1
Question
If u=f(r), where r2=x2+y2+z2, then prove that:
∂2u∂x2+∂2u∂y2+∂2u∂z2=f′′(r)+2rf(r)
∂2u∂x2+∂2u∂y2+∂2u∂z2=f′′(r)+2rf(r)
Open in App
Solution
r2=x2+y2+z2,v=f(r)∂U∂x=∂U∂r⋅∂r∂x=f′(r)xr
similarly ∂U∂y=f′(r)yπ
∂U∂z=f′(π)zr
∂2U∂x2=∂∂x(f(r)xr)
=xrf′′(r)xπ+f′(r)(1r−xr2∂r∂x)
=f′′(r)x2r2+f′(r)(1r−x2r3)
∂2v∂y2=f′′(r)x2r2+f′(r)(1r−y2r3)
∂2v∂z2=f′′(r)x2r2+f′(r)(1r−z2r3)
thus RHS=f′′(r)+f′(r)2r
r2=x2+y2+z2,v=f(r)
∂U∂x=∂U∂r⋅∂r∂x
=f′(r)xr
similarly ∂U∂y=f′(r)yπ
∂U∂z=f′(π)zr
∂2U∂x2=∂∂x(f(r)xr)
=xrf′′(r)xπ+f′(r)(1r−xr2∂r∂x)
=f′′(r)x2r2+f′(r)(1r−x2r3)
∂2v∂y2=f′′(r)x2r2+f′(r)(1r−y2r3)
∂2v∂z2=f′′(r)x2r2+f′(r)(1r−z2r3)
thus RHS=f′′(r)+f′(r)2r
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
