Partial Differential Equations (Math 3303)

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Chapter 2First Order Equations

A first-order partial differential equation in two independent variables has the general form

F (x, y, u, ux, uy) = 0,

where F is a given function of its arguments, and u = u(x, y) is an unknown function of the inde-pendent variables x and y which lie in some given domain Ω ⊆ R2. First-order equations have manyapplications such as the transport of material in a fluid flow and propagation of wavefronts in optics.

The main result for first-order PDEs is the fact that the general solution of such equations canbe obtained by solving systems of ordinary differential equations. This is not true for higher orderequations or for systems of first-order equations. However first-order equations have the advantageof providing a conceptual basis that can be utilized for higher-order equations.

In this chapter we are concerned with linear and quasi-linear first-order partial differential equa-tions. We will study solving such equations by using change of variables and the Lagrange methodof characteristics and its generalizations.

2.1 Linear First-order Equations

A first-order PDE is said to be linear if it has the form

A(x, y)ux +B(x, y)uy + C(x, y)u = G(x, y).

It is assumed that A, B, C, and G have continuous first-order derivatives with respect to x and y insome region Ω of the xy−plane. Moreover it is assumed that at least one of the coefficient functionsA and B is not identically zero on Ω.

Example. ux + uy = 1 is a first-order linear PDE.

Remarks.

(1) The linear equationA(x, y)ux +B(x, y)uy + C(x, y)u = G(x, y)

can be written asL(u) = G(x, y),

where L is the differential operator

L = A∂

∂x+B

∂

∂y+ C.

1

(2) The operator L is linear; that is, for each pair of functions u1, u2, and each pair of constantsc1, c2, we have

L(c1u1 + c2u2) = c1L(u1) + c2L(u2).

Definition. (Solution of linear equation)

(1) A solution of L(u) = G on a region Ω is a function u = φ(x, y) with continuous first-order partialderivatives in Ω such that L(φ) = G.

(2) If u = φ(x, y) is a solution of L(u) = G, then the corresponding surface in the xyu space is calledan integral surface.

(3) The general solution of L(u) = 0 is a relation involving an arbitrary function such that eachchoice for the arbitrary function yields a solution of L(u) = 0.

(4) The general solution of L(u) = G is a solution which includes all solutions of L(u) = G.

Example. Show that u(x, y) = e−yf(x− y) is the general solution of ux + uy + u = 0.

Solution.

How to obtain the general solution of L(u) = G

Theorem 1. (General solution of L(u) = G)Let uh be the general solution of L(u) = 0 and let up be any particular solution of L(u) = G. Thenu = uh + up is the general solution of L(u) = G.

Proof.

2

Example 1. Find the general solution of the linear equation ux + u = x.

Solution.

Example 2. Find the general solution of ux + uy + u = 0.

Solution.

3

How to obtain the general solution of L(u) = 0

In the above examples we have seen how we can solve special cases of homogeneous linear partialdifferential equations. In this section we will study a general method to solve homogeneous linearpartial differential equations.

Canonical form of first-order linear equation

Consider a homogeneous linear partial differential equation

L(u) = A(x, y)ux +B(x, y)uy + C(x, y)u = 0.

We look for a transformationξ = ξ(x, y), η = η(x, y)

that transforms L(u) = 0 to a simple equation. We assume that the Jacobian

∂(ξ, η)

∂(x, y)=

∣∣∣∣ξx ηxξy ηy

∣∣∣∣is not zero. This guarantee the existence of the inverse transformation

x = x(ξ, η), y = y(ξ, η).

Definition. (Non-singular transformation)A transformation

ξ = ξ(x, y), η = η(x, y)

is called non-singular in a domain Ω if the Jacobian J =∂(ξ, η)

∂(x, y)= ξxηy − ξyηx 6= 0 of at any point

(x, y) ∈ Ω.

Now consider u(x, y) = u(ξ, η). Using the chain rule we obtain

ux = uξξx + uηηx,

uy = uξξy + uηηy.

Substituting into L(u) = 0 we obtain

(Aξx +Bξy)uξ + (Aηx +Bηy)uη + Cu = 0.

