A specific method of characteristics problem


by ericm1234
Tags: characteristics, method, specific
ericm1234
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#1
Feb5-13, 08:50 PM
P: 67
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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Simon Bridge
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#2
Feb5-13, 09:28 PM
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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 in to 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_...ntial_equation
Chestermiller
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#3
Feb5-13, 09:42 PM
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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.

ericm1234
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#4
Feb6-13, 12:46 PM
P: 67

A specific method of characteristics problem


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.
Chestermiller
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Feb6-13, 02:15 PM
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Quote Quote by ericm1234 View Post
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.
ericm1234
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#6
Feb6-13, 04:06 PM
P: 67
I'm lost. Could you walk me through this?
Chestermiller
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Feb6-13, 06:11 PM
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Quote Quote by ericm1234 View Post
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
ericm1234
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#8
Feb6-13, 06:28 PM
P: 67
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?
Chestermiller
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Feb6-13, 07:57 PM
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Quote Quote by ericm1234 View Post
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].
ericm1234
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#10
Feb7-13, 04:36 AM
P: 67
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. :(
Chestermiller
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Feb7-13, 08:43 AM
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Quote Quote by ericm1234 View Post
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


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