MATH 343: PARTIAL DIFFERENTIAL EQUATIONS Second Midterm Exam (with solutions) 04 April 2000 Tuesday 08.40-10.30

QUESTIONS:

1. Find the solution of the following Cauchy problem: z_{,xx} − 4 z_{,xy} = e^{y+4x} , where z = z_{,x} = 0 on the y-axis. Solution: D (D − 4D′)z = e^{y+4x}. The general Solution. Homogenous solution z_{h} = z_{1} +z_{2} where D z_{1} = 0 and (D − 4D′) z_{2} = 0 with the solutions z_{1} = f_{1}(y) and z_{2} = f_{2}(y + 4x), hence z_{h} = f_{1}(y) + f_{2}(y + 4x) where f_{1} and f_{2} are some differentiable functions of their arguments. Particular solution is obtained as follows: Since y + 4x = constant curves are characteristics of equation we let z_{p} = e^{y+4x} ϕ(x, y) where ϕ satisfies ϕ_{xx} − 4ϕ_{xy} + 4ϕ_{x} − 16ϕ_{y} = 1 . Any particular solution of this equation will be enough for our purