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
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
\[ \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0, \qquad 0 < x < 1,\; 0 < y < 1, \] 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
Show 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.
Updated On: Jul 21, 2026
- \(\dfrac{7}{6}\)
- \(\dfrac{5}{6}\)
- \(\dfrac{1}{6}\)
- \(\dfrac{1}{2}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Try the simplest harmonic function that could fit.
The boundary values \(x\), \(y\), \(1+x\), \(1+y\) are all linear in \(x\) and \(y\), so try \(T(x,y) = x+y\), since any function of the form \(ax+by+c\) automatically satisfies Laplace's equation, as its second derivatives in \(x\) and \(y\) are both zero.
Step 2: Check T(x,y) = x+y against every boundary condition.
On \(y=0\): \(T(x,0) = x+0 = x\), matching \(T(x,0)=x\).
On \(x=0\): \(T(0,y) = 0+y = y\), matching \(T(0,y)=y\).
On \(y=1\): \(T(x,1) = x+1 = 1+x\), matching \(T(x,1)=1+x\).
On \(x=1\): \(T(1,y) = 1+y\), matching \(T(1,y)=1+y\).
So \(T(x,y)=x+y\) matches all four boundary edges exactly, not just at the corners.
Step 3: Use uniqueness of the Dirichlet problem.
The Laplace equation on a bounded region, together with prescribed values on the whole boundary, has exactly one solution (this is the Dirichlet problem uniqueness result). Since \(T(x,y)=x+y\) is harmonic and matches the boundary data everywhere on the boundary of the unit square, it must be the solution to this problem; there is no other function to consider.
Step 4: Evaluate at the given point.
\[ T\left(\frac{1}{2}, \frac{1}{3}\right) = \frac{1}{2}+\frac{1}{3} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}. \]
Final Answer:
\(T(1/2,1/3) = 5/6\). Options like \(7/6\), \(1/6\), or \(1/2\) come from mis-adding the fractions or guessing a wrong form for \(T\); the direct sum \(x+y\) evaluated here gives exactly \(5/6\). \[ \boxed{\dfrac{5}{6}} \]
The boundary values \(x\), \(y\), \(1+x\), \(1+y\) are all linear in \(x\) and \(y\), so try \(T(x,y) = x+y\), since any function of the form \(ax+by+c\) automatically satisfies Laplace's equation, as its second derivatives in \(x\) and \(y\) are both zero.
Step 2: Check T(x,y) = x+y against every boundary condition.
On \(y=0\): \(T(x,0) = x+0 = x\), matching \(T(x,0)=x\).
On \(x=0\): \(T(0,y) = 0+y = y\), matching \(T(0,y)=y\).
On \(y=1\): \(T(x,1) = x+1 = 1+x\), matching \(T(x,1)=1+x\).
On \(x=1\): \(T(1,y) = 1+y\), matching \(T(1,y)=1+y\).
So \(T(x,y)=x+y\) matches all four boundary edges exactly, not just at the corners.
Step 3: Use uniqueness of the Dirichlet problem.
The Laplace equation on a bounded region, together with prescribed values on the whole boundary, has exactly one solution (this is the Dirichlet problem uniqueness result). Since \(T(x,y)=x+y\) is harmonic and matches the boundary data everywhere on the boundary of the unit square, it must be the solution to this problem; there is no other function to consider.
Step 4: Evaluate at the given point.
\[ T\left(\frac{1}{2}, \frac{1}{3}\right) = \frac{1}{2}+\frac{1}{3} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}. \]
Final Answer:
\(T(1/2,1/3) = 5/6\). Options like \(7/6\), \(1/6\), or \(1/2\) come from mis-adding the fractions or guessing a wrong form for \(T\); the direct sum \(x+y\) evaluated here gives exactly \(5/6\). \[ \boxed{\dfrac{5}{6}} \]
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