Some Aspects of Solutions of Partial Differential Equations

Some Aspects of Solutions of Partial Differential Equations K. Sakthivel Department of Mathematics Indian Institute of Space Science & Technology(IIST) Trivandrum - 695 547, Kerala Sakthivel@iist.ac.in Periyar University, Salem February 22, 2013 K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 1 / 24

Plan of the Talk 1 Formulation of Diffusion Model 2 Solution by Fourier Method 3 Nonlinear PDEs - Conservation Equations - Method of Characteristics 4 Weak Solutions for PDEs 5 Weak Solutions by Variational Methods K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 2 / 24

Formulation of Diffusion Model Consider A motionless liquid filling a straight tube or pipe A chemical substance, say die, which is diffusing through the liquid Let u(z, t) be the concentration of the die at position z of the pipe at time t. On the region x to x + x, the mass of the dye is x+ x x+ x M(t) = u(z, t)dz, and so dm u = (z, t)dz. x dt x t Fick s Law: Flux goes from region of higher concentration to the region of lower concentration. The rate of motion is propositional to the concentration gradient. dm dt = x+ x x u (z, t)dz = Net Change of Concentration t = k u x u (x + x, t) k (x, t). x K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 3 / 24

Formulation of Diffusion Model... Now, x+ x x u t (z, t)dz = k[ u x where k proportionality constant. u (x + x, t) (x, t)], x Applying Mean Value Theorem (!!!) on x < ζ < x + x, we get u [ u t (ζ, t) = k u ] (x + x, t) x x (x, t) / x. Taking x 0, we have u t (x, t) = k 2 u (x, t). x 2 K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 4 / 24

Formulation of Diffusion Model... Now, x+ x x u t (z, t)dz = k[ u x where k proportionality constant. u (x + x, t) (x, t)], x Applying Mean Value Theorem (!!!) on x < ζ < x + x, we get u [ u t (ζ, t) = k u ] (x + x, t) x x (x, t) / x. Taking x 0, we have u t (x, t) = k 2 u (x, t). x 2 In the multidimensional case, we get u t = k u K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 4 / 24

Formulation of Diffusion Model... Now, x+ x x u t (z, t)dz = k[ u x where k proportionality constant. u (x + x, t) (x, t)], x Applying Mean Value Theorem (!!!) on x < ζ < x + x, we get u [ u t (ζ, t) = k u ] (x + x, t) x x (x, t) / x. Taking x 0, we have u t (x, t) = k 2 u (x, t). x 2 In the multidimensional case, we get u t = k u + f (x, t), where f is a source" or sink" of the dye. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 4 / 24

Fourier Transform and its Properties Let f L 1 (R). The Fourier transform of f is defined as f (ω) := F [f (x)] = 1 2π e iωx f (x)dx. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 5 / 24

Fourier Transform and its Properties Let f L 1 (R). The Fourier transform of f is defined as Inverse Fourier Transform: f (ω) := F [f (x)] = 1 2π f (x) := F 1 [ f (ω)] = 1 2π e iωx f (x)dx. e iωx f (ω)dω. Some Properties: If f is a continuous piecewise differentiable function with f, f x, f xx L 1 (R) and f (x) = 0, then lim x F [f x (x)] = iωf [f (x)] and F [f xx (x)] = ω 2 F [f (x)]!!!. Fourier transform is linear, that is, f, g, L 1 (R) and a, b R, then F [af + bg] = af [f ] + bf [g]. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 5 / 24

Fourier Transform and its Properties... Note that F [f (x)g(x)] F [f (x)]f [g(x)], whereas F [(f g)(x)] = F [f (x)]f [g(x)], where (f g) is the convolution" of f and g defined as (f g)(x) := 1 2π f (x y)g(y)dy = (g f )(x). K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 6 / 24

Fourier Transform and its Properties... Note that F [f (x)g(x)] F [f (x)]f [g(x)], whereas F [(f g)(x)] = F [f (x)]f [g(x)], where (f g) is the convolution" of f and g defined as (f g)(x) := 1 2π 1-D Heat Conduction Model (in an infinite rod): f (x y)g(y)dy = (g f )(x). u t (x, t) = ku xx (x, t), < x <, 0 < t < u(x, 0) = φ(x), < x <. Applying Fourier transform F [u t (x, t)] = kf [u xx (x, t)], F [u(x, 0)] = F [φ(x)] K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 6 / 24

1-D Heat Conduction Model Recall that F [u t (x, t)] = = t 1 2π e [ 1 2π iξx u (x, t)dx t ] e iξx u(x, t)dx Taking φ(ξ) = F [φ(x)], we get the following ODE": dû dt (t) = kξ2 û(t) with data û(0) = φ(ξ). = û (ξ, t). t K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 7 / 24

