Linear vs. Nonlinear First Order Differential...
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Linear vs. Nonlinear First Order Differential Equations Linear Nonlinear Form y′ + p(t)y = g(t), y(t0)=y0 y′ = F(t, y), y(t0)=y0 Theorem If p and g are continuous on an open interval I: α <t< β containing the point t0, then there exists a unique function y = ϕ(t) that satisfies the differential equation for each t in I and that also satisfies the initial condition. If f and ∂f/∂y are continuous in some rectangle α <t< β, γ <y< δ containing the point (t0,y0), then in some interval t0 –h<t<t0 +h contained in α <t< β there is a unique solution y = ϕ(t) of the initial value problem. Weaker result None If f is continuous, then a solution exists, but it may not be unique Method Integrating factor If separable, separate. Otherwise…? Discontinuity The solution can only be discontinuous where p and g are discontinuous. Even then, it still may exist even at points were either p or g are discontinuous. The solution can be discontinuous at points other than where f and ∂f/∂y are discontinuous. Moreover, there may be nothing in the differential equation that indicates where these additional discontinuities exist. Completeness of solutions All solutions can be generated by varying a constant in a single expression A general solution may be produced in the form of an algebraic expression, but even then it is possible that not all solutions may be produced by varying the constant Explicit solution There is always a way to solve for y explicitly It may not be possible to solve for y explicitly
Transcript of Linear vs. Nonlinear First Order Differential...
Linearvs.NonlinearFirstOrderDifferentialEquations
Linear Nonlinear
Form y′+p(t)y=g(t),y(t0)=y0 y′=F(t,y),y(t0)=y0
Theorem IfpandgarecontinuousonanopenintervalI:α<t<βcontainingthepointt0,thenthereexistsauniquefunctiony=ϕ(t)thatsatisfiesthedifferentialequationforeachtinIandthatalsosatisfiestheinitialcondition.
Iffand∂f/∂yarecontinuousinsomerectangleα<t<β,γ<y<δcontainingthepoint(t0,y0),theninsomeintervalt0–h<t<t0+hcontainedinα<t<βthereisauniquesolutiony=ϕ(t)oftheinitialvalueproblem.
Weakerresult
None Iffiscontinuous,thenasolutionexists,butitmaynotbeunique
Method Integratingfactor Ifseparable,separate.Otherwise…?
Discontinuity Thesolutioncanonlybediscontinuouswherepandgarediscontinuous.Eventhen,itstillmayexistevenatpointswereeitherporgarediscontinuous.
Thesolutioncanbediscontinuousatpointsotherthanwherefand∂f/∂yarediscontinuous.Moreover,theremaybenothinginthedifferentialequationthatindicateswheretheseadditionaldiscontinuitiesexist.
Completenessofsolutions
Allsolutionscanbegeneratedbyvaryingaconstantinasingleexpression
Ageneralsolutionmaybeproducedintheformofanalgebraicexpression,buteventhenitispossiblethatnotallsolutionsmaybeproducedbyvaryingtheconstant
Explicitsolution
Thereisalwaysawaytosolveforyexplicitly
Itmaynotbepossibletosolveforyexplicitly
