热传导偏微分方程的推导

Deduce the parabolic equation: ∂ T ‾ ∂ t ‾ = ∂ 2 T ‾ ∂ x ‾ 2 \frac{\partial \overline{T}}{\partial\overline{t}}=\frac{\partial^2 \overline{T}}{\partial\overline{x}^2} ∂t∂T=∂x2∂2T

Your deduction can be based on heat conduction process, or you can select other physical process if you need a high grade.

h e a t f l u x i n = − K A ∂ T ∂ x ∣ x , t ( k i s a p h y s i c s c o n s t a n t ) heat~flux~in=-KA\frac{\partial T}{\partial x}|_{x,t}(k~is~a~physics~constant) heat flux in=−KA∂x∂T∣x,t(k is a physics constant)
h e a t f l u x i n = − K A ∂ T ∂ x ∣ x + Δ t , t heat~flux~in=-KA\frac{\partial T}{\partial x}|_{x+\Delta t,t} heat flux in=−KA∂x∂T∣x+Δt,t
c h a n g e o f h e a t p e r t : c v A Δ x ∂ T ∂ t change ~of~heat~per~t:c_vA\Delta x\frac{\partial T}{\partial t} change of heat per t:cvAΔx∂t∂T
W e c o u l d w r i t e t h e b a l a n c e e q u a t i o n : c v A Δ x ∂ T ∂ t = − K A ∂ T ∂ x ∣ x , t + K A ∂ T ∂ x ∣ x + Δ t , t We~could~write~the~balance~equation: c_vA\Delta x\frac{\partial T}{\partial t}=-KA\frac{\partial T}{\partial x}|_{x,t} +KA\frac{\partial T}{\partial x}|_{x+\Delta t,t} We could write the balance equation:cvAΔx∂t∂T=−KA∂x∂T∣x,t+KA∂x∂T∣x+Δt,t
∂ T ∂ t = k c v 1 Δ x ( ∂ T ∂ x ∣ x + Δ t , t − ∂ T ∂ x ∣ x , t ) \frac{\partial T}{\partial t}=\frac{k}{c_v}\frac{1}{\Delta x}(\frac{\partial T}{\partial x}|_{x+\Delta t,t}-\frac{\partial T}{\partial x}|_{x,t}) ∂t∂T=cvkΔx1(∂x∂T∣x+Δt,t−∂x∂T∣x,t)
t a k e t h e l i m i t : ∂ T ∂ t = k c v ∂ 2 T ∂ x 2 take~the~limit:\frac{\partial T}{\partial t}=\frac{k}{c_v}\frac{\partial^2 T}{\partial x^2} take the limit:∂t∂T=cvk∂x2∂2T
s e l e c t n o n d i m e n s i o n a l v a r i a b l e s x ‾ t ‾ T ‾ x ‾ = x L T ‾ = t t s c a l e T ‾ = T T 0 select~nondimensional~variables~\overline{x} ~\overline{t}~\overline{T} \overline{x}=\frac{x}{L}~~~~\overline{T}=\frac{t}{t_scale}~~\overline{T}=\frac{T}{T_0} select nondimensional variables x t Tx=Lx T=tscalet T=T0T
∂ T ∂ t = ∂ T ‾ T 0 ∂ t ‾ t s c a l e = T 0 t s c a l e ∂ T ‾ ∂ t ‾ ∂ 2 T ∂ x 2 = ∂ 2 T ‾ T 0 ∂ ( x ‾ L ) 2 = T 0 L 2 ∂ 2 T ‾ ∂ x ‾ 2 T 0 t s c a l e ∂ T ‾ ∂ t ‾ = T 0 L 2 ∂ 2 T ‾ ∂ x ‾ 2 ∗ k c v ∴ ∂ T ‾ ∂ t ‾ = ∂ 2 T ‾ ∂ x ‾ 2 f o r t s c a l e = c v L 2 k \frac{\partial T}{\partial t}=\frac{\partial \overline{T}T_0}{\partial \overline{t}t_scale}=\frac{T_0}{t_scale}\frac{\partial \overline{T}}{\partial\overline{t}}\\ \frac{\partial^2 T}{\partial x^2}=\frac{\partial^2 \overline{T}T_0}{\partial (\overline{x}L)^2}=\frac{T_0}{L^2}\frac{\partial^2 \overline{T}}{\partial\overline{x}^2}\\ \frac{T_0}{t_scale}\frac{\partial \overline{T}}{\partial\overline{t}}=\frac{T_0}{L^2}\frac{\partial^2 \overline{T}}{\partial\overline{x}^2}*\frac{k}{c_v}\\ \\ \therefore \frac{\partial \overline{T}}{\partial\overline{t}}=\frac{\partial^2 \overline{T}}{\partial\overline{x}^2}~for~t_{scale}=\frac{c_vL^2}{k} ∂t∂T=∂ttscale∂TT0=tscaleT0∂t∂T∂x2∂2T=∂(xL)2∂2TT0=L2T0∂x2∂2TtscaleT0∂t∂T=L2T0∂x2∂2T∗cvk∴∂t∂T=∂x2∂2T for tscale=kcvL2
One of my familiar process is Brownian motion, resulting from the random movements and collisions of the particles which is closely related to the very beginning of random process. Also, other related physical process is diffusion function:

The equation is usually written as:
∂ Φ ( r , t ) ∂ t = ▽ [ D ( Φ , r ) ▽ Φ ( r , t ) ] \frac{\partial{\Phi(r,t)}}{\partial t}=\bigtriangledown [D(\Phi,r)\bigtriangledown \Phi(r,t)] ∂t∂Φ(r,t)=▽[D(Φ,r)▽Φ(r,t)]
where ϕ(r, t) is the density of the diffusing material at location r and time t and D(ϕ, r) is the collective diffusion coefficient for density ϕ at location r; and ∇ represents the vector differential operator. If the diffusion coefficient depends on the density then the equation is nonlinear, otherwise it is linear.

The equation above applies when the diffusion coefficient is isotropic; in the case of anisotropic diffusion, D is a symmetric positive definite matrix, and the equation is written (for three dimensional diffusion) as:
∂ Φ ( r , t ) ∂ t = ∑ i = 1 3 ∑ j = 1 3 ∂ ∂ t [ D i j ( ϕ , r ) ∂ Φ ( r , t ) ∂ x j ] \frac{\partial{\Phi(r,t)}}{\partial t}=\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial{}}{\partial t}[D_{ij}(\phi,r)\frac{\partial{\Phi(r,t)}}{\partial x_j}] ∂t∂Φ(r,t)=i=1∑3j=1∑3∂t∂[Dij(ϕ,r)∂xj∂Φ(r,t)]
If D is constant, then the equation reduces to the following linear differential equation:
∂ Φ ( r , t ) ∂ t = D ▽ 2 Φ ( r , t ) \frac{\partial{\Phi(r,t)}}{\partial t}=D\bigtriangledown ^2\Phi(r,t) ∂t∂Φ(r,t)=D▽2Φ(r,t)
which is identical to the heat equation I deduced above.

Reference: Diffusion_equation