Modeling using conservation laws. Let u(x, t) = density (heat, momentum, probability,...) so that. u dx = amount in region R Ω. R

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1 Modeling using conservation laws Let u(x, t) = density (heat, momentum, probability,...) so that u dx = amount in region R Ω. R

2 Modeling using conservation laws Let u(x, t) = density (heat, momentum, probability,...) so that u dx = amount in region R Ω. R A quantity is conserved if it can be gained or lost only by flow through domain boundaries, because of sources and sinks in the domain.

3 Modeling using conservation laws Let u(x, t) = density (heat, momentum, probability,...) so that u dx = amount in region R Ω. R A quantity is conserved if it can be gained or lost only by flow through domain boundaries, because of sources and sinks in the domain. Model ingredients: The flow or flux is a vector field J(x, t), so that J ˆndA is flow across infinitesimal area ˆndA. Q(x, t) is the rate of inflow at point x

4 Deriving a PDE for a conserved quantity Conservation of u(x, t) on any region R Ω implies d u u dx = dt t dx = J ˆndx + Q(x, t)dx. R R (recall notation: R is boundary of R, ˆn is outward normal vector) R R

5 Deriving a PDE for a conserved quantity Conservation of u(x, t) on any region R Ω implies d u u dx = dt t dx = J ˆndx + Q(x, t)dx. R R (recall notation: R is boundary of R, ˆn is outward normal vector) Use divergence theorem to turn boundary integral into integral on region R, ( u ) t + J Q dx =. R R R

6 Deriving a PDE for a conserved quantity Conservation of u(x, t) on any region R Ω implies d u u dx = dt t dx = J ˆndx + Q(x, t)dx. R R (recall notation: R is boundary of R, ˆn is outward normal vector) Use divergence theorem to turn boundary integral into integral on region R, ( u ) t + J Q dx =. R Since this is true for any subregion R, integrand is zero: u t R + J = Q. conservation form/continuity equation/transport equation R

7 Boundary conditions for conserved quantities Flux-type boundary conditions specify flow F (x) : Ω R through physical boundary.

8 Boundary conditions for conserved quantities Flux-type boundary conditions specify flow F (x) : Ω R through physical boundary. Flux is typically modeled as a function of u and its derivatives J = J(u, u,...); boundary condition becomes J(u, u,...) ˆn = F (x), x Ω.

9 Boundary conditions for conserved quantities Flux-type boundary conditions specify flow F (x) : Ω R through physical boundary. Flux is typically modeled as a function of u and its derivatives J = J(u, u,...); boundary condition becomes J(u, u,...) ˆn = F (x), x Ω. Example: for heat diffusion, Fourier s law says J = D u. Thus D u ˆn = F (x), x Ω,

10 Boundary conditions for conserved quantities Flux-type boundary conditions specify flow F (x) : Ω R through physical boundary. Flux is typically modeled as a function of u and its derivatives J = J(u, u,...); boundary condition becomes J(u, u,...) ˆn = F (x), x Ω. Example: for heat diffusion, Fourier s law says J = D u. Thus D u ˆn = F (x), x Ω, More specifically, if boundary in insulating, get Neumann boundary condition u ˆn =. Remark: Dirichlet boundary condition u = U(x), guarantee flux is zero at boundary. x Ω will not

11 Example: transport and traffic flow Suppose that u = u(x, t) is transported at velocity c, so that (one dimensional) scalar flux is J = cu.

12 Example: transport and traffic flow Suppose that u = u(x, t) is transported at velocity c, so that (one dimensional) scalar flux is J = cu. Then u t + J = Q becomes (assuming Q = ) u t + cu x =. (linear transport equation)

13 Example: transport and traffic flow Suppose that u = u(x, t) is transported at velocity c, so that (one dimensional) scalar flux is J = cu. Then u t + J = Q becomes (assuming Q = ) u t + cu x =. (linear transport equation) Traffic flow: speed can be modeled as a decreasing function of density c = c mu, so J = u(c mu); conservation law becomes u t + c u x m(u 2 ) x =. (nonlinear transport equation)

14 Example: Diffusion with a source Random motions of particles (and other things) leads to diffusion. Means that net flow has a direction toward regions of less density.

