Solve the first order hyperbolic equation

[andrey21]

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?

 

[hunt_mat]

I do this in a slightly different way, my characteristic equations are:
<br /> \dot{x}=3,\quad\dot{t}=2x,\quad \dot{u}=2u<br />
Where the dot denotes differentiation to the characteristic variable, s. I write the cauchy data as:
<br /> t(s=0)=0,\quad x(s=0)=r,\quad u(s=0)=2r+4<br />
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

 

[andrey21]

Thank you hunt_mat. I shall read through ur notes. Is the method I was using incorrect for this question then?

 

[hunt_mat]

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.

 

[andrey21]

So from what u have said, will the quotient of the characteristics be

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

 

[hunt_mat]

I think so, yes. To be perfectly honest, I was never really happy doing things this way.

 

[andrey21]

Are u referring to the method I have adopted? Sorry for the confusion, I'm just anxious to answer this question.

 

[hunt_mat]

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.

 

[andrey21]

Ok hunt_mat I appreciate your help, now here's where I get stuck:

Integrating du/dx=2u/3?

 

[hunt_mat]

So you have computed the characteristics which are given as 3t-x^{2}=k where k is a constant, you know that u is a constant on these characteristics, and you know that characteristics curve passes through the point (t,x)=(0,r), what will u be at this point?

 

[andrey21]

I'm a little confused how u got 3t-x²=k?

But will u(r,0) =2r+4

 

[hunt_mat]

Because I solved the equation:
<br /> \frac{dt}{dx}=\frac{2x}{3}<br />
You are right with the second part.

 

[andrey21]

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?

 

[hunt_mat]

Now you need to compute K, you know that the characteristic passes through the point (t,x)=(0,r), 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 (t,x)=(0,r), u=2r+4, so all you have to do is substitute for r.

 

[andrey21]

So establishing K:

t = x² /3 +K

0= r² /3 +K

K=-r² /3

Therefore:

t = x² /3 - r² /3

Correct??

 

[hunt_mat]

Good so far. Now do the same thing for u and you should end up with a function with x's and r's in, and now you know an equation for r.

 

[andrey21]

So the next step is to do:

du/dx where u = 2r+4

so u = 2rx +4x

Is this correct??

 

[hunt_mat]

No, you have an equation for u of the form:
<br /> \frac{du}{dx}=\frac{2u}{3}<br />
Solve that. You will have a constant to compute, that is when you use the initial condition you were talking about.

 

[andrey21]

Ok this is the part I can't get past:

du/dx= 2u/3

I'm nt sure how to solve the above:

 

[hunt_mat]

You integrate it like any other equation:
<br /> \log u-\log u_{0}=\frac{x^{2}}{3}<br />
Now we know that at the point (t,x)=(0,r), the solution is u_{0}=2r+4, 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...

 

[andrey21]

So substituting u₀=2r+4 into:

log_u-log_u0 = x²/3

log_u-log(2r+4)=x²/3

 

[hunt_mat]

Not quite, You need to evaluate everything at the point (t,x)=(0,r), I note that you have done nothing about te x...

 

[hunt_mat]

It would have been better to have written the solution as:
<br /> \log u=\frac{x^{2}}{3}+c<br />
Then to find c, just evaluate the equation at the point (t,x)=(0,r)

 

[andrey21]

Oh yes I see:

logu=r²/3 +c
Is this correct now?

 

[hunt_mat]

Yes, you also know what value of u at this point as well u=2r+4 and therefore
<br /> \log (2r+4)=\frac{r^{2}}{3}+c<br />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.

 

[andrey21]

So

c= log (2r+4) - r²/3

So

log u = x²/3 + log (2r+4) - r²/3

 

[hunt_mat]

Yey, and now you know what r in terms of t and x are, substitute these in and you will have your solution.

 

[andrey21]

So:

t = x² /3 - r² /3

r² /3 = x² /3-t

r² = x² -3t

r = SQRT (x² -3t)

 

[hunt_mat]

Good so far, remember there is an associated \pm along with the answer. You have to decide which it is, but this shouldn't be too hard.

 

[andrey21]

So :

r = \pmSQRT (x² -3t)

What is the next step?

 

[hunt_mat]

Plug it in and see what sign causes you trouble (I think it'll be the - sign that will be the troublesome one). Look for large values of x to find the right sign. Then you're done, you've got the right solution!

 

[andrey21]

Which equation shall I plug the r value into?

Is it:

log u = x²/3 + log (2r+4) - r²/3

 

[hunt_mat]

Yes.