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.
  • #36
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:
prob3.jpg

solution:
shocks.jpg


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...
 
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  • #37
The characteristic curves from the system of equations:
[tex]\frac{dt}{ds}=1,\quad\frac{dx}{ds}=u,\quad\frac{du}{ds}=...[/tex]
Are given by dividing the differential equations by each other to obtain:
[tex]\frac{dx}{dt}=u [/tex]
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.
 
  • #38
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.
charsshock.jpg
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 ?
 
  • #39
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 [itex]x-ut=\textrm{constant}[/itex] and will therefore be straight lines, as you have drawn. For the region where [itex]0<x<1[/itex] the solution was [itex]u=x/(1+t)[/itex], and therefore you solve the equation:
[tex]\frac{dx}{dt}=\frac{x}{1+t}[/tex], solve this to get the shape of the characteristics in that region. What do you get?
 
  • #40
I should add that once we have done this, we can calculate when the shock happens and the path of the shock.
 
  • #41
It should be [itex]x = K (1+t)[/itex] So a straight line, for some constant K. If I use the range x is in I get that
[itex]-1<t<\frac{1}{K}+1[/itex].

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!
 
  • #42
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 [itex]t_{0}[/itex] and [itex]t[/itex], we get the equation:
[tex]x=\frac{1+t}{1+t_{0}}[/tex]
Where [itex]t_{0}[/itex] defines the characteristic. Now we expect that neighbouring characteristics to intersect when a shock forms, so:
[tex]x(t,t_{0})\approx x(t,t_{0})=\frac{\partial x}{\partial t_{0}}\delta t_{0}[/tex]
So the point where the shock forms is when [tex]\frac{\partial x}{\partial t_{0}}=0[/tex], so can you compute this?
 
  • #43
Having checked the calculation, I realize this is the wrong solution, you should do the same thing for [tex]\frac{dx}{dt}=\frac{x-1}{t}[/tex], what value of t does this vanish for?
 
  • #44
So x vanishes for t = 1/C... So the x intercept. If C is the "sigma" this confuses me even further...
 
  • #45
I think it shows that [tex]x=1+\frac{t}{t_{0}}[/tex], then:[tex]\frac{\partial x}{\partial t_{0}}=-\frac{t}{t_{0}^{2}}[/tex] which shows that [itex]t=0[/itex] is the time when the shock forms and that implies that the shock forms when [itex]x=1[/itex].
 
  • #46
I did not notice I deleted that part. I got [itex]x = 1 - Ct[/itex] (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!
 
  • #47
If you send me a PM, I will send you my characteristics notes.
 
<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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