15 Example: Diffusion with a source Random motions of particles (and other things) leads to diffusion. Means that net flow has a direction toward regions of less density. Simplest model: Fick/Fourier law J = D u.

16 Example: Diffusion with a source Random motions of particles (and other things) leads to diffusion. Means that net flow has a direction toward regions of less density. Simplest model: Fick/Fourier law J = D u. Sources Q(x, t) created by, for example heat production or chemical reactions.

17 Example: Diffusion with a source Random motions of particles (and other things) leads to diffusion. Means that net flow has a direction toward regions of less density. Simplest model: Fick/Fourier law J = D u. Sources Q(x, t) created by, for example heat production or chemical reactions. The conservation equation becomes u t = D u + Q = D u + Q, (diffusion equation)

18 Example: wave equation Conserved quantity is momentum density ρu t (x, t), where ρ = mass density

19 Example: wave equation Conserved quantity is momentum density ρu t (x, t), where ρ = mass density Momentum flux occurs because of force imbalance between parts of material; simplest model is J = σ u.

20 Example: wave equation Conserved quantity is momentum density ρu t (x, t), where ρ = mass density Momentum flux occurs because of force imbalance between parts of material; simplest model is J = σ u. Momentum conservation leads to where c 2 = σ/ρ. (u t ) t = c 2 u = c 2 u, (wave equation)

21 Steady state equations Often dynamical processes settle down ; for PDEs this means the time derivative can be ignored.

22 Steady state equations Often dynamical processes settle down ; for PDEs this means the time derivative can be ignored. For a conservation law with flux J(u) and time independent source term Q, a steady state solution solves J(u) = Q. Interpretation: the amount flowing into a region in space equals the amount flowing out.

23 Steady state equations Often dynamical processes settle down ; for PDEs this means the time derivative can be ignored. For a conservation law with flux J(u) and time independent source term Q, a steady state solution solves J(u) = Q. Interpretation: the amount flowing into a region in space equals the amount flowing out. Example (diffusion with a source): Flux is given by Fick s law J = D u, and Q(x, y) is a prescribed source term.

24 Steady state equations Often dynamical processes settle down ; for PDEs this means the time derivative can be ignored. For a conservation law with flux J(u) and time independent source term Q, a steady state solution solves J(u) = Q. Interpretation: the amount flowing into a region in space equals the amount flowing out. Example (diffusion with a source): Flux is given by Fick s law J = D u, and Q(x, y) is a prescribed source term. Steady state u = u(x, y) solves D u = u = Q(x, y). If Q, get Poisson s equation; if Q, get Laplace s equation.

25 Using conserved and dissipated quantities Salient fact: solutions of PDEs have a lot of detail, but...

26 Using conserved and dissipated quantities Salient fact: solutions of PDEs have a lot of detail, but... We can t always know everything about them (especially nonlinear equations) Even if we could, often hard to see the essential aspects.

27 Using conserved and dissipated quantities Salient fact: solutions of PDEs have a lot of detail, but... We can t always know everything about them (especially nonlinear equations) Even if we could, often hard to see the essential aspects. Idea: use coarse-grained quantities to study solutions qualitatively. Some of these are inspired by physics (energy, entropy), whereas others are completely abstract.

28 Conserved and dissipated quantities A functional F [u] maps u to the real numbers, e.g. F [u] = u(x) dx or F [u] = u 2 dx. Ω In our case, these are often quantities of physical interest (mass, energy, momentum) Ω

29 Conserved and dissipated quantities A functional F [u] maps u to the real numbers, e.g. F [u] = u(x) dx or F [u] = u 2 dx. Ω In our case, these are often quantities of physical interest (mass, energy, momentum) Let u(x, t) : Ω [, ) R be a solution of some PDE, and suppose F [u] has the form F [u(t)] = f (u, u x,...)dx. so that F can be regarded as depending on t. Ω Ω

