g2esebyy7

2022-04-23

Solution of nonlinear BVP is symmetric where derivative is 0:

For the following nonlinear BVP,

$y{}^{\u2033}\left(x\right)+y{\left(x\right)}^{2}=0,\text{}y\left(0\right)=y\left(1\right)=0$ ,

I wish to show that if

${y}^{\prime}\left({x}_{0}\right)=0$ , then $y\left(x\right)=y({x}_{0}-x)$ .

That is, the solution is symmetric about this point.

I've already shown that the nontrivial solution is unique, so it should be enough to show that$z\left(x\right)=y({x}_{0}-x)$ satisfies the BVP where y is a nontrivial solution but I can't seem to do this.

For the following nonlinear BVP,

I wish to show that if

That is, the solution is symmetric about this point.

I've already shown that the nontrivial solution is unique, so it should be enough to show that

Jonas Dickerson

Beginner2022-04-24Added 22 answers

Step 1

You make a little error, the reflection of x at$x}_{0$ has the formula

$\tau \left(x\right)=2{x}_{0}-x$ .

Then at the reflection point you have$\tau \left({x}_{0}\right)={x}_{0}$ , so it is indeed a fixed point of the reflection.

You have now a point${x}_{0},\text{}{y}_{0}=y\left({x}_{0}\right)$ and ${y}^{\prime}\left({x}_{0}\right)=0$

Next you consider with the corrected reflection$z\left(x\right)=y(2{x}_{0}-x)$ and get likewise $z\left({x}_{0}\right)=y\left({x}_{0}\right)={y}_{0}$ and ${z}^{\prime}\left({x}_{0}\right)=-{y}^{\prime}\left({x}_{0}\right)=0$ .

Additionally z is a solution of the DE. By uniqueness$z=y$ follows.

To progress further you probably have to use next that any solution is concave, so has at most two roots.

You make a little error, the reflection of x at

Then at the reflection point you have

You have now a point

Next you consider with the corrected reflection

Additionally z is a solution of the DE. By uniqueness

To progress further you probably have to use next that any solution is concave, so has at most two roots.

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