Solve the first order hyperbolic equation

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  • #1
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Solve the first order hyperbolic equation

3 du/dx + 2x du/dt =2u

With initial condition: u(x,0) = 2x+4



My attempt at a solution

I usually adopt the method of characteristics:

dx/a = dt/b = du/c


So from the above:

a=3, b=2x and c=2u

am I on the right track here?
 

Answers and Replies

  • #2
hunt_mat
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I do this in a slightly different way, my characteristic equations are:
[tex]
\dot{x}=3,\quad\dot{t}=2x,\quad \dot{u}=2u
[/tex]
Where the dot denotes differentiation to the characteristic variable, [itex]s[/itex]. I write the cauchy data as:
[tex]
t(s=0)=0,\quad x(s=0)=r,\quad u(s=0)=2r+4
[/tex]
Then just re-arrange to get rid of s and r and that will be your solution. I posted some notes on this in https://www.physicsforums.com/showthread.php?t=467445
 
  • #3
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Thank you hunt_mat. I shall read through ur notes. Is the method I was using incorrect for this question then?
 
  • #4
hunt_mat
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Not an incorrect method, you can get the solution that way, I just think that the solution method that I presented is far easier to implement and you know if you're making a mistake or not.
 
  • #5
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So from what u have said, will the quotient of the characteristics be

dt/dx = 2x/3 du/dx= 2u/3
 
  • #6
hunt_mat
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I think so, yes. To be perfectly honest, I was never really happy doing things this way.
 
  • #7
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Are u referring to the method I have adopted? Sorry for the confusion, I'm just anxious to answer this question.
 
  • #8
hunt_mat
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So far, I think you're doing fine, but I haven't much experience with the method you're adopting. I have a book with it in though, so I will keep helping you with this method.
 
  • #9
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Ok hunt_mat I appreciate your help, now heres where I get stuck:

Integrating du/dx=2u/3?
 
  • #10
hunt_mat
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So you have computed the characteristics which are given as [itex]3t-x^{2}=k[/itex] where [itex]k[/itex] is a constant, you know that [itex]u[/itex] is a constant on these characteristics, and you know that characteristics curve passes through the point [itex](t,x)=(0,r)[/itex], what will [itex]u[/itex] be at this point?
 
  • #11
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I'm a little confused how u got 3t-x2=k?

But will u(r,0) =2r+4
 
  • #12
hunt_mat
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Because I solved the equation:
[tex]
\frac{dt}{dx}=\frac{2x}{3}
[/tex]
You are right with the second part.
 
  • #13
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Yes sorry its because I had it in the form:

t = x2 /3 +K

Ok so given u, where do I go next?

I still need to solve du/dx correct?
 
  • #14
hunt_mat
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Now you need to compute K, you know that the characteristic passes through the point [itex](t,x)=(0,r)[/itex], so you can use that to compute K. Then you need to integrate your equation for u and use that fact that at the point [itex](t,x)=(0,r)[/itex], [itex]u=2r+4[/itex], so all you have to do is substitute for r.
 
  • #15
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So establishing K:

t = x2 /3 +K

0= r2 /3 +K

K=-r2 /3

Therefore:

t = x2 /3 - r2 /3

Correct??
 
  • #16
hunt_mat
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Good so far. Now do the same thing for [itex]u[/itex] and you should end up with a function with x's and r's in, and now you know an equation for r.
 
  • #17
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So the next step is to do:

du/dx where u = 2r+4

so u = 2rx +4x

Is this correct??
 
  • #18
hunt_mat
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No, you have an equation for u of the form:
[tex]
\frac{du}{dx}=\frac{2u}{3}
[/tex]
Solve that. You will have a constant to compute, that is when you use the initial condition you were talking about.
 
  • #19
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Ok this is the part I can't get past:

du/dx= 2u/3

I'm nt sure how to solve the above:
 
  • #20
hunt_mat
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You integrate it like any other equation:
[tex]
\log u-\log u_{0}=\frac{x^{2}}{3}
[/tex]
Now we know that at the point [itex](t,x)=(0,r)[/itex], the solution is [itex]u_{0}=2r+4[/itex], so insert this into the equation and you will be left with an equation with lots of x's and r's, well you know an equation for r...
 
  • #21
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So substituting u0=2r+4 into:

logu-logu0 = x2/3

logu-log(2r+4)=x2/3
 
  • #22
hunt_mat
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Not quite, You need to evaluate everything at the point [itex](t,x)=(0,r)[/itex], I note that you have done nothing about te x...
 
  • #23
hunt_mat
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It would have been better to have written the solution as:
[tex]
\log u=\frac{x^{2}}{3}+c
[/tex]
Then to find c, just evaluate the equation at the point [itex](t,x)=(0,r)[/itex]
 
  • #24
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Oh yes I see:

logu=r2/3 +c



Is this correct now?
 
  • #25
hunt_mat
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Yes, you also know what value of u at this point as well [itex]u=2r+4[/itex] and therefore
[tex]
\log (2r+4)=\frac{r^{2}}{3}+c
[/tex]So now you know what c is in terms of r, and you know what r is in terms of t and x, you can write down the solution.
 

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