By Hans Sagan
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Extra info for Integral and Differential Calculus : An Intuitive Approach
Example text
These solutions are valid for any C 1 -functions F , G : R → R. Remark 1. If F and G are of class C 2 , then the functions are solutions of the wave equation. ♦ Remark 2. The reason why the example is placed here is that one latently applies the chain rule in a very simple version when we calculate the partial derivative. However, this cannot be clearly seen due to all the other messages in the example. 5 Let Pn (x, y, z) be an homogeneous polynomial of degree n in three variables. Consider Pn as a function of the spherical coordinates (r, θ, ϕ).
6) When f (r, t) = tα exp − r2 ∂f = αtα−1 exp − ∂t 4t r2 , we get 4t + tα exp − r2 4t · r2 = 4t2 αtα−1 + 1 2 α−2 r2 r t exp − 4 4t , and accordingly (3) r2 ∂f = ∂t αr2 tα−1 + 1 4 α−2 r2 r t exp − 4 4t . Furthermore, ∂f r2 = tα exp − ∂r 4t · − 1 r2 r = − tα−1 r exp − 2t 2 4t , so r2 1 ∂f r2 = − tα−1 r3 exp − ∂r 2 4t and ∂ ∂r (4) r2 ∂f ∂r r2 r2 3 1 = − tα−1 r2 exp − + r2 tα−2 exp − 2 4t 4 4t 2 1 r 3 = − tα−1 r2 + r4 tα−2 exp − . com 61 Calculus 2c-2 Partial derivatives of higher order 3 By comparison we see that (3) and (4) only equals each other when α = − , corresponding to 2 the fact that only f (r, t) = t−3/2 exp − r2 4t , t > 0, of the given set of functions are solutions of r2 ∂ ∂f = ∂t ∂r r2 ∂f ∂r .
4 Prove in each of the following cases that the given function f satisfies the given differential equation everywhere in its domain. In some of the cases there occur some constants α, β, γ; check if these can be chosen freely. Notice that the variables are not x, y or z in all cases. x2 + y 2 , defined in R2 \ {0}, fulfils the differential equation 1) Prove that the function ln ∂2f ∂2f + 2 = 0. 2 ∂x ∂y 2) Prove that the function eαx cos(αy), defined in R2 , fulfils the differential equation ∂2f ∂2f + = 0.