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2.2.2: Initial Value Problem of Cauchy

( \newcommand{\kernel}{\mathrm{null}\,}\)

Consider again the quasilinear equation

() a1(x,y,u)ux+a2(x,y,u)uy=a3(x,y,u).

Let
Γ:  x=x0(s), y=y0(s), z=z0(s), s1ss2, <s1<s2<+
be a regular curve in R3 and denote by C the orthogonal projection of Γ onto the (x,y)-plane, i. e.,
$$
\mathcal{C}:\ \ x=x_0(s),\ \ y=y_0(s).
\]

Initial value problem of Cauchy: Find a C1-solution u=u(x,y) of () such that u(x0(s),y0(s))=z0(s), i. e., we seek a surface S defined by z=u(x,y) which contains the curve Γ.

Cauchy initial value problem
Figure 2.2.2.1: Cauchy initial value problem

Definition. The curve Γ is said to be non-characteristic if
$$
x_0'(s)a_2(x_0(s),y_0(s))-y_0'(s)a_1(x_0(s),y_0(s))\not=0.
\]

Theorem 2.1. Assume a1, a2, a2C1 in their arguments, the initial data x0, y0, z0C1[s1,s2] and Γ is non-characteristic.

Then there is a neighborhood of C such that there exists exactly one solution u of the Cauchy initial value problem.

Proof. (i) Existence. Consider the following initial value problem for the system of characteristic equations to ():
x(t)=a1(x,y,z)y(t)=a2(x,y,z)z(t)=a3(x,y,z)
with the initial conditions
x(s,0)=x0(s)y(s,0)=y0(s)z(s,0)=z0(s).
Let x=x(s,t), y=y(s,t), z=z(s,t) be the solution, s1ss2, |t|<η for an η>0. We will show that this set of curves, see Figure 2.2.2.1, defines a surface. To show this, we consider the inverse functions s=s(x,y), t=t(x,y) of x=x(s,t), y=y(s,t) and show that z(s(x,y),t(x,y)) is a solution of the initial problem of Cauchy. The inverse functions s and t exist in a neighborhood of t=0 since
det(x,y)(s,t)|t=0=|xsxtysyt|t=0=x0(s)a2y0(s)a10,
and the initial curve Γ is non-characteristic by assumption.

Set
u(x,y):=z(s(x,y),t(x,y)),
then u satisfies the initial condition since
u(x,y)|t=0=z(s,0)=z0(s).
The following calculation shows that u is also a solution of the differential equation ().
a1ux+a2uy=a1(zssx+zttx)+a2(zssy+ztty)=zs(a1sx+a2sy)+zt(a1tx+a2ty)=zs(sxxt+syyt)+zt(txxt+tyyt)=a3
since 0=st=sxxt+syyt and 1=tt=txxt+tyyt.

(ii) Uniqueness. Suppose that v(x,y) is a second solution. Consider a point (x,y) in a neighborhood of the curve (x0(s),y(s)), s1ϵss2+ϵ, ϵ>0 small. The inverse parameters are s=s(x,y), t=t(x,y), see Figure 2.2.2.2.

 Uniqueness proof

Figure 2.2.2.2: Uniqueness proof

Let
A:  x(t):=x(s,t), y(t):=y(s,t), z(t):=z(s,t)
be the solution of the above initial value problem for the characteristic differential equations with the initial data
x(s,0)=x0(s), y(s,0)=y0(s), z(s,0)=z0(s).
According to its construction this curve is on the surface S defined by u=u(x,y) and u(x,y)=z(s,t). Set
ψ(t):=v(x(t),y(t))z(t),
then
ψ(t)=vxx+vyyz=xxa1+vya2a3=0
and
ψ(0)=v(x(s,0),y(s,0))z(s,0)=0
since v is a solution of the differential equation and satisfies the initial condition by assumption. Thus, ψ(t)0, i. e.,
v(x(s,t),y(s,t))z(s,t)=0.
Set t=t, then
v(x,y)z(s,t)=0,
which shows that v(x,y)=u(x,y) because of z(s,t)=u(x,y).

Remark. In general, there is no uniqueness if the initial curve Γ is a characteristic curve, see an exercise and Figure 2.2.2.3, which illustrates this case.

