Heat Equation with Non-Zero Temperature Boundaries — Question 4

Question 4 – Solution

Strategy. The stationary gradient and the natural storage norm both follow from the variable coefficients.

Step 1: Integrate constant stationary flux. The stationary equation is $((1+x)S')'=0$, so $(1+x)S'=C$. The endpoint difference gives $C=2/\ell$, hence $$\boxed{S=1+2\log(1+x)/\ell,\qquad j_S=-2/\ell.}$$ For $1+2x$, the quantity $((1+x)u_x)'=2$ is nonzero, so it is not stationary. Equal endpoint values do not make different interior profiles solve the same PDE.

Step 2: Transform and solve the full problem. Write $U(s,t)=u(e^s-1,t)$ on $0<s<\ell$. Then $u_x=U_s/(1+x)$ and $((1+x)u_x)_x=U_{ss}/(1+x)$. The transformed equation is $U_t=U_{ss}$, with endpoints one and three. Subtracting $1+2s/\ell$ gives one sine mode. With $\lambda=\pi^2/\ell^2$, $$\boxed{u=S+e^{-\lambda t}\sin(\pi\log(1+x)/\ell).}$$ The chain-rule identities verify the original PDE, and the sine gives zero endpoint correction and exactly the prescribed initial field.

Step 3: Check storage against boundary flux. Since $dx/(1+x)=ds$, $$H(t)=2\ell+\frac{2\ell}{\pi}e^{-\lambda t},\qquad H'=-\frac{2\pi}{\ell}e^{-\lambda t}.$$ Also $j=-U_s=-2/\ell-(\pi/\ell)e^{-\lambda t}\cos(\pi s/\ell)$. Its left value minus its right value equals the displayed derivative. The unweighted integral of $u$ is not the stored quantity in this model.

Step 4: Derive the weighted error dissipation. Multiplying $v_t/(1+x)=((1+x)v_x)_x$ by $v$ and integrating, with zero error at both ends, gives $$\frac 12\frac d{dt}\int_0^1\frac{v^2}{1+x}\,dx =-\int_0^1(1+x)v_x^2\,dx.$$ For this mode the squared norm is $(\ell/2)e^{-2\lambda t}$ and the right-hand integral is $(\pi^2/(2\ell))e^{-2\lambda t}$, agreeing with the identity. The weight is the storage coefficient; dropping it changes the energy calculation.