Does anyone know how this works ? I have dψ = (x^2 -cy^2 )dy And my physics teacher says it is (or can be) a harmonic function (Δψ = 0) Can anyone explain ?


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Question Number 206273 by EmGent last updated on 10/Apr/24

$Does anyone know how this works ?$ $I have d ψ = (x^{2} - {c y}^{2}) d y$ $And my physics teacher says it is (or can$ $be) a harmonic function (Δ ψ = 0)$ $Can anyone explain ?$
	Answered by aleks041103 last updated on 11/Apr/24
	$Since ∃dψ, then dψ = (∂ψ/∂x)dx + (∂ψ/∂y)dy = (x^2 −cy^2 )dy ⇒ { ((∂_x ψ = 0)),((∂_y ψ = x^2 −cy^2 )) :} it is obvious that ∂_x ψ,∂_y ψ are continuous. ⇒ψ∈C^1 Also it is easily seen that ∃∂_x ^( 2) ψ,∂_x ∂_y ψ,∂_y ∂_x ψ,∂_y ^( 2) ψ. So, ψ is twice differentiable and is once continously differentiable. Therefore Schwartz rule applies: (∂^2 ψ/(∂x∂y)) = (∂^2 ψ/(∂y∂x)) BUT (∂^2 ψ/(∂x∂y)) = (∂/∂x)((∂ψ/∂y))=(∂/∂x)(x^2 −cy^2 )=2x (∂^2 ψ/(∂y∂x)) = (∂/∂y)((∂ψ/∂x))=(∂/∂y)(0)=0 BUT 2x≠0 ⇒ ∄ψ: dψ=(x^2 −cy^2 )dy Therefore, such ψ cannot be harmonic, since such ψ doesn′t exist.$
	$S i n c e \exists d ψ, t h e n$ $d ψ = \frac{\partial ψ}{\partial x} d x + \frac{\partial ψ}{\partial y} d y = (x^{2} - {c y}^{2}) d y$ $\Rightarrow {\begin{cases} \partial_{x} ψ = 0 \\ \partial_{y} ψ = x^{2} - {c y}^{2} \end{cases}$ $i t i s o b v i o u s t h a t \partial_{x} ψ, \partial_{y} ψ a r e c o n t i n u o u s .$ $\Rightarrow ψ \in C^{1}$ $A l s o i t i s e a s i l y s e e n t h a t \exists \partial_{x}^{2} ψ, \partial_{x} \partial_{y} ψ, \partial_{y} \partial_{x} ψ, \partial_{y}^{2} ψ .$ $S o, ψ i s t w i c e d i f f e r e n t i a b l e a n d i s o n c e$ $c o n t i n o u s l y d i f f e r e n t i a b l e .$ $T h e r e f o r e S c h w a r t z r u l e a p p l i e s :$ $\frac{\partial^{2} ψ}{\partial x \partial y} = \frac{\partial^{2} ψ}{\partial y \partial x}$ $B U T$ $\frac{\partial^{2} ψ}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{\partial ψ}{\partial y}) = \frac{\partial}{\partial x} (x^{2} - {c y}^{2}) = 2 x$ $\frac{\partial^{2} ψ}{\partial y \partial x} = \frac{\partial}{\partial y} (\frac{\partial ψ}{\partial x}) = \frac{\partial}{\partial y} (0) = 0$ $B U T 2 x \neq 0$ $\Rightarrow ∄ ψ : d ψ = (x^{2} - {c y}^{2}) d y$ $T h e r e f o r e, s u c h ψ c a n n o t b e h a r m o n i c,$ $s i n c e s u c h ψ {d o e s n}^{'} t e x i s t .$
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