Question

Consider the Laplace equation
\[ \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0, \qquad 0 < x < 1,\; 0 < y < 1, \] with the boundary conditions
\[ T(x,0) = x, \qquad T(0,y) = y \] \[ T(x,1) = 1+x, \qquad T(1,y) = 1+y. \] Then the value of \(T\left(\dfrac{1}{2}, \dfrac{1}{3}\right)\) is equal to

• A. \(\dfrac{7}{6}\)
• B. \(\dfrac{5}{6}\)
• C. \(\dfrac{1}{6}\)
• D. \(\dfrac{1}{2}\)

Hint:

Try T(x,y) = x+y; it is harmonic and matches all four boundary conditions exactly, so by uniqueness of the Dirichlet problem it is the solution.

Answer 1

The Correct Option is B

Instead of directly guessing the answer, build it by comparing $T$ to the plane function $x+y$ using the maximum principle for harmonic functions.

Define $w(x,y) = T(x,y) - (x+y)$. Since $x+y$ is harmonic (linear functions always satisfy Laplace's equation, as their second partial derivatives are zero), and $T$ is harmonic by the problem statement, the difference $w$ is also harmonic on the square, being a difference of two harmonic functions.

Now check what $w$ looks like on each edge of the square, using the given boundary values of $T$:

1. On $y=0$: $w(x,0) = T(x,0) - (x+0) = x - x = 0$.
2. On $x=0$: $w(0,y) = T(0,y) - (0+y) = y - y = 0$.
3. On $y=1$: $w(x,1) = T(x,1) - (x+1) = (1+x) - (x+1) = 0$.
4. On $x=1$: $w(1,y) = T(1,y) - (1+y) = (1+y) - (1+y) = 0$.

So $w$ is harmonic on the whole square and equals $0$ on all four edges of the boundary. By the maximum principle, a harmonic function attains both its maximum and minimum on the boundary of the region. Since $w$ is $0$ everywhere on the boundary, its maximum and minimum are both $0$, which forces $w(x,y)=0$ at every point inside the square too.

This means $T(x,y) - (x+y) = 0$ for all $(x,y)$ in the square, so $T(x,y) = x+y$ everywhere.

Let's summarize:

• Writing $w = T-(x+y)$ turns the problem into checking a harmonic function with all zero boundary data.
• The maximum principle says such a $w$ must be identically zero inside the region, not just on the boundary.
• So $T(x,y)=x+y$ exactly, and $T(1/2,1/3) = 1/2+1/3 = 5/6$.

So the value of $T(1/2, 1/3)$ is $5/6$.

\[\boxed{T(1/2,1/3) = \dfrac{5}{6}}\]