← 2023 Paper 2

UPSC 2023 Maths Optional Paper 2 Q3b — Step-by-Step Solution

15 marks · Section A

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

Question

Prove that u(x,y)=ex(xcos⁡y−ysin⁡y)u(x,y)=e^x(x\cos y-y\sin y) is harmonic. Find its conjugate harmonic function v(x,y)v(x,y) and express the corresponding analytic function f(z)f(z) in terms of zz.

Technique

Verify Laplace; integrate one CR equation, then differentiate to pin down the integration “constant” (function of xx).

Solution

Part 1 — Harmonic check

Step 1 — First partials.

ux=ex(xcos⁡y−ysin⁡y)+excos⁡y=ex[(x+1)cos⁡y−ysin⁡y],u_x=e^{x}(x\cos y-y\sin y)+e^{x}\cos y=e^{x}\bigl[(x+1)\cos y-y\sin y\bigr], uy=ex(−xsin⁡y−sin⁡y−ycos⁡y)=−ex[(x+1)sin⁡y+ycos⁡y].u_y=e^{x}(-x\sin y-\sin y-y\cos y)=-e^{x}\bigl[(x+1)\sin y+y\cos y\bigr].

Step 2 — Second partials.

uxx=ex[(x+1)cos⁡y−ysin⁡y]+excos⁡y=ex[(x+2)cos⁡y−ysin⁡y].u_{xx}=e^{x}\bigl[(x+1)\cos y-y\sin y\bigr]+e^{x}\cos y=e^{x}\bigl[(x+2)\cos y-y\sin y\bigr]. uyy=∂∂y{−ex[(x+1)sin⁡y+ycos⁡y]}=−ex[(x+1)cos⁡y+cos⁡y−ysin⁡y]=−ex[(x+2)cos⁡y−ysin⁡y].u_{yy}=\frac{\partial}{\partial y}\bigl\{-e^{x}\bigl[(x+1)\sin y+y\cos y\bigr]\bigr\}=-e^{x}\bigl[(x+1)\cos y+\cos y-y\sin y\bigr]=-e^{x}\bigl[(x+2)\cos y-y\sin y\bigr].

Step 3 — Verify Laplace.

uxx+uyy=ex[(x+2)cos⁡y−ysin⁡y]−ex[(x+2)cos⁡y−ysin⁡y]=0.u_{xx}+u_{yy}=e^{x}\bigl[(x+2)\cos y-y\sin y\bigr]-e^{x}\bigl[(x+2)\cos y-y\sin y\bigr]=0.

So uu is harmonic on R2\mathbb{R}^{2}.

Part 2 — Find the harmonic conjugate vv

The Cauchy–Riemann equations give vy=uxv_y=u_x and vx=−uyv_x=-u_y.

Step 4 — Integrate vy=uxv_y=u_x w.r.t. yy.

v=∫ex[(x+1)cos⁡y−ysin⁡y]dy=ex ⁣ ⁣∫(x+1)cos⁡y dy−ex ⁣ ⁣∫ysin⁡y dy.v=\int e^{x}\bigl[(x+1)\cos y-y\sin y\bigr]dy=e^{x}\!\!\int(x+1)\cos y\,dy-e^{x}\!\!\int y\sin y\,dy.

The first is (x+1)sin⁡y(x+1)\sin y. For the second, ∫ysin⁡y dy=−ycos⁡y+sin⁡y\int y\sin y\,dy=-y\cos y+\sin y, so −∫ysin⁡y dy=ycos⁡y−sin⁡y-\int y\sin y\,dy=y\cos y-\sin y. Combining,

v=ex[(x+1)sin⁡y+ycos⁡y−sin⁡y]+g(x)=ex[xsin⁡y+ycos⁡y]+g(x).v=e^{x}\bigl[(x+1)\sin y+y\cos y-\sin y\bigr]+g(x)=e^{x}\bigl[x\sin y+y\cos y\bigr]+g(x).

Step 5 — Determine g(x)g(x) from vx=−uyv_x=-u_y.

vx=ex[xsin⁡y+ycos⁡y]+exsin⁡y+g′(x)=ex[(x+1)sin⁡y+ycos⁡y]+g′(x).v_x=e^{x}\bigl[x\sin y+y\cos y\bigr]+e^{x}\sin y+g'(x)=e^{x}\bigl[(x+1)\sin y+y\cos y\bigr]+g'(x).

But −uy=ex[(x+1)sin⁡y+ycos⁡y]-u_y=e^{x}\bigl[(x+1)\sin y+y\cos y\bigr]. Equating, g′(x)=0g'(x)=0, so g(x)=Cg(x)=C (constant). Take C=0C=0 (constants in vv shift ff by an imaginary constant; conventional to set C=0C=0):

  v(x,y)=ex(xsin⁡y+ycos⁡y).  \boxed{\;v(x,y)=e^{x}\bigl(x\sin y+y\cos y\bigr).\;}

Part 3 — Express f(z)f(z) in terms of zz

f(z)=u+iv=ex[(xcos⁡y−ysin⁡y)+i(xsin⁡y+ycos⁡y)].f(z)=u+iv=e^{x}\bigl[(x\cos y-y\sin y)+i(x\sin y+y\cos y)\bigr].

Group:

xcos⁡y+ixsin⁡y=x(cos⁡y+isin⁡y)=xeiy,x\cos y+ix\sin y=x(\cos y+i\sin y)=xe^{iy}, −ysin⁡y+iycos⁡y=iy(cos⁡y+isin⁡y)=iy eiy.-y\sin y+iy\cos y=iy(\cos y+i\sin y)=iy\,e^{iy}.

Therefore

f(z)=ex(xeiy+iyeiy)=exeiy(x+iy)=ex+iy(x+iy)=z ez.f(z)=e^{x}\bigl(x e^{iy}+iy e^{iy}\bigr)=e^{x}e^{iy}(x+iy)=e^{x+iy}(x+iy)=z\,e^{z}.

Answer

  f(z)=z ez.  \boxed{\;f(z)=z\,e^{z}.\;}
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