← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

Harmonic functions and harmonic conjugate · Complex Analysis · asked 7× in 13 yrs · Read the full method →

Question

Show that the function $v(x,y)=\ln(x^2+y^2)+x+y$ is harmonic. Find its conjugate harmonic function $u(x,y)$. Also, find the corresponding analytic function $f(z)=u+iv$ in terms of $z$.

Technique

Verify $\nabla^2 v=0$; reconstruct $u$ via Cauchy–Riemann; identify $f(z)$ by recognising $\ln(x^2+y^2)$ and $\arctan(y/x)$ as $\Re$ and $\Im$ of $\ln z$.

Solution

Step 1 — Check $v$ is harmonic: $\nabla^2 v=v_{xx}+v_{yy}=0$

$v_x=\dfrac{2x}{x^2+y^2}+1$, $v_{xx}=\dfrac{2(x^2+y^2)-2x\cdot 2x}{(x^2+y^2)^2}=\dfrac{2(y^2-x^2)}{(x^2+y^2)^2}$.

$v_y=\dfrac{2y}{x^2+y^2}+1$, $v_{yy}=\dfrac{2(x^2+y^2)-2y\cdot 2y}{(x^2+y^2)^2}=\dfrac{2(x^2-y^2)}{(x^2+y^2)^2}$.

$v_{xx}+v_{yy}=\dfrac{2(y^2-x^2)+2(x^2-y^2)}{(x^2+y^2)^2}=0$ ✓.

$v$ is harmonic (on the punctured plane $\mathbb C\setminus\{0\}$).

Step 2 — Find harmonic conjugate $u$ via Cauchy–Riemann

For $f=u+iv$ to be analytic: $u_x=v_y$ and $u_y=-v_x$.

$u_x=v_y=\dfrac{2y}{x^2+y^2}+1$.

Integrate w.r.t. $x$:

Hmm — actually $\int dx/(x^2+y^2)=(1/y)\arctan(x/y)$, so $\int 2y/(x^2+y^2)\,dx=2\arctan(x/y)$. (Assume $y>0$; choose principal branch.)

Differentiate w.r.t. $y$ and match $u_y=-v_x$:

Required: $u_y=-v_x=-\dfrac{2x}{x^2+y^2}-1$.

Match: $-\dfrac{2x}{x^2+y^2}+g'(y)=-\dfrac{2x}{x^2+y^2}-1\;\Rightarrow\;g'(y)=-1\;\Rightarrow\;g(y)=-y+C$.

So $u=2\arctan(x/y)+x-y+C$. Take $C=0$:

Equivalently (using $\arctan(x/y)=\pi/2-\arctan(y/x)$ in the relevant branch): $u=-2\arctan(y/x)+x-y+\pi$, which differs by a constant. The conventional form is

(Choice differs by constant of integration. We use this form below.)

Step 3 — Analytic function $f(z)=u+iv$ in terms of $z$

Note $\ln(x^2+y^2)=2\ln|z|=2\,\Re(\ln z)$, and $\arctan(y/x)=\arg z=\Im(\ln z)$.

So $\ln z=\ln|z|+i\arg z$, giving:

• $\Re(\ln z)=\ln|z|=\tfrac{1}{2}\ln(x^2+y^2)$
• $\Im(\ln z)=\arg z=\arctan(y/x)$

Thus $2i\ln z=2i\ln|z|-2\arg z=-2\arctan(y/x)+i\ln(x^2+y^2)$.

So $-2\arctan(y/x)=\Re(2i\ln z)$ and $\ln(x^2+y^2)=\Im(2i\ln z)$.

Now $f(z)=u+iv$:

(Check the second bracket: $(x-y)+i(x+y)=x+iy+i(x+iy)\cdot$ wait, $i(x+iy)=ix-y$, so $z+iz=(x+iy)+(ix-y)=(x-y)+i(x+y)$ ✓.)

Answer