University:
Georgia Southern University
Course:
MATH 3230 | Ordinary Differential Equations
Academic year:

2024
Views:

227

Pages:

19
Author:

Jalen W.

Definition Consider the multivariable function f (x, y ). The domain of f (x, y ) is the set of all points (a, b) such that f (a, b) is defined. Algebra I Polynomials are defined everywhere. I Exponentials are defined everywhere. I Rational functions are defined whenever the denominator is not zero. I Odd radicals (third root, fifth root, etc.) are defined everywhere. I Even radicals (square root, fourth root, etc.) are defined everywhere where the radicand is nonnegative. ) I In general, (a, b) is in the domain of gf (x,y (x,y ) if I (a, b) is in the domain of f (x, y ), I (a, b) is in the domain of g (x, y ), I and g (a, b) 6= 0. Example Find the domain of f (x, y ) = y 2 − x. Shade the regions of the plane that lie in the domain of f (x, y ). Example −3 . Shade the regions of the plane Find the domain of f (x, y ) = yy −x that lie in the domain of f (x, y ). Example p Find the domain of f (x, y ) = y 2 − x. Shade the regions of the plane that lie in the domain of f (x, y ). Example Find the domain of f (x, y ) = log(y 2 − x). Shade the regions of the plane that lie in the domain of f (x, y ). Trigonometry I Sine is defined everywhere. I Cosine is defined everywhere. I Tangent is undefined at (2n+1)π for all n ∈ Z. 2 I Secant is undefined at (2n+1)π for all n ∈ Z. 2 I Cosecant is undefined at nπ for all n ∈ Z. I Cotangent is undefined at nπ for all n ∈ Z. Example Find the domain of f (x, y ) = csc(y − x). Shade the regions of the plane that lie in the domain of f (x, y ). Example p Find the domain of f (x, y ) = sec( x 2 + y 2 ). Shade the regions of the plane that lie in the domain of f (x, y ). Partial Derivative For f (x, y ) we define the following partial derivatives. ∂f ∂ I ∂x = ∂x [f (x, y )]: Treat y like it is a constant and take the derivative with respect to x just as you would in Calculus 1 (product rule, quotient rule, chain rule, etc.). ∂ ∂f I ∂y = ∂y [f (x, y )]: Treat x like it is a constant and take the derivative with respect to y just as you would in Calculus 1 (product, rule, quotient rule, chain rule, etc.). Example p Let f (x, y ) = sec( x 2 + y 2 ). Find ∂f ∂x and ∂f ∂y . On Finding Solutions to IVPs How do we know whether an IVP is solvable, i.e. when do we know that a solution exists? Every ODE has infinitely many solutions, but what about IVPs? Our intuition should tell us that since we are imposing this extra condition (say, y (x0 ) = y0 ) that this limits the possibilities to solutions that go through (x0 , y0 ). Is this enough to guarantee that there is only one solution? Existence and Uniqueness Theorem, Picard -Lindelof Theorem Consider the IVP dy dx = f (x, y ), y (x0 ) = y0 . This IVP is guaranteed to have a unique solution if BOTH conditions below are satisfied: 1. f (x, y ) is continuous at (x0 , y0 ); 2. ∂f ∂y is continuous at (x0 , y0 ). Continuity for Multivariable Functions Note: Continuity is hard for multivariable functions because limits are hard for multivariable functions. Here are some things you can take for granted. 1. Every function I give you will be continuous on the interior of its domain (i.e. not on the boundary). 2. Functions are discontinuous on outside of their domains. 3. Functions are discontinuous on their boundaries. Example Does the IVP have a unique solution? dy y −x = , y (0) = 0 dx x −1 Example Does the IVP have a unique solution? dy y −x = , y (1) = 0 dx x −1 Example Does the IVP have a unique solution? p dy = y 2 − x 2 , y (1) = 0 dx Example Does the IVP have a unique solution? p dy = 4 − y 2 − x 2 , y (2) = 0 dx Example Does the IVP have a unique solution? p dy = 4 − y 2 − x 2 , y (1) = 1 dx

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