You are right Wrobel. I overlooked that U is negative (the assumption). Sorry.
The result now is consistent with an example that I have.
u" + 4u = x^2 , u(0)=u(1)=0
Solution: u(x) = (2x^2 + sin(1-2x)/sin(1) - 1)/8
Lower bound solution F=4, G=1
U" + 4U = 1 , U(0)=U(1)=0
Solution...
Sorry to come back again to this thread.
If I understand correctly, the mapping \mathcal F : W \rightarrow W where W=\{u\in C[0,1]\mid U\le u\le 0\} required the statement
if h\le 0 then Ph=x\int_0^1d\xi\int_0^\xi h(s)ds-\int_0^xd\xi\int_0^\xi h(s)ds \le 0 .
Also in the equation
-U''=F...
Wow! What a solution. Thank you very much Wrobel.
I need some time to properly understand the solution. Hope you don't mind if I ask again in case I do have problem understanding the argument.
:smile:
Thanks. Interesting suggestion wrobel. So to which DE should I compare my equation ?
Although I would prefer f(x) and g(x) to be general, I would still be content if f(x) is a monotonic increasing function since I'll be solving later on a specific DE with f(x) known.
I have a BVP of the form u" + f(x)u = g(x) , u(0)=u(1)= 0
where f(x) and g(x) are positive functions.
I suspect that u(x) < 0 in the domain 0 < x < 1. How do I go proving this.
I have try proving by contradiction. Assuming first u > 0 but I can't deduce that u" > 0 which contradict that u has...
Looks like you have a boundary value problem not initial value problem. If your DE is a linear constant coefficient, I think you still can solve it with Laplace transform.
I got the expression from Prof. J.L. Butcher note, an authoritative person in Runge-Kutta method. That term is related to a rooted tree.
Just google rooted tree Runge-Kutta for detail.
When deriving the Runge-Kutta Method to solve y'=f(x) we need to use Taylor expansion. Hence we need to differentiate the function many times.
y'(x)=f(y(x))
y''(x)=f'(y(x))y'(x) = f'(y(x))f(y(x))
y''' = f''(y(x))(f(y(x)),y'(x)) + f'(y(x))f'(y(x))y'(x)
I can understand the second...
Assuming that your system can be written as the matrix form \dot{X}=AX+F(t). , e.g. X=[x1 x2]t etc
Then the general solution for this equation should be (if I'm not mistaken)
X(t)=e^{At}C+e^{At}\int_0^t e^{-As}F(s) ds
Your particular solution in this case is then
X_p(t)=e^{At}\int_0^t...
I get this example from >>help eval
and add a semicolon.
for n = 1:12
eval(['M' num2str(n) ' = magic(n)']);
end
The above commands display all 12 magic square. How do I suppress the output? I only want matlab to assign the variables not display them.
Given n by n matrices A, B, C. I know how to solve the Sylvester equation
AX + XB + C = 0
using the matlab command >> X=lyap(A,B,C)
But how do we solve the extended Sylvester equation
AX + XB + CXD + E = 0 ?
Either numerical or analytical method I'm willing to learn.
From what I understand the command format won't effect the computation. It is only for display purposes. I show you another session in the default format.
>> 1/3-1/2
ans =
-0.1667
>> ans + 1/6
ans =
-2.7756e-017
>> -1/6
ans =
-0.1667
>> -1/6+1/6
ans =...
Thank you for all those information.
When we were first introduced to matlab, they said matlab can be used as a calculator. Now I think my desktop scientific calculator or Window Accessories calculator can do a much better job for simple arithmetic calculation.
What puzzled me is that when...
This really surprised me.
>> format rat
>> 1/3-1/2+1/6
ans =
-1/36028797018963968
Even school student knows that the answer is 0.
Even format short does not give a correct answer.
>> format short
>> 1/3-1/2+1/6
ans =
-2.7756e-017
The general solution is a linear combination of two linearly independent solutions y1(x) and y2(x).
p.s. I think something is not right with your notation e^(xt) .
I want to determine whether u=-x^3_1-x_1-\sqrt{3}x_2 is a stabilizing control for the system
\begin{array}{cc}\dot{x}_1=x_2\\ \dot{x}_2=x^3_1+u\end{array}
with cost functional
\frac{1}{2}\int^{\infty}_0 x^2_1 +x^2_2+u^2 \ dt.
After looking at some examples, I understand that I have to find...
I have gone through the paper again but cannot extract new information about the terminal point other than what I have already written.
But I see there is a sentence which claim that this example is for linear unstable system.
Why is it unstable? Will it effect the computation?
Re: Y'' + py'+ qy = 0 explain why the value of y''(a) is determined by the values of
The DE is linear so it must have a general solution which is a linear combination of two linearly independent solutions y1(x) and y2(x).
y(x)=c1y1(x) + c2y2(x) .
c1 and c2 can be determined uniquely...
This is the part that really confuse me, the terminal point, because so far I have been doing by just following examples.
Some problem have specific fixed end. Whilst others are free and yet some have infinite time.
So I'm not fully understand what I'm doing here whether it is fixed end...
Thanks Pyrrhus. Probably thats my mistake.
My arguement why the control u is a scalar because in the equation \dot{x}=Ax+Bu , B is a column vector. The only way we can compute Bu is when u is a scalar.
Bu=\left(\begin{array}{cc}0\\u\end{array}\right).
The initial condition x(0) is...
I will be following this thread. :smile:
For a diffusion equation, the first derivative is wrt to time.
\frac{\partial u(x,t)}{\partial t}= \frac{\partial^2 u(x,t)}{\partial x^2}