First Order Partial Differential Equations Method of characteristics Web Lecture WI2607-2008 H.M. Schuttelaars May 15, 2018 Delft Institute of Applied Mathematics

Contents Linear First Order Partial Differential Equations Derivation of the Characteristic Equation Examples (solved using Maple) Quasi-Linear Partial Differential Equations Nonlinear Partial Differential Equations Derivation of Characteristic Equations Example May 15, 2018

Contents Linear First Order Partial Differential Equations Derivation of the Characteristic Equation Examples (solved using Maple) Quasi-Linear Partial Differential Equations Nonlinear Partial Differential Equations Derivation of Characteristic Equations Example May 15, 2018

Contents Linear First Order Partial Differential Equations Derivation of the Characteristic Equation Examples (solved using Maple) After this lecture: you can recognize a linear first order PDE you can write down the corresponding characteristic equations you can parameterize the initial condition and solve the characteristic equation using the initial condition, either analytically or using Maple May 15, 2018

First Order Linear Partial Differential Equations Definition of a first order linear PDE: May 15, 2018

First Order Linear Partial Differential Equations Definition of a first order linear PDE: May 15, 2018

First Order Linear Partial Differential Equations Definition of a first order linear PDE: This is the directional derivative of u in the direction <a,b> May 15, 2018

First Order Linear Partial Differential Equations Example May 15, 2018

First Order Linear Partial Differential Equations Plot the direction field: May 15, 2018

First Order Linear Partial Differential Equations Plot the direction field: t x May 15, 2018

First Order Linear Partial Differential Equations Plot the direction field: t x May 15, 2018

First Order Linear Partial Differential Equations Direction field: Through every point, a curve exists that is tangent to <a,b> everywhere. t x May 15, 2018

First Order Linear Partial Differential Equations Direction field: Through every point, a curve exists that is tangent to <a,b> everywhere: 1) Take points (0.5,0.5), (-0.1,0.5) and (0.2,0.01) X X X May 15, 2018

First Order Linear Partial Differential Equations Direction field: Through every point, a curve exists that is tangent to <a,b>everywhere: 1) Take points (0.5,0.5), (-0.1,0.5) and (0.2,0.01) 2) Now draw the lines through those points that are tangent to <a,b> for all points on the lines. May 15, 2018

First Order Linear Partial Differential Equations Zooming in on the line through (0.5,0.5), tangent to <a,b> for all x en t on the line: Direction field: May 15, 2018

First Order Linear Partial Differential Equations Zooming in on the line through (0.5,0.5), tangent to <a,b> for all x en t on the line: Direction field: Parameterize these lines with a parameter s May 15, 2018

First Order Linear Partial Differential Equations SHORT INTERMEZZO May 15, 2018

First Order Linear Partial Differential Equations SHORT INTERMEZZO Parameterization of a line in 2 dimensions Parameter representation of a circle May 15, 2018

First Order Linear Partial Differential Equations Or in 3 dimensions Parameter representation of a helix May 15, 2018

First Order Linear Partial Differential Equations Or in 3 dimensions NOW BACK TO THE CHARACTERISTIC BASE CURVES Parameter representation of a helix May 15, 2018

First Order Linear Partial Differential Equations Parameterize these lines with a parameter s: Direction field: s=0.1 s=0.02 s=0 For example: s=0: (x(0),t(0)) = (0.5,0.5) changing s results in other points on this curve s=0.04 s=0.01 May 15, 2018

First Order Linear Partial Differential Equations Parameterize these lines with a parameter s: Its tangent vector is: Direction field: s=0.1 s=0.02 s=0 s=0.04 s=0.01 May 15, 2018

First Order Linear Partial Differential Equations Parameterize these lines with a parameter s, x=x(s), t=t(s). Its tangent vector is given by On the curve: May 15, 2018

First Order Linear Partial Differential Equations Parameterize these lines with a parameter s, x=x(s), t=t(s). Its tangent vector is given by On the curve: May 15, 2018

First Order Linear Partial Differential Equations Parameterize these lines with a parameter s, x=x(s), t=t(s). Its tangent vector is given by On the curve: May 15, 2018

