ODE: Assignment-3

1. (T) A surface z = y 2 − x2 in the shape of a saddle is lying outdoors in a rainstorm. Find the
paths along which raindrops will run down the surface.
Solution:
A curve on the surface is determined by a curve y = y(x) on the xy-plane. The raindrop will
take the path where z decreases at maximum rate. We know that for a real valued differentiable
function f (x, y), f will have maximum increase rate in direction ∇f and maximum decrease
rate in direction −∇f (This comes from the fact that directional derivative of f in direction
v, |v| = 1, is given by (∇f ).v).
So the required curve in xy-plane will have slope −∇f = (2x, −2y). So its differential equation
is dy/dx = −2y/2x. Solving we get xy = c. Thus the curve on the surface is the intersection
of the saddle z = x2 − y 2 with hyperbolic cylinder xy = c.

2. (T) Does f (x, y) = xy 2 satisfies Lipschitz condition (LC) on any rectangle [a, b] × [c, d]? What
about on an infinite strip [a, b] × R?
[A function f (x, y) is said to satisfy Lipschitz condition on a domain D ⊆ R2 , if there exists
L > 0 such that |f (x, y1 ) − f (x, y2 )| ≤ L|y1 − y2 | for all (x, y1 ), (x, y2 ) ∈ D.]
Solution:
On closed rectangle [a, b] × [c, d], the partial derivative fy is continuous and hence bounded and
hence f satisfies LC. Alternatively,
|f (x, y1 ) − f (x, y2 )|
= |x||y1 + y2 | ≤ max {|a|, |b|} × 2 max {|c|, |d|}
|y1 − y2 |

On the vertical strip, |x| is bounded but |y1 + y2 | can be made arbitrarily large for large choices
of y1 and y2 . So f does not satisfy LC there.

3. (T) Let (x0 , y0 ) be an arbitrary point in the plane and consider the initial value problem (IVP)

y 0 = y 2 , y(x0 ) = y0 .

Explain why Picard theorem guarantees that this problem has a unique solution on some
interval |x − x0 | ≤ h. Since f (x, y) = y 2 and ∂f /∂y are continuous on the entire plane, it is
tempting to conclude that this solution is valid for all x. But considering the solutions through
the points (0, 0) and (0, 1), show that this consideration is sometime true and sometime false,
and that therefore the inference is not legitimate.
[Remark: Compare the above with the fact that if f is continuous and Lipschitz on [a, b] × R,
then the IVP y 0 = f (x, y), y(x0 ) = y0 , x0 ∈ [a, b] has solution over [a, b]. Simmons book
Theorem B in chapter ‘The Existence and Uniqueness of Solutions’.]
Solution:
Since f (x, y) = y 2 and ∂f /∂y are continuous on the entire plane, they are continuous on any
closed rectangle containing (x0 , y0 ). Hence Picard theorem guarantees unique solution on some
interval |x − x0 | ≤ h.
 1 1
Solving the equation we get y = − x+c . Initial condition y(0) = 1 gives us y = 1−x which is
valid for (−∞, 1). For initial condition y(0) = 0, we cant find value of c. But we observe that
y(x) = 0 satisfies the equation with y(0) = 0. So it is valid for all R.

4. (T) Consider the IVP y 0 = 2 sin(3xy), y(0) = y0 . Show that it has unique solution in (−∞, ∞).
Solution:
It suffices to show that it has unique solution on every interval [−L, L]. This is because if we
have a unique solution on [−L1 , L1 ] and a unique solution on [−L2 , L2 ] with L2 > L1 , then by
uniqueness part the two solution has to agree on the smaller interval [−L1 , L1 ].
Now fix L. Define R = [−L, L] × [y0 − b, y0 + b for some large b > 0. Note that the function
f (x, y) = 2 sin(3xy) satisfies |f ≤ 2 and |fy | ≤ 6L on the rectangle R. So by Picard theorem,
unique solution exist on the interval [−h, h] where h = min {L, b/2}. We can choose b > 2L so
that h = L. Thus we get a unique solution on [−L, L].

