunt_mat]

I was hoping you would just solve the problem. If u_{0}(0,x)=0, then t(s)=s and du/ds=0 which then implies u(\sigma ,s)=0, then dx/ds=0, which then implies x=\sigma. So we have solved the problem when u_{0}(0,x)=0. The next case to tackle is when u_{0}(0,x)=x. How do you think we should proceed?

 

[Leb]

Following the same idea, would it not just be x = \sigma t + \sigma ? (And I expanded the solution in my previous post). Sorry, I thought I solved the problem in the previous post.

 

[hunt_mat]

It's good to get these things done, step by step rather than ploughing headlong into things. So what is sigma in this case?

 

[Leb]

OK \sigma = x / (1+t)

 

[hunt_mat]

In terms of u?

 

[Leb]

I assume you are asking for this: u(0,\sigma) = \sigma ?

Otherwise, maybe this u(\sigma,t) = F(\sigma) = u(x,t) = F(x/(1+t)) ?

 

[hunt_mat]

Really it's u=\sigma 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)?

 

[Leb]

Really it's u=\sigma 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 \sigma = 1 which will give the characteristic x = \alpha t + 1, where \alpha = u and in range from 0 to 1. \alpha = x-1/t so u = x-1/t

 

[hunt_mat]

The function u_{0}(0,x) 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:
\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, 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.

 

[Leb]

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) ?

 

[hunt_mat]

You weren't wrong, I just pointed out the usual process that someone would do when solving these characteristics problems.

 

[Leb]

I am still not sure if this all helped (not that your explaining is bad, but me being daft). I am now looking at shocks and weak solutions. I (think I) know how to mechanically get weak solutions, but when it comes to drawing the characteristics for the weak solution - I'm in trouble.

For example here:

solution:

I can sort of understand why it is straight lines before -1 and after 1 (from below). Because I think I should be looking for the gradient (1/c(σ)) and c(σ) = ρ(x,t), so the 1/0 gives and infinite gradient (i.e. a straight line), but what about the non straight lines ? How to I look at
1+t/(1+x) ? I would probably try and set t first, so the -1<x<sqrt would hold (for example, t = 1) and then just pick any x from those values (in this case x is in (-1;1).

I am also not sure where the "fan" (chars from 0 to 1 point to (1,1)) originates (it's not in the weak solution, or at least I cannot see it...

 

[hunt_mat]

The characteristic curves from the system of equations:
\frac{dt}{ds}=1,\quad\frac{dx}{ds}=u,\quad\frac{du}{ds}=...
Are given by dividing the differential equations by each other to obtain:
\frac{dx}{dt}=u
From this you just solve the equation to obtain x=f(t,a) where a is the parameter that decides which characteristic you're on.

 

[Leb]

Maybe you could tell me about how you think (your thought process) when you are drawing the characteristics (especially the sloping ones) ? I have circled the ones that maybe will make life easier for both of us.

Because, however I approach it, I get nonsensical line.

Let's say for the first one from the left, I see that it t=0 at say x=-3/4, so If I plug it into the 1+x/(1+t) and invert it, I get gradient 4 (which seems to be a good approximation, as it is really steep). So σ = -3/4 (the value of sigma will aways be the value of x at t=0, right ?) Using the equation for characteristic x=ρt+σ. What does it mean when the characteristic reaches the path of the shock ? Is it where breaking occurs ?

 

[hunt_mat]

To compute the characteristics as I have mentioned, I would solve a differential equation I wrote down in the previous post of mine, in the region where u is constant, the characteristics are going to be of the form x-ut=\textrm{constant} and will therefore be straight lines, as you have drawn. For the region where 0&lt;x&lt;1 the solution was u=x/(1+t), and therefore you solve the equation:
\frac{dx}{dt}=\frac{x}{1+t}, solve this to get the shape of the characteristics in that region. What do you get?

 

[hunt_mat]

I should add that once we have done this, we can calculate when the shock happens and the path of the shock.

 

[Leb]

It should be x = K (1+t) So a straight line, for some constant K. If I use the range x is in I get that
-1&lt;t&lt;\frac{1}{K}+1.

I think I am wasting your time by now. I think it could be high school maths that's my problem. I have no idea where I get K from and no idea how this leads to shock time and path...I'll just try to waffle something in the exam and solve as much as possible "mechanically" If I have not understood this simple subset of PDE's I never will.

I appreciate you taking time to guide me through this, thank you so much!

 

[hunt_mat]

The K in your equation specifies which characteristic you're on. Now you can calculate the characteristics we're not in a good position to compute the shock position. Now if we integrate between t_{0} and t, we get the equation:
x=\frac{1+t}{1+t_{0}}
Where t_{0} defines the characteristic. Now we expect that neighbouring characteristics to intersect when a shock forms, so:
x(t,t_{0})\approx x(t,t_{0})=\frac{\partial x}{\partial t_{0}}\delta t_{0}
So the point where the shock forms is when \frac{\partial x}{\partial t_{0}}=0, so can you compute this?

 

[hunt_mat]

Having checked the calculation, I realize this is the wrong solution, you should do the same thing for \frac{dx}{dt}=\frac{x-1}{t}, what value of t does this vanish for?

 

[Leb]

So x vanishes for t = 1/C... So the x intercept. If C is the "sigma" this confuses me even further...

 

[hunt_mat]

I think it shows that x=1+\frac{t}{t_{0}}, then:\frac{\partial x}{\partial t_{0}}=-\frac{t}{t_{0}^{2}} which shows that t=0 is the time when the shock forms and that implies that the shock forms when x=1.

 

[Leb]

I did not notice I deleted that part. I got x = 1 - Ct (I did it by differentiating, and equating the logs.

Are we still talking about the same problem as in the picture ? Because it says the shock forms at t=1, x=1.

Anyway, never mind. I understand nothing. Thanks for your help anyway!

 

[hunt_mat]

If you send me a PM, I will send you my characteristics notes.