First-Order PDEs

MATH 467 Partial Differential Equations

J. Robert Buchanan

Department of Mathematics

Spring 2016

Objectives

In this lesson we will learn:

▶ solutions to linear and quasilinear first-order partial

differential equations, ▶ applications of first-order PDEs as mathematical models of

traffic flow, structured populations, etc.

Transport Equations

The general form of a first-order scalar PDE is

u_{t} + ∇ · f (x, t, u, ∇u) = g(x, t, u)

where ▶ ∇ · f denotes the divergence of f , and ▶ ∇u denotes the gradient of u.

Simple Case

Let c be a constant and consider

u_{t} + c u_{x} = 0
⟨u_{x}, u_{t}⟩ · ⟨c, 1⟩ = 0
D_{⟨c,1⟩}u(x, t) = 0.

Remarks:

▶ This can be interpreted as stating the directional derivative

of u(x, t) in the direction of vector ⟨c, 1⟩ is 0. ▶ Function u(x, t) is constant along lines of the form

x − c t = k. ▶ Function u(x, t) ≡ f (k) where f is an arbitrary differentiable

function.

Let c be a constant and consider

u_{t} + c u_{x} = 0
⟨u_{x}, u_{t}⟩ · ⟨c, 1⟩ = 0
D_{⟨c,1⟩}u(x, t) = 0.

Remarks:

▶ This can be interpreted as stating the directional derivative

of u(x, t) in the direction of vector ⟨c, 1⟩ is 0. ▶ Function u(x, t) is constant along lines of the form

x − c t = k. ▶ Function u(x, t) ≡ f (k) where f is an arbitrary differentiable

function.

Let c be a constant and consider

u_{t} + c u_{x} = 0
⟨u_{x}, u_{t}⟩ · ⟨c, 1⟩ = 0
D_{⟨c,1⟩}u(x, t) = 0.

Remarks:

▶ This can be interpreted as stating the directional derivative

of u(x, t) in the direction of vector ⟨c, 1⟩ is 0. ▶ Function u(x, t) is constant along lines of the form

x − c t = k. ▶ Function u(x, t) ≡ f (k) where f is an arbitrary differentiable

function.

Let c be a constant and consider

u_{t} + c u_{x} = 0
⟨u_{x}, u_{t}⟩ · ⟨c, 1⟩ = 0
D_{⟨c,1⟩}u(x, t) = 0.

Remarks:

▶ This can be interpreted as stating the directional derivative

of u(x, t) in the direction of vector ⟨c, 1⟩ is 0. ▶ Function u(x, t) is constant along lines of the form

x − c t = k. ▶ Function u(x, t) ≡ f (k) where f is an arbitrary differentiable

function.

Confirmation

Let u(x, t) = f (x − c t) then

u_{t} = −c f′(x − c t)
u_{x} = f′(x − c t)

and

u_{t} + c u_{x} = −c f′(x − c t) + c f′(x − c t) = 0.

Functions of the form u(x, t) = f (x − c t) are referred to as the general solution to this simple PDE.

Confirmation

Let u(x, t) = f (x − c t) then

u_{t} = −c f′(x − c t)
u_{x} = f′(x − c t)

and

u_{t} + c u_{x} = −c f′(x − c t) + c f′(x − c t) = 0.

Functions of the form u(x, t) = f (x − c t) are referred to as the general solution to this simple PDE.

Confirmation

Let u(x, t) = f (x − c t) then

u_{t} = −c f′(x − c t)
u_{x} = f′(x − c t)

and

u_{t} + c u_{x} = −c f′(x − c t) + c f′(x − c t) = 0.

Functions of the form u(x, t) = f (x − c t) are referred to as the general solution to this simple PDE.

Initial Conditions

If the value of u(x, t) along the line where t = 0 is specified, then an initial condition of the form u(x, 0) = ϕ(x) further specifies the solution.

u(x, t) = f (x − c t) (general solution)
u(x, 0) = f (x) = ϕ(x) (initial condition)
u(x, t) = ϕ(x − c t) (solution to IVP)

Initial Conditions

If the value of u(x, t) along the line where t = 0 is specified, then an initial condition of the form u(x, 0) = ϕ(x) further specifies the solution.

u(x, t) = f (x − c t) (general solution)
u(x, 0) = f (x) = ϕ(x) (initial condition)
u(x, t) = ϕ(x − c t) (solution to IVP)

General First-Order Linear PDE

The solution technique used previously can be extended to the general first-order linear PDE of the form

a(x, y)u_{x} + b(x, y)u_{y} + c(x, y)u = d(x, y),

for (x, y) ∈ D ⊂ ℝ^{2}.