5. (T) Given
2
0 ey − 1
y = , y(−2) = 1.
1 − x2 y 2
Find an interval on which solution exist.
Solution:
y2
e −1
Here our function f is defined by f = 1−x 2 y 2 and x0 = −2, y0 = 1. Thus we need to pick a

rectangle R which is centered at (−2, 1). In this rectangle we need to have f, fy continuous and
so we certainly have to choose R so small that it contains no points at which the denominator
1 − x2 y 2 vanishes. The exact choice of the rectangle is up to you.
We choose a = 1/2, b = 1/4 so that

R = [x0 − a, x0 + a] × [y0 − b, y0 + b] = [−5/2, −3/2] × [3/4, 5/4],

so that R is disjoint from the hyperbolas xy = ±1.

On R, we have x2 ≥ 9/4, y 2 ≥ 9/16 and therefore |1 − x2 y 2 | ≥ (81/64) − 1 = 17/64 > 1/4.
2
Also |ey − 1| ≤ e9/16 < e < 3. Thus |f (x, y)| ≤ 3 × 4 = 12 = M on R. This is a legitimate
(but non-optimal) bound.
Since R is disjoint from xy = ±1, clearly fy will be continuous on R. So by Picard theorem
unique solution will exist on [−2 − h, −2 + h] for h = min {a, b/M } = min {1/2, 1/48} = 1/48.

6. (T) Consider the ode y 0 = x22xy

−y 2
. Solve it. Sketch the solutions. Verify Picard theorem for
initial values in R − {(x, y) : x = y 2 }. What is your solution passing through (1, 0)?
2 2

Solution:
1
Comparing with M dx + N dy = 0, we have M = 2xy, N = x2 − y 2 . So M
(My − Nx ) = 2/y. So
R
integrating factor is e− 1/ydy = 1/y 2 . We get solution x2 + y 2 = cy.
(Also we can solve it as homogeneous equation.)
Solution curves are circles with centre on the y-axis and touching the x axis at the origin.
 The function f (x, y) = x22xy
−y 2
and fy is continuous on D = R2 − {(x, y) : x2 = y 2 }. So Picard
theorem tells us: given any (x0 , y0 ) ∈ D there passes through a unique solution curve.
Given initial condition (x0 , y0 ), x0 6= 0 there is circle as above passing though that point.
For point (x0 , 0), x0 6= 0 we can not find a circle like that. But we observe that y(x) = 0 is also a
solution of the equation and so this must be the unique solution passing through (x0 , 0), x0 6= 0.

7. A function f (x, y) is said to satisfy Lipschitz condition on a domain D ⊆ R2 , if there exists

L > 0 such that |f (x, y1 ) − f (x, y2 )| ≤ L|y1 − y2 | for all (x, y1 ), (x, y2 ) ∈ D.
(i) Show that if f (x, y) satisfies Lipschitz condition (LC)with respect to y on a rectangle D,
then for each fixed x, the resulting function of y is continuous function of y.
(ii) Let f (x, y) = y + [x]. Then how that f satisfies LC on R2 but not continuous on R2 .
(iii) Let f (x, y) = xy. Then show that f is continuous on R2 but not LC on R2 .
Solution:
(i) Follows from definition.
(ii) Let f (x, y) = y + [x]. Clearly |f (x, y1 ) − f (x, y2 )| = |y1 − y2 |, so LC is satisfied on the entire
plane. But f is not continuous for any integral x.
|f (x,y1 )−f (x,y2 )|
(iii) Let f (x, y) = xy. It is continuous on entire plane, being polynomial. But |y1 −y2 |
=
|x| can be made arbitrarily large on R2 . So LC not satisfied on R2 .

