Differentiation theory Flashcards
(41 cards)
When is F(t,y) autonomous?
If in the expression F(t,y) the variable t does not occur explicitly.
C^n-function:
Function y:R–>R is n times differentiable and the n-th derivative is continuous
n-th order linear differential equation form:
a_ny^(n)+a_(n-1)y^(n-1)+…+a_oy=b or y’=ay+b
If a and b in y’=ay+b depend on t, we assume that they depend continuously on t, which implies:
all the solutions of differential equations will be C^1-functions
General solution set of y’:
y’=b(t)
{y=B(t)+k|k∈R}
General solution set of y’:
y’=ay
{y=ke^(at)|k∈R}
General solution set of y’:
y’=ay+b
{y(t)=-b/a + ke^(at)|k∈R}
General solution set of y’:
y’=a(t)y
{y=ke^(A(t))|k∈R}
A(t)=∫a(s)ds
General solution set of y’:
y’=a(t)y+b(t)
{y(t)=e^(A(t))(∫e^(-A(s))b(s)ds+c)
Consider initial value problem: y’=F(t,y), y(t0)=y0.
If F is a continuous function in t and y, then:
there exists an open interval I⊆R containing t0 and a C^1-function y:I–>R such that y(t0)=y0 and y’(t)=F(t,y(t)) for all t in I; y is a solution of the initial value problem.
Note: If F is a C^1-function, then this solution is, given a fixed domain I, unique
Euler’s solution. Choose appropriate number of iterations n. Then:
h=(T-t0)/n. Define ti=t_(i-1)+h=t0+ih. Then [t0,T] is divided in n equally sized subsegments. Define subsequently yi=y_(i-1)+hF(t_(i-1,y_(i-1)). Then yn is the required approximation of ^y(T)
Let f:R–>R be a function with lim f(h)=0 (h->0) f is called convergent of order O(h^n) if:
there exist ε>0 and k>0 such that for all h in (0, ε): |f(h)|<Kh^n
A first order differential equation y’=F(t,y) is called separable if:
F(t,y) is the product of a function of y and a function of t: F(t,y)=g(y)*h(t)
How to solve separable equations:
1: F(t,y)=g(y)*h(t) –> y’/g(y)=h(t) we don’t have to bother that this might cause division by zero; if g(y0)=0, then y=y0 is a constant solution. Other solutions do not intersect these constant solutions, so we may assume g(y)≠0, for all y
2: Denote z=1/g, so z(y)y’=h(t). Let Z and H be antiderivatives of z and h. Then Z’(y(t))=z(y(t))y’(t)
3: Integrating both sides: Z(y)=H(t)+k
4: Then in initial value problem, determine the value of k and rewrite for y into an explicit solution
A linear n-th order differential equation is called homogeneous if:
it equates a linear combination of y,y’,… to zero; a_ny^(n)+a_(n-1)y^(n-1)+…+a0y=0. Advantage of homogenous equations: aggregating solutions to each other or multiplying them with scalars lead to new solutions
If y^ and ybar are solutions of the same homogeneous differential equation, then:
for all λ and μ in R, λy^+μybar are solutions of it as well
Let Sl be the general solution set of a linear differential equation, let y^ be a particular solution of the equation and let Sh be the general solution set of the corresponding homogeneous differential equation. Then:
Sl={y^+y|y∈Sh}
|α+βi|
=√(α^2+β^2) (distance to zero/length)
Euler’s formula
e^(βi)=cos(β)+isin(β)
For all z in C, we have:
|e^z|=e^(re(z))>0
Consider the homogeneous linear differential equation: y^(n)+a_(n-1)y^(n-1)+…+a0y=0 for functions ai:R–>R (i=0,…,n-1)
Solutions with complex codomain are also allowed, then we can write:
y(t)=u(t)+iv(t) for u,v: R–>R and note: If y(t) is a solution, then so are its real and imaginary parts u(t) and v(t), and vice versa.
For complex number λ=a+bi, the function
y(t)=e^(λt)=e^(at)(cos(bt)+isin(bt)) is a solution if:
λ satisfies the characteristic equation: λ^n+a_(n-1)λ^(n-1)+…+a0=0
High order equations can be written as first-order systems:
y(t)=[y y’ … y^(n-1)]=[y1 y2 … yn] and the system becomes y’=Ay, where A =
[0 1 0 …..,
0 0 1 …….,
……………..,
-a0 -a1 -a2 … -a(n-1)]
Linear second order differential equation solution sets for:
1: characteristic function has two different real roots λ1 and λ2
2: characteristic function has one root λ
3: characteristic function has complex roots λ1=a+bi and λ2=a-bi
1: y(t)=k1e^(λ1t)+k2e^(λ2t) if D>0
2: y(t)=k1e^(λt)+k2te^(λt) if D=0
3: y(t)=e^(at)(k1cos(bt)+k2sin(bt)) if D<0 (k1=c1+c2, k2=(c1-c2)i)
