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. 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

 

[Chestermiller]

Can you solve the following ODE, subject to the initial condition u = 0 at t = 0?

\frac{Du}{Dt}=f(t,x_0+at)

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]

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]

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

 

[Chestermiller]

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]

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(\tau, \xi) with \xi = x-at and \tau = t
which leads to, after partial derivatives, ∂u/∂\tau = f(t,x)

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

 

[Chestermiller]

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(\tau, \xi) with \xi = x-at and \tau = t
which leads to, after partial derivatives, ∂u/∂\tau = f(t,x)

Now integrate and either get u=\tau 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/∂\tau = f(t,x). Now, you substitute x=\xi+at to get

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

Now you take any value of \xi, say \xi=-5 at time t = 0 (x = -5), and start integrating with respect to t (at constant \xi=-5), 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:

u=0 for t\leqslant 4/a

u=t-4/a for 4/a\leqslant t\leqslant 6/a

u=2/a for 6/a\leqslant t

You can do this for different values of \xi.

 

[ericm1234]

I have to say I still do not understand several aspects. What exactly is meant by the subscript \xi and why is "u=0 until x=-1"

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

 

[Chestermiller]

I have to say I still do not understand several aspects. What exactly is meant by the subscript \xi and why is "u=0 until x=-1"

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

The subscript \xi means "holding \xi constant." So, the derivative with a subscript is the partial of u with respect to t holding \xi 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