First Order Linear Partial Differential Equations Its tangent vector is given by On the curve: IN WORDS: THE PDE REDUCES TO AN ODE ON THE CHARACTERISTIC CURVES May 15, 2018

First Order Linear Partial Differential Equations The PDE reduces to an ODE on the characteristic curves. The characteristic equations (that define the characteristic curves) read: May 15, 2018

First Order Linear Partial Differential Equations The PDE reduces to an ODE on the characteristic curves. The characteristic equations (that define the characteristic curves) read: One can solve for x(s) and t(s) without solving for u(s). May 15, 2018

First Order Linear Partial Differential Equations The PDE reduces to an ODE on the characteristic curves. The characteristic equations (that define the characteristic curves) read: One can solve for x(s) and t(s) without solving for u(s). Gives the characteristic base curves May 15, 2018

First Order Linear Partial Differential Equations The equations for the characteristic base were solved to get the base curves in the example: Solving May 15, 2018

First Order Linear Partial Differential Equations The equations for the characteristic base were solved to get the base curves in the example: Solving gives May 15, 2018

First Order Linear Partial Differential Equations This parameterisation, i.e., was plotted for (0.5,0.5) (x(0),t(0)) = (-0.1,0.5) (0.2,0.01) by varying s! May 15, 2018

First Order Linear Partial Differential Equations To solve the original PDE, u(x,t) has to be prescribed at a certain curve C =C (x,t). May 15, 2018

First Order Linear Partial Differential Equations To solve the original PDE, u(x,t) has to be prescribed at a certain curve C =C (x,t). The corresponding system of ODE’s has to be solved such that u(x,t) has the prescribed value at this curve C . May 15, 2018

First Order Linear Partial Differential Equations The corresponding system of ODE’s has to be solved such that u(x,t) has the prescribed value at this curve C . As a first step, parameterize the initial curve C with the parameter τ: x=x(τ), t=t(τ) and u=u(τ). May 15, 2018

First Order Linear Partial Differential Equations The corresponding system of ODE’s has to be solved such that u(x,t) has the prescribed value at this curve C . As a first step, parameterize the initial curve C with the parameter τ: x=x(τ), t=t(τ) and u=u(τ). Next, the family of characteristic curves, determined by the points on C , may be parameterized by x=x(s, τ), t=t(x, τ) and u=u(s, τ), with the initial conditions prescribed for s=0. May 15, 2018

First Order Linear Partial Differential Equations As a first step, parameterize the initial curve C with the parameter τ: x=x(τ), t=t(τ) and u=u(τ). Next, the family of characteristic curves, determined by the points on C , may be parameterized by x=x(s, τ), t=t(x, τ) and u=u(s, τ), with the initial conditions prescribed for s=0. This gives the solution surface May 15, 2018

First Order Linear Partial Differential Equations EXAMPLE 1 May 15, 2018

First Order Linear Partial Differential Equations Consider with The corresponding PDE reads: May 15, 2018

First Order Linear Partial Differential Equations Consider with Parameterize this initial curve with parameter l: May 15, 2018

First Order Linear Partial Differential Equations Consider with Parameterize this initial curve with parameter l: Solve the characteristic equations with these initial conditions. May 15, 2018

First Order Linear Partial Differential Equations Consider with The (parameterized) solution reads: May 15, 2018

First Order Linear Partial Differential Equations Visualize the solution for various values of l: May 15, 2018

First Order Linear Partial Differential Equations When all values of l and s are considered, we get the solution surface: May 15, 2018

First Order Linear Partial Differential Equations EXAMPLE 2 May 15, 2018

First Order Linear Partial Differential Equations PDE: Initial condition: May 15, 2018

First Order Linear Partial Differential Equations PDE: Initial condition: Char eqns: May 15, 2018

First Order Linear Partial Differential Equations PDE: Initial condition: Char eqns: Parameterised initial condition: May 15, 2018

First Order Linear Partial Differential Equations Char eqns: Parameterized initial condition: Parameterized solution: May 15, 2018

