# Solve the first order hyperbolic equation

1. Jul 4, 2011

### 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?

2. Jul 4, 2011

### hunt_mat

I do this in a slightly different way, my characteristic equations are:
$$\dot{x}=3,\quad\dot{t}=2x,\quad \dot{u}=2u$$
Where the dot denotes differentiation to the characteristic variable, $s$. I write the cauchy data as:
$$t(s=0)=0,\quad x(s=0)=r,\quad u(s=0)=2r+4$$
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. Jul 4, 2011

### andrey21

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

4. Jul 4, 2011

### 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.

5. Jul 4, 2011

### andrey21

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

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

6. Jul 4, 2011

### hunt_mat

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

7. Jul 4, 2011

### andrey21

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

8. Jul 4, 2011

### 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.

9. Jul 4, 2011

### andrey21

Ok hunt_mat I appreciate your help, now heres where I get stuck:

Integrating du/dx=2u/3?

10. Jul 4, 2011

### 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?

11. Jul 4, 2011

### andrey21

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

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

12. Jul 4, 2011

### hunt_mat

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

13. Jul 4, 2011

### 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?

14. Jul 4, 2011

### 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.

15. Jul 4, 2011

### andrey21

So establishing K:

t = x2 /3 +K

0= r2 /3 +K

K=-r2 /3

Therefore:

t = x2 /3 - r2 /3

Correct??

16. Jul 5, 2011

### 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.

17. Jul 5, 2011

### andrey21

So the next step is to do:

du/dx where u = 2r+4

so u = 2rx +4x

Is this correct??

18. Jul 5, 2011

### hunt_mat

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

19. Jul 5, 2011

### andrey21

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

du/dx= 2u/3

I'm nt sure how to solve the above:

20. Jul 5, 2011

### hunt_mat

You integrate it like any other equation:
$$\log u-\log u_{0}=\frac{x^{2}}{3}$$
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...