A Boundary Value Problem

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 a maximum in the domain. Or because my conjecture is wrong :cry:
 
In the theory of elliptic PDE there are comparison theorems, you can try such a type argument here. Perhaps I describe details later
 
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.
 
Well. First consider a boundary value problem
$$-u''(x)=h(x)\in C[0,1],\quad u(0)=u(1)=0.$$ (We can replace the spaces ##C^k[0,1]## with the Sobolev or Holder spaces , they are also suitable. Actually I will reason very rough , it is only to illustrate the general idea. ) Solving this problem we obtain a bounded operator ##P:C[0,1]\to C^2[0,1]## defined by the formula
$$u(x)=Ph=x\int_0^1d\xi\int_0^\xi h(s)ds-\int_0^xd\xi\int_0^\xi h(s)ds.$$
It is easy to see that if ##h\ge 0## then ##Ph\ge 0##.

In your case the equation is as follows $$-u''=f(x)u-g(x),\quad f,g>0.\qquad (*)$$ Assume that ##f,g\in C[0,1]## then introduce constants
##F=\max_{x\in[0,1]}f(x),\quad G=\max_{x\in[0,1]}g(x).##
Assume that the constants ##F,G## are such that the problem $$-U''=F U-G,\quad U(0)=U(1)=0$$
has a solution ##U(x)<0,\quad x\in(0,1)##. Then the problem (*) has a solution ##\tilde u(x)## such that ##U\le \tilde u\le 0##. Indeed,
consider an operator $$\mathcal F(u)=P(f(x)u-g(x)).$$
This operator takes the set
$$W=\{u\in C[0,1]\mid U\le u\le 0\}$$ to itself. Moreover, The operator ##\mathcal F## is a compact operator in ##W## with respect to ##C[0,1]## topology. By the Schauder fixed point theorem we get a fixed point of the operator ##\mathcal F##. This fixed point is the solution ##\tilde u##.
 
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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:
 
Sorry to come back again to this thread.
If I understand correctly, the mapping [itex]\mathcal F : W \rightarrow W [/itex] where [itex] W=\{u\in C[0,1]\mid U\le u\le 0\} [/itex] required the statement
if [itex]h\le 0 [/itex] then [itex] Ph=x\int_0^1d\xi\int_0^\xi h(s)ds-\int_0^xd\xi\int_0^\xi h(s)ds \le 0 .[/itex]

Also in the equation
[itex] -U''=F U-G,\quad U(0)=U(1)=0, [/itex]
[itex]G=\min_{x\in[0,1]}g(x) [/itex] instead of maximum value.
 
from this:
It is easy to see that if h≥0h\ge 0 then Ph≥0Ph\ge 0.
it follows that if ##v\le V## then ##Pv\le PV## and particularly
the mapping F:W→W\mathcal F : W \rightarrow W where W={u∈C[0,1]∣U≤u≤0} W=\{u\in C[0,1]\mid U\le u\le 0\} required the statement
if h≤0h\le 0 then Ph=x∫10dξ∫ξ0h(s)ds−∫x0dξ∫ξ0h(s)ds≤0. 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′′=FU−G,U(0)=U(1)=0, -U''=F U-G,\quad U(0)=U(1)=0,
G=minx∈[0,1]g(x)G=\min_{x\in[0,1]}g(x) instead of maximum value.
O, for me that is a hardest point in whole the argument, I every time confuse in it:)
taking into account that ##f,F>0## we see that if ##0\ge u\ge U## then ## f(x)u-g(x)\ge FU-g(x)## that is clear. Then it must be ## FU-g(x)\ge FU-G ## so that ##-g\ge- G,\quad g\le G##. It seems everything has been written ok
 
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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: U(x)=-sin(1-x)sin(x)/2cos(1)

:smile:
 

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