For $M>0, c\in\mathbb{R}$ consider the boundary value problem $$ \begin{cases}y''+cy'+f(y)=0,\quad~ -M<x<M\\ y(-M)=1,y(M)=0,\qquad 0\leqslant y\leqslant 1\end{cases}\tag{Eqn(M,c)}\label{Eq} $$ where $f$ is some non-linearity which is Lipschitz and $f(0)=f(1)=0$.

For each pair $(M,c)$ there is a unique solution $y(\cdot,M,c)$ which is monotonous.

(I can assume this fact.)

Show that, for fixed $M$, the function $c\mapsto y$ solution of \ref{Eq} is continuous and decreasing.

Hint: Consider two solution $(y_1,c_1)$ and $(y_2,c_2)$ with $c_2>c_1$ and plug $y_1$ into Eqn(M,$c_2$).

Concerning the continuity

Isn't it true that for fixed $M$ the solutions depend continuously on $c$ (isn't that a general thing)? And doesn't this already imply the continuity of the given function?

Concerning the decrease

If I plug $y_1$ into Eqn(M,$c_2$), what I get is $$ y_1''+c_2y_1'+f(y_1)\neq 0 $$

(This cannot be equal to $0$, since $y_1$ is the unique solution to Eqn(M,$c_1$)...)

To be honest, I do not see how to use the given hint.


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