8. (T) What does Picard theorem says about existence and uniqueness of solution of the IVP
y 0 = (3/2)y 1/3 , y(0) = 0? Show that it has uncountably many solutions.
Solution:
Here f (x, y) = (2/3)y 1/3 is continuous on the plane. So Picard theorem (Peano existence) tells
us that it has at least one solution. But fy is not continuous in any rectangle containing (0, 0)
and also f does not satisfy Lipschitz condition on any rectangle containing (0, 0). So we can
not say anything about uniqueness of the solution from the theorem.
Solving the equation we get y 2 = x3 . Also y(x) = 0 satisfies the IVP. Moreover, y(x) = (x−a)3/2
for x ≥ a and y(x) = 0 for x ≤ a also satisfies the IVP for any a ≥ 0 (just need check derivative
at x = a exists and equal to 0). Thus we get uncountably many solutions.
√ √
9. Consider the IVP y 0 = y + 1, y(0) = 0, x ∈ [0, 1]. Show that f (x, y) = y + 1 does not
satisfy Lipschitz condition in any rectangle containing origin, but still the solution is unique.
(Remark: It is fact that if an IVP, with f is continuous (not necessarily Lipschitz), has more
than one solution, then it has uncountably many solutions. This is known as Kneser’s Theorem.
The previous exercise illustrates this phenomenan.)
Solution:
Consider any rectangle R = [0, a] × [0, d] containing origin We have

√ √ √
|f (x, y1 ) − f (x, y2 )| | y1 − y2 |
= = 1/ δ, for y1 = δ > 0, y2 = 0.
|y1 − y2 | |y1 − y2 |
 For δ arbitrary small, we can make |f (x,y|y11)−f (x,y2 )|
−y2 |
arbitrarily large on R. Hence f does not
satisfy Lipschitz condition in any rectangle containing origin.
√ √
Let g1 (x), g2 (x) be two solutions of the IVP. Consider z(x) = ( g1 − g2 )2 . Then z 0 (x) =
− √gz(x)
√
1 g2
≤ 0. Thus z(x) is a decreasing function. Further z(x) is non negative and z(0) = 0.
Then z(x) = 0 for all x ≥ 0. Hence g1 = g2 .

10. (T) (i) Let f (x, y) be continuous on the closed rectangle R : |x − x0 | ≤ a, |y − y0 | ≤ b. Show
0
that y is a solution of the initial value
Z x problem y = f (x, y), y(x0 ) = y0 iff


y(x) = y0 + f t, y(t) dt.
x0
Z x

(ii) Let |f (x, y)| ≤ M on the closed rectangle R and yn (x) = y0 + f t, yn−1 (t) dt, with
x0
y0 (x) = y0 . Use induction to show that yn+1 (x) is well defined for I : |x − x0 | ≤ h, where
h = min{a, b/M }; that is |yn (x) − y0 | ≤ b for x ∈ I.
(Remark: The sequence of functions yn (x) are called Picard’s Iterates. Precisely because of
this step, the solution exist in possibly smaller interval in Picard theorem.)
Solution:
(i) Let y(x) is the solution to y 0 = f (x, y), y(x0 ) = y0 . Then y 0 = f x, y(x) , y(x0 ) = y0 .


Rx

Integrating from x0 to x we get y(x) − y0 = x0 f t, y(t) dt.
Z x


Conversely, let y(x) = y0 + f t, y(t) dt. Then y(x0 ) = y0 and from fundamental theorem
x0
of integral calculus, y 0 = f x, y(x) = f (x, y).


Rx

(ii) For n = 0, y0 (x) ≡ y0 and the relation is obvious. For n = 1, |y1 (x)−y0 | = | x0 f t, y0 (t) dt| ≤
Rx

x0
|f t, y0 (t) | dt ≤ M h ≤ b. Let it be true for n = m and so |ym (x) − y0 | ≤ b. So for


a ≤ x ≤ b, (x, ym (x)) lies in the rectangle R and hence |f x, ym (x) | ≤ M . Therefore,
Rx

|ym+1 − y0 | ≤ x0 |f t, ym (t) | dt ≤ M h ≤ b. Hence proved.

11. Use Picard’s method of successive approximation to solve the following initial value problems
and compare these results with the exact solutions:
√ √
(i) (T) y 0 = 2 x, y(0) = 1 (ii) y 0 + xy = x, y(0) = 0 (iii) y 0 = 2 y/3, y(0) = 0
Solution:
Rx

Picard iteration is yn+1 (x) = y0 + x0 f t, yn (t) dt with y0 (x) ≡ y0 .
Rx√
(i) y0 = 1, yn (x) = 1 + 2 0 t dt = 1 + (4/3)x3/2 , n ≥ 1 (since f is independent of y). Here
yn (x) (n ≥ 1) coincides with the exact solution.
(ii) For exact solution
dy x2
= x dx =⇒ − ln(1 − y) = +C
1−y 2
Using y(0) = 0 we find C = 0. So,
2 /2 2 /2
1 − y = e−x =⇒ y = 1 − e−x .