Note: symbol y is used in place of t as in the previous example. This change of variable is made so that t can be used as a parameter in the solution procedure to be outlined next.

Vector Field (1 of 2)

Suppose (x, y) ≡ (x(t), y(t)) where t is a parameter, then

\frac{d}{dt}u(x, y) = u_{x}x′(t) + u_{y}y′(t)

and we can think of

a(x, y)u_{x} + b(x, y)u_{y} + c(x, y)u = d(x, y) as
u_{x}x′(t) + u_{y}y′(t) = −c(x, y)u + d(x, y) where
\frac{dx}{dt} = a(x, y)
\frac{dy}{dt} = b(x, y).

Remark: the parametric curve (x(t), y(t)) is tangent to the vector field ⟨a(x, y), b(x, y)⟩ for all (x, y) ∈ D.

Vector Field (1 of 2)

Suppose (x, y) ≡ (x(t), y(t)) where t is a parameter, then

\frac{d}{dt}u(x, y) = u_{x}x′(t) + u_{y}y′(t)

and we can think of

a(x, y)u_{x} + b(x, y)u_{y} + c(x, y)u = d(x, y) as
u_{x}x′(t) + u_{y}y′(t) = −c(x, y)u + d(x, y) where
\frac{dx}{dt} = a(x, y)
\frac{dy}{dt} = b(x, y).

Remark: the parametric curve (x(t), y(t)) is tangent to the vector field ⟨a(x, y), b(x, y)⟩ for all (x, y) ∈ D.

Vector Field (2 of 2)

If \frac{dx}{dt} = a(x, y) and \frac{dy}{dt} = b(x, y) then

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b(x, y)}{a(x, y)}.

Suppose the implicit form of the solution of this ODE is ϕ(x, y) = k where k is an arbitrary constant.

Along the curves defined by ϕ(x, y) = k the general first-order linear PDE

a(x, y)u_{x} + b(x, y)u_{y} = −c(x, y)u + d(x, y)

becomes

u_{x} + u_{y} \frac{b(x, y)}{a(x, y)}

=

−c(x, y)u + d(x, y)
a(x, y)

u_{x} + u_{y} \frac{dy}{dx} = −c(x, y)u + d(x, y)

a(x, y)
\frac{du}{dx} = \frac{−c(x, y)u + d(x, y)}{a(x, y)}.

Ordinary Differential Equation

\frac{du}{dx} = \frac{−c(x, y)u + d(x, y)}{a(x, y)}

is an ordinary differential equation for u as a function of variable x (since y ≡ y(x)) and arbitrary constant k.

Solving this ODE yields u ≡ u(x, k) ≡ u(x, y) since k = ϕ(x, y).

Characteristics

▶ The ordinary differential equations

\frac{dy}{dx} = \frac{b(x, y)}{a(x, y)}
\frac{du}{dt} = −c(x, y)u + d(x, y)

are called the characteristic equations. ▶ The solution curves (x(t), y(t), u(x(t), y(t))) to the

characteristic equations are called the characteristic curves.

▶ The curves in the xy-plane of the form (x(t), y(t)) are called characteristics.

▶ The ordinary differential equations

\frac{dy}{dx} = \frac{b(x, y)}{a(x, y)}
\frac{du}{dt} = −c(x, y)u + d(x, y)

are called the characteristic equations. ▶ The solution curves (x(t), y(t), u(x(t), y(t))) to the

characteristic equations are called the characteristic curves.

▶ The curves in the xy-plane of the form (x(t), y(t)) are called characteristics.

▶ The ordinary differential equations

\frac{dy}{dx} = \frac{b(x, y)}{a(x, y)}
\frac{du}{dt} = −c(x, y)u + d(x, y)

are called the characteristic equations. ▶ The solution curves (x(t), y(t), u(x(t), y(t))) to the

characteristic equations are called the characteristic curves.

▶ The curves in the xy-plane of the form (x(t), y(t)) are called characteristics.