To simplify the last equation we may choose η such that

Aηx +Bηy = 0.

A solution η(x, y) of the last equation can be accomplished as follows. Assume A(x, y) 6= 0 andconsider the ordinary differential equation

dy

dx=B(x, y)

A(x, y).

Let η(x, y) = k be the general solution of this ODE with ηy 6= 0. Then

ηx + ηydy

dx= 0.

4

Usingdy

dx=B(x, y)

A(x, y), we obtain

Aηx +Bηy = 0.

Therefore, we can choose η(x, y) to be any solution of the ordinary differential equation

dy

dx=B(x, y)

A(x, y).

Let us take ξ = x. Then∂(ξ, η)

∂(x, y)=

∣∣∣∣1 ηx0 ηy

∣∣∣∣ = ηy 6= 0.

Thus, the transformationξ = x, η = η(x, y),

where η(x, y) = k is a solution of the ODEdy

dx=B(x, y)

A(x, y)with ηy 6= 0, is invertible.

Definition. (Canonical form of first-order linear equation)The equation

A(ξ, η)uξ + C(ξ, η)u = G(ξ, η)

is the canonical form of the first-order linear equation

A(x, y)ux +B(x, y)uy + C(x, y)u = G(x, y).

Now we can solve the simple equation

A(ξ, η)uξ + C(ξ, η)u = 0.

It is separable; that is it can be written as

uξu

= −CA.

Integrating both sides with respect to ξ, we obtain

u(ξ, η) = f(η) exp(−∫C(ξ, η)

A(ξ, η)dξ),

where f is an arbitrary function.

Theorem 2. (Solution of homogeneous linear PDEs)Let

L(u) = A(x, y)ux +B(x, y)uy + C(x, y)u = 0

be a first-order linear partial differential equation with A(x, y) 6= 0. Assume that ξ = x and

η(x, y) = k is the general solution of the ordinary differential equationdy

dx=B(x, y)

A(x, y)with ηy 6= 0.

Then the general solution of L(u) = 0 is given by

u(x, y) = f(η) exp(−∫C(ξ, η)

A(ξ, η)dξ),

where f is an arbitrary function.

5

Example 1. Solve the partial differential equation

x2ux − xyuy + yu = 0.

Solution.

Example 2. Find the general solution of the partial differential equation

xux + yuy = 2u.

Solution.

6

Example 3. Find the general solution of the partial differential equation

2ux + 3uy = x2.

Solution.

Example 4. Solve the partial differential equation

3ux + 6uy + u = x+ 2ey.

Solution.

7

Exercises

(1) Hold one independent variable constant and integrate with respect to the remaining variableto obtain the general solution for each of the following equations. Verify that your answer iscorrect by substituting into the equation.

(a) uy = x2 + y2

(b) ux = sin

(x

y

)(c) ux + xu = x3 + 3xy

(2) Consider the first-order equation Aux + Buy = 0, where A and B are constants. Assume asolution of the form u = f(αx + βy), where f is an arbitrary function and α and β areconstants. Substitute into the differential equation to determine suitable values of α and β.

(3) Show that if A, B, and C are constants, then the general solution of the homogeneous first-orderlinear equation Aux+Buy +Cu = 0 is given by u = f(Bx−Ay)e−Cx/A, where f is an arbitraryfunction.

(4) Use exercise (3) to obtain the general solution of the following equations:

(a) 3ux − 4uy = x2

(b) ux − 3uy = sinx+ cos y

(c) 5ux + 4uy + u = x2 + 1 + 2e3y

(d) ux + 2uy − 5u = cosx+ y2 + 1

(e) ux − auy = emx cos(by), where a, b, m are constants.

(5) Make the change of independent variables ξ = lnx, η = ln y and reduce the given differentialequation to one with constant coefficients. Then obtain the general solution.

(a) 4xux − 2yuy = 0

(b) 2xux + 3yuy = lnx

(c) xux − 7yuy = x2y

(d) 8xux − 5yuy + 4u = x2 cosx

(e) axux + byuy + cu = x2 + y2, where a, b, c are constants.

