Intuition The one-sentence pictureHeat flows from hot to cold, and the rate at which a tiny slab heats up depends on how curved its temperature profile is. A bump (concave-down) cools; a dip (concave-up) warms. That "curvature drives change in time" is exactly u t = α u x x u_t = \alpha\, u_{xx} u t = α u xx
A thin insulated rod lying along the x x x
u ( x , t ) u(x,t) u ( x , t ) temperature at position x x x t t t ρ \rho ρ c c c k k k
We want a PDE governing u ( x , t ) u(x,t) u ( x , t ) parabolic because the discriminant of A u x x + 2 B u x t + C u t t A u_{xx}+2Bu_{xt}+Cu_{tt} A u xx + 2 B u x t + C u tt B 2 − A C = 0 B^2-AC=0 B 2 − A C = 0 C = 0 C=0 C = 0 B = 0 B=0 B = 0
Take a slab between x x x x + Δ x x+\Delta x x + Δ x A A A
Worked example Step-by-step energy balanceStep 1 — Energy stored in the slab.
E ( t ) = ∫ x x + Δ x ρ c u ( s , t ) A d s . E(t)=\int_x^{x+\Delta x} \rho c\, u(s,t)\,A\,ds. E ( t ) = ∫ x x + Δ x ρ c u ( s , t ) A d s . Why? Energy density ρ c u \rho c u ρ c u A d s A\,ds A d s
Step 2 — Rate of change of stored energy.
d E d t = ∫ x x + Δ x ρ c u t A d s . \frac{dE}{dt}=\int_x^{x+\Delta x}\rho c\,u_t\,A\,ds. d t d E = ∫ x x + Δ x ρ c u t A d s . Why? Differentiate under the integral (the limits are fixed in space).
Step 3 — Net heat flowing IN. Flux enters the left face, leaves the right face:
net in = A q ( x , t ) − A q ( x + Δ x , t ) . \text{net in}=A\,q(x,t)-A\,q(x+\Delta x,t). net in = A q ( x , t ) − A q ( x + Δ x , t ) . Why? q ( x ) q(x) q ( x ) + x +x + x + A q ( x ) +Aq(x) + A q ( x ) + A q ( x + Δ x ) +Aq(x+\Delta x) + A q ( x + Δ x )
Step 4 — Equate (conservation).
∫ x x + Δ x ρ c u t A d s = A ( q ( x , t ) − q ( x + Δ x , t ) ) . \int_x^{x+\Delta x}\rho c\,u_t\,A\,ds = A\big(q(x,t)-q(x+\Delta x,t)\big). ∫ x x + Δ x ρ c u t A d s = A ( q ( x , t ) − q ( x + Δ x , t ) ) .
Step 5 — Divide by A Δ x A\Delta x A Δ x Δ x → 0 \Delta x\to 0 Δ x → 0
Left side mean value → ρ c u t \rho c\, u_t ρ c u t q ( x ) − q ( x + Δ x ) Δ x → Δ x → 0 − ∂ q ∂ x . \frac{q(x)-q(x+\Delta x)}{\Delta x}\;\xrightarrow{\Delta x\to0}\;-\frac{\partial q}{\partial x}. Δ x q ( x ) − q ( x + Δ x ) Δ x → 0 − ∂ x ∂ q . Why? That is minus the definition of the derivative of q q q ρ c u t = − ∂ q ∂ x . \rho c\, u_t = -\frac{\partial q}{\partial x}. ρ c u t = − ∂ x ∂ q .
Step 6 — Insert Fourier's law q = − k u x q=-k u_x q = − k u x ρ c u t = − ∂ ∂ x ( − k u x ) = k u x x ( k constant ) . \rho c\, u_t = -\frac{\partial}{\partial x}\big(-k u_x\big)=k\,u_{xx}\quad(k\text{ constant}). ρ c u t = − ∂ x ∂ ( − k u x ) = k u xx ( k constant ) .
