Cho các phát biểu sau: (a) Vinyl axetat có phản ứng trùng hợp. (b) Glucozơ bị oxi hóa bởi nước brom tạo thành axit gluco… – Olm

phương trình dạng toán tử : (widehat{H})(Psi) = E(Psi)

Toán tử Laplace: (bigtriangledown)2 = (frac{partial^2}{partial x^2})+ (frac{partial^2}{partial y^2})+(frac{partial^2}{partial z^2})

thay vào từng bài cụ thể ta có :

a.sin(x+y+z)

(bigtriangledown)2 f(x,y,z) = ( (frac{partial^2}{partial x^2})+ (frac{partial^2}{partial y^2})+(frac{partial^2}{partial z^2}))sin(x+y+z)

=(frac{partial^2}{partial x^2})sin(x+y+z) + (frac{partial^2}{partial y^2})sin(x+y+z) + (frac{partial^2}{partial z^2})sin(x+y+z)

=(frac{partial}{partial x})cos(x+y+z) + (frac{partial}{partial y})cos(x+y+z) + (frac{partial}{partial z})cos(x+y+z)

= -3.sin(x+y+z)

(Rightarrow) sin(x+y+z) là hàm riêng. với trị riêng bằng -3.

b.cos(xy+yz+zx)

(bigtriangledown)2 f(x,y,z) = ( (frac{partial^2}{partial x^2})+ (frac{partial^2}{partial y^2})+(frac{partial^2}{partial z^2}))cos(xy+yz+zx)

=(frac{partial^2}{partial x^2})cos(xy+yz+zx) +(frac{partial^2}{partial y^2})cos(xy+yz+zx) + (frac{partial^2}{partial z^2})cos(xy+yz+zx)

=(frac{partial}{partial x})(y+z).-sin(xy+yz+zx) + (frac{partial}{partial y})(x+z).-sin(xy+yz+zx) + (frac{partial}{partial z})(y+x).-sin(xy+yz+zx)

=- ((y+z)2cos(xy+yz+zx) + (x+z)2cos(xy+yz+zx) + (y+x)2cos(xy+yz+zx))

=-((y+z)2+ (x+z)2 + (x+z)2).cos(xy+yz+zx)

(Rightarrow) cos(xy+yz+zx) không là hàm riêng của toán tử laplace.

c.exp(x2+y2+z2)

(bigtriangledown)2 f(x,y,z) = ((frac{partial^2}{partial x^2})+ (frac{partial^2}{partial y^2})+ (frac{partial^2}{partial z^2}))exp(x2+y2+z2) =(frac{partial^2}{partial x^2})exp(x2+y2+z2)+(frac{partial^2}{partial y^2})exp(x2+y2+z2) +(frac{partial^2}{partial z^2})exp(x2+y2+z2) =(frac{partial}{partial x})2x.exp(x2+y2+z2)+(frac{partial}{partial y})2y.exp(x2+y2+z2)+(frac{partial}{partial z})2z.exp(x2+y2+z2) =2.exp(x2+y2+z2) +4×2.exp(x2+y2+z2)+2.exp(x2+y2+z2) +4y2.exp(x2+y2+z2)+2.exp(x2+y2+z2) +4z2.exp(x2+y2+z2) =(6+4×2+4y2+4z2).exp(x2+y2+z2)(Rightarrow)exp(x2+y2+z2) không là hàm riêng của hàm laplace.d.ln(xyz)(bigtriangledown)2 f(x,y,z) = ((frac{partial^2}{partial x^2})+ (frac{partial^2}{partial y^2})+ (frac{partial^2}{partial z^2}))ln(xyz) =(frac{partial^2}{partial x^2})ln(xyz)+(frac{partial^2}{partial y^2})ln(xyz)+(frac{partial^2}{partial z^2})ln(x+y+z) =(frac{partial}{partial x})yz.(frac{1}{xyz}) + (frac{partial}{partial y})xz.(frac{1}{xyz}) + (frac{partial}{partial z})xy.(frac{1}{xyz}) =(frac{partial}{partial x})(frac{1}{x}) + (frac{partial}{partial y})(frac{1}{y})+(frac{partial}{partial z})(frac{1}{z}) = – (frac{1}{x^2})- (frac{1}{y^2})- (frac{1}{z^2})(Rightarrow) ln(xyz) không là hàm riêng của hàm laplace.