{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture14 - 13.4 Cauchy-Riemann Equations Laplaces Equation...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
13.4 Cauchy-Riemann Equations. Laplace’s Equation. The Cauchy-Riemann equations are a system of two linear first-order par- tial differential equations which provide a criterion for differentiability of a complex function: Theorem 2 If a complex function f ( x + iy ) = u ( x, y ) + iv ( x, y ) is holomor- phic in U C , then the first partial derivatives of u and v exist and they satisfy the Cauchy-Riemann equations u x = v y , u y = v x , in U. (609) Theorem 3 If two real-valued continuous functions u, v : R 2 R have continuous first partial derivatives that satisfy the Cauchy-Riemann equations (609) in some domain U C , then the complex function f ( x + iy ) := u ( x, y )+ iv ( x, y ) is holomorphic in U . Remark: valid under weaker conditions (Looman, 1923; Menchoff, 1936) Sketch of Proofs: Theorem 2: For a point z 0 U , we know that the derivative of f is given by f ( z 0 ) = lim z z 0 f ( z ) f ( z 0 ) z z 0 (610) (and that this limit exists). We choose two particular paths z z 0 such that f ( x 0 + iy 0 ) = lim x x 0 u ( x, y 0 ) u ( x 0 , y 0 ) x x 0 + i lim x x 0 v ( x, y 0 ) v ( x 0 , y 0 ) x x 0 , f ( x 0 + iy 0 ) = i lim y y 0 u ( x 0 , y ) u ( x 0 , y 0 ) y y 0 + lim y y 0 v ( x 0 , y ) v ( x 0 , y 0 ) y y 0 . From these equations, we conclude that the first partial derivatives of u, v exist at ( x 0 , y 0 ) and also that u x ( x 0 , y 0 ) = v y ( x 0 , y 0 ) , u y ( x 0 , y 0 ) = v x ( x 0 , y 0 ) . (611) This is true for any point z 0 = x 0 + iy 0 U . Remark: We also conclude from this proof that f ( x 0 + iy 0 ) = u x ( x 0 , y 0 )+ iv x ( x 0 , y 0 ) = iu y ( x 0 , y 0 )+ v y ( x 0 , y 0 ) . (612) 103
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Theorem 3: Choose z 0 U . Because U is open, it contains an open ball around z 0 = ( x 0 , y 0 ), and therefore we may find a second point z = ( x, y ) U such that the line segment which connects the two points lies entirely in U . Because of the differentiability of u and v , we may apply the mean value theorem in R 2 . It guarantees the existence of points m 1 , m 2 R 2 on the line segment connecting the points ( x 0 , y 0 ) and ( x, y ), such that u ( x, y ) u ( x 0 , y 0 ) = ( x x 0 ) u x ( m 1 ) + ( y y 0 ) u y ( m 1 ) , (613) v ( x, y ) v ( x 0 , y 0 ) = ( x x 0 ) v x ( m 2 ) + ( y y 0 ) v y ( m 2 ) . (614)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}