(6) Find the general solution of the following equations:

(a) xyux − x2yuy + yu = 0

(b) yux − xuy = 0

(c) (x+ a)ux + (y + b)uy + cu = 0, where a, b, c are constants.

(d) x2ux + y2uy = 2xy

(e) x2ux + y2uy = (x+ y)u

2.2 Quasilinear First-order Equations

A first-order PDE F (x, y, u, ux, uy) = 0 is called quasilinear if it is linear in ux and uy. The generalform of quasilinear first-order PDE is

P (x, y, u)ux +Q(x, y, u)uy = R(x, y, u).

Remark. The linear first-order PDEs are special cases of quasilinear first-order PDEs.

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The Method of Lagrange

Consider the system of first-order ordinary differential equations

dx

P (x, y, u)=

dy

Q(x, y, u)=

du

R(x, y, u). (1)

This system is called subsidiary equations. If P 6= 0, then the subsidiary equations is equivalent tothe system

dy

dx=Q(x, y, u)

P (x, y, u),

du

dx=R(x, y, u)

P (x, y, u). (2)

Similarly, if Q 6= 0 or R 6= 0, then the subsidiary equations may be written on the equivalent form

dx

dy=P (x, y, u)

Q(x, y, u),

du

dy=R(x, y, u)

Q(x, y, u)(3)

ordx

du=P (x, y, u)

R(x, y, u),

dy

du=Q(x, y, u)

R(x, y, u), (4)

respectively. The advantage of the subsidiary system is it avoids the distinguishing between depen-dent and independent variables.

The general solution of system (2) has the form

y = y(x, c1, c2), u = u(x, c1, c2),

where c1, c2 are arbitrary (dependent) constants. If these equations are solved for c1 and c2, thenthe general solution of the subsidiary equations (1) can be written in the form

v(x, y, u) = c1, w(x, y, u) = c2.

Each relation v = c1, w = c2 is called an integral of the subsidiary equations.If v and w are functionally independent is some region Ω in the xyu space; that is, the Jacobians

∂(v, w)

∂(x, y),

∂(v, w)

∂(x, u),

∂(v, w)

∂(y, u)

are not all zero at any point of Ω, then the general solution of the quasilinear equation is given by

F (v, w) = 0,

where F is an arbitrary function. The solution F (v, w) = 0 can be written in the alternative forms

w = f(v) or v = g(w),

where f and g are arbitrary functions.

Remark. The two solutions of the subsidiary equations can be obtained by solving any one of thesystems (2), (3) or (4).

Theorem 3. (Solution of quazilinear equation)Let v and w be two functionally independent solutions of the subsidiary equation (1) in a domain

Ω ⊆ R3. Let F (v, w) be an arbitrary function with continuous first-order derivatives. Then theequation

F (v(x, y, u), w(x, y, u)) = 0

defines u implicitly as a function of x and y and this function is a solution of the equation

P (x, y, u)ux +Q(x, y, u)uy = R(x, y, u).

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Example 1. Solve the equationxux + yuy + u = 0.

Solution.

Exercise: Show that u defined by F(yx, xu)

= 0 is a solution of xux + yuy + u = 0.

Example 2. Find the general solution of the equation

xuux + yuuy + x2 + y2 = 0.

Solution.

10

Exercise: Show that u defined by F(yx, u2 + x2 + y2

)= 0 is a solution of

xuux + yuuy + x2 + y2 = 0.

The Method of Multipliers

A useful technique for integrating a system of first-order equations is the method of multipliers.

Proposition 1. Ifa

b=c

d, then

λa+ µc

λb+ µd=a

b=c

d,

for arbitrary values of multipliers λ, µ.

Proof.

Corollary 1. Ifa

b=c

d=e

f, then

λa+ µc+ νe

λb+ µd+ νf=a

b=c

d=e

f,

for arbitrary values of multipliers λ, µ, ν.

Proof. Exercise

Remark. Applying Corollary 1 to the subsidiary system

dx

P (x, y, u)=

dy

Q(x, y, u)=

du

R(x, y, u)

we obtainλdx+ µdy + νdu

λP + µQ+ νR=

dx

P (x, y, u)=

dy

Q(x, y, u)=

du

R(x, y, u).

