S = cvln(pVγ).
In conclusion, the first law of thermodynamics is
de = T dS − pdV. (1.28)
This means that the change of internal energy can be due to the heat T dS exchange with external world and the work −pdV exerted from outside. For ideal polytropic gases, using the ideal gas law and the constitutive relation, plus the first law of thermodynamics, we can choose p, V as independent variables and express
T = pV
R , e = cv
RpV, S = cvln(pVγ).
*Homework
1. Express thermo variables in terms of e, V for ideal polytropic gases..
1.6 Existence and Uniqueness
In this section, we shall state but without proof the existence and uniqueness theorems. We also give examples and counter-examples regarding to the existence and uniqueness. Finally, we give application of these fundamental theorems.
Existence
Theorem 1.3 (Local existence theorem). Suppose f (t, y) is continuous in a neighborhood of (t0, y0).
Then the initial value problem
y0(t) = f (t, y), y(t0) = y0
has a solutiony(·) existing on a small interval (t0− , t0+ ) for some small number > 0.
This theorem states that there exists an interval (may be small) where a solution does exist. The solution may not exist for all t. Let us see the following example.
Examples Consider the initial value problem
y0 = y2 y(0) = y0 By the method of separation of variable,
dy y2 = dt
Z y y0
dy y2 = t
−y−1+ y0−1= t y(t) = y0
1 − ty0
.
When y0< 0, the solution does exist in (1/y0, ∞). But when y0 > 0, the solution can only exist in (−∞, 1/y0). The solution blows up when t → 1/t0:
lim
t→1/y0
y(t) = ∞.
The maximal interval of existence is (−∞, 1/y0) when y0 > 0 and is (1/y0, ∞) when y0 < 0.
In the local existence theorem, it only states that the solution exists in a small region. If the solution does have a limit at the end, say t1, of this interval, we can solve the equation again to extend this function. One can show that this extended function also satisfies the differential equation at t1and beyond. Eventually, we can find the maximal interval of existence. If the solution remains bounded whenever it exists, then we can always find globally exists if y(·) stays bounded whenever it exists. We have the following corollary.
Corollary 1.1. Consider the ODE: y0 = f (t, y). Suppose f (t, y) is continuous on R × R and assume a solution stays bounded as long as it exists, then this solution exists for all time.
Proof. Suppose the maximal interval of existence is (t0, t1). The assumption that y(t) remains bounded in (t0, t1) plus f (t, y) is continuous imply that limt→t1y(t) exists (why?). Then we can extend y(·) beyond t1by the local existence theorem. This contradicts to the hypothesis that (t0, t1) is the maximal interval of existence.
Homeworks Find the maximal interval of existence for the problems below.
1. y0 = 1 + y2, y(0) = y0
2. y0 = y3, y(0) = y0
3. y0 = ey, y(0) = y0
4. y0 = y ln y, y(0) = y0 > 0.
Uniqueness
Theorem 1.4 (Uniqueness). Assume that f and ∂f /∂y are continuous in a small neighborhood of (t0, y0). Suppose y1(t) and y2(t) are two solutions that solve the initial value problem
y0 = f (t, y), y(t0) = y0 on an interval(t0− , t0+ ) for some > 0. Then
y1(t) = y2(t), for all t ∈ (t0− , t0+ ).
In other word, no two solutions can pass through the same point in the t-y plane.
1.6. EXISTENCE AND UNIQUENESS 35 Application 1. Reduce high order equation to first-order system The above existence and uniqueness theorems also hold for general first-order ODE system:
y0 = f (t, y)
where f : R × Rn → Rn is a Lipschitz function. This means that: given initial data (t0, y0) ∈ R × Rn, there exists a unique solutiony : (t0− , t0+ ) → Rnwithy(t0) = y0. This theorem can be applied to high-order equations too. Indeed, any high-order equation can be transformed to an equivalent first-order system. Namely, the general n-th order equation
y(n)= f (t, y, y0, · · · , y(n−1)) (1.29) is equivalent to the following system
y10 = y2 y20 = y3
...
yn0 = f (t, y1, y2, · · · , yn)
(1.30)
We need n conditions to determine a unique solution for the first-order system (1.30). Likely, we need n conditions to determine a unique solution for the nth-order differential equations (1.29).