Example

Consider the initial value problem:

2u_{x} + 3u_{y} − 4u = 0
u(x, 0) = sin x.

1. Find the general solution to the PDE using the method of characteristics.

2. Find the solution to the IVP.

Solution (1 of 2)

2u_{x} + 3u_{y} − 4u = 0

In this PDE we can let a(x, y) = 2 and b(x, y) = 3, thus

\frac{dy}{dx} = \frac{3}{2} ⟹ y = \frac{3}{2}x + k̂ ⟺ 2y − 3x = k.

Comment: the characteristics for this PDE are straight lines of the form 2y − 3x = k.

Think of u(x, y) = u(x, y(x)) so that

\frac{du}{dx} = u_{x} + u_{y} \frac{dy}{dx} = u_{x} + \frac{3}{2}u_{y} = 2u

which implies

u(x, y) = f (k)e^{2x} = f (2y − 3x)e^{2x}

where f is an arbitrary differentiable function.

2u_{x} + 3u_{y} − 4u = 0

In this PDE we can let a(x, y) = 2 and b(x, y) = 3, thus

\frac{dy}{dx} = \frac{3}{2} ⟹ y = \frac{3}{2}x + k̂ ⟺ 2y − 3x = k.

Comment: the characteristics for this PDE are straight lines of the form 2y − 3x = k.

Think of u(x, y) = u(x, y(x)) so that

\frac{du}{dx} = u_{x} + u_{y} \frac{dy}{dx} = u_{x} + \frac{3}{2}u_{y} = 2u

which implies

u(x, y) = f (k)e^{2x} = f (2y − 3x)e^{2x}

where f is an arbitrary differentiable function.

Solution (2 of 2)

In order to satisfy the initial condition

u(x, 0) = sin x
f (−3x)e^{2x} = sin x
f (−3x) = e^{−2x} sin x
f (z) = −e^{2z/3} sin(z/3).

Illustration

Thus the solution to the IVP is u(x, y) = −e^{2(2y−3x)/3} sin((2y−3x)/2)e^{2x} = −e^{4y/3} sin (y − \frac{3}{2}x). u(x, y) = −e^{4y/3} sin (y − \frac{3}{2}x)

Example

Find the general solution to the first-order linear PDE: −y u_{x} + x u_{y} = 0.

If we let a(x, y) = −y and b(x, y) = x then the characteristics satisfy the ODE

\frac{dy}{dx} = −\frac{x}{y}
y dy = −x dx
∫ y dy = − ∫ x dx
\frac{1}{2}y^{2} = −\frac{1}{2}x^{2} + k̂
x^{2} + y^{2} = k.

In this example the characteristics are circles of radius \sqrt{k} for k ≥ 0.

Solution (2 of 2)

Assuming u(x, y) = u(x, y(x)) then

\frac{du}{dx} = u_{x} + u_{y} \frac{dy}{dx}
= u_{x} − \frac{x}{y}u_{y}
\frac{du}{dx} = 0
u = f (k)
u(x, y) = f (x^{2} + y^{2})

where f is an arbitrary differentiable function.

Example

Solve the initial value problem:

2x y u_{x} + u_{y} − u = 0
u(x, 0) = x,

for x > 0 and y > 0.

Solution (1 of 2)

Let a(x, y) = 2x y and b(x, y) = 1 then

\frac{dy}{dx} = \frac{1}{2x y}
y dy = \frac{1}{2x} dx
∫ y dy = ∫ \frac{1}{2x} dx
\frac{1}{2}y^{2} = \frac{1}{2} lnx + k̂
y^{2} − lnx = k.

These are the characteristics of the solution.

Solution (2 of 2)

If u(x, y) = u(x, y(x)) then

\frac{du}{dx} = u_{x} + u_{y} \frac{dy}{dx}
= u_{x} + \frac{1}{2x y} u_{y}
\frac{du}{dx} = \frac{u}{2x y} = \frac{u}{2x\sqrt{k + lnx}}
\frac{1}{u} du = \frac{1}{2x\sqrt{k + lnx}} dx
∫ \frac{1}{u} du = ∫ \frac{1}{2x\sqrt{k + lnx}} dx
lnu = \sqrt{k + lnx} + lnf (k)
u(x, y) = f (k)e^{\sqrt{k+lnx}} = f (y^{2} − lnx)e^{y}

where f is an arbitrary differentiable function. Since u(x, 0) = x then

x = f (− lnx) ⟺ f (x) = e^{−x} ⟹ u(x, y) = e^{−y^{2}+lnx}e^{y} = x e^{y−y^{2}}.

