A specific method of characteristics problem

In summary: That's why u jumps up to t-4/a. And once your line crosses over into the region x > +1, the integrand drops back to zero, and u stays at +2/a for all subsequent times. This is one way of looking at the solution. A more traditional way of looking at the solution would involve the use of Green's functions.
  • #1
ericm1234
73
2
In my previous PDE class we did nothing with method of chars. Now I am assigned one in a higher class, but all the examples I find online are not helpful for a particular problem I have.
du/dt+a*du/dx=f(t,x)

with u(0,x)=0 and f(t,x)=1 if -1<=x<=1, and f(t,x)= 0 otherwise

Differences between this and typical examples I see in my PDE book and online:
1. I rarely see non-homogeneous ones.
2. Initial condition is NEVER equal to 0 in any example I've ever seen.
3. The solution I'm supposed to "show" is piecewise defined in 5 parts. I've seen a couple examples whose solution is defined into 2 parts, sometimes. But 5?

Please be gentle.
 
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  • #2
1. I rarely see non-homogeneous ones.
2. Initial condition is NEVER equal to 0 in any example I've ever seen.
3. The solution I'm supposed to "show" is piecewise defined in 5 parts. I've seen a couple examples whose solution is defined into 2 parts, sometimes. But 5?

1. If you write out the equation for each regeon, you'll see that f(t,x) is a constant in each case.

2. shouldn't matter - just plug the zero into the general solution like you would any other number.

3. makes no difference - you just have more steps to do. Do it the same way - just divide the DE into the different regions.

What is the actual problem you have to solve

You've seen:
http://en.wikipedia.org/wiki/Method..._of_first-order_partial_differential_equation
 
  • #3
Can you solve the following ODE, subject to the initial condition u = 0 at t = 0?

[tex]\frac{Du}{Dt}=f(t,x_0+at)[/tex]

where x0 is a constant.
 
  • #4
Obviously I'm missing something fairly rudimentary.

Simon:
I wrote the problem I have to solve.

du/dt+a*du/dx=f(t,x)

with u(0,x)=0 and f(t,x)=1 if -1<=x<=1, and f(t,x)= 0 otherwise

If I write out the equation for each region, I get 2 equations.

Chestermiller:

When you use capital D, is that the same as an ordinary derivative?

I really don't see how to proceed.
 
  • #5
ericm1234 said:
Obviously I'm missing something fairly rudimentary.

Simon:
I wrote the problem I have to solve.

du/dt+a*du/dx=f(t,x)

with u(0,x)=0 and f(t,x)=1 if -1<=x<=1, and f(t,x)= 0 otherwise

If I write out the equation for each region, I get 2 equations.

Chestermiller:

When you use capital D, is that the same as an ordinary derivative?

I really don't see how to proceed.

Yes, that is the same as an ordinary derivative. I wanted to use d's, but you had already used d's in your original equation, where you properly should have been using partials. The solution to the equation I gave is u(t,x0+at). So you have an initial value problem, starting out at various locations x0. It is like following the time rate of change of a particle that is moving with velocity a. This is how the method of characteristics works.
 
  • #6
I'm lost. Could you walk me through this?
 
  • #7
ericm1234 said:
I'm lost. Could you walk me through this?

Do you know how to make a change of variables from t and x, to t and (x-at)?

Chet
 
  • #8
Do I 'know how'? I guess not really.
Ok, let me put my attempt to 'solve' this.

problem:
∂u/∂t+a*∂u/∂x=f(t,x)
with f(t,x)=1 if -1<=x<=1 and f= 0 otherwise
u(0,x)=0

For whatever reason, (I read this a long time ago) I set u(x,t)=U([itex]\tau[/itex], [itex]\xi[/itex]) with [itex]\xi[/itex] = x-at and [itex]\tau[/itex] = t
which leads to, after partial derivatives, ∂u/∂[itex]\tau[/itex] = f(t,x)

Now integrate and either get u=[itex]\tau[/itex] for -1<=x<=1 or get 0 for x values other than that.
So what am I missing?
 