First Order Linear Partial Differential Equations Char eqns: Initial condition: May 15, 2018

First Order Linear Partial Differential Equations Char curves: Initial condition: May 15, 2018

First Order Linear Partial Differential Equations This gives the solution surface, parameterised by s and l. May 15, 2018

First Order Linear Partial Differential Equations This gives the solution surface, parameterised by s and l. This is the solution of the original PDE With initial condition May 15, 2018

First Order Linear Partial Differential Equations Recipe to solve the PDE To solve the partial differential equation explicitly, u(x,t) must be given at a certain curve C . As a first step, parameterize this curve with parameter l. May 15, 2018

First Order Linear Partial Differential Equations Write down the characteristic equations, note that x, t and u now depend on both the parameters s and l. May 15, 2018

First Order Linear Partial Differential Equations Solve the characteristic equations, using the conditions on the curve C . Take s=0 at this curve. Hence x, t and u are obtained as functions of s and l. May 15, 2018

First Order Linear Partial Differential Equations Solve the characteristic equations, using the conditions on the curve C . Take s=0 at this curve. Hence x, t and u are obtained as functions of s and l. To get u in terms of x and t, at least in the neighbourhood of C , explicit expressions for s and l are needed: May 15, 2018

First Order Linear Partial Differential Equations Solve the characteristic equations, using the conditions on the curve C . Take s=0 at this curve. Hence x, t and u are obtained as functions of s and l. To get u in terms of x and t, at least in the neighbourhood of C , explicit expressions for s and l are needed: Implicit Function Theorem: possible in a neighborhood of C if May 15, 2018

First Order Linear Partial Differential Equations Solve the characteristic equations, using the conditions on the curve C . Take s=0 at this curve. Hence x, t and u are obtained as functions of s and l. To get u in terms of x and t, at least in the neighbourhood of C , explicit expressions for s and l are needed: (Implicit Function Theorem) possible in a neighborhood of C if otherwise: no solution of infinitely many solutions May 15, 2018

First Order Linear Partial Differential Equations EXAMPLE 3 (using Maple) May 15, 2018

First Order Linear Partial Differential Equations Consider the PDV with initial condition May 15, 2018

First Order Linear Partial Differential Equations Consider the PDV with initial condition Visualize the IC using May 15, 2018

First Order Linear Partial Differential Equations Consider the PDV with initial condition Write down the characteristic equations May 15, 2018

First Order Linear Partial Differential Equations Consider the PDV with initial condition Solve the characteristic equations using dsolve: May 15, 2018

First Order Linear Partial Differential Equations Consider the PDV with initial condition Solve the characteristic equations using dsolve: May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to vizualize the solution surface: Note that in this example we can easily find s and l in terms of x and t and hence get the solution in terms of x and t: s = t l = x-t May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to plot the base characteristics: May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to plot the base characteristics: May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to vizualize the solution surface: May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to vizualize the solution surface: May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to vizualize the solution surface: May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to animate the solution in time: May 15, 2018

First Order Linear Partial Differential Equations Use the solution sol to animate the solution in time: May 15, 2018

First Order Linear Partial Differential Equations EXAMPLE 4 (using Maple) May 15, 2018

First Order Linear Partial Differential Equations Consider the PDV with initial condition May 15, 2018

First Order Linear Partial Differential Equations Consider the PDV with initial condition Write down the characteristic equations May 15, 2018

First Order Linear Partial Differential Equations Using the same commands, we can find: solution surface May 15, 2018

First Order Linear Partial Differential Equations Using the same commands, we can find: animation of the solution solution surface May 15, 2018

Conclusions You are able to: you can recognize a linear first order PDE you can write down the corresponding characteristic equations you can parameterize the initial condition and solve the characteristic equation using the initial condition, either analytically or using Maple May 15, 2018

Conclusions You are able to: you can recognize a linear first order PDE you can write down the corresponding characteristic equations you can parameterize the initial condition and solve the characteristic equation using the initial condition, either analytically or using Maple Next lecture: quasi-linear first order partial differential equations May 15, 2018