Solution (2 of 2)

If u(x, y) = u(x, y(x)) then

\frac{du}{dx} = u_{x} + u_{y} \frac{dy}{dx}
= u_{x} + \frac{1}{2x y} u_{y}
\frac{du}{dx} = \frac{u}{2x y} = \frac{u}{2x\sqrt{k + lnx}}
\frac{1}{u} du = \frac{1}{2x\sqrt{k + lnx}} dx
∫ \frac{1}{u} du = ∫ \frac{1}{2x\sqrt{k + lnx}} dx
lnu = \sqrt{k + lnx} + lnf (k)
u(x, y) = f (k)e^{\sqrt{k+lnx}} = f (y^{2} − lnx)e^{y}

where f is an arbitrary differentiable function. Since u(x, 0) = x then

x = f (− lnx) ⟺ f (x) = e^{−x} ⟹ u(x, y) = e^{−y^{2}+lnx}e^{y} = x e^{y−y^{2}}.

u(x, y) = x e^{y−y^{2}}

Homework

▶ Read Section 2.1 ▶ Exercises: 1–5

Quasilinear and Semilinear PDEs

A first-order PDE is called quasilinear if it has the form

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0

(assuming u ≡ u(x, y) for (x, y) ∈ D ⊂ ℝ^{2}). A first-order PDE is called semilinear if it has the form

a(x, y)u_{x} + b(x, y)u_{y} − c(x, y, u) = 0.

We will generalize the method of characteristics in order to solve quasilinear PDEs.

Method of Lagrange

Solutions of the quasilinear PDE

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0

can be expressed implicitly as φ(x, y, z) = 0 when φ(x, y, z) is a solution to the PDE

a(x, y, z)φ_{x} + b(x, y, z)φ_{y} + c(x, y, z)φ_{z} = 0.

Proof (1 of 3)

▶ Suppose u ≡ u(x, y) is a solution to

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0.

▶ Define φ(x, y, z) = u(x, y) − z, then

φ_{x} = u_{x}, φ_{y} = u_{y}, φ_{z} = −1.

▶ Substitute into the expression

a(x, y, z)φ_{x} + b(x, y, z)φ_{y} + c(x, y, z)φ_{z}

= a(x, y, u(x, y))u_{x} + b(x, y, u(x, y))u_{y} + c(x, y, u(x, y))(−1)
= 0,

when z = u(x, y) or equivalently when φ(x, y, z) = 0.

Proof (1 of 3)

▶ Suppose u ≡ u(x, y) is a solution to

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0.

▶ Define φ(x, y, z) = u(x, y) − z, then

φ_{x} = u_{x}, φ_{y} = u_{y}, φ_{z} = −1.

▶ Substitute into the expression

a(x, y, z)φ_{x} + b(x, y, z)φ_{y} + c(x, y, z)φ_{z}

= a(x, y, u(x, y))u_{x} + b(x, y, u(x, y))u_{y} + c(x, y, u(x, y))(−1)
= 0,

when z = u(x, y) or equivalently when φ(x, y, z) = 0.

Proof (2 of 3)

▶ Suppose φ(x, y, z) is a solution of

a(x, y, z)φ_{x} + b(x, y, z)φ_{y} + c(x, y, z)φ_{z} = 0

and let (x_{0}, y_{0}, z_{0}) be a point on the surface φ(x, y, z) = 0 for which φ_{z}(x_{0}, y_{0}, z_{0}) ≠ 0.

▶ Near point (x_{0}, y_{0}, z_{0}) the surface φ(x, y, z) = 0 is the graph of some function, call it u(x, y).

φ(x, y, u(x, y)) = 0

▶ Differentiate the equation above.

