Terminology — Question 8

Question 8 – Solution

Strategy. Subtract the lifting from every part of the problem, including the differential equation and the initial condition.

Step 1: Compute the transformed full problem. Since $\ell_{xx}=0$, $$\boxed{v_t-\kappa v_{xx}=-\ell_t =-\left(1-\frac{x}{L}\right)A'(t)-\frac{x}{L}B'(t).}$$ The endpoints are $v(0,t)=v(L,t)=0$, and $v(x,0)=f(x)-\ell(x,0)$. The endpoint conditions are homogeneous Dirichlet. The transformed PDE is linear, but generally nonhomogeneous. Its initial condition is homogeneous only if $f=\ell(\cdot,0)$.

Step 2: Determine when the source vanishes. For each $t$, the displayed affine function of $x$ vanishes throughout $(0,L)$ if and only if both its endpoint limits vanish. Thus $A'(t)=B'(t)=0$ for all $t$, or equivalently $\boxed{A,B\text{ are constant in time}}$. Nonzero constant boundary values can therefore produce a homogeneous transformed PDE, although the original boundary data were nonhomogeneous.

Step 3: Inspect time-dependent endpoint values. For $A=t,B=2t$, $\ell=t(1+x/L)$, so $$v_t-\kappa v_{xx}=-(1+x/L),\qquad v(0,t)=v(L,t)=0,\qquad v(x,0)=0.$$ The boundary forcing has moved into the interior equation; it has not been removed from the full problem. In particular, $v=0$ would fail this PDE. The stated value compatibility holds, but no corner-smooth existence is claimed for these data; higher compatibility would require additional checks.

Step 4: Track the freedom in the lifting. Let $\widetilde v=u-\widetilde\ell=v-q$. Direct differentiation gives $$\widetilde v_t-\kappa\widetilde v_{xx} =-\ell_t-q_t+\kappa q_{xx},\qquad \widetilde v(x,0)=f(x)-\ell(x,0)-q(x,0).$$ The endpoints still vanish. For example, $q=x(L-x)e^{-t}$ is a nonzero smooth choice with zero endpoints. Although the source and transformed initial data change, $u=v+\ell=\widetilde v+\widetilde\ell$ is identical. Each lifting gives an equivalent formulation when every datum is transformed consistently.