# 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 Related Differential Equations News on Phys.org
wrobel
In the theory of elliptic PDE there are comparison theorems, you can try such a type argument here. Perhaps I describe details later

• matematikawan
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.

wrobel
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
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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• matematikawan
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. 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
$-U''=F U-G,\quad U(0)=U(1)=0,$
$G=\min_{x\in[0,1]}g(x)$ instead of maximum value.

wrobel
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
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) 