设u=f(r).r=根号下x^+y^,其中f是二阶可微函数,且f(1)=1,f '(1)=1 u对x的二次偏导+对y的二次偏导=0.求f(r)

u=f(r),r=√(x2+y2)
∂r/∂x=x/√(x2+y2),∂r/∂y=y/√(x2+y2)
∂2r/∂x2=[√(x2+y2)-x2/√(x2+y2)]/(x2+y2)=y2/[(x2+y2)√(x2+y2)]
∂2r/∂y2=[√(x2+y2)-y2/√(x2+y2)]/(x2+y2)=x2/[(x2+y2)√(x2+y2)]
∂u/∂x=∂f/∂r*∂r/∂x,∂u/∂y=∂f/∂r*∂r/∂y
∂2u/∂x2=∂[∂f/∂r*∂r/∂x]/∂x=∂[∂f/∂r]/∂x*∂r/∂x+∂f/∂r*∂[∂r/∂x]/∂x=∂2f/∂r2*(∂r/∂x)2+∂f/∂r*∂2r/∂x2
∂2u/∂y2=∂[∂f/∂r*∂r/∂y]/∂y=∂[∂f/∂r]/∂y*∂r/∂y+∂f/∂r*∂[∂r/∂y]/∂y=∂2f/∂r2*(∂r/∂y)2+∂f/∂r*∂2r/∂y2
∂2u/∂x2+∂2u/∂y2=[∂2f/∂r2*(∂r/∂x)2+∂f/∂r*∂2r/∂x2]+[∂2f/∂r2*(∂r/∂y)2+∂f/∂r*∂2r/∂y2]
=∂2f/∂r2*[(∂r/∂x)2+(∂r/∂y)2]+∂f/∂r*[∂2r/∂x2+∂2r/∂y2]
=∂2f/∂r2*{(x2+y2)/[√(x2+y2)]2}+∂f/∂r*{(x2+y2)/[(x2+y2)√(x2+y2)]}
=∂2f/∂r2*1+∂f/∂r*[1/√(x2+y2)]
=∂2f/∂r2+∂f/∂r*1/r
∂2u/∂x2+∂2u/∂y2=∂2f/∂r2+∂f/∂r*1/r=0 => r*∂2f/∂r2+∂f/∂r=0
∫r*(∂2f/∂r2)dr+∫(∂f/∂r)dr=∫0dr=C
∫rd(∂f/∂r)+∫(∂f/∂r)dr =C
r*(∂f/∂r)-∫(∂f/∂r)dr+∫(∂f/∂r)dr=C
r*(∂f/∂r)=C => (∂f/∂r)=f'(r)=C/r
∫f'(r)dr=∫C/r*dr => ∫df(r)=C∫dlnr
f(r)=Clnr+C1
f(1)=1,=> f(1)=Cln1+C1=0+C1=C1=1
f'(1)=1,=> f’(1)=C/1=C=1
f(r)=lnr+1=ln[√(x2+y2)]+1