Application 2 Let us apply the existence and uniqueness to the qualitative study of the autonomous system in one dimension. For instance, let consider a smooth f (y) which has the property (i) the only zeros of f are 0 and K, (ii) f (y) > 0 for 0 < y < K. The logistic model: y0 = f (y) :=
ry(1 − y/K), is one such example. The constant states 0 and K naturally partition the domain R into three regions: I1 = (−∞, 0), I2 = (0, K) and I3 = (K, ∞). By the uniqueness theorem, no solution can cross these two constant states. Thus, starting y(0) ∈ (0, K), the trajectory y(t) stays in (0, K) for all t because it cannot cross these two constant solutions. So, the solution stays bounded and thus exists for all time. The limit limt→∞y(t) must exist because the function y(·) monotonically increases and stays bounded above. Let us call limt→∞y(t) = ¯y ∈ [0, K]. Then
t→∞lim y0(t) = lim
t→∞f (y(t)) = f (¯y).
We claim that f (¯y) = 0. Suppose not, then we have f (¯y) > 0 because f (y) > 0 for y ∈ (0, K).
We choose > 0 so that f (¯y) − > 0. With this , there exists M > 0 such that f (y(t)) > f (¯y) − for all t > M . Thus,
y(t) − y(M ) = Z t
M
f (y(s)) ds > (f (¯y) − )(t − M ) → ∞ as t → ∞.
This is a contradiction. Thus, we get f (¯y) = 0. But the only constant states are 0 and K. It has to be K because 0 < y(0) < y(t) for all t > 0. This shows that when y(0) ∈ (0, K), we have y(t) → K as t → ∞. This is asymptotic stability result. We will see more applications of the uniqueness theorem in the subsequent chapters.
Remarks.
1. The initial value problem may not have a unique solution. Let us see the following problem:
y0= 2y1/2, y(0) = 0 By the method of separation of variable,
dy 2√
y = dt,
√y = t − C With the initial condition y(0) = 0, we get C = 0. Hence
y(t) = t2
is a solution. On the other hand, we know y(t) ≡ 0 is also a solution. We should be careful here. The portion y(t) = t2for t < 0 is not a solution because y0< 0 for t < 0. This portion does not satisfy the equation y0= 2√
y > 0. Therefore, one solution is y(t) =
0 for − ∞ < t < 0 t2 for t ≥ 0.
We have known that y(t) ≡ 0 is another solution. In fact, there are infinite many solutions passing through (0, 0):
y(t) =
0 for − ∞ < t < C (t − C)2 for t ≥ C,
with parameter C ≥ 0 being arbitrary.
It is important to notice two things. (1) The ODE associated with the family of parabolas y = (t − C)2 is y02 = 4y, see the subsection 1.5.2. It contains two branches: y0 = ±2√
y.
The solutions also contain two branches. The branch y = (t − C)2, t ≥ C satisfies y0 = 2√ y, while the branch y = (t − C)2, t ≤ C satisfies y0 = −2√
y. (2) The curve y(t) ≡ 0 is the envelop of both families of parabolas.
2. You can find non-uniqueness examples easily from the envelop of a family of curves. In fact, suppose the family of curve Ψ(x, y, C) = 0 is the solution of some ODE: F (x, y, y0) = 0.
Suppose ((x(C), y(C)) is the envelop of this family of curves. Then at C, both Ψ(x, y, C) = 0 and the envelop (x(C), y(C)) are the solution curves of the ODE: F (x, y, y0) = 0 at (x(C), y(C)).
3. For vector field V(x, y) = (u(x, y), v(x, y)), its integral curves do not intersect if V 6= 0.
More precisely, if (u(x0, y0), v(x0, y0)) 6= (0, 0), then the integral curve through (x0, y0) is unique. To show this, if u(x0, y0) 6= 0, then the integral curve of (u, v) satisfies
dy
dx = v(x, y)
u(x, y) = f (x, y).
