设u=f(v)在(0,十∞)具有二阶连续导数,且u=f(1n√X^2+y^2+z^2)满足方程d^2u/dx^2+d^2u/dy^2+d^2u/dz^2=(x^2+y^2+z^2)^-3/2.若f(1)=-2/e,f'(1)=1/e,求f(v)的具体表达式

Question

设u= f(v)在(0,十∞)具有二阶连续导数,且u=f(1n√X^ 2+y^2+z^2)满足方程d^2u/dx^2+ d^2u/dy^2+ d^2u/dz^2=(x^2+ y^2+ z^2 )^-3/2 .若f(1)=-2/e,f'(1)=1/e,求f(v)的具体表达式

Answer 1

v=1n√(x^ 2+y^2+z^2),w=x^ 2+y^2+z^2=e^(2v)
u=f(1n√(x^ 2+y^2+z^2))=f(1n(x^2+y^2+z^2)/2)=f(lnw/2)
du/dx=f'(v)(x/w),d^2u/dx^2=f''(v)(x/w)^2+f'(v)(w-2x^2)/w^2
同样：d^2u/dy^2=f''(v)(y/w)^2+f'(v)/(w-2y^2)/w^2 d^2u/dz^2=f''(v)(z/w)^2+f'(v)/(w-2z^2)/w^2
代入得：f''(v)(x/w)^2+f'(v)(w-2x^2)/w^2+
+f''(v)(y/w)^2+f'(v)/(w-2y^2)/w^2+f''(v)(z/w)^2+f'(v)/(w-2z^2)/w^2=w^（-3/2）
即：f''(v)/w+f'(v)/w==w^(-3/2）
代入得：f''(v)+f'(v)=e^(-v)
上述方程的通解为：f(v)=e^(-v)(C1-v)+C2
将f(1)=-2/e,f'(1)=1/e代入可得C1,C2.自己作吧