The Archive
This is the last problem from the summer series.
To see problems and solutions in the fall series, which runs from October 13 through December 15 visit The Fall 2025 Featured Problem Series
Problem
Our final problem for this summer’s Featured Problem of the Week series comes from Penn State Math 411, an upper level ordinary differential equations class. The methods used in the solution are taught in a sophomore-level course in differential equations, such as Penn State Math 250 or 251, but applying the tools to assemble a solution requires more mathematical sophistication than can be expected of a sophomore.
This week’s problem was selected to illustrate the rich behavior of nonlinear differential equations.
Solve the second-order nonlinear differential equation with initial conditions and , by consideration of three cases for the initial conditions
,
, and
Solution
The first step in the solution of is to rewrite it as It follows that where is a constant determined by the initial conditions. This first-order differential equation satisfies the conditions of the Existence and Uniqueness Theorem for first-order differential equations, as discussed at the end of this write-up, thus it is guaranteed to have a unique local solutions satisfying the initial condition .
Equation is a separable first order equation that is equivalent to where is an arbitrary constant, provided that . To carry out the integration on the left three case, which correspond to the case in the statement of the problem, are considered:
.
In this case, . Hence is an equilibrium of . Consequently if , then the solution is for .
If , then becomes This yields the solution The initial condition is used to find . The result is Observe that if , then gives for all , hence gives the solution for all initial values of for this case.
The right hand side of is undefined at , if . It follows that where . Since is equivalent to , and the function in is twice differentiable, this is the solution of the second-order nonlinear differential equation whenever .
The solution has the following asymptotic behavior In summary, if the system starts in equilibrium it stays in equilibrium for all . If the system starts to the left of equilibrium () it will converge to equilibrium as . Finally, if the system starts to the right of equilibrium () it will diverge to infinity in a finite amount of time. Thus the equilibrium is semi-stable.
This classification of the equilibrium can be deduced from a qualitative analysis of . However, a qualitative analysis would not lead to the conclusion that the solution blows up in a finite amount of time whenever the system starts to the right of equilibrium.
In this case, . Set The equilibrium points of are . Thus if , then for all is the solution to the initial value problem.
Suppose that . Then becomes where the equality is a consequence of a straightforward partial fraction decomposition. The integral on the right is evaluated term by term, and after simplification, gives This is equivalent to where .
To solve for , it is shown that the absolute value on the left can be dropped. First, note, as shown at the end of this write-up, there exists a closed interval centered at , such that satisfies the initial value problem on . Thus is a differentiable function on the interval, and consequently continuous there. Hence the functions and are both continuous on . Further, the right hand side of is finite for . Therefore for . By the quotient rule for continuous functions is continuous on . In addition, the right hand side of is nonzero for . Thus for . If a continuous function does not vanish on an interval, then it does not change signs on the interval. Hence the sign of determines the sign of on . Consequently Equation is evaluated at , to find in terms of the initial conditions. With the help of and , becomes This is solved for in terms of the hyperbolic tangent. At this point an observation is in order. The right hand side of is defined everywhere except for satisfying The range of the hyperbolic tangent is . Therefore if , is defined on In particular, if , then gives that for . Hence gives the solution for all possible initial values. On the other hand, if , then is undefined at Note that if and that . Set . The solution to the initial value problem in this case is It follows from the equivalence of and , and that is a twice differentiable function, that is the solution of the second-order nonlinear differential equation whenever .
The asymptotic behavior of the solution is now examined.For , and for or Thus is an unstable equilibrium and is a stable equilibrium. As in Case a, this much could have been deduced from a qualitative analysis of . However, the fact that the solution blows up in finite amount of time if the system starts to the right of would not have been revealed by a qualitative analysis.
In this case, . Set There are no equilibrium solutions of in this case. Consequently the solution is given by for all . Hence where . The range of the arctangent is . Therefore the domain of is and . From the bounds on , one can conclude that therefore as it must. The solution to the initial value problem and consequently , since it is a twice differentiable function, for any satisfying is The asymptotics of the solution in this case are straight forward. For all initial conditions that is the solution blows up in a finite amount of time for all . A qualitative analysis of would have led to the conclusion that the solution diverges regardless of the initial value, but not that the solution would do so in finite time for all initial values.
This concludes the discussion of the solution of the initial value problem .
A brief discussion of the Existence and Uniqueness Theorem as it applies to follows.
Showing that the initial value problem where C is a constant, satisfies the conditions of the Existence and Uniqueness Theorem for first-order differential equations may appear redundant, at least with regard to the existence of solutions, since solutions have been exhibited in all cases, which proves solutions exist. However, the existence of a differentiable solution to the initial value problem was used in the course of the solution in Case b. Hence the argument in that case is not complete without this step. Further, showing uniqueness proves that no solutions have been overlooked.
First, recall the theorem.
In the problem at hand, is given by The right hand side does not depend on . Hence for fixed , is a continuous function of for . Next consider as a function of for fixed restricted to a closed and bounded interval . For . where the inequality follows from the triangle inequality. Set . Then for , Thus is Lipshitz continuous in on any closed and bound interval with Lipshitz constant which does not depend on . It follows that for restricted to that the initial value problem satisfies Theorem .