MATH 343: PARTIAL DIFFERENTIAL EQUATIONS First Midterm Exam: 02 March 2000 Thursday 08.40-10.30

QUESTIONS ans SOLUTIONS: (25)1.a. If there exists one integrating factor of a Pfaffian differential equation prove that there exist infinetely many of them. Solution: See Sneddon page 23 theorem 6. 1.b. Verify that yzdx + (x^{2}y − zx)dy + (x^{2}z − xy)dz = 0 is integrable and find its primitive. Solution: X = (yz, x^{2}y−zx, x^{2}z−xy). Hence, curlX = 2 (0, −xz+y, xy−z). Then X · curlX = 0 identically. This means that given Pfafian differential equation is integrable. Then we can search for its primitive. There are several ways of finding its primitive. The easisest way is by inspection.

yzdx + (x^{2}y − xz)dy + (x^{2}z − xy)dz =  (1)
z(ydx − xdy) + x^{2}ydy + x^{2}zdz − xydz =  (2)
−zx^{2}d(\frac{y}{x}) + x^{2}ydy + x^{2}zdz − ydz  (3)

dividing by x^{2} one reduces the above equation to

−zd\frac{y}{x} + ydy + zdz − \frac{y}{x}dz = 0

or

\frac{zy}{x} + \frac{y^{2} + z^{2}}{2} = c

Hence the primitive of the above Pfaffian differential equation is ϕ = \frac{zy}{x} +

\frac{y^{2}+z^{2}}{2}

(25)2. Let u(x, y, z), v(x, y, z), and w(x, y, z) be three differentiable functions. Prove that a necessary and sufficient condition that there exists a relation F (u, v, w) = 0 , not involving x, y, and z explicitly is that

\frac{∂(u, v, w)}{∂(x, y, z)} = 0

. Solution: The condition necessary. Let F (u, v, w) = 0. Differentiate this equation with respect x,y, and z we get

u_{x}F_{u} + v_{x}F_{v} + w_{x}F_{w} = 0,  (4)
u_{y}F_{u} + v_{y}F_{v} + w_{y}F_{w} = 0,  (5)
u_{z}F_{u} + v_{z}F_{v} + w_{z}F_{w} = 0.  (6)

Since F is arbitrary the F_{u}, F_{v}, F_{w} are simultanouly vanishing. Hence to have a nontrivial solution of the above linear equations the coeffient matrix must have zero determinant which is equivalent to the condition \frac{∂(u,v,w)}{∂(x,y,z)} = 0.

This condition is also sufficient. From u = U(x, y, z) and v = V (x, y, z) we may solve (for instance) x and y interms of u, v, and z. Interting all these into w = W (x, y, z) we obtain a relation among the variables u, v, w, and z i.e., F (u, v, w, z). Differentiating this equation with respect to x, y, and z, we obtain.

u_{x}F_{u} + v_{x}F_{v} + w_{x}F_{w} = 0,  (7)
u_{y}F_{u} + v_{y}F_{v} + w_{y}F_{w} = 0,  (8)
u_{z}F_{u} + v_{z}F_{v} + w_{z}F_{w} + F_{z} = 0.  (9)

Solving F_{u} and F_{v} in the first two equations above we obtain

F_{u} = \frac{v_{x}w_{y} − w_{x}v_{y}}{v_{y}u_{x} − v_{x}u_{y}} F_{w}  (10)
F_{v} = \frac{u_{x}w_{y} − w_{x}u_{y}}{v_{y}u_{x} − v_{x}u_{y}} F_{w}  (11)

Inserting these into the last equation we obtain

(u_{z}(v_{x}w_{y} − w_{x}v_{y}) + v_{z}(u_{x}w_{y} − w_{x}u_{y}) + w_{z}(v_{y}u_{x} − v_{x}u_{y}))F_{w} +  (12)
(v_{y}u_{x} − v_{x}u_{y}))F_{z} = 0  (13)

