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Lecture Notes in Differential Equations - Bruce E. Shapiro

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226 LESSON 26. GENERAL EXISTENCE THEORY*<br />

to a system.<br />

We create a system by def<strong>in</strong><strong>in</strong>g the variables<br />

Then the differential equation becomes<br />

which we can rewrite as<br />

We then def<strong>in</strong>e functions f, and g,<br />

so that our system can be written as<br />

x 1 = y<br />

x 2 = y ′ (26.62)<br />

x ′ 2 + 4t 3 x 2 + x 3 1 = s<strong>in</strong> t (26.63)<br />

x ′ 2 = s<strong>in</strong> t − x 3 1 − 4t 3 x 2 (26.64)<br />

f(x 1 , x 2 ) = s<strong>in</strong> t − x 3 1 − 4t 3 x 2<br />

g(x 1 , x 2 ) = x 2<br />

(26.65)<br />

x ′ 1 = f(x 1 , x 2 )<br />

x ′ 2 = g(x 1 , x 2 )<br />

(26.66)<br />

with <strong>in</strong>itial condition<br />

x 1 (0) = 1<br />

x 2 (0) = 1<br />

(26.67)<br />

It is common to def<strong>in</strong>e a vector x = (x 1 , x 2 ) and a vector function<br />

F(x) = (f(x 1 , x 2 ), g(x 1 , x 2 )) (26.68)<br />

Then we have a vector <strong>in</strong>itial value problem<br />

}<br />

x ′ (t) = F(x) = (s<strong>in</strong> t − x 3 1 − 4t 3 x 2 , x 2 )<br />

x(0) = (1, 1)<br />

(26.69)<br />

S<strong>in</strong>ce the set of all differentiable functions on R 2 is a vector space, our<br />

theorem on vector spaces applies. Even though we proved theorem (26.50)<br />

for first order equations every step <strong>in</strong> the proof still works when y and f<br />

become vectors. On any closed rectangle surround<strong>in</strong>g the <strong>in</strong>itial condition<br />

F and ∂F/∂x i is bounded, cont<strong>in</strong>uous, and differentiable. So there is a<br />

unique solution to this <strong>in</strong>itial value problem.

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