# Homework Help: Using the Maximum Principle

1. Feb 27, 2010

### Vuldoraq

Hi, Please could someone help me with the following? Sorry for the extra long post!

1. The problem statement, all variables and given/known data
a) Using the Maximum/Minimum principle, find the maximum and minimum points of the solution of the following mixed problem;

$$u_t - c^2u_{xx} = 0, \ \ 0 < x < l, \ \ t>0$$
$$u(0,x) = x(x-l) \ \ 0< x< l$$
$$u(t,0) = 0 = u(t,l) \ \ t > o$$

and

b) Let u, belonging to the set of continuously differentiable once and continuously differentiable twice functions, be the solution to the following mixed problem;

$$u_t - u_{xx} =0, \ \ 0 < x < 2, \ \ 0 < t < T$$
$$u(0,x) = 0$$
$$u(t,0) = f(t)$$
$$u(t,2) = 0$$
where $$0 \le f(t) \le 1$$

Prove that,

$$0 \leq u(t,x) \leq 1 - \frac{x}{2}$$

hint: Write the initial boundary value problem for $$w(t,x) = u(t,x) - v(t,x)$$ where $$v(t,x) = 1 - x/2$$ and use the maximum principle.
2. Relevant equations

None?

3. The attempt at a solution
I'm really stuck on these. For a start I don't really understand the Maximum principle, so any information would be most helpful. I think it means that if you can find a maximum on the boundary of a function, then you have found the maximum for the whole space. The same is true for the minimum.

So is the boundary defined by our boundary/initial conditions? And how would you then find the maximum/minimum on that boundary?

Here is what I have for part a)

The only boundary that gives us any information is the one where $$u(0,x) = x(x-l) \ \ 0< x< l$$. The minimum of this boundary must be zero, since x lies between zero and l. The maximum can be found by differentiating once and setting equal to zero. This gives x=l/2. Plugging this back into the original boundary equation gives me a maximum of l/4. Therefore we have,

$$0 \le u(x,t) \le \frac{l}{4}$$

Is this the right thing to do? It doesn't seem very rigorous to me, so I think that I have done something wrong.

For part b)

We get an equation for w,

$$w_t - w_{xx} = 0$$
$$w(0,x) = \frac{x}{2} - 1$$
$$w(t,0) = f(t) -1$$
$$w(t,2) = 0$$

I find the minimum on the boundary of w to be minus one, because the lowest of f(t)-1 is when f(t)=0. I also get the maximum of w to be zero, corresponding to when x=2 or f(t)=1. Therefore we have the following,

$$w(t,x) \ge -1$$
$$w(t,x) \le 0$$

since w = u-v we have,

$$u-v \ge -1$$
$$u-v \le 0$$

which implies,

$$u \ge v - 1 = -x/2$$
$$u \le v = 1-x/2$$

Therfore I get, $$-x/2 \le u(t,x) \le 1-x/2$$ which is clearly wrong!

Please could someone help me make sense of all this? I have spent hours and hours trying to work it out and I must be missing something obvious. Thanks for your time!

Last edited: Feb 27, 2010
2. Feb 28, 2010

### Vuldoraq

Anyone have any ideas? Please say if I haven't explained myself properly.