Heat Equation with Non-Zero Temperature Boundaries — Question 8

Question 8 – Solution

Strategy. A convenient lifting can solve the PDE as well as its boundary traces; a less convenient lifting is still valid if every transformed term is retained.

Step 1: Verify the exact nonaffine solution. The field $\boxed{u=e^t\cosh x}$ has $u_t=u_{xx}=e^t\cosh x$. Its two endpoint values and its initial trace are exactly the prescribed ones. With lifting $P$, the remaining field is zero, with zero source and initial data.

Step 2: Transform all parts under the affine lifting. Write $p(x)=1+(\cosh 1-1)x$, so $\ell=e^tp$. Since $\ell_{xx}=0$, $$v_t-v_{xx}=-e^tp(x),\quad v(0,t)=v(1,t)=0,\quad v(x,0)=\cosh x-p(x).$$ The zero field fails both the interior source equation and the initial trace. The affine lifting satisfies the boundary values but is not itself a heat solution.

Step 3: Solve the transformed modal equations. Direct integration gives $$2\int_0^1p(x)\sin(k_nx)\,dx=\frac{2D_n}{k_n},\qquad 2\int_0^1\cosh x\sin(k_nx)\,dx=\frac{2k_nD_n}{1+k_n^2}.$$ Thus the initial coefficient is $c_n=-2D_n/[k_n(1+k_n^2)]$. The modal equation is $v_n'+k_n^2v_n=-(2D_n/k_n)e^t$, $v_n(0)=c_n$. Since $(1+k_n^2)c_n=-2D_n/k_n$, its solution is $\boxed{v_n(t)=c_ne^t}$. No additional decaying term is needed for these data.

Step 4: Reconstruct the same physical temperature. The coefficients $c_n=O(n^{-3})$ are summable. Their sine series equals the smooth zero-endpoint function $\cosh x-p(x)$ uniformly on $[0,1]$. Consequently $$u=\ell+e^t\sum_{n\ge 1}c_n\sin(k_nx) =e^t p+e^t(\cosh x-p)=e^t\cosh x.$$ This identity verifies the reconstructed PDE directly without claiming absolute uniform convergence of every differentiated sine series at the boundary. Different liftings change the auxiliary source and initial data together; retaining both transformations preserves the original solution.