Due to the condition the last equation reduces to (v_{y}u_{x} − v_{x}u_{y}))F_{z} = 0. The coefficient of F_{z} cannot vanish, becouse it implies H(u, v) = 0. Hence F_{z} = 0 which completes the proof. (25)3.a. Find the integral curves of the equation ydx − xdy + dz = 0 on the surface y = xz. Solution: Please see last comment of Sneddon in page 25: The Pfaffian differential equation ydx − xdy + dz = 0 is not integrable (X · curlX ≠ 0). Although not integrable , it still has solutions. Consider the surface simultanously. That is

ydx − xdy + d(y/x) = y(1 − \frac{1}{x^{2}} )dx − x(1 − \frac{1}{x^{2}} )dy = 0  (14)

two types of curves i.x = ±1 and y = ±z ii.ydx − xdy = 0 and dz = 0 or y = cx, x = c. 3.b. Find the integral curvesof the equation

\frac{dx}{x^{2}(y^{3} − z^{3})} = \frac{dy}{y^{2}(z^{3} − x^{3})} = \frac{dz}{z^{2}(x^{3} − y^{3})}

. Solution: One of the easiest question: The integral curve is given by

u = x^{2} + y^{2} + z^{2} = c_{1}, v = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = c_{2}  (15)

(25)4.a. Find a surface orthogonal to the sphere x^{2} + y^{2} + z^{2} + −2ax = 0 and passing through the line y = x, z = h, 0 < h < a. Here a and h are constants.

Solution: Problem reduces to finding the solution of the quasi-linear first order pde (x − a)p + yq = z with the initial conditions x = t, y = t, z = h.

\frac{dx}{x − a} = \frac{dy}{y} = \frac{dz}{z}  (16)

These equations has solutions \frac{x−a}{y} = c_{1} and \frac{z}{y} = c_{2}. On the initial curve one has \frac{t−a}{t} = c_{1} and \frac{h}{t} = c_{2}. Finding t from the second one and inserting it into the first equation we obtain

h(1 − c_{1}) = ac_{2}  (17)

This leads to the following surface (a plane passing throgh the center of the sphere)

h(y − x + a) = az  (18)

4.b. Obtain the solution of the Cauchy problem: Find the integral surface of the equation xu_{,x} + yu_{,y} + zu_{,z} = nu which contains the surface x = t, y = s, z = st, (s, t) ∈ R^{2} . Solution: Using the Lagrange method we find that

\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z} = \frac{du}{nu}  (19)

whic has solution given by

\frac{y}{x} = c_{1}, \frac{z}{x}, \frac{u}{x^{n}} = c_{3}  (20)

i: x = t, y = s, z = st, u = c. Solving t and s we find

u = c(\frac{xy}{z})^{n}  (21)

ii: x = t, y = s, z = st, z = t + s, st. Solving t and s we find that

z^{n−1} u + (xy)^{n−1} (x + y + z) = 0  (22)

(25)5. Using the method of characteristics solve the following two problems (a). (x − y)p + (y − x − z)q = z, initial curve : z_{0}(s) = 1, x_{0}(s) = s, y_{0}(s) = s^{2}. Solution: We have the following set of ODEs

x′ = x − y, y′ = y − x − z, z′ = z  (23)

Solution satisfying also the initial conditions is given by

x = −e^{t} + \frac{1}{2} (1 + s + s^{2}) + \frac{1}{2} (1 + s − s^{2}) e^{2t}  (24)
y = \frac{1}{2}(1 + s + s^{2}) + \frac{1}{2} (1 + s − s^{2}) e^{2t}  (25)
z = e^{t}  (26)

To find the integral surface first find t and s from any of these three equations and insert it the last one. We obtain

e^{t} = z, s = −\frac{1}{2} + \frac{1}{2}(x + y = z − \frac{y − x − z}{z^{2}} )  (27)

Equation of the surface is lengthy , i am not displaying it here the above explanation is suficient. (b). pq = 1, initial curve : x_{0}(s) = 0, y_{0}(s) = s^{2}, z_{0}(s) = s. Solution: We find the following set of ODEs

x′ = q, y′ = q, z′ = 2pq = 2, p′ = 0, q′ = 0  (28)

with the initial conditions x_{0}(s) = 0, y_{0}(s)s^{2}, x_{0}(s) = s. We complete the initial data by using the equations

p_{0}(s) q_{0}(s) = 1, z′_{0} = p_{0} x′_{0} + q_{0} y′_{0}  (29)

solving these we find that p_{0} = 2s, q_{0} = \frac{1}{2s}. The complete solution of the problem is as follows

x = \frac{t}{2s}, y = 2st + s^{2}, z = 2t + s  (30)

which correspond to the integral surface z^{2} − y(x + 1).