Exercises — Heat equation (parabolic) 1D — derivation from Fourier's law

(a) ✅ This is the heat equation: one time derivative $u_t$, curvature $u_{xx}$, positive diffusivity $\alpha >0$. (b) ❌ Two time derivatives $u_{tt}$ — this is the wave equation; it oscillates, does not smooth. (c) ❌ The backward heat equation. The minus sign makes curvature drive the wrong way (hills grow, valleys deepen) — it is wildly unstable and runs heat backward in time. (d) ❌ No time at all — this is the Laplace equation, the steady state.

Apply $u_t=\alpha u_{xx}=2\times (-3)=-6$. Since $u_t<0$ the point is cooling at $6$ K/s. A bump ($u_{xx}<0$) always cools — its neighbours are colder, so heat leaks out.

Step 1 — curvature. $u_{xx}={\partial }_x^2[20+5\sin \pi x]=-5{\pi }^2\sin \pi x$. (Why twice? Because the equation feeds on $u_{xx}$.) Step 2 — evaluate. At $x=\frac{1}{4}$, $\sin (\pi /4)=\frac{\sqrt{2}}{2}$, so $u_{xx}=-5{\pi }^2\cdot \frac{\sqrt{2}}{2}=-\frac{5{\pi }^2\sqrt{2}}{2}$. Step 3 — time rate. $u_t=\alpha u_{xx}=2\cdot (-\frac{5{\pi }^2\sqrt{2}}{2})=-5{\pi }^2\sqrt{2}\approx -69.79$ K/s. Cooling (on the rising flank of a bump).

Step 1 — time derivative. $u_t=-4\alpha e^{-4\alpha t}\cos 2x$ (chain rule on the exponential). Step 2 — space derivatives. Why differentiate in space twice? Because the equation compares $u_t$ against the curvature $u_{xx}$, and curvature is the second spatial derivative — so we must go two steps deep. First step: $u_x=-2e^{-4\alpha t}\sin 2x$ (derivative of $\cos 2x$ is $-2\sin 2x$; the $-2$ is the chain-rule factor from the inner $2x$). Second step: $u_{xx}=-4e^{-4\alpha t}\cos 2x$ (derivative of $-2\sin 2x$ is $-2\cdot 2\cos 2x=-4\cos 2x$). Each differentiation of the sine/cosine pulls out another factor of the wavenumber $2$, which is why the constant grows to $4=2^2$. Step 3 — compare. $\alpha u_{xx}=-4\alpha e^{-4\alpha t}\cos 2x=u_t$. ✅ The decay rate $4\alpha$ is exactly $\alpha \times (\text{wavenumber}{)}^2=\alpha \cdot 2^2$.

$\alpha =\frac{k}{\rho c}=\frac{400}{8960\times 385}=\frac{400}{3449600}\approx 1.16\times 10^{-4}{\text{m}}^2/\text{s}.$ Copper conducts brilliantly yet stores lots of energy, so $\alpha$ is moderate, not huge.

Step 1 — plug in. $u_t=-\alpha n^2{\pi }^2{e}^{-\alpha n^2{\pi }^{2}t}\sin n\pi x$ and $u_{xx}=-n^2{\pi }^2{e}^{-\alpha n^2{\pi }^{2}t}\sin n\pi x$, so $u_t=\alpha u_{xx}$ ✅. What is ${\lambda }_n$? Notice the whole solution is $(\text{shape }\sin n\pi x)\times e^{-{\lambda }_{n}t}$ with the number ${\lambda }_n=\alpha n^2{\pi }^2$ sitting in the exponent. We call ${\lambda }_n$ the decay rate because $e^{-{\lambda }_{n}t}$ is a shrinking exponential: the larger ${\lambda }_n$, the faster the factor $e^{-{\lambda }_{n}t}$ collapses toward zero. So ${\lambda }_n$ literally measures "how quickly this mode dies." Step 2 — compare rates. $\frac{{\lambda }_3}{{\lambda }_1}=\frac{\alpha 3^2{\pi }^2}{\alpha 1^2{\pi }^2}=\frac{3^2}{1^2}=9$. Conclusion: the $n=3$ mode decays 9 times faster. Why physically? Higher $n$ = tighter wiggles = sharper curvature $u_{xx}$; sharper curvature drives faster smoothing. The heat equation erases fine detail first.

Step 1 — steady means no time change. $u_t=0\Rightarrow \alpha u_{xx}=0\Rightarrow u_{xx}=0$. Step 2 — integrate twice. $u_{xx}=0\Rightarrow u_{\infty }=a+bx$ (a straight line — the only curvature-free profile). Step 3 — fit ends. $u_{\infty }(0)=a=100$; $u_{\infty }(1)=100+b=40\Rightarrow b=-60$. Answer: $u_{\infty }(x)=100-60x$. At $x=\frac{1}{2}$, $u_{\infty }=70$ K.

Step 1 — chain rule in time. $\frac{\partial u}{\partial t}=\frac{\partial u}{\partial \tau }\frac{d\tau }{dt}=u_{\tau }\cdot \frac{\alpha }{L^2}$. Step 2 — chain rule in space (twice). $\frac{\partial u}{\partial x}=u_{\xi }\cdot \frac{1}{L}$, so $\frac{{\partial }^{2}u}{\partial x^2}=u_{\xi \xi }\cdot \frac{1}{L^2}$. Step 3 — substitute. $u_{\tau }\frac{\alpha }{L^2}=\alpha u_{\xi \xi }\frac{1}{L^2}$. Cancel $\alpha /L^2$: $u_{\tau }=u_{\xi \xi }$. Meaning: every rod is the same rod once you measure time in units of $L^2/\alpha$ (the natural diffusion time). This is why bigger objects take $\sim L^2$ longer to equilibrate — a headline of Separation of Variables for the Heat Equation.