Multiple solutions

Figure 2.2.2.3: Multiple solutions

Examples

Example 2.2.2.1:

Consider the Cauchy initial value problem
ux+uy=0
with the initial data
x0(s)=s, y0(s)=1, z0(s) is a given C1-function.
These initial data are non-characteristic since y0a1x0a2=1. The solution of the associated system of characteristic equations
x(t)=1, y(t)=1, u(t)=0
with the initial conditions
x(s,0)=x0(s), y(s,0)=y0(s), z(s,0)=z0(s)
is given by
x=t+x0(s), y=t+y0(s), z=z0(s),
i. e.,
x=t+s, y=t+1, z=z0(s).
It follows s=xy+1, t=y1 and that u=z0(xy+1) is the solution of the Cauchy initial value problem.

Example 2.2.2.2:

A problem from kinetics in chemistry. Consider for x0, y0 the problem
ux+uy=(k0ek1x+k2)(1u)
with initial data
u(x,0)=0, x>0, and u(0,y)=u0(y), y>0.
Here the constants kj are positive, these constants define the velocity of the reactions in consideration, and the function u0(y) is given. The variable x is the time and y is the height of a tube, for example, in which the chemical reaction takes place, and u is the concentration of the chemical substance.

In contrast to our previous assumptions, the initial data are not in C1. The projection C1C2 of the initial curve onto the (x,y)-plane has a corner at the origin, see Figure 2.2.2.4.

alt

Figure 2.2.2.4: Domains to the chemical kinetics example

The associated system of characteristic equations is
x(t)=1, y(t)=1, z(t)=(k0ek1x+k2)(1z).
It follows x=t+c1, y=t+c2 with constants cj. Thus the projection of the characteristic curves on the (x,y)-plane are straight lines parallel to y=x. We will solve the initial value problems in the domains Ω1 and Ω2, see Figure 2.2.2.4, separately.

(i) The initial value problem in Ω1. The initial data are
x0(s)=s, y0(s)=0, z0(0)=0, s0.
It follows
x=x(s,t)=t+s, y=y(s,t)=t.
Thus
z(t)=(k0ek1(t+s)+k2)(1z), z(0)=0.
The solution of this initial value problem is given by
z(s,t)=1exp(k0k1ek1(s+t)k2tk0k1ek1s).
Consequently
u1(x,y)=1exp(k0k1ek1xk2yk0k1ek1(xy))
is the solution of the Cauchy initial value problem in Ω1. If time x tends to , we get the limit
$$
\lim_{x\to\infty} u_1(x,y)=1-e^{-k_2y}.
\]

(ii) The initial value problem in Ω2. The initial data are here

x0(s)=0, y0(s)=s, z0(0)=u0(s), s0.
It follows
x=x(s,t)=t, y=y(s,t)=t+s.
Thus
z(t)=(k0ek1t+k2)(1z), z(0)=0.
The solution of this initial value problem is given by
z(s,t)=1(1u0(s))exp(k0k1ek1tk2tk0k1).
Consequently
u2(x,y)=1(1u0(yx))exp(k0k1ek1xk2xk0k1)
is the solution in Ω2.

If x=y, then
u1(x,y)=1exp(k0k1ek1xk2xk0k1)u2(x,y)=1(1u0(0))exp(k0k1ek1xk2xk0k1).
If u0(0)>0, then u1<u2 if x=y, i. e., there is a jump of the concentration of the substrate along its burning front defined by x=y.

Remark. Such a problem with discontinuous initial data is called Riemann problem. See an exercise for another Riemann problem.

The case that a solution of the equation is known

Here we will see that we get immediately a solution of the Cauchy initial value problem if a solution of the homogeneous linear equation
a1(x,y)ux+a2(x,y)uy=0
is known.

Let
x0(s), y0(s), z0(s), s1<s<s2
be the initial data and let u=ϕ(x,y) be a solution of the differential equation. We assume that
ϕx(x0(s),y0(s))x0(s)+ϕy(x0(s),y0(s))y0(s)0
is satisfied. Set
g(s)=ϕ(x0(s),y0(s))
and let s=h(g) be the inverse function.

The solution of the Cauchy initial problem is given by u0(h(ϕ(x,y))).

This follows since in the problem considered a composition of a solution is a solution again, see an exercise, and since
$$
u_0\left(h(\phi(x_0(s),y_0(s))\right)=u_0(h(g))=u_0(s).
\]

Example 2.2.2.3:

Consider equation
ux+uy=0
with initial data
x0(s)=s, y0(s)=1, u0(s) is a given function.
A solution of the differential equation is ϕ(x,y)=xy. Thus
ϕ((x0(s),y0(s))=s1
and
u0(ϕ+1)=u0(xy+1)
is the solution of the problem.

Contributors and Attributions


This page titled 2.2.2: Initial Value Problem of Cauchy is shared under a not declared license and was authored, remixed, and/or curated by Erich Miersemann.

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