1-D Heat Conduction Model Recall that F [u t (x, t)] = = t 1 2π e [ 1 2π iξx u (x, t)dx t ] e iξx u(x, t)dx Taking φ(ξ) = F [φ(x)], we get the following ODE": dû dt (t) = kξ2 û(t) with data û(0) = φ(ξ). = û (ξ, t). t Solution to the transformed ODE: F [u(x, t)] = û(t) = φ(ξ)e ktξ2. Note that: (f g)(x) = F 1[ F [f (x)]f [g(x)] ] and F [e a2 x 2 ] = 1 2a e ξ2 /4a 2. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 7 / 24

Solution of 1-D Heat Conduction Model Taking inverse Fourier transform u(x, t) = F 1 [ φ(ξ)e ktξ2 ] = 1 F 1[ ] F [φ(x)]f [e x 2 /4kt ] 2kt = φ(x) [ 1 ] e x 2 /4kt 2kt = 1 4πkt φ(y)e (x y)2 /4kt dy. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 8 / 24

Solution of 1-D Heat Conduction Model Taking inverse Fourier transform u(x, t) = F 1 [ φ(ξ)e ktξ2 ] = 1 F 1[ ] F [φ(x)]f [e x 2 /4kt ] 2kt = φ(x) = 1 4πkt [ 1 ] e x 2 /4kt 2kt φ(y)e (x y)2 /4kt dy. Fundamental Solution (or Heat Kernel) of 1-D Heat Equation: Multidimensional case: Φ(x, t) = 1 4πt e x 2 /4t, k = 1, t > 0. 1 Ψ(x, t) = /4t (4πt) d/2 e x 2, x R d, t > 0, where x = (x 1, x 2,, x d ) and x = x 2 1 + x 2 2 + + x 2 d. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 8 / 24

Properties of Fundamental Solution For t > 0, the function Ψ(x, t) > 0 is an infinitely differentiable function of x and t. (Recall Smoothing Property) Ψ t (x, t) = Ψ(x, t), x R d, t > 0 and Ψ(x, t)dx = 1, t > 0. R d For any continuous and bounded function g(x) : lim Ψ(x, t)g(x)dx = g(0). t 0 R d K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 9 / 24

Properties of Fundamental Solution For t > 0, the function Ψ(x, t) > 0 is an infinitely differentiable function of x and t. (Recall Smoothing Property) Ψ t (x, t) = Ψ(x, t), x R d, t > 0 and Ψ(x, t)dx = 1, t > 0. R d For any continuous and bounded function g(x) : lim Ψ(x, t)g(x)dx = g(0). t 0 R d Since Ψ(x, t) is the probability density of a Gaussian random variable, Ψ(x, t)dx = 1, for all t > 0!!!. R d Moreover, by Lebesgue dominated convergence theorem Ψ(x, t)g(x)dx R d 1 = (4πt) d/2 e x 2 /4t g(x)dx = 1 R d π d/2 e y 2 g(y 4t)dx R d 1 = g(0)dx = g(0) as t 0. (π) d/2 e y 2 R d K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 9 / 24

Conservation Equations - Traffic Flow Model Consider a stretch of highway on which cars are moving from left to right. Assume that no exit or entrance of ramps. u(x, t) density of cars at x f (x, t) flux of cars at x (cars per minute passing the point x) Change in the number of cars in [a, b] = d dt b Using flux, change in the number of cars in [a, b] = = f (a, t) f (b, t) = b a a u(x, t)dx f (x, t)dx, x by fundamental theorem of calculus. The last two integrals yield b a u (x, t)dx = t The interval length [a, b] is arbitrary, b a f (x, t)dx. x u t (x, t) + f x (x, t) = 0. (Conservation Equation) K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 10 / 24

Traffic Flow Model The amount of cars passing a given point is generally a function of density u, that is, f (u) : for example, f (u) = u 2 (quadratic flow rate). From conservation equation and chain rule: u t + 2uu x = 0. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 11 / 24

Traffic Flow Model The amount of cars passing a given point is generally a function of density u, that is, f (u) : for example, f (u) = u 2 (quadratic flow rate). From conservation equation and chain rule: Consider the model u t + 2uu x = 0. u t + 2uu x = 0, < x <, 0 < t < with the initial density of cars!!! 1 x 0 u(x, 0) = u 0 (x) = 1 x 0 < x < 1 0 x 1 K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 11 / 24

Solution by Method of Characteristics Let x(t) be a curve in x t plane and u(x(t), t) is the function of x and t. How does u change from the perspective of x(t)? By chain rule du dx = u x dt dt + u t; but taking traffic flow model into account So dx dt u t + 2uu x = 0 { dx dt = 2u du dt = 0 = 2u x = 2u(x, t)t + x 0 Characteristic equation Along such characteristic u is constant. From the initial density of cars, the characteristic curve starting from (x 0, 0) is x = 2u(x 0, 0)t + x 0 K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 12 / 24

