Question #264536

Uxx + Uyy =0 convert the situation equation into its Canonical form and find out its general solution


1
Expert's answer
2021-11-23T08:37:12-0500

this is the canonical form of elliptic equation, 2-D Laplace equation:

uxx+uyy=0u_{xx} + u_{yy}=0


we have uxx=uyy=au_{xx} =- u_{yy}=a

where a is constant


general solution:

u(x,y)=a4(x2y2)u(x, y) = \frac{a}{4}(x^2-y^2)


solution for rectangle with sides L and H:


for boundary conditions:

u(0,y)=g(y),u(L,y)=0,u(x,0)=0,u(x,H)=0u(0,y)=g(y),u(L,y)=0,u(x,0)=0,u(x,H)=0


u(x,y)=Bnsinh(nπ(xL)H)sin(nπyH)u(x,y)=\sum B_n sinh(\frac{n\pi (x-L)}{H})sin(\frac{n\pi y}{H})


Bn=2Hsinh(nπ(L)H)0Hg(y)sin(nπyH)B_n=\frac{2}{Hsinh(\frac{n\pi (-L)}{H})}\int^H_0 g(y)sin(\frac{n\pi y}{H})


for boundary conditions:

u(0,y)=0,u(L,y)=0,u(x,0)=0,u(x,H)=f(x)u(0,y)=0,u(L,y)=0,u(x,0)=0,u(x,H)=f(x)


u(x,y)=Bnsinh(nπyL)sin(nπxL)u(x,y)=\sum B_n sinh(\frac{n\pi y}{L})sin(\frac{n\pi x}{L})


Bn=2Lsinh(nπHL)0Lf(x)sin(nπxL)B_n=\frac{2}{Lsinh(\frac{n\pi H}{L})}\int^L_0 f(x)sin(\frac{n\pi x}{L})


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