Laplace's Equation — Question 3

Question 3 – Solution

Strategy. Boundary polynomials can identify an exact harmonic field, but its interior stationary point must be classified rather than assumed to be an extremum.

Step 1: Reconstruct and verify the polynomial. A general quadratic $aX^2+bXY+cY^2+dX+eY+f$ is harmonic precisely when $c=-a$. The top and bottom traces force $a=1$, $b=2$, $d=e=0$, and $f=0$. The side traces then agree: $$\boxed{u=X^2+2XY-Y^2.}$$ At the corners its values are $1/2,-1/2,-1/2,1/2$, in the order $(0,0),(1,0),(0,1),(1,1)$, whether computed from horizontal or vertical edges. The difference of two continuous harmonic solutions has zero boundary data; its maximum and minimum are zero.

Step 2: Classify the critical point. The gradient is $(2X+2Y,2X-2Y)$, so its sole zero is $(X,Y)=(0,0)$. The Hessian has eigenvalues $2\sqrt 2$ and $-2\sqrt 2$, hence the center is a saddle. In particular $u(X,0)=X^2$ while $u(0,Y)=-Y^2$. The maximum principle prohibits a nonconstant harmonic field’s interior maximum or minimum, not a saddle.

Step 3: Locate the actual extrema. On the bottom, $X^2-X-1/4$ decreases on $[-1/2,1/2]$, with range $[-1/2,1/2]$; the top has the same range. On the left, $1/4-Y-Y^2$ decreases over the allowed interval, also with that range, and the right behaves symmetrically. Therefore $$\boxed{\max u=\tfrac 12\text{ at }(0,0),(1,1),\qquad \min u=-\tfrac 12\text{ at }(1,0),(0,1).}$$ No interior point attains either value by the strong maximum principle.

Step 4: Draw the zero set and check flux. Solving $X^2+2XY-Y^2=0$ gives the two lines $Y=(1\pm\sqrt 2)X$, clipped to the square. The horizontal regions are positive and the vertical ones negative. Direct integration of $-\partial_nu$ gives flux $-1$ on each vertical edge and $+1$ on each horizontal edge; the sum is zero, also following from $\Delta u=0$.