30 Conserved and dissipated quantities A functional F [u] maps u to the real numbers, e.g. F [u] = u(x) dx or F [u] = u 2 dx. Ω In our case, these are often quantities of physical interest (mass, energy, momentum) Let u(x, t) : Ω [, ) R be a solution of some PDE, and suppose F [u] has the form F [u(t)] = f (u, u x,...)dx. so that F can be regarded as depending on t. Time evolution of F [u] may be categorized as: If df /dt = for all u, then F is called conserved, If df /dt for all u, then F is called dissipated. Ω Ω

31 Example: energy in the wave equation Let u solve u tt = u xx, u(, t) = = u(l, t).

32 Example: energy in the wave equation Let u solve Energy functional is conserved. u tt = u xx, E(t) = u(, t) = = u(l, t). 1 2 u2 t u2 x dx

33 Example: energy in the wave equation Let u solve u tt = u xx, u(, t) = = u(l, t). Energy functional 1 E(t) = 2 u2 t u2 x dx is conserved. Fist differentiate under integral sign: de dt = u t u tt + u x u xt dx,

34 Example: energy in the wave equation Let u solve u tt = u xx, u(, t) = = u(l, t). Energy functional 1 E(t) = 2 u2 t u2 x dx is conserved. Fist differentiate under integral sign: Integrate by parts de dt = de dt = u xu t x=l x= + u t u tt + u x u xt dx, u t u tt u xx u t dx, Then use equation and boundary conditions to get de/dt =.

35 Example: energy in the wave equation Let u solve u tt = u xx, u(, t) = = u(l, t). Energy functional 1 E(t) = 2 u2 t u2 x dx is conserved. Fist differentiate under integral sign: Integrate by parts de dt = de dt = u xu t x=l x= + u t u tt + u x u xt dx, u t u tt u xx u t dx, Then use equation and boundary conditions to get de/dt =. If u(x, ) = = u t (x, ) initially, does the solution remain zero?

36 Example: energy in the wave equation Let u solve u tt = u xx, u(, t) = = u(l, t). Energy functional 1 E(t) = 2 u2 t u2 x dx is conserved. Fist differentiate under integral sign: Integrate by parts de dt = de dt = u xu t x=l x= + u t u tt + u x u xt dx, u t u tt u xx u t dx, Then use equation and boundary conditions to get de/dt =. If u(x, ) = = u t (x, ) initially, does the solution remain zero? Yes, since E() =, E(t), thus u x. Using boundary conditions gives u(x, t).

37 Example: energy in the wave equation Let u solve u tt = u xx, u(, t) = = u(l, t). Energy functional 1 E(t) = 2 u2 t u2 x dx is conserved. Fist differentiate under integral sign: Integrate by parts de dt = de dt = u xu t x=l x= + u t u tt + u x u xt dx, u t u tt u xx u t dx, Then use equation and boundary conditions to get de/dt =. If u(x, ) = = u t (x, ) initially, does the solution remain zero? Yes, since E() =, E(t), thus u x. Using boundary conditions gives u(x, t). Converse also true: if u(x, ) initially, then solution never dies out.

38 Example: heat equation Suppose u solves the diffusion equation u t = u xx, u(, t) = = u(l, t).

39 Example: heat equation Suppose u solves the diffusion equation u t = u xx, u(, t) = = u(l, t). Then the quantity is dissipated, F (t) = 1 2 u2 x dx

40 Example: heat equation Suppose u solves the diffusion equation u t = u xx, u(, t) = = u(l, t). Then the quantity is dissipated, since df dt = F (t) = u x u xt dx = 1 2 u2 x dx u xx u t dx = u 2 xx dx,

41 Example: heat equation Suppose u solves the diffusion equation u t = u xx, u(, t) = = u(l, t). Then the quantity is dissipated, since df dt = F (t) = u x u xt dx = 1 2 u2 x dx u xx u t dx = u 2 xx dx, One interpretation: arclength of x-cross sections of u can be approximated 1 + uxdx u2 x dx. Since df /dt, arclength diminishes and spatial oscillations die away.