φ_{x} + φ_{z}u_{x} = 0 ⟺ φ_{x} = −φ_{z}u_{x}
φ_{y} + φ_{z}u_{y} = 0 ⟺ φ_{y} = −φ_{z}u_{y}

Proof (3 of 3)

▶ So far we have φ(x, y, z) = 0 when z = u(x, y) where

φ(x, y, z) is a solution of

a(x, y, z)φ_{x} + b(x, y, z)φ_{y} + c(x, y, z)φ_{z} = 0

▶ We also know φ_{x} = −φ_{z}u_{x} and φ_{y} = −φ_{z}u_{y} where φ_{z} ≠ 0. ▶ Substitute into the PDE above.

−a(x, y, u)φ_{z}u_{x} − b(x, y, u)φ_{z}u_{y} + c(x, y, z)φ_{z} = 0
a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0

So u(x, y) solves the quasilinear PDE.

Integral Surface

Suppose u(x, y) solves the quasilinear PDE:

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0

for (x, y) ∈ D, then the surface

S = {(x, y, u(x, y)) : (x, y) ∈ D}

is called an integral surface. The vector ⟨u_{x}, u_{y}, −1⟩ is normal to the integral surface for all (x, y) ∈ D.

Since

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0
⟨u_{x}, u_{y}, −1⟩ · ⟨a(x, y, u), b(x, y, u), c(x, y, u)⟩ = 0

then vector ⟨a(x, y, u), b(x, y, u), c(x, y, u)⟩ is perpendicular to the normal to the integral surface, i.e., tangent to the integral surface.

Integral Surface

Suppose u(x, y) solves the quasilinear PDE:

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0

for (x, y) ∈ D, then the surface

S = {(x, y, u(x, y)) : (x, y) ∈ D}

is called an integral surface. The vector ⟨u_{x}, u_{y}, −1⟩ is normal to the integral surface for all (x, y) ∈ D.

Since

a(x, y, u)u_{x} + b(x, y, u)u_{y} − c(x, y, u) = 0
⟨u_{x}, u_{y}, −1⟩ · ⟨a(x, y, u), b(x, y, u), c(x, y, u)⟩ = 0

then vector ⟨a(x, y, u), b(x, y, u), c(x, y, u)⟩ is perpendicular to the normal to the integral surface, i.e., tangent to the integral surface.

Characteristic System

The vector ⟨a(x, y, u), b(x, y, u), c(x, y, u)⟩ defines a vector field in xyu-space. The integral curves (x(t), y(t), u(t)) defined by the characteristic system of ODEs

\frac{dx}{dt} = a(x, y, u)
\frac{dy}{dt} = b(x, y, u)
\frac{du}{dt} = c(x, y, u)

are called the characteristic curves. The projections of the characteristic curves in the xy-plane will be called characteristics.

Constructing a Solution

Suppose Γ is a non-characteristic curve in xyu-space and we construct a family of characteristic curves through points of Γ.

Solution Surface

The union of all the characteristic curves through points on Γ is an solution (integral) surface to the quasilinear PDE.

Example

Find the solution of

y u_{x} − x u_{y} − e^{u} = 0

that passes through the curve Γ = {(x, y, u) = (s, sin s, 0) : s ∈ ℝ}.

Remarks:

▶ Γ is parameterized by s for clarity, this parameter will not

be used for any other purpose. ▶ The solution u must satisfy the condition

u(x, sin x) = 0

for all x.

Solution (1 of 5)

The characteristic system for this example is

\frac{dx}{dt} = y, \frac{dy}{dt} = −x, \frac{du}{dt} = e^{u}.

From the last equation, −e^{−u} = t − C where C is a constant. From the second equation,

\frac{dy}{dt} = −x = \frac{d}{dt} [\frac{dx}{dt}] = \frac{d^{2}x}{dt^{2}} ⟺ x′′(t) + x = 0. Thus x(t) = A cos t + B sin t and consequently y(t) = x′(t) = −A sin t + B cos t where A and B are constants.

Solution (1 of 5)

The characteristic system for this example is

\frac{dx}{dt} = y, \frac{dy}{dt} = −x, \frac{du}{dt} = e^{u}.

From the last equation, −e^{−u} = t − C where C is a constant. From the second equation,

\frac{dy}{dt} = −x = \frac{d}{dt} [\frac{dx}{dt}] = \frac{d^{2}x}{dt^{2}} ⟺ x′′(t) + x = 0. Thus x(t) = A cos t + B sin t and consequently y(t) = x′(t) = −A sin t + B cos t where A and B are constants.

