跳至主要内容

2.6 Exact Differential Equations

THM​

Let M ,N ,My,NxM_y ,N_xare continuous in R={(x,y)∣ α<x<β, γ<y<δR = \{ (x ,y) |\ \alpha < x< \beta ,\ \gamma <y <\delta

Then the equ(∗)equ(*) is exact in RR ⟺ My(x,y)=Nx(x,y)\Longleftrightarrow \ M_y(x ,y) = N_x(x ,y)at each point in RR

proof​

  • ⟹\Longrightarrow

Suppose (*) is an exact differential equation ,∃ ψ s.t. ∂ψ∂x=M(x,y) , ∂ψ∂y=N(x,y)\exist \ \psi \ s.t.\ \frac{\partial \psi}{\partial x} = M(x ,y) \ ,\ \frac{\partial \psi}{\partial y} = N(x ,y)

then ,My(x,y)=∂ψ∂x∂y=∂ψ∂x∂yM_y(x ,y) = \frac{\partial \psi}{\partial x\partial y} = \frac{\partial \psi}{\partial x \partial y}, MyM_yand NxN_xare continuous ,ψxy , ψyx\psi_{xy} \ ,\ \psi_{yx}are continuous.

Hence ,∂2ψ∂y∂x=∂2ψ∂x∂y\frac{\partial^2 \psi}{\partial y \partial x} = \frac{\partial^2 \psi}{\partial x \partial y} ,My=NxM_y = N_x

  • ⟸\Longleftarrow

Good Find a ψ(x) s.t. ψx=M , ψy=N\psi(x) \ s.t. \ \psi_x = M\ ,\ \psi_y = N, Integrating ψx=M w.r.t x\psi_x = M \ w.r.t \ x

ψ(x,y)=∫x0xM(s,y) ds+h(y)=Q(x,y)+h(y)\psi(x ,y) = \int_{x_0}^{x}M(s ,y)\ ds + h(y) = Q(x ,y) + h(y) ,where Q(x,y)=∫x0xM(s,y) dsQ(x ,y) = \int_{x_0}^{x} M(s ,y)\ dsand x0x_0is some specified constant with α<x<β\alpha < x < \beta

Differentiating Q(x,y)+h(y)w.r.tyQ(x ,y) + h(y) \quad w.r.t \quad y

ψy(x,y)=∂Q∂y(x,y)+h′(y)=N(x,y)\psi_y(x ,y) = \frac{\partial Q}{\partial y}(x ,y) + h'(y) = N(x ,y), Then h′(y)=N(x,y)−∂Q∂y(x,y)h'(y) = N(x ,y) - \frac{\partial Q}{\partial y}(x ,y)

To show N(x,y)−∂Q∂y(x,y)N(x ,y) - \frac{\partial Q}{\partial y}(x ,y)only depends on y

∂∂x(N(x,y)−∂Q∂y(x,y))=Nx−∂2Q∂y∂x=Nx−My=0\frac{\partial}{\partial x}(N(x ,y) - \frac{\partial Q}{\partial y}(x ,y)) = N_x - \frac{\partial^2Q}{\partial y \partial x} = N_x - M_y = 0