Step 7 — Define diffusivity α = k ρ c \alpha=\dfrac{k}{\rho c} α = ρ c k u t = α u x x \boxed{\,u_t=\alpha\,u_{xx}\,} u t = α u xx
Intuition Why "parabolic" / infinite speedClassifying A u t t + 2 B u x t + C u x x Au_{tt}+2Bu_{xt}+Cu_{xx} A u tt + 2 B u x t + C u xx 0 ⋅ u t t + 0 + ( − α ) u x x + u t = 0 0\cdot u_{tt}+0+(-\alpha)u_{xx}+u_t=0 0 ⋅ u tt + 0 + ( − α ) u xx + u t = 0 B 2 − A C = 0 B^2-AC=0 B 2 − A C = 0 parabolic . Physically this means smoothing/diffusion with infinite signal speed (a disturbance is felt everywhere instantly, unlike the wave equation).
Worked example Will this point heat or cool?At an instant a rod has u ( x , t ) = 20 + 5 sin ( π x ) u(x,t)=20+5\sin(\pi x) u ( x , t ) = 20 + 5 sin ( π x ) α = 2 \alpha=2 α = 2 u t u_t u t x = 1 / 2 x=1/2 x = 1/2 Step 1: u x x = ∂ x 2 [ 20 + 5 sin π x ] = − 5 π 2 sin π x u_{xx}=\partial_x^2[20+5\sin\pi x]=-5\pi^2\sin\pi x u xx = ∂ x 2 [ 20 + 5 sin π x ] = − 5 π 2 sin π x Why? Differentiate twice; constants vanish.
Step 2: At x = 1 / 2 x=1/2 x = 1/2 sin ( π / 2 ) = 1 \sin(\pi/2)=1 sin ( π /2 ) = 1 u x x = − 5 π 2 u_{xx}=-5\pi^2 u xx = − 5 π 2 Step 3: u t = α u x x = 2 ( − 5 π 2 ) = − 10 π 2 ≈ − 98.7 u_t=\alpha u_{xx}=2(-5\pi^2)=-10\pi^2\approx-98.7 u t = α u xx = 2 ( − 5 π 2 ) = − 10 π 2 ≈ − 98.7 Interpretation: the peak of the bump cools rapidly — consistent with intuition.
Worked example Check a known solutionShow u = e − α π 2 t sin ( π x ) u=e^{-\alpha\pi^2 t}\sin(\pi x) u = e − α π 2 t sin ( π x ) u t = α u x x u_t=\alpha u_{xx} u t = α u xx Step 1: u t = − α π 2 e − α π 2 t sin π x u_t=-\alpha\pi^2 e^{-\alpha\pi^2 t}\sin\pi x u t = − α π 2 e − α π 2 t sin π x Why? Chain rule on the exponential.
Step 2: u x = π e − α π 2 t cos π x u_x=\pi e^{-\alpha\pi^2 t}\cos\pi x u x = π e − α π 2 t cos π x u x x = − π 2 e − α π 2 t sin π x u_{xx}=-\pi^2 e^{-\alpha\pi^2 t}\sin\pi x u xx = − π 2 e − α π 2 t sin π x Step 3: α u x x = − α π 2 e − α π 2 t sin π x = u t . \alpha u_{xx}=-\alpha\pi^2 e^{-\alpha\pi^2 t}\sin\pi x=u_t. α u xx = − α π 2 e − α π 2 t sin π x = u t .
Recall Predict before you check
If you double α \alpha α Forecast…
A linear initial profile u = a + b x u=a+bx u = a + b x u t u_t u t Forecast…
Verify: (1) Faster — u t u_t u t α \alpha α u x x = 0 ⇒ u t = 0 u_{xx}=0\Rightarrow u_t=0 u xx = 0 ⇒ u t = 0
Common mistake "Heat equation should haveu t t u_{tt} u tt Why it feels right: the wave equation u t t = c 2 u x x u_{tt}=c^2u_{xx} u tt = c 2 u xx Fix: energy storage uses ρ c u \rho c u ρ c u t t t one time derivative. This breaks time-reversal symmetry (heat spreads, never un-spreads).