In this way related differential equations can be formed, some of which may be easy to integrate. Inparticular if λ, µ, ν are chosen such that

λP + µQ+ νR = 0,

thenλdx+ µdy + νdu = 0.

Now if there exist a function v such that

dv = λdx+ µdy + νdu,

then v(x, y, u) = c1 is an integral of the subsidiary equations.

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Example 1. Find the general solution of the equation

uux + yuy = x.

Solution.

Example 2. Find the general solution of the equation

(y − x)ux + (y + x)uy =x2 + y2

u.

Solution.

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Exercises

Find the general solution of each of the following PDEs:

(a) ux + xuy = u

(b) xux + yuy = nu, where n is a constant.

(c) (x+ u)ux + (y + u)uy = 0

(d) xux + yuy = y

(e) (x+ y)(ux − uy) = u

(f) yux − xuy = x3y + xy3

(g) ux − 2uy = 3x2 sin(y + 2x)

(h) cos(y)ux + cos(x)uy = cosx cos y

(i) (x+ y)ux + (y + u)uy = u

(j) (x2 − y2 − u2)ux + 2xyuy = 2xu

(k) (xy3 − 2x4)ux + (2y4 − x3y)uy = 9u(x3 − y3)

Answers

(a) u(x, y) = exf(2y − x2)

(b) u = xnf(y/x).

(c) u = f

(y + u

x+ u

)(d) u = y + f(y/x)

(e) u = ex/(x+y)f(x+ y)

(f) u =1

2x2(x2 + y2) + f(x2 + y2)

(g) u = x3 sin(2x+ y) + f(2x+ y)

(h) u = sin y + f(sinx− sin y)

(i) u = esin−1(x/√x2+y2)f(x2 + y2) + 1− x2 − y2

(j) x2 + y2 + u2 = yf(u/y)

(k)u2

(x3 + y3)3= f

(x6(x3 + y3)

y6

)

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2.3 Cauchy Problem for Quasilinear First-order Equations

A fundamental problem in the study of differential equations is to solve an initial value problem; thatis, a differential equation subject to an initial condition. In the case of partial differential equation thebasic initial value problem is to determine an integral surface that passes through a given curve in thexyu−space. The problem will also be called Cauchy problem in honor of the French mathematicianAugustin Louis Cauchy (1789− 1857).

Definition. (Cauchy problem)Let Γ be a curve in R3 described parametrically by the equations

Γ : x = x0(s), y = y0(s), u0(s), s ∈ [a, b],

where x0(s), y0(s), and u0(s) are functions with continuous first-order derivatives on [a, b]. The initialvalue problem (Cauchy problem) for a quasilinear equation

P (x, y, u)ux +Q(x, y, u)uy = R(x, y, u)

is to find a function u = u(x, y) defined in a domain Ω of R2 such that:

(a) u = u(x, y) is a solution of

P (x, y, u)ux +Q(x, y, u)uy = R(x, y, u)

in Ω.

(b) On the curve Γ, u is equal to the given function u0(s); that is

u(x0(s), y0(s)) = u0(s), s ∈ [a, b].

The curve Γ will be called the initial curve and the condition

u(x0(s), y0(s)) = u0(s), s ∈ [a, b],

will be called the initial condition.

Example 1. Solve the Cauchy problem

yux − xuy = 0, u(x, 0) = x4.

Solution.

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Example 2. Solve the Cauchy problem

ux + xuy = u, u = sin y, x = 0.

Solution.

Existence and uniqueness of solution

Theorem 4. (Existence and uniqueness Theorem)Consider the quasilinear equation

P (x, y, u)ux +Q(x, y, u)uy = R(x, y, u),

where P, Q, R are continuous functions with continuous first-order partial derivatives in a domainΩ ⊆ R3. Let

Γ : x = x0(s), y = y0(s), u = u0(s), s ∈ [a, b],

be an initial smooth curve in Ω. If

J = P (x0, y0, u0)dy0ds−Q(x0, y0, u0)

dx0ds6= 0, s ∈ [a, b],

then there exists a unique solution u = u(x, y) defined in a neighborhood of the initial curve Γ, whichsatisfies the quasilinear equation

P (x, y, u)ux +Q(x, y, u)uy = R(x, y, u)

and the initial conditionu(x0(s), y0(s)) = u0(s), s ∈ [a, b].

