# Boundary Value Problem

## Main Question or Discussion Point

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
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.

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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• 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.

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) 