Now we calculate the Picard iterates. Here f (x, y) = x(1 − y) and y0 = 0. Thus y1 (x) =
Rx Rx
0
t(1 − 0) dt = x2 /2. Using y1 , we get y2 (x) = 0 t(1 − t2 /2) dt = x2 /2 − (x2 /2)2 /2. y3 (x) =
 x2 /2 − (x2 /2)2 /2 + (x2 /2)3 /3!. By induction, we get yn (x) = nm=1 (−1)m−1 (x2 /2)m /m!. Thus
P

as n → ∞, yn (x) → − ∞ 2 m −x2 /2
P
m=0 (−x /2) /m! + 1 = 1 − e , which is the exact solution.
√
(iii) Here y0 = 0 and f (x, y) = 2 y/3. If we take y0 (x) ≡ y0 = 0, then yn (x) = 0, n ≥ 1. Here
yn (x), ∀n coincides with the analytical solution y(x) = 0. The other solution y(x) = (x/3)2 is
not reachable from here.
Note: However, if we start with y0 (x) = 1, then
 5/2  9/4
2 2 2 4 7/4
y1 (x) = x, y2 (x) = x3/2 , y3 (x) = x
3 3 3 7
 17/8  1/2
2 4
y4 (x) = x15/8
3 7
Clearly, yn (x) = an xbn where a1 = 2/3, a2 = (2/3)5/2 , a3 = (2/3)9/4 (4/7), · · · and bn = (2n −
1)/2n−1 . The sequence bn → 2 and an is a decreasing sequence bounded below. Hence, yn (x) →
Ax2 . To find we substitute in the integral relation and find
√
Ax2 = 2/3 Ax2 /2 =⇒ A = 1/32 =⇒ yn (x) → (x/3)2 .

12. Consider the initial value problem (IVP) xy 0 − y = 0, y(x0 ) = y0 . Solve it for different values
of x0 and y0 . Does the result contradict Picard theorem ?
Solution:
We have xdy − ydx = 0. Dividing by x2 , we have d(y/x) = 0 Integrating we get y = cx for
arbitrary c. If x0 6= 0, then we have unique solution for any y0 . If x0 = 0 and y0 = 0 then initial
condition is satisfied for any c and so there are infinite solutions. If x0 = 0 and y0 6= 0, there
is no solution.
Here f (x, y) = y/x which is not even defined on y-axis. So Picard theorem does not apply there.
At other points conditions of Picard theorem is satisfied and also we have unique solution.

13. Solve y 0 = (y − x)2/3 + 1. Show that y = x is also a solution. What can be said about the
uniqueness of the initial value problem consisting of the above equation with y(x0 ) = y0 , where
(x0 , y0 ) lies on the line y = x.
Solution:
Put u = y − x =⇒ u0 = u2/3 . Solving we get y = x + [(x + C)/3]3 . Also y = x is a
solution by direct verification. If y(x0 ) = y0 and x0 = y0 , then C = −x0 . Thus the solutions
y = x + [(x − x0 )/3]3 and y = x both satisfy the initial conditions y(x0 ) = y0 with x0 = y0 .
Clearly the solution to the IVP is nonunique.

14. Discuss the existence and uniqueness of the solution of the initial value problem

(x2 − 2x)y 0 = 2(x − 1)y, y(x0 ) = y0 .

Solution:
 Here f (x, y) = 2(x − 1)y/(x2 − 2x) and ∂f /∂y = 2(x − 1)/(x2 − 2x). The existence and
uniqueness theorem guarantees the existence of unique solution in the vicinity of (x0 , y0 ) where
f and ∂f /∂y are continuous and bounded. Thus, existence of unique solution is guaranteed at
all x0 for which x0 (x0 − 2) 6= 0. Hence, unique solution exists when x0 6= 0, 2.
When x0 = 0 or x0 = 2, nothing can be said using the existence and uniqueness theorem.
However, since the equation is separable, we can find the general solution to be y = Cx(x − 2).
Using initial condition we get y0 = Cx0 (x0 −2). Clearly the IVP has no solution if x0 (x0 −2) = 0
and y0 6= 0. If x0 (x0 − 2) = 0 and y0 = 0 then y = αx(x − 2) is a solution to the IVP for any
real α. Hence, in summary
(i) No solution for x0 = 0 or x0 = 2 and y0 6= 0;
(ii) Infinite number of solutions for x0 = 0 or x0 = 2 and y0 = 0;
(iii) Unique solution for x0 6= 0, 2.

