Q32E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the result of Problem 31 to prove that all solutions to the equation\({\bf{y'' + }}{{\bf{y}}^{\bf{3}}}{\bf{ = 0}}\)remain bounded. [Hint: Argue that \(\frac{{{{\bf{y}}^{\bf{4}}}}}{{\bf{4}}}\) is bounded above by the constant appearing in Problem 31.]

Step-by-Step Solution

Answer

The solution is\({\bf{y = }}\sqrt[{\bf{4}}]{{{\bf{4c}}}}\)and bounded.

Step 1: Find the value of F

Suppose that y(t) solves \({\bf{f(y) =  - }}{{\bf{y}}^{\bf{3}}}\) where f is a continuous function independent of y’. We know that \(\frac{{{\bf{y'(t}}{{\bf{)}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - F(y) = constant}}\) where F(y) is an antiderivative of (y) and then\({\bf{F(y) = }}\int {{\bf{ - }}{{\bf{y}}^{\bf{3}}}{\bf{ =  - }}\frac{{{{\bf{y}}^{^{\bf{4}}}}}}{{\bf{4}}}{\bf{ + C}}} \).

Step 2: Prove solutions remain bounded

Now, 

 

\(\begin{array}{c}\frac{{{{{\bf{(y')}}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + }}\frac{{{{\bf{y}}^{\bf{4}}}}}{{\bf{4}}}{\bf{ = c}}\\\frac{{{{\bf{y}}^{\bf{4}}}}}{{\bf{4}}}{\bf{ = c - }}\frac{{{{{\bf{(y')}}}^{\bf{2}}}}}{{\bf{2}}}\\\frac{{{{\bf{y}}^{\bf{4}}}}}{{\bf{4}}} \le c\\{{\bf{y}}^{\bf{4}}} \le {\bf{4c}}\\{\bf{y = }}\sqrt[{\bf{4}}]{{{\bf{4c}}}}\end{array}\)

 

Thus, all the solutions are bounded.