A specific method of characteristics problem

ericm1234

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.

Simon Bridge

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

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

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

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

I'm lost. Could you walk me through this?

Chestermiller

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

Chet

ericm1234

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

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

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

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