15. (T) Consider the IVP y 0 = x−y, y(0) = 1. Show that for Euler method, yn = 2(1−h)n −1+nh
where h is the step size. (xn = nh with x0 = 0, y0 = y(0) = 1). Deduce that if we take h = 1/n,
then the limit of yn converges to actual value of y(1).
Solution:
The inductive formula of Euler method is

yn = yn−1 + hf (xn−1 , yn−1 ) = yn−1 + h(xn−1 − yn−1 ) = (1 − h)yn−1 + h2 (n − 1).

(Using xn = nh.)
We now use induction to prove the required formula for yn . Clearly it is true for n = 0. Assume
the formula is true for n. Then yn+1 = (1 − h)yn + h2 n = 2(1 − h)n+1 − 1 + (n + 1)h.
Taking h = 1/n, we have xn = 1. Thus approximate value of y(1) is given by yn = 2(1 − 1/n)n
which converges to 2e−1 .
Exact solution of the equation is y = 2e−x − 1 + x. So y(1) = 2e−1 .

16. Use Euler method and step size .1 on the IVP y 0 = x+y 2 , y(0) = 1 to calculate the approximate
value for the solution y(x) when x = .1, .2, .3. Is your answer for y(.3) is higher or lower than
the actual value ?
Solution:
We have x0 = 0, y0 = 1. Using the Euler iterative formula with h = .1 (see previous exercise),
we get y1 = 1.1, y2 = 1.231, y3 = 1.403.
Using graphical method, we see that the solution curve through (0, 1) is convex. So Euler
method approximate value is lower than actual value.

17. Verify that y = x2 sin x and y = 0 are both solution of the initial value problem (IVP)

x2 y 00 − 4xy 0 + (x2 + 6)y = 0, y(0) = y 0 (0) = 0.

Does it contradict uniqueness of solution of IVP?

 Solution: It is easy to verify that they satisfies the equation. For second order ode y 00 +
p(x)y 0 +q(x)y = r(x), with initial condition y(x0 ) = a, y 0 (x0 ) = b, the existence and uniqueness
theorem assets unique solution when p, q, r are continuous on an interval containing x0 . Here
p(x) = −4/x and q(x) = (x2 + 6)/2 are not continuous at x = 0.

18. (T) (i)The differential equation of the form y = xy 0 + f (y 0 ) is called a Clairaut equation. Show
that the general solution of this equation is the family of straight lines y = cx + f (c). In
addition to these show that it has a special solution given by f 0 (p) = −x where p = y 0 . This
special solution which does not (in general) represent one of the straight lines y = cx + f (c), is
a singular solution.
(Hint. Differentiate the given equation w.r.t. x.)
(Recall: A General Solution of an n-th order differential equation is one that involves n arbitrary
constants. A singular solution of a differential equation is a solution that is not obtainable by
specifying the values of the arbitrary constant in general solution)
(T) (ii) Solve the equation: y 02 − xy + y = 0.
Solution:
(i)The given equation is y = xp + f (p), p = y 0 . Differentiating with respect to x, we get
dp dp
dx
(x + f 0 (p)) = 0. For dx = 0 implies that p = const = c1 . Further integrating, we have
y = c1 x + c2 . Substituting in the given equation we have c1 x + c2 = xc1 + f (c1 ), i.e. c2 = f (c1 ).
Hence the general solution is
y = c1 x + c2 = c1 x + f (c1 ).

(ii) The given equation is y = xy 0 − y 02 which is in Clairaut form with f (p) = −p2 . General
solution is the family of st lines y = mx + m2 . For singular solution, −2p = f 0 (p) = −x.
Putting y 0 = p = x/2 in the given equation, we have singular solution y = x2 /2 − x2 /4 = x2 /4.