Terminology — Question 4

Question 4 – Solution

Strategy. At a corner, two different traces must describe the same value and the same time derivative.

Step 1: Match the values. Continuity at $(0,0)$ and $(1,0)$ requires $$\boxed{A(0)=f(0)=0,\qquad B(0)=f(1)=0.}$$ If these equalities fail, even a continuous solution on the entire closed rectangle is impossible. Interior smoothness alone does not imply these equalities.

Step 2: Match the PDE to the time traces. The left boundary trace gives $u_t(0,0)=A'(0)$ and the initial trace gives $u_{xx}(0,0)=f''(0)$. The PDE therefore requires $A'(0)=f''(0)$. The right corner gives the analogous identity. Since $f''=-2$, $$\boxed{A'(0)=B'(0)=-2.}$$ These conditions use the stated corner regularity; they cannot be imposed silently on a class that has no continuous $u_{xx}$ at the corners.

Step 3: Diagnose homogeneous endpoint data. For $A=B=0$, both value conditions hold, but $A'(0)=B'(0)=0\ne-2$. Thus no solution exists in the corner-smooth class specified here. This is a regularity obstruction at $t=0$, not a proof of nonexistence in a weaker class. A solution may be continuous up to the initial edge and smooth for $t>0$ without its second spatial derivative extending continuously through the initial corners. No such weaker solution is needed for the contradiction.

Step 4: Construct a compatible full example. Take $A(t)=B(t)=-2t$ and $$\boxed{u(x,t)=x(1-x)-2t.}$$ Then $u_t=u_{xx}=-2$, $u(x,0)=f(x)$ and $u(0,t)=u(1,t)=-2t$. This polynomial has all required corner derivatives and proves existence for the modified problem. Matching a finite list of necessary compatibility conditions would not alone prove existence for arbitrary data; the explicit field supplies the sufficiency argument in this case.