Solution of nonlinear BVP is symmetric where derivative is 0:For the following nonlinear BVP,y″(x)+y(x)2=0, y(0)=y(1)=0,I wish to show that ify′(x0)=0, then y(x)=y(x0−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(x)=y(x0−x) satisfies the BVP where y is a nontrivial solution but I can't seem to do this.

Question

Solution of nonlinear BVP is symmetric where derivative is 0:For the following nonlinear BVP,y″(x)+y(x)2=0, y(0)=y(1)=0,I wish to show that ify′(x0)=0, then y(x)=y(x0−x).That is, the solution is symmetric about this point.I&#039;ve already shown that the nontrivial solution is unique, so it should be enough to show that z(x)=y(x0−x) satisfies the BVP where y is a nontrivial solution but I can&#039;t seem to do this.

Jonas Dickerson · Accepted Answer

Step 1You make a little error, the reflection of x at x0 has the formulaτ(x)=2x0−x.Then at the reflection point you have τ(x0)=x0, so it is indeed a fixed point of the reflection.You have now a point x0, y0=y(x0) and y′(x0)=0Next you consider with the corrected reflection z(x)=y(2x0−x) and get likewise z(x0)=y(x0)=y0 and z′(x0)=−y′(x0)=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.

Solution of nonlinear BVP is symmetric where derivative is 0: For the following nonlinear BVP, N

Answered question

Answer & Explanation

New Questions in Calculus 2