If

J = P (x0, y0, u0)dy0ds−Q(x0, y0, u0)

dx0ds

= 0,

then the Cauchy problem has either no solution at all, or it has infinitely many solutions.

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Example 1. Solve the equation ux = 1 subject to the initial condition u(0, y) = y.

Solution.

Example 2. Determine whether the Cauchy problem

yux − xuy = 0,

Γ : x = cos s, y = sin s, u = sin s,

has a unique solution, infinitely many solutions, or no solution

Solution.

16

Example 3. Show that the Cauchy problem

ux + uy = 1, u(x, x) = x

has infinitely many solutions.

Solution.

Alternative method for solving Cauchy problem

Let Γ be a given smooth curve defined parameterically by

x = x(s), y = y(s), u = u(s), s ∈ [a, b].

Let v = c1 and w = c2 be two functionally independent integral of the subsidiary equations. Writedown the equations

v(x(s), y(s), u(s)) = c1, w(x(s), y(s), u(s)) = c2.

Eliminating s from these equations gives a functional relation

F (c1, c2) = 0

between c1 and c2. Then the solution of the Cauchy problem is given by

F (v(x, y, u), w(x, y, u)) = 0.

Example 1. Find the integral surface of the equation

(y + xu)ux + (x+ yu)uy = u2 − 1

which passes through the parabola

x = s, y = 1, u = s2.

17

Solution.

Example 2. Solve the Cauchy problem

uux + uuy = y + x, u = y2, x = 1.

Solution.

18

Exercises

(1) Find the solution of each of the following Cauchy problem:

(a) ux + uy = u; u = cosx, y = 0.

(b) u(ux − uy) = y − x; x = 1, u = y2.

(c) (x+ u)ux + (y + u)uy = 0; x = 1− s, y = 1 + s, u = s.

(d) xux − yuy = 0; x = y = u = s.

(e) x2ux + y2uy = u2; x = s, y = 2s, u = 1.

(f) xuux + yuuy + xy = 0; xy = 1, u = 4.

(g) 2xuux + 2yuuy = u2 − x2 − y2; x+ y + u = 0, x2 + y2 + u2 = 1.

(h) (u+ y)ux + (u− x)uy = x− y; x = s, y = 2s, u = 0.

(i) x(x2 + y2)ux + 2y2(xux + yuy − u) = 0; x2 + y2 = a2, u = b, where a, b are constants.

(j) (x2 + y2)ux + 2xyuy = xu; x = a, y2 + u2 = a2, where a is a constant.

(k) x(y − u)ux + y(u− x)uy = u(x− y); x = y = u = s.

(l) u(x+ u)ux − y(y + u)uy = 0; x = 1, y = s, u =√s.

(m) secx ux + uy = cot y; u(0, y) = sin y.

(2) Consider the equation yux−xuy = 0, (y > 0). Check for each of the following initial conditionswhether the problem is solvable. If it is solvable, find a solution. If it is not, explain why.

(a) u(x, 0) = x2.

(b) u(x, 0) = x.

(c) u(x, 0) = x, x > 0.

(3) Determine whether each of the following Cauchy problems has a unique solution, infinitely manysolutions, or no solution. If it has solutions, then find them.

(a) (y + u)ux + yuy = x− y, u(x, 1) = x+ 1.

(b) (y2 − u2)ux − xyuy = xu, u(0, y) = −y.

2.4 The method of characteristics

The method of characteristics is a method to solve Cauchy problems for first-order partial differentialequations. This method was developed in the middle of the nineteenth century by Hamilton.

The characteristics method is based on generating the solution surface from a one-parameterfamily of curves that intersect a given curve in space.

Consider the general first-order quasilinear equation

P (x, y, u)ux +Q(x, y, u)uy = R(x, y, u) (5)

and write the initial condition parameterically:

Γ : x = x0(s), y = y0(s), u = u0(s), s ∈ [a, b].

The curve Γ will be called the initial curve.

