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The existence theorem The existence theorem (p. 66) basically says that if f(t, y) is continuous “near” (t 0, y 0 ), then the differential equation has a solution “near” time t 0. Most of the functions we’ll see in this class are continuous (at least, most of the time!).
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Section 1.5
EXISTENCE AND UNIQUENESS OF SOLUTIONS
The eternal existential question
If we are given any old initial-value problem
• Does there have to be a solution?• If so, could there be more than one solution?
(Think of questions like “does 2x5 - 10x + 5 = 0 have a solution? if so, how many?” We can show that there is a solution between x=-1 and x=1, but we can’t factor the polynomial to find it, and we don’t know how many there are.)
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dydt= f (t, y), y(t0 ) = y0
The existence theorem
The existence theorem (p. 66) basically says that if f(t, y) is continuous “near” (t0, y0), then the differential equation
has a solution “near” time t0.
Most of the functions we’ll see in this class are continuous (at least, most of the time!).
€
dydt= f (t, y), y(t0 ) = y0
Formal statement of the existence theorem
• Check out the theorem on p. 66.• The statement “there exists an > 0” means that
there is some positive value the variable can take on so that the statement becomes true.
• The theorem does not tell us how large that value is.
Example
This is a continuous function, so solutions will exist “near” any point (t0, y0 ).
HOWEVER, there is no guarantee that the solution you have for (t0, y0 ) will be valid for times far from t0.
For example, the general solution is y(t)=tan(t+c), so a solution of the IVP y(0)=0 is y(t) = tan(t), which is undefined at t = pi/2.
This situation is called lack of extendability. The existence theorem doesn’t guarantee extandability.
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dydt= f (t, y) =1+ y2
Uniqueness of solutionsOK, so in most reasonable situations, at least one
solution to an IVP will exist.Did I say at least one????Does this mean there can be more than one????
YES.
If the function f(t, y) and its partial derivative are continuous at (t0, y0), a solution exists and is unique near (t0, y0). Otherwise, there might be more than one solution!
Try dy/dt = 3y2/3, y(0)=0, to see how this can look.€
∂f ∂y
Questions
• What is the difference between the Existence Theorem and the Uniqueness Theorem? In particular, how do their hypotheses differ?
• What does this tell us about how solution curves “look?” (There’s a particularly nice relationship for equilibrium solutions.)
• How can these theorems help us to spot problems in numerical solutions?
• Exercises: p. 73: 2, 13