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.