Solution (1 of 5)

The characteristic system for this example is

\frac{dx}{dt} = y, \frac{dy}{dt} = −x, \frac{du}{dt} = e^{u}.

From the last equation, −e^{−u} = t − C where C is a constant. From the second equation,

\frac{dy}{dt} = −x = \frac{d}{dt} [\frac{dx}{dt}] = \frac{d^{2}x}{dt^{2}} ⟺ x′′(t) + x = 0. Thus x(t) = A cos t + B sin t and consequently y(t) = x′(t) = −A sin t + B cos t where A and B are constants.

The characteristic curves of the solution can be parameterized as

x = A cos t + B sin t
y = B cos t − A sin t
u = − ln(C − t).

The non-characteristic curve Γ is parameterized as

x = s
y = sin s
u = 0.

For each point on Γ we want to find a characteristic curve which passes through the point when t = 0 (arbitrary choice).

The characteristic curves of the solution can be parameterized as

x = A cos t + B sin t
y = B cos t − A sin t
u = − ln(C − t).

The non-characteristic curve Γ is parameterized as

x = s
y = sin s
u = 0.

For each point on Γ we want to find a characteristic curve which passes through the point when t = 0 (arbitrary choice).

When t = 0,

x(0) = s = A
y(0) = sin s = B
u(0) = 0 = − lnC ⟺ C = 1

The characteristic curve intersecting Γ at t = 0 can be parameterized as

x(t) = s cos t + sin s sin t
y(t) = sin s cos t − s sin t
u(t) = − ln(1 − t).

These equations also produce the integral surface for the solution of the quasilinear PDE.

x = s cos t + sin s sin t
y = sin s cos t − s sin t
u = − ln(1 − t).

We can eliminate s and t from the equations. Multiply the first equation by cos t and the second equation by − sin t and add them together.

x cos t − y sin t = s cos^{2} t + sin s cos t sin t − sin s cos t sin t + s sin^{2} t
= s

Multiply the first equation by sin t and the second equation by cos t and add them together.

x sin t + y cos t = s cos t sin t + sin s sin^{2} t + sin s cos^{2} −s cos t sin t
= sin s

Combining the results produces:

sin(x cos t − y sin t) = x sin t + y cos t.

We managed to eliminate parameter s. Using u = − ln(1 − t) ⟺ t = 1 − e^{−u} we can eliminate parameter t as well.

The solution to the quasilinear PDE can be expressed in implicit form as

sin (x cos [1 − e^{−u}] − y sin [1 − e^{−u}])
= x sin [1 − e^{−u}] + y cos [1 − e^{−u}] .

Illustration Example

Consider the quasilinear first-order PDE

x u_{x} + y u_{y} − sec u = 0.

Find the solution passing through the parametric curve Γ given by Γ = {(x, y, u) = (s^{2}, sin s, 0) : s ∈ ℝ}.

Solution (1 of 3)

The characteristic system is:

\frac{dx}{dt} = x ⟹ x(t) = Ae^{t}
\frac{dy}{dt} = y ⟹ y(t) = Be^{t}
\frac{du}{dt} = sec u ⟹ sin u = t + C

where A, B, and C are constants. Suppose the characteristics pass through Γ at t = 0 (arbitrary choice).

x(0) = A = s^{2}
y(0) = B = sin s
u(0) = 0 = 0 + C ⟹ C = 0

\frac{dx}{dt} = x ⟹ x(t) = Ae^{t}
\frac{dy}{dt} = y ⟹ y(t) = Be^{t}
\frac{du}{dt} = sec u ⟹ sin u = t + C

where A, B, and C are constants. Suppose the characteristics pass through Γ at t = 0 (arbitrary choice).

x(0) = A = s^{2}
y(0) = B = sin s
u(0) = 0 = 0 + C ⟹ C = 0

\frac{dx}{dt} = x ⟹ x(t) = Ae^{t}
\frac{dy}{dt} = y ⟹ y(t) = Be^{t}
\frac{du}{dt} = sec u ⟹ sin u = t + C

where A, B, and C are constants. Suppose the characteristics pass through Γ at t = 0 (arbitrary choice).

x(0) = A = s^{2}
y(0) = B = sin s
u(0) = 0 = 0 + C ⟹ C = 0

\frac{dx}{dt} = x ⟹ x(t) = Ae^{t}
\frac{dy}{dt} = y ⟹ y(t) = Be^{t}
\frac{du}{dt} = sec u ⟹ sin u = t + C

where A, B, and C are constants. Suppose the characteristics pass through Γ at t = 0 (arbitrary choice).

x(0) = A = s^{2}
y(0) = B = sin s
u(0) = 0 = 0 + C ⟹ C = 0

Solution (2 of 3)

The characteristic curves take the form:

x = s^{2}e^{t}, y = e^{t}sin s, u = arcsin t.

