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UPSC 2026 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Laplace equation: Dirichlet/Neumann, separation of variables · PDEs · asked 5× in 14 yrs · Read the full method →

Question

Two-dimensional harmonic equation in plane polar coordinates $(r, \theta)$ takes the form:

Deduce that it has solutions of the form $(Ar^n + Br^{-n})e^{\pm in\theta}$, where $A$, $B$, $n$ are constants. Determine $V$ if it satisfies the two-dimensional harmonic equation in the region $0 \leq r \leq a$, $0 \leq \theta \leq 2\pi$ and satisfies the conditions:

(i) $V$ remains finite as $r \to 0$,

(ii) $V = \displaystyle\sum_{n} c_n\cos(n\theta)$ on $r = a$.

Technique

Separation of variables $V = R(r)\Theta(\theta)$ turns Laplace’s equation on the disc into an angular oscillator plus a Cauchy–Euler radial equation $r^2R'' + rR' - n^2R = 0$, whose trial solution $R = r^m$ gives $m = \pm n$ — that is exactly the printed form. Two things then do all the work: single-valuedness in $\theta$ forces $n$ to be an integer, and $n = 0$ is a repeated indicial root, so it carries the separate pair $\{1,\ \log r\}$ that the printed form does not contain. Finiteness at the origin kills $r^{-n}$ and $\log r$; matching Fourier coefficients on $r = a$ fixes the rest.

Solution

Step 1 — Separate the variables

Look for a non-trivial product solution

Substituting into $V_{rr} + \dfrac1r V_r + \dfrac1{r^2}V_{\theta\theta} = 0$,

Multiply through by $\dfrac{r^2}{R\Theta}$ (legitimate wherever $R\Theta \neq 0$) and separate:

The left side depends on $r$ alone and the right on $\theta$ alone, so each equals a common constant. Call it $\lambda$:

Step 2 — The radial equation is Cauchy–Euler; deduce $r^{\pm n}$

Equation (2) is homogeneous of Cauchy–Euler (equidimensional) type: substituting the trial solution $R = r^m$ gives

Write $n = \sqrt{\lambda}$, so that $\lambda = n^2$ and $m = \pm n$.

• If $n \neq 0$ the indicial roots $m = n$ and $m = -n$ are distinct, and the two independent radial solutions are $r^{n}$ and $r^{-n}$:

• With $\lambda = n^2$, equation (1) reads $\Theta'' + n^2\Theta = 0$, whose independent solutions are

Multiplying,

which is the form asserted in the question. $\qquad\blacksquare$

Replacing $n$ by $-n$ merely interchanges $A \leftrightarrow B$ and the two exponentials, so no generality is lost by taking $n \ge 0$.

Step 3 — The exceptional case $n = 0$ (which the printed form omits)

If $n = 0$ the indicial equation $m^2 = 0$ has a repeated root, and $r^{n}, r^{-n}$ both collapse to the single solution $1$. The second solution is obtained from (2) directly: with $\lambda = 0$,

Likewise (1) becomes $\Theta'' = 0$, giving $\Theta = C_0 + D_0\theta$. So the $n=0$ family is

not covered by $(Ar^n + Br^{-n})e^{\pm in\theta}$. This matters here, because the boundary data in (ii) contains the term $n = 0$.

Step 4 — Single-valuedness forces $n \in \mathbb{Z}$ and kills the $\theta$-linear term

The point $(r,\theta)$ and the point $(r,\theta + 2\pi)$ are the same physical point of the disc. A potential must therefore be single-valued:

For the $n \neq 0$ family this requires $e^{in(\theta + 2\pi)} = e^{in\theta}$, i.e.

(Note this rules out complex $\lambda$ automatically: a non-real $n$ would give a factor $e^{-(\operatorname{Im}n)\theta}$, which is not $2\pi$-periodic. In particular $\lambda \ge 0$.)

For the $n = 0$ family, single-valuedness forces $D_0 = 0$, removing the $\theta$ term.

Hence, taking $n = 0,1,2,\dots$ and forming real combinations $\cos n\theta,\ \sin n\theta$ of $e^{\pm in\theta}$, the general single-valued solution on the punctured disc $0 < r \le a$ is

Step 5 — Impose condition (i): finiteness as $r \to 0$

As $r \to 0^{+}$,

while $r^{n} \to 0$ and the constant stays bounded. Since the terms of (3) are mutually orthogonal in $\theta$, each Fourier mode must be separately bounded, and boundedness of $V$ near the origin therefore forces

So

Condition (i) is not decorative — it is what makes the problem well-posed. Without it, $\log(r/a)$ and $\bigl(a^{n}r^{-n} - a^{-n}r^{n}\bigr)\cos n\theta$ are harmonic on $0 < r \le a$ and vanish identically on $r = a$; they could be added to any solution without disturbing (ii). Finiteness at the origin is exactly the condition that removes them.

Step 6 — Impose condition (ii): match Fourier coefficients on $r = a$

Take the index set in (ii) to be $n = 0,1,2,\dots$, the $n=0$ term being the constant $c_0$ (since $\cos 0 = 1$); assume $\sum_n |c_n| < \infty$ so the boundary series converges uniformly. Setting $r = a$ in (4),

Both sides are uniformly convergent Fourier series on $[0,2\pi]$, so their coefficients agree. Extracting them by orthogonality — multiply by $1$, by $\cos m\theta$, by $\sin m\theta$ and integrate over $[0,2\pi]$, using

— gives

Hence $A_n' = 0$ for every $n$ (no sine modes: the data is even in $\theta$, and so therefore is $V$), and $A_n = c_n/a^{n}$.

Step 7 — The solution, its validity, and uniqueness

It really is harmonic. For $0 \le r \le \rho < a$ we have $\bigl|c_n (r/a)^n\bigr| \le |c_n|(\rho/a)^n$, and after $k$ term-by-term differentiations each term is bounded by $|c_n|\,n^{k}\rho^{-k}(\rho/a)^{n}$; since $\rho/a < 1$ and $\sum |c_n| < \infty$, these majorants are summable. By the Weierstrass $M$-test the differentiated series converge uniformly on every closed sub-disc, so term-by-term differentiation is valid. Each term $r^{n}\cos n\theta$ satisfies the equation by Step 2, hence so does $V$.

It is the only such solution. $V$ is harmonic on the open disc and continuous on the closed disc (the boundary series converges uniformly), and by the maximum principle for harmonic functions the difference of two such solutions is harmonic with zero boundary values, hence identically zero. So the answer is unique.

Sanity value at the centre. $V(0,\theta) = c_0$, which is precisely the mean of the boundary data $\frac{1}{2\pi}\int_0^{2\pi}\sum_n c_n\cos n\theta\,d\theta = c_0$ — the mean-value property of harmonic functions. $\qquad\blacksquare$

Remark (closed integral form). Since $\dfrac{a^2-r^2}{a^2-2ar\cos(\theta-\phi)+r^2} = 1 + 2\sum_{n\ge1}\left(\dfrac{r}{a}\right)^{n}\cos n(\theta-\phi)$, the same answer can be written as the Poisson integral

which is a useful cross-check but is not needed for the marks.

Answer