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Solutions HW 5
1 (a) Separation of variables yields the solution
x(t) =
1 2 −^ t
Equation (1.1) is defined for t 6 = 1/2, but this consideration does not determine the interval of existence. The graph of the solution (1.1) con- sists of two hyperbolas. Only one of them contains the initial point (t 0 , x 0 ) = (0, 2). The hyperbola that contains the initial point is defined for −∞ < t < 1 /2, which is the interval of existence. (The hyperbola for t > 1 /2 does not contain the initial point and so is not a solution curve.) (b) The solution is x(t) =
t + (^12)
As in (a), the graph of x consists of two hyperbolas. The interval of exis- tence is − 1 / 2 < t < ∞—the interval over which the hyperbola containing the initial condition is defined. (c) By inspection, the equilibrium solution x(t) = 0 is a solution of the differential equation that satisfies the initial condition x 0 = 0. The interval of existence is the entire real line. (d) The general solution is x(t) =
x(0)
− t
which has a singularity at t = 1/10 if x(0) = 10. (e) As in (a), the graph of the general solution (1.2 consists of two hyperbo- las. The one containing the initial condition x(0) = x 0 > 0 is defined for −∞ < t < 1 /x 0 , which is the interval of existence. (f) If x(0) < 0, then the general solution is still given by Eq. (2), but the interval of existence is 1/x 0 < t < ∞.
(g) There is a singularity at t = δ if x 0 = 1/δ.
6 (a) The solution is defined implicitly by 2
x = t + 2 and it exists for all − 2 ≤ t < ∞ (b) The initial slope exists only if x 0 ≥ 0. (c) The implicit solution is 2
x = t + 2
x 0 and it exists for all − 2
x 0 ≤ t < ∞. Therefore the singularity exists only for negative values of t. (d) no
1 In (a), (b) and (d), f and ∂f /∂x are defined and continuous everywhere. Thus the solution of every initial value problem is unique. In (c), we have that f and ∂f /∂x are not defined at x = 0. Thus, the theorem says nothing about the existence of solutions for initial conditions of the form x(t 0 ) = 0.
2 Rewrite the equation as x′^ = −kx + g(t). Then, in order to get a unique solution to an initial value problem, Theorem 3.3 requires that f (t, x) = −kx+ g(t) and ∂f /∂x be continuous in a region containing the initial point. Since g(t) is assumed to be continuous and ∂f /∂x = −k, with k a constant, these hypotheses hold for every initial condition.
3 (a) Since f (t, x) = x^1 /^3 , we have ∂f /∂x = 13 x−^2 /^3 , which is undefined if x = 0. (b) Clearly x = 0 is a solution. The separation of variables yields the solu- tions x 1 , 2 = ± (^29)
6 t^3 /^2. These solutions can be shifted in time to get
x(t) =
6(t − t 0 )^3 /^2 , t ≥ t 0 > 0 0 , t < t 0 ,
which is a whole family of solutions of the initial value problem. (c) Since f (x) = x^1 /^3 is continuous everywhere, existence and uniqueness is determined by the continuity properties of the partial derivative ∂f /∂x. If S is a sufficiently small box, then ∂f /∂x is defined and continuous on S, so Theorem 3.3 guarantees a unique solution. (d) ∂f /∂x = 13 x−^2 /^3 , which is finite on any region S that does not contain the origin. If x 0 6 = 0, then we may always choose S in such a way that it contains the initial point (t 0 , x 0 ) but avoids the line x = 0.
(e) x(t) = 271
[
6(t − t0) + (27x 0 )^2 /^3
] 3 / 2
5 f (t, x) = x^2 and ∂f /∂x = 2x are both continuous functions for all x. Hence, by Theorem 3.3 we have a unique solution through any initial point. The interval of existence of this solution has nothing to do with the question of uniqueness or existence of the solution.
