← 2018 Paper 2
UPSC 2018 Maths Optional Paper 2 Q1c — Step-by-Step Solution 10 marks · Section A
Harmonic functions and harmonic conjugate · Complex Analysis · asked 7× in 14 yrs · Read the full method →
Question
Prove that the function u ( x , y ) = ( x − 1 ) 3 − 3 x y 2 + 3 y 2 u(x,y)=(x-1)^3-3xy^2+3y^2 u ( x , y ) = ( x − 1 ) 3 − 3 x y 2 + 3 y 2 f ( z ) f(z) f ( z ) z z z
Technique
Verify Laplace’s equation; recover the conjugate by integrating the Cauchy–Riemann equations; assemble f ( z ) f(z) f ( z ) f ′ ( z ) = u x − i u y f'(z)=u_x-iu_y f ′ ( z ) = u x − i u y y = 0 , x = z y=0,x=z y = 0 , x = z
Solution
Step 1 — Show u u u
Compute the first and second partials of u = ( x − 1 ) 3 − 3 x y 2 + 3 y 2 u=(x-1)^3-3xy^2+3y^2 u = ( x − 1 ) 3 − 3 x y 2 + 3 y 2
u x = 3 ( x − 1 ) 2 − 3 y 2 , u x x = 6 ( x − 1 ) = 6 x − 6 , u_x=3(x-1)^2-3y^2,\qquad u_{xx}=6(x-1)=6x-6, u x = 3 ( x − 1 ) 2 − 3 y 2 , u xx = 6 ( x − 1 ) = 6 x − 6 , u y = − 6 x y + 6 y , u y y = − 6 x + 6. u_y=-6xy+6y,\qquad u_{yy}=-6x+6. u y = − 6 x y + 6 y , u y y = − 6 x + 6. Hence
∇ 2 u = u x x + u y y = ( 6 x − 6 ) + ( − 6 x + 6 ) = 0. \nabla^2u=u_{xx}+u_{yy}=(6x-6)+(-6x+6)=0. ∇ 2 u = u xx + u y y = ( 6 x − 6 ) + ( − 6 x + 6 ) = 0. So u u u R 2 \mathbb R^2 R 2 u u u ✓ \checkmark ✓
Step 2 — Find the harmonic conjugate v v v
We seek v ( x , y ) v(x,y) v ( x , y ) f = u + i v f=u+iv f = u + i v
v y = u x , v x = − u y . v_y=u_x,\qquad v_x=-u_y. v y = u x , v x = − u y . From the first,
v y = u x = 3 ( x − 1 ) 2 − 3 y 2 = 3 x 2 − 6 x + 3 − 3 y 2 . v_y=u_x=3(x-1)^2-3y^2=3x^2-6x+3-3y^2. v y = u x = 3 ( x − 1 ) 2 − 3 y 2 = 3 x 2 − 6 x + 3 − 3 y 2 . Integrate with respect to y y y x x x
v = ∫ ( 3 x 2 − 6 x + 3 − 3 y 2 ) d y = ( 3 x 2 − 6 x + 3 ) y − y 3 + g ( x ) , v=\int(3x^2-6x+3-3y^2)\,dy=(3x^2-6x+3)y-y^3+g(x), v = ∫ ( 3 x 2 − 6 x + 3 − 3 y 2 ) d y = ( 3 x 2 − 6 x + 3 ) y − y 3 + g ( x ) , i.e. v = 3 x 2 y − 6 x y + 3 y − y 3 + g ( x ) v=3x^2y-6xy+3y-y^3+g(x) v = 3 x 2 y − 6 x y + 3 y − y 3 + g ( x ) g g g x x x
Now impose the second CR equation. We have − u y = 6 x y − 6 y -u_y=6xy-6y − u y = 6 x y − 6 y
v x = 6 x y − 6 y + g ′ ( x ) . v_x=6xy-6y+g'(x). v x = 6 x y − 6 y + g ′ ( x ) . Setting v x = − u y v_x=-u_y v x = − u y
6 x y − 6 y + g ′ ( x ) = 6 x y − 6 y ⟹ g ′ ( x ) = 0 ⟹ g ( x ) = C ( const ) . 6xy-6y+g'(x)=6xy-6y\ \Longrightarrow\ g'(x)=0\ \Longrightarrow\ g(x)=C\ (\text{const}). 6 x y − 6 y + g ′ ( x ) = 6 x y − 6 y ⟹ g ′ ( x ) = 0 ⟹ g ( x ) = C ( const ) . Therefore the harmonic conjugate is
v ( x , y ) = 3 x 2 y − 6 x y − y 3 + 3 y + C . \boxed{\,v(x,y)=3x^2y-6xy-y^3+3y+C.\,} v ( x , y ) = 3 x 2 y − 6 x y − y 3 + 3 y + C . Step 3 — Build f ( z ) = u + i v f(z)=u+iv f ( z ) = u + i v z z z
Use the Milne–Thomson method: for an analytic f f f
f ′ ( z ) = u x − i u y . f'(z)=u_x-i\,u_y. f ′ ( z ) = u x − i u y . Substitute y = 0 , x = z y=0,\ x=z y = 0 , x = z u x − i u y u_x-iu_y u x − i u y
u x − i u y ∣ y = 0 , x = z = ( 3 ( z − 1 ) 2 − 0 ) − i ( 0 ) = 3 ( z − 1 ) 2 = 3 z 2 − 6 z + 3. u_x-iu_y\big|_{y=0,\,x=z}=\big(3(z-1)^2-0\big)-i\,(0)=3(z-1)^2=3z^2-6z+3. u x − i u y y = 0 , x = z = ( 3 ( z − 1 ) 2 − 0 ) − i ( 0 ) = 3 ( z − 1 ) 2 = 3 z 2 − 6 z + 3. Integrate:
f ( z ) = ∫ ( 3 z 2 − 6 z + 3 ) d z = z 3 − 3 z 2 + 3 z + const . f(z)=\int(3z^2-6z+3)\,dz=z^3-3z^2+3z+\text{const}. f ( z ) = ∫ ( 3 z 2 − 6 z + 3 ) d z = z 3 − 3 z 2 + 3 z + const . Recognizing the pattern,
z 3 − 3 z 2 + 3 z = ( z − 1 ) 3 + 1 , z^3-3z^2+3z=(z-1)^3+1, z 3 − 3 z 2 + 3 z = ( z − 1 ) 3 + 1 , so up to an additive constant
Answer
f ( z ) = ( z − 1 ) 3 + i C ′ , e.g. f ( z ) = ( z − 1 ) 3 ( taking the constant = 0 ) . \boxed{\,f(z)=(z-1)^3+iC',\qquad\text{e.g. }f(z)=(z-1)^3\ (\text{taking the constant }=0).\,} f ( z ) = ( z − 1 ) 3 + i C ′ , e.g. f ( z ) = ( z − 1 ) 3 ( taking the constant = 0 ) .