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Diffusion equation • One-dimensional diffusion equation is ∂u ∂ 2 u − = 0, −∞ < x < ∞, t > 0. ∂t ∂x2 • Introduce unit heat source at time t = 0 at the origin x = 0. • We wish to solve ut − uxx = δ(x)δ(t), −∞ < x < ∞, 0≤t • Initial condition is u(x, 0− ) = 0. • Here δ is the Dirac delta function APDEs (EMAT32110) 2007-08 3.20 Diffusion equations: similarity Fundamental solution • Observe that L.H.S. does NOT involve x and t • Can translate the solution to ﬁnd the result for a source located at x = ξ and turned on at t = τ • Suppose u = F (x, t) is the solution of ut − uxx = δ(x)δ(t). • Consider ut − uxx = δ(x − ξ)δ(t − τ ), u(x, τ − ) = 0 • Its solution is then u = F (x − ξ, t − τ ). APDEs (EMAT32110) 2007-08 3.21 Diffusion equations: similarity Similarity • Assume we have a solution u = F (x, t) • Question: it is possible to ﬁnd a second solution u = G(x, t)? ¯ ¯ • Set x = βx, t = γt. • Deﬁne G as G(x, t) ≡ αF (βx, γt) • If G is a solution it must satisfy Gt − Gxx = δ(x − ξ)δ(t − τ ), u(x, τ − ) = 0 • Substituting the derivatives gives, β2 β Ft − ¯ Fxx = ¯ x ¯ δ(¯)δ(t), F (¯, 0−) = 0 x γ α APDEs (EMAT32110) 2007-08 3.22 Diffusion equations: similarity Similarity • G(x, t) can be a solution only if β = α and γ = α2 • G(x, t) must be of the form G(x, t) = αF (αx, α2 t) • Have we found a new soluion ? • NO! The solution of the diffusion equation is unique. • We have discovered similarity structure of F , i.e. αF (αx, α2 t) = F (x, t). • This property implies that F must be of the form 1 x 1 x2 1 x F (x, t) = √ f √ , or √ g √ , or h √ ,... t t t t x t APDEs (EMAT32110) 2007-08 3.23 Diffusion equations: similarity Reduction to an ODE • Each choice of F (x, t) reduces the original PDE to an ODE ! • Solution of an ODE will give the same result for F . • Let’s try 1 x F (x, t) = √ f (ζ), ζ= √ . t t • Substituting into PDE leads to ζ 1 f ′′ + f ′ + f = 0 2 2 • The solution of the second order linear ODE is ζ −ζ 2 /4 2 /4 2 /4 f = Ae e−s ds + Be−ζ , A, B − constants. APDEs (EMAT32110) 2007-08 3.24 Diffusion equations: similarity Finding the constants • Constants A and B are determined by taking the total heat content H(t) in a bar. • The total heat is just an integral of the temperature, i.e. ∞ H(t) = F (x, t)dx −∞ ∞ ∞ A x B 2 /4t = √ g √ dx + √ e−x dx. t −∞ t t −∞ • Integrating by parts gives 2 g(ζ) = + O(ζ −3 ) as |ζ| → ∞ |ζ| ∞ 1 • It implies that √ gdx is undounded. t −∞ • The total heat must be ﬁnite, so set A = 0. APDEs (EMAT32110) 2007-08 3.25 Diffusion equations: similarity Heaviside function • Since A = 0, therefore, B 2 F = √ e−x /4t , t > 0. t • Have to choose B to satisfy the homogeneous problem. • Differentiate H w.r.t. t: ∞ dH = Ft (x, t)dx dt −∞ = Fx (∞, t) − Fx (−∞, t) + δ(t) = δ(t) • H(t) is the Heaviside function, i.e. 1 if t >0 H(t) = 0 if t > 0. APDEs (EMAT32110) 2007-08 3.26 Diffusion equations: similarity General solution • Using properties of the Heaviside function, we obtain ∞ B 2 1= √ e−x /4t dx −∞ t • Integrating gives 1 B= √ . 2 π • The fundamental solution of a diffusion equation is 1 −x2 /4t F (x, t) = √ e . 4πt • More generally, the solution is 1 2 /4(t−τ ) F (x − ξ, t − τ ) = e−(x−ξ) . 4π(t − τ ) APDEs (EMAT32110) 2007-08 3.27 Diffusion equations: similarity

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diffusion equation, boundary conditions, diﬀusion equation, initial condition, Partial Differential Equations, diffusion coefficient, fundamental solution, heat equation, boundary condition, time step

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posted: | 1/24/2011 |

language: | English |

pages: | 8 |

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