Common mistake Forgetting the minus sign in Fourier's law.Why it feels right: "flux proportional to gradient" sounds clean. Fix: without the minus, heat would flow toward hotter regions, violating thermodynamics. The minus makes the final equation + α u x x +\alpha u_{xx} + α u xx u t = − α u x x u_t=-\alpha u_{xx} u t = − α u xx
α = k \alpha=k α = k Why it feels right: k k k Fix: the PDE's time scale is set by diffusivity α = k / ( ρ c ) \alpha=k/(\rho c) α = k / ( ρ c ) k k k ρ c \rho c ρ c α \alpha α
Recall Explain to a 12-year-oldImagine a metal ruler, hot in the middle, cold at the ends. Heat is like a crowd that always pushes from a crowded (hot) place to an emptier (cold) place. How fast a tiny spot's temperature changes depends on whether it's a "hill" or a "valley" on the temperature map: hills lose heat to lower neighbours and cool; valleys gain heat and warm. The equation u t = α u x x u_t=\alpha u_{xx} u t = α u xx
Mnemonic Remember the form"Time-change equals C urvature × D iffusivity" → u t = α u x x \;u_t=\alpha u_{xx} u t = α u xx "flux flows down-hill" → q = − k u x q=-k u_x q = − k u x
State Fourier's law of heat conduction (1D) and explain the minus sign. q = − k u x q=-k\,u_x q = − k u x ; minus because heat flows from hot to cold, i.e. opposite to the temperature gradient.
What two physical principles derive the heat equation? Fourier's law (flux ∝ −gradient) and conservation of energy on a control slab.
Write the 1D heat equation and define α \alpha α u t = α u x x u_t=\alpha u_{xx} u t = α u xx with thermal diffusivity
α = k / ( ρ c ) \alpha=k/(\rho c) α = k / ( ρ c ) , units m²/s.
Why is the heat equation called parabolic? Its second-order classification gives discriminant
B 2 − A C = 0 B^2-AC=0 B 2 − A C = 0 ; physically it is diffusive/smoothing.
In the derivation, what does ρ c u \rho c\,u ρ c u Thermal energy density (J/m³): density × specific heat × temperature.
Why only one time derivative (not u t t u_{tt} u tt Energy storage and Fourier flux are each first order in time, so the balance gives a single
u t u_t u t .
Physical meaning of u x x > 0 u_{xx}>0 u xx > 0 Profile is concave up (a dip);
u t > 0 u_t>0 u t > 0 so the point warms up.
Steady-state temperature profile shape in 1D (no source)? Linear, since
u x x = 0 u_{xx}=0 u xx = 0 .
Does doubling α \alpha α Speeds it up;
u t u_t u t scales with
α \alpha α .
Heat eq with internal source f f f ρ c u t = k u x x + f \rho c\,u_t = k u_{xx}+f ρ c u t = k u xx + f .
Stored energy E over slab
dE/dt = integral rho c u_t
Net heat in = A q x minus q x+dx
Heat equation u_t = alpha u_xx
Intuition Hinglish mein samjhoDekho, heat equation ka core idea bahut simple hai. Ek patli rod lo, aur uske har point ka temperature u ( x , t ) u(x,t) u ( x , t ) Fourier's law — heat flux q = − k u x q=-k u_x q = − k u x energy conservation — kisi chhoti slab mein energy tabhi badhegi jab net heat andar aaye.
Ab ek chhota slab [ x , x + Δ x ] [x, x+\Delta x] [ x , x + Δ x ] ∫ ρ c u A d s \int \rho c\,u\,A\,ds ∫ ρ c u A d s Δ x → 0 \Delta x\to 0 Δ x → 0 ρ c u t \rho c\,u_t ρ c u t − ∂ q / ∂ x -\partial q/\partial x − ∂ q / ∂ x ρ c u t = k u x x \rho c\,u_t = k u_{xx} ρ c u t = k u xx u t = α u x x u_t=\alpha u_{xx} u t = α u xx α = k / ( ρ c ) \alpha=k/(\rho c) α = k / ( ρ c )
Samajhne ka shortcut: u x x u_{xx} u xx curvature . Agar point ek "hill" (bump) hai to neighbours thande hain, garmi bahar jaati hai, point thanda hota hai (u t < 0 u_t<0 u t < 0 u t = α × u_t = \alpha\times u t = α × parabolic hai, diffusion type.
Yeh important kyun hai? Yahi equation thermal engineering, finance (Black-Scholes bhi diffusion hai!), aur chemistry mein diffusion sab jagah aati hai. Exam mein dhyan rakho: minus sign mat bhoolo Fourier law mein, aur α = k / ( ρ c ) \alpha=k/(\rho c) α = k / ( ρ c ) k k k