Step 1 — linearity. The heat equation is linear, so a sum of solutions is a solution; each term already satisfies it (rates $\alpha {\pi }^2$ and $9\alpha {\pi }^2$ from Problem 3.1). Step 2 — check term one. For $u_1=3e^{-\alpha {\pi }^{2}t}\sin \pi x$: $u_{1,t}=-3\alpha {\pi }^2{e}^{-\alpha {\pi }^{2}t}\sin \pi x$; and $u_{1,xx}=-3{\pi }^2{e}^{-\alpha {\pi }^{2}t}\sin \pi x$, so $\alpha u_{1,xx}=u_{1,t}$ ✅. Step 3 — check term two explicitly. For $u_2=2e^{-9\alpha {\pi }^{2}t}\sin 3\pi x$, do not hand-wave — run the same two derivatives. Time: $u_{2,t}=-9\alpha {\pi }^2\cdot 2e^{-9\alpha {\pi }^{2}t}\sin 3\pi x$ (the exponent $-9\alpha {\pi }^2$ drops out front). Space, twice: $u_{2,x}=2\cdot 3\pi e^{-9\alpha {\pi }^{2}t}\cos 3\pi x$, then $u_{2,xx}=-2\cdot 9{\pi }^2{e}^{-9\alpha {\pi }^{2}t}\sin 3\pi x$ (each $x$-derivative of $\sin 3\pi x/\cos 3\pi x$ pulls out a factor $3\pi$, so two of them give $(3\pi {)}^2=9{\pi }^2$). Multiply by $\alpha$: $\alpha u_{2,xx}=-18\alpha {\pi }^2{e}^{-9\alpha {\pi }^{2}t}\sin 3\pi x=u_{2,t}$ ✅. The rate $9\alpha {\pi }^2$ is exactly the $n=3$ case of ${\lambda }_n=\alpha n^2{\pi }^2$. Step 4 — evaluate at $x=\frac{1}{2}$, $t=0$. $\sin (\pi /2)=1$, $\sin (3\pi /2)=-1$: $u=3(1)(1)+2(1)(-1)=3-2=1$ K. This superposition is exactly a two-term Fourier series.

Why this exact shape? A point-burst has no built-in length scale, so the profile can only spread by the natural diffusion width $\sqrt{4\alpha t}$ (from Problem 4.1, lengths scale as $\sqrt{\alpha t}$). Two consequences are baked in: the width in the exponent must be $4\alpha t$, and the prefactor must be $1/\sqrt{4\pi \alpha t}$ so that the total heat $\int udx$ stays constant as the bump flattens and widens. We now verify it directly — no ansatz gymnastics, just the two derivatives the equation asks for. Step 1 — time derivative. Write $u=C{t}^{-1/2}{e}^{-x^2/(4\alpha t)}$ with the constant $C=\frac{1}{\sqrt{4\pi \alpha }}$. Differentiating the product ($t^{-1/2}$ gives $-\frac{1}{2}{t}^{-3/2}$; the exponent $-x^2/(4\alpha t)$ has $t$-derivative $+x^2/(4\alpha t^2)$): $u_t=C\left[-\frac{1}{2}{t}^{-3/2}+t^{-1/2}\cdot \frac{x^2}{4\alpha t^2}\right]e^{-x^2/(4\alpha t)}=u\left(-\frac{1}{2t}+\frac{x^2}{4\alpha t^2}\right).$ Step 2 — space derivatives. $u_x=u\cdot \left(-\frac{x}{2\alpha t}\right)$, and differentiating again (product of $u$ and $-x/(2\alpha t)$): $u_{xx}=u\left[{\left(-\frac{x}{2\alpha t}\right)}^2-\frac{1}{2\alpha t}\right]=u\left(\frac{x^2}{4{\alpha }^2{t}^2}-\frac{1}{2\alpha t}\right).$ Step 3 — multiply by $\alpha$. $\alpha u_{xx}=u\left(\frac{x^2}{4\alpha t^2}-\frac{1}{2t}\right)=u_t$. ✅ Edge case $t\to 0^{+}$. Watch the shape: the width $\sqrt{4\alpha t}\to 0$, so the bump becomes infinitely narrow, while the prefactor $1/\sqrt{4\pi \alpha t}\to \infty$, so it becomes infinitely tall — yet the area under it stays fixed at $1$. This limiting object is the Dirac delta $\delta (x)$: "all the heat piled at a single point." So the Gaussian is the solution whose initial condition is a point-burst $u(x,0)=\delta (x)$, and for any $t>0$ it is a smooth spread-out bump. This is why it is called the fundamental solution / heat kernel — the building block from which Fourier Series Solutions and general initial data are assembled.

Step 1 — differentiate the energy. $\frac{dE}{dt}={\int }_0^{1}2u{u}_{t}dx={\int }_0^{1}2u(\alpha u_{xx})dx$ (used the PDE). Step 2 — integrate by parts. ${\int }_0^{1}u{u}_{xx}dx=[u{u}_x{]}_0^1-{\int }_0^1(u_x{)}^{2}dx$. Step 3 — kill the boundary term. Insulated ends give $u_x=0$ at $x=0,1$, so $[u{u}_x{]}_0^1=0$. Step 4 — conclude. $\frac{dE}{dt}=-2\alpha {\int }_0^1(u_x{)}^{2}dx\le 0$, since $\alpha >0$ and $(u_x{)}^2\ge 0$. Meaning: the "size" of the temperature profile can only shrink or hold steady — a rigorous face of diffusion smooths, never sharpens. This is the energy-method cousin of Conservation Laws and Continuity Equation.