Solution by Method of Characteristics... Case 1: For x 0 0, u(x 0, 0) = 1, the characteristics are x = 2t + x 0 or t = 1 2 (x x 0). Case 2: For 0 < x 0 < 1 u(x 0, 0) = 1 x 0, the characteristics are x = 2(1 x 0 )t + x 0 or t = 1 x x 0. 2 1 x 0 Case 3: For 1 x 0 <, u(x 0, 0) = 0, the characteristics are vertical lines given by x = x 0. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 13 / 24

Solution by Method of Characteristics... Case 1: For x 0 0, u(x 0, 0) = 1, the characteristics are x = 2t + x 0 or t = 1 2 (x x 0). Case 2: For 0 < x 0 < 1 u(x 0, 0) = 1 x 0, the characteristics are x = 2(1 x 0 )t + x 0 or t = 1 x x 0. 2 1 x 0 Case 3: For 1 x 0 <, u(x 0, 0) = 0, the characteristics are vertical lines given by x = x 0. Observations: Note that characteristics run together starting at t = 1/2. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 13 / 24

Solution by Method of Characteristics... Case 1: For x 0 0, u(x 0, 0) = 1, the characteristics are x = 2t + x 0 or t = 1 2 (x x 0). Case 2: For 0 < x 0 < 1 u(x 0, 0) = 1 x 0, the characteristics are x = 2(1 x 0 )t + x 0 or t = 1 x x 0. 2 1 x 0 Case 3: For 1 x 0 <, u(x 0, 0) = 0, the characteristics are vertical lines given by x = x 0. Observations: Note that characteristics run together starting at t = 1/2. So except for short time t = 1/2, the solution is not continuous. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 13 / 24

Solution by Method of Characteristics... Case 1: For x 0 0, u(x 0, 0) = 1, the characteristics are x = 2t + x 0 or t = 1 2 (x x 0). Case 2: For 0 < x 0 < 1 u(x 0, 0) = 1 x 0, the characteristics are x = 2(1 x 0 )t + x 0 or t = 1 x x 0. 2 1 x 0 Case 3: For 1 x 0 <, u(x 0, 0) = 0, the characteristics are vertical lines given by x = x 0. Observations: Note that characteristics run together starting at t = 1/2. So except for short time t = 1/2, the solution is not continuous. When characteristics run together, we have a phenomenon of Shock Waves" (discontinuous solutions) K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 13 / 24

Solution by Method of Characteristics... Case 1: For x 0 0, u(x 0, 0) = 1, the characteristics are x = 2t + x 0 or t = 1 2 (x x 0). Case 2: For 0 < x 0 < 1 u(x 0, 0) = 1 x 0, the characteristics are x = 2(1 x 0 )t + x 0 or t = 1 x x 0. 2 1 x 0 Case 3: For 1 x 0 <, u(x 0, 0) = 0, the characteristics are vertical lines given by x = x 0. Observations: Note that characteristics run together starting at t = 1/2. So except for short time t = 1/2, the solution is not continuous. When characteristics run together, we have a phenomenon of Shock Waves" (discontinuous solutions) Shock waves occurs due to the flux that grows very large as a function of density u. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 13 / 24

Notion of Weak Solution for PDEs Test Functions : C 0 () Let R d be a bounded domain. The set of all functions beloging to C () and compactly supported in. Support of a function ϕ(x) : supp(ϕ) = {x : ϕ(x) 0}. Example (Compact Support) Suppose d = 1. { ϕ(x) = e 1 x 2 1 if x < 1. 0 elsewhere. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 14 / 24

Notion of Weak Solution for PDEs Test Functions : C 0 () Let R d be a bounded domain. The set of all functions beloging to C () and compactly supported in. Support of a function ϕ(x) : supp(ϕ) = {x : ϕ(x) 0}. Example (Compact Support) Suppose d = 1. { ϕ(x) = e 1 x 2 1 if x < 1. 0 elsewhere. Example (Weak Differentiability): Suppose d = 1 and = ( 1, 1). The function u(x) = x, x is not differentiable in the classical sense. But it has a weak derivative!!! { 1 if x < 0 Du = 1 if x > 0. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 14 / 24

Weak Differentiability... For any ϕ C 0 (), integrating by parts, we get: 1 1 = u(x)dϕ(x)dx 0 = = 1 0 1 1 u(x)dϕ(x)dx + 1 1 0 u(x)dϕ(x)dx Du(x)ϕ(x)dx + uϕ 0 1 Du(x)ϕ(x)dx + uϕ 1 Du(x)ϕ(x)dx [u(0)]ϕ(0). Note that ϕ(1) = ϕ( 1) = 0 and [u(0)] = u(0 + ) u(0 ) = 0 since u is continuous at x = 0. 0 1 0 K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 15 / 24