We can eliminate parameters s and t from the characteristics curves. Since t = sin u then

y = e^{sin u} sin s ⟺ sin s = y e^{− sin u} ⟹ s = arcsin (y e^{− sin u}) .

Consequently the implicit form of the integral surface can be expressed as

x = [arcsin (y e^{− sin u})]^{2} e^{sin u}.

x = [arcsin (y e^{− sin u})]^{2} e^{sin u} Find the solution to the first-order, quasi linear PDE with side condition.

u_{x} + u u_{y} = 6x
u(0, y) = 3y

Remark: since no non-characteristic curve Γ is mentioned we are free to create one. A natural one to use would be Γ = {(x, y, u) = (0, s, 3s) : s ∈ ℝ}.

Find the solution to the first-order, quasi linear PDE with side condition.

u_{x} + u u_{y} = 6x
u(0, y) = 3y

Remark: since no non-characteristic curve Γ is mentioned we are free to create one. A natural one to use would be Γ = {(x, y, u) = (0, s, 3s) : s ∈ ℝ}.

Solution (1 of 3)

Re-write the PDE as

u_{x} + u u_{y} − 6x = 0

which has the characteristic system:

\frac{dx}{dt} = 1 ⟹ x = t + A
\frac{dy}{dt} = u = 3t^{2} + 6A t + B ⟹ y = t^{3} + 3A t^{2} + B t + C
\frac{du}{dt} = 6x = 6t + 6A ⟹ u = 3t^{2} + 6A t + B.

Solution (1 of 3)

Re-write the PDE as

u_{x} + u u_{y} − 6x = 0

which has the characteristic system:

\frac{dx}{dt} = 1 ⟹ x = t + A
\frac{dy}{dt} = u = 3t^{2} + 6A t + B ⟹ y = t^{3} + 3A t^{2} + B t + C
\frac{du}{dt} = 6x = 6t + 6A ⟹ u = 3t^{2} + 6A t + B.

Solution (1 of 3)

Re-write the PDE as

u_{x} + u u_{y} − 6x = 0

which has the characteristic system:

\frac{dx}{dt} = 1 ⟹ x = t + A
\frac{dy}{dt} = u = 3t^{2} + 6A t + B ⟹ y = t^{3} + 3A t^{2} + B t + C
\frac{du}{dt} = 6x = 6t + 6A ⟹ u = 3t^{2} + 6A t + B.

Solution (1 of 3)

Re-write the PDE as

u_{x} + u u_{y} − 6x = 0

which has the characteristic system:

\frac{dx}{dt} = 1 ⟹ x = t + A
\frac{dy}{dt} = u = 3t^{2} + 6A t + B ⟹ y = t^{3} + 3A t^{2} + B t + C
\frac{du}{dt} = 6x = 6t + 6A ⟹ u = 3t^{2} + 6A t + B.

Solution (2 of 3)

For each point on Γ find the characteristic curve which passes through the point when t = 0 (arbitrary choice).

x(0) = 0 = A
y(0) = s = C
u(0) = 3s = B

Consequently the characteristic curves are in parametric form:

x(t) = t
y(t) = t^{3} + s(3t + 1)
u(t) = 3t^{2} + 3s.

Now eliminate s and t so as to write the solution in implicit form. Taking the parametric equations for x and u we can solve for s and t to obtain

s = \frac{u}{3} − x^{2}t  =
x.

Substituting into the equation for y we get

y = \frac{u}{3} + x u − x^{2} − 3x^{3}

which can be solved explicitly for

u(x, y) = \frac{3(2x^{3} + x^{2} + y)}{3x + 1}.

Illustration Homework

▶ Read Section 2.2 ▶ Exercises: 6–8