  • #9
ericm1234 said:
Do I 'know how'? I guess not really.
Ok, let me put my attempt to 'solve' this.

problem:
∂u/∂t+a*∂u/∂x=f(t,x)
with f(t,x)=1 if -1<=x<=1 and f= 0 otherwise
u(0,x)=0

For whatever reason, (I read this a long time ago) I set u(x,t)=U([itex]\tau[/itex], [itex]\xi[/itex]) with [itex]\xi[/itex] = x-at and [itex]\tau[/itex] = t
which leads to, after partial derivatives, ∂u/∂[itex]\tau[/itex] = f(t,x)

Now integrate and either get u=[itex]\tau[/itex] for -1<=x<=1 or get 0 for x values other than that.
So what am I missing?


Yes. You're very close. So you have ∂u/∂[itex]\tau[/itex] = f(t,x). Now, you substitute [itex]x=\xi+at[/itex] to get

[tex](\frac{\partial u}{\partial t})_\xi=f(t,\xi+at)[/tex]

Now you take any value of [itex]\xi[/itex], say [itex]\xi=-5[/itex] at time t = 0 (x = -5), and start integrating with respect to t (at constant [itex]\xi=-5[/itex]), subject to the initial condition u = 0. So, in this example, u(x,t) = 0 until
x = -5 + at = -1.
That would be t = 4/a. After that u(t,x) = u(t,-5+at) = (t - 4/a) until x = -5 + at = +1
That would be at t = 6/a. After that u(t,x)=u(t,-5+at) = 2/a for all subsequent times along the line x = -5 + at.

So, integrating along the line x = -5 + at, you have:

[itex]u=0[/itex] for [itex]t\leqslant 4/a[/itex]

[itex]u=t-4/a[/itex] for [itex]4/a\leqslant t\leqslant 6/a[/itex]

[itex]u=2/a[/itex] for [itex]6/a\leqslant t[/itex]

You can do this for different values of [itex]\xi[/itex].
 
  • #10
I have to say I still do not understand several aspects. What exactly is meant by the subscript [itex]\xi[/itex] and why is "u=0 until x=-1"

Further, this does not appear to be the solution we're supposed to show. :(
 
  • #11
ericm1234 said:
I have to say I still do not understand several aspects. What exactly is meant by the subscript [itex]\xi[/itex] and why is "u=0 until x=-1"

Further, this does not appear to be the solution we're supposed to show. :(

The subscript [itex]\xi[/itex] means "holding [itex]\xi[/itex] constant." So, the derivative with a subscript is the partial of u with respect to t holding [itex]\xi[/itex] constant. The initial condition on u is u=0 for all x.

You are integrating with respect to t along a line

x - at=constant.

The constant is the value of x at t = 0.

Anywhere you start integrating to the left of x = -1 at time = 0, the integrand f(t,x) is equal to zero. You will find that, throughout the entire region x < -1, the solution for u is going to be u = 0 for all t. But, once your line crosses over into the region -1<x<+1, your integrand f(t,x) suddenly jumps up to 1 (and stays there until your line crosses out of the region again at x = 1).

I'm afraid I'm not explaining things very well. Hopefully, I said enough to give you the gist of what is happening. If not, all I can say is I did my best.

Chet
 

1. What is the method of characteristics and how does it work?

The method of characteristics is a mathematical technique used to solve partial differential equations. It involves finding a set of curves, called characteristics, along which the solution to the equation is constant. By determining the values of the solution at specific points along these curves, the solution can be found for the entire domain.

2. When is the method of characteristics most commonly used?

The method of characteristics is commonly used to solve hyperbolic partial differential equations, such as those found in fluid dynamics and acoustics. It is also useful for solving equations with initial or boundary conditions that are not defined on a specific boundary, but rather along a characteristic curve.

3. What are the advantages of using the method of characteristics?

One advantage of the method of characteristics is that it can be used to solve equations with complex or variable coefficients. It also allows for the use of non-uniform grids, which can provide more accurate solutions. Additionally, the method can be applied to both linear and nonlinear equations.

4. What are some limitations of the method of characteristics?

The method of characteristics may not be suitable for solving equations with discontinuous solutions, as it relies on smooth characteristic curves. It also requires a high level of mathematical understanding and can be computationally intensive, making it less practical for some applications.

5. How does the method of characteristics compare to other numerical methods?

The method of characteristics is a specific type of numerical method, and its effectiveness depends on the nature of the problem being solved. For certain types of equations, it may be more efficient and accurate than other methods, but for others, alternative techniques may be more suitable.

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