1.7. *NUMERICAL METHODS: FIRST ORDER DIFFERENCE EQUATIONS 37 The function f (x, y) is well-defined in a neighborhood of (x0, y0) because u(x0, y0) 6= 0.
By the uniqueness theorem, the solution y(x) of y0 = f (x, y) with y(x0) = y0 is unique.
Therefore, the integral curve is unique there. If on the other hand, u(x0, y0) = 0, then v(x0, y0) 6= 0, we solve
dx
dy = u(x, y) v(x, y) instead.
However,the integral curves can “intersect” at those critical points where V(x, y) = (0, 0).
For instance, the integral curves of V(x, y) = (−x, −y) point to (0, 0). The integral curve of V = (−x, y) are xy = C. As C = 0, the corresponding integral curve is x = 0 or y = 0.
They intersect at (0, 0).
4. In the example of application 2, we cannot obtain the rate of convergence for y(t) → K as t → ∞. However, if we know that f0(K) 6= 0 (in fact, f0(K) < 0), then we can get that y(t) → K at exponential rate. This means that
|y(t) − K| ≤ Cef0(K)t
as t → ∞. A concrete is the logistic model, where f (y) = ry(1 − y/K) and f0(K) = −r.
For y ∼ K, the Taylor expansion of f gives f (y) ∼ −r(y − K). The equation y0 = f (y) ∼
−r(y − K) leads to y(t) ∼ O(1)e−rt.
On the other hand, if f0(K) = 0, then f (y) is of high order near y = K. In this case, we can not have exponential convergence, as you can see from this simple example: y0 = (y − K)2 with y(0) < K.
1.7 *Numerical Methods: First Order Difference Equations
1.7.1 Euler method
Consider the first order equation
y0= f (t, y).
If the solution is smooth (this is what we would expect), we may approximate the derivative y0(t) by a finite difference
y0(t) ∼ y(t + h) − y(t)
h .
Thus, we choose a time step size h. Let us denote t0+ nh = tnand t0is the initial time. We shall approximate y(tn) by yn. For tn < t < tn+1, y(t) is approximated by a linear function. Thus, we approximate y0 = f (t, y) by
yn+1− yn
h = f (tn, yn). (1.31)
This is called the Euler method. It approximates the solution by piecewise linear function. The approximate solution yn+1 can be computed from yn. If we refine the mesh size h, we would
expect the solution get closer to the true solution. To be more precise, let us fix any time t. Let us divide [0, t] into n subintervals evenly. Let h = t/n be the step size. We use Euler method to construct yn. The convergence at t means that yn → y(t) as n → ∞ (with nh = t fixed, hence h → 0).
Homework
1. Use Euler method to compute the solution for the differential equation y0= ay
where a is a constant. Find the condition on h such that the sequence yn so constructed converges as n → ∞ and nh = t is fixed.
1.7.2 First-order difference equation
This subsection is a computer project to study the discrete logistic map:
yn+1= ρyn 1 −yn
k
. (1.32)
It is derived from the Euler method for the logistic equation.
yn+1− yn
h = ryn 1 −yn
K
,
with ρ = 1 + rh and k = K(1 + rh)/rh. We use the following normalization: xn= yn/k to get xn+1 = ρxn(1 − xn) := F (xn). (1.33) This mapping (F : xn 7→ xn+1) is called the logistic map. The project is to study the behaviors of this logistic map by computer simulations.
Iterative map In general, we consider a function F : R → R. The mapping xn+1= F (xn), n = 0, 1, 2, · · · ,
is called an iterative map. We denote the composition F ◦ F by F2.
A point x∗is called a fixed point (or an equilibrium) of the iterative map F if it satisfies F (x∗) = x∗
A fixed point x∗ is called stable if we start the iterative map from any x0 close to x∗, the sequence {Fn(x0)} converges to x∗. A fixed point x∗is called unstable if we start the iterative map from any x0arbitrarily close to x∗, the sequence {Fn(x0)} cannot converge to x∗. The goal here is to study the behavior (stable, unstable) of a fixed point as we vary the parameter ρ.
1.8. HISTORICAL NOTE 39