Sobolev Space H 1 () H 1 () := {v L 2 (), v L 2 ()}. The space H 1 () is a separable Hilbert space equipped with inner product (u, v) H 1 () = uvdx + u vdx. Norm on H 1 (): u 2 H 1 () = Space H 1 0 () := {u H1 () : u = 0}. u 2 dx + u 2 dx. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 16 / 24

Weak Solution for Poisson Equation Let R d be a bounded domain with smooth boundary. } u(x) = f (x) in u(x) = 0 (1) where f (x) is the external force; x = (x 1, x 2,..., x d ) and is the Laplace operator = 2 x 2 1 + 2 x 2 2 +... + 2 xd 2. Weak and Classical solutions : Assume u C 2 () and ϕ C0 (). Multiplying Poisson equation by ϕ and integrating on, u(x)ϕ(x)dx = f (x)ϕ(x)dx. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 17 / 24

Weak and Classical solutions... Green s identity: u u(x)ϕ(x)dx = ν (x)ϕ(x)ds + u ϕdx where ν is the unit normal vector outward to. Further, u ϕdx = u ϕ ν ds u ϕ(x)dx We get the following identities : u ϕdx = f (x)ϕ(x)dx (2) u ϕ(x)dx = f (x)ϕ(x)dx (3) Thus, if u C 2 () is a solution of Poisson equation then for any ϕ C0 () the above identities hold. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 18 / 24

Weak and Classical solutions... Conversely for any ϕ C0 (), u C2 () satisfies (3), we also get ( u f )ϕ(x)dx = 0. Since ϕ is arbitrary, u f = 0 in, i.e, u is a solution of Poisson equation. Note : The integral (2) makes sense if u H 1 () whereas for (3) we only need u L 2 (). Weak Solution: A function u H 1 () is said to be a weak solution of (1), if the following identity holds: u ϕdx = f ϕdx for any ϕ C0 () or H1 0 () since C 0 () is dense in H1 0 (). K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 19 / 24

Optimization Result from Linear Algebra Consider a system Ax = b, where A is n n matrix, symmetric and positive definite and b R n. Recall that for x, y R n x y =< x, y >= x T y = n x i y i. i=1 A vector x R n is a solution of Ax = b iff it is global minimizer of φ(x) = 1 2 < Ax, x > < b, x >. Assume that φ has a global minimizer!!. For any y R n, ɛ R the variation Φ(ɛ) := φ(x + ɛy) achieves its minimum at ɛ = 0. Φ (0) = d dɛ φ(x + ɛy) ɛ=0 = 0. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 20 / 24

Optimization Result from Linear Algebra... Now Φ (ɛ) = d [ ] 1 < A(x + ɛy), x + ɛy > < b, x + ɛy > dɛ 2 = < A(x + ɛy), y > < b, y > But Φ (0) = 0 implying that < Ax, y > < b, y >=< Ax b, y >= 0. Since y is arbitrary, x is a solution of Ax = b. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 21 / 24

Variational Approach for Poisson Equation Consider a functional J[v] = 1 v 2 dx fvdx. 2 If u H0 1() is an extremal of the functional J[v] in H1 0 (), then u is a weak solution of the Dirichlet problem (1). Suppose u H0 1 () is an extremum of J[v](need a proof!!), then for any ϕ H0 1 () and ɛ R, F(ɛ) := J[u + ɛϕ] = 1 (u + ɛϕ) 2 dx f (u + ɛϕ)dx. 2 But F achieves its minimum at ɛ = 0, F (0) = dj dɛ (u + ɛϕ) ɛ=0 = 0. Note that F (ɛ) = ( u + ɛ ϕ) ϕdx f ϕdx So F (0) = 0 u ϕdx = f ϕdx. Therefore, K.Sakthivel u(iist) is a weak solution Solutions of Partial of the Differential Poisson Equationsequation. Periyar University, Salem 22 / 24

Uniqueness of Weak Solutions Suppose u 1 and u 2 are two weak solutions of the Poisson equation. For any ϕ H0 1 (), we have u 1 ϕdx = f ϕdx and u 2 ϕdx = f ϕdx Taking u := u 1 u 2, we arrive at u ϕdx = 0, Choosing ϕ = u, u 2 dx = 0. Applying the Poincare inequality: u 2 dx C() so that ϕ H 1 0 (). u 2 dx, u 2 dx = 0 u = 0 a.e. in. K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 23 / 24

Thank You! K.Sakthivel (IIST) Solutions of Partial Differential Equations Periyar University, Salem 24 / 24