19

A solution of equation (5) defines an integral surface u = u(x, y) in the (x, y, u)−space. Thenormal to this surface at a point A is the vector −→n = (ux, uy,−1). Let −→v = (P,Q,R) be a vector.Then the equation (5) can be interpreted as the condition that at each point A of the integral surface,the vector −→v is tangent to the integral surface. Thus, the quasilinear equation (5) can be written as

(P,Q,R) · (ux, uy,−1) = 0.

In other words, the vector direction −→v = (P,Q,R) is tangential to the integral surface at each point.The direction −→v = (P,Q,R) at any point on the surface is called the characteristic direction. Aspace curve

(x(t), y(t), u(t))

whose tangent at every point coincides with the characteristic direction is called a characteristiccurve.

If the parametric equations of this characteristic curve are

x = x(t), y = y(t), u = u(t),

then the tangent vector to this curve is

(dx

dt,dy

dt,du

dt

)which must be equal to −→v = (P,Q,R).

Therefore, the system of ordinary differential equations of the characteristic curve is given by theequations

dx

dt= P (x, y, u),

dy

dt= Q(x, y, u),

du

dt= R(x, y, u), (6)

defines curves lying on the solution surface. This system is called the system of characteristic equa-tions or, for short, the characteristic equations.

In fact, there are only two independent ordinary differential equations in the system (6); therefore,its solutions consist of a two-parameter family of curves in (x, y, u)−space. The projection on u = 0of a characteristic curve on the (x, y)−plane is called a characteristic ground curve.

In order to determine a characteristic curve we need an initial condition. We shall require theinitial point to lie on the initial curve Γ. Since each curve (x(t), y(t), u(t)) emanates from a differ-ent point Γ(s), we shall explicitly write the curves in the form (x(t, s), y(t, s), u(t, s)). The initialconditions are written as:

x(0, s) = x0(s), y(0, s) = y0(s), u(0, s) = u0(s). (7)

Notice that we selected the parameter t such that the characteristic curve is located on Γ whent = 0. One may, of course, select any other parameterization. We also notice that, in general, theparameterization (x(t, s), y(t, s), u(t, s)) represents a surface in R3.

Note that the characteristic equations (6) in the nonparametric form are

dx

P (x, y, u)=

dy

Q(x, y, u)=

du

R(x, y, u). (8)

Thus, we have the subsidiary equations for (5).

Remarks.

(1) The above geometrical interpretation can be generalized for higher-order partial differentialequations. However, it is not easy to visualize geometrical arguments that have been describedfor the case of three space dimensions.

(2) The geometrical interpretation is more complicated for the case of nonlinear partial differentialequations, because the normals to possible solution surfaces through a point do not lie in aplane. The tangent planes no longer intersect along one straight line, but instead, they envelopealong a curved surface.

20

Example 1. Solve the equationux + uy = 2

subject to the initial condition u(x, 0) = x2.

Solution.

Example 2. Solve the equation ux = 1 subject to the initial condition u(0, y) = g(y).

Solution.

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Example 3. Determine whether the Cauchy problem

ux + yuy = 2u,

Γ : x = 1, y = s, u = s,

has a unique solution, infinitely many solutions, or no solution

Solution.

Exercises

(1) Solve the initial value problem

ux + uy = u2, u(s, 0) = s2.

(2) Solve the Cauchy problem

xux + (y + x2)uy = u, u(2, s) = s− 4.

(3) Find the integral surface of the equation

ux + uy = u2

passing through the curve

Γ : x = s, y = −s, u = s, s ∈ R.

(4) Find the solution of the equationuux + uy = 1

for the following initial conditions:

22

(a) u(x, 0) = x.

(b) u(x, 0) = x2.

(5) Consider the equation

uux + uy =1

2u.

(a) Show that there is a unique integral surface in a neighborhood of the curve

Γ : (x, y, u) = (s, 0, sin s), s ∈ R.

(b) Find the parametric representation

x = x(t, s), y = y(t, s), u = u(t, s)

of the integral surface S for initial condition of part (a).

(c) Find an integral surface S1 of the same PDE passing through the initial curve

Γ1 : (x, y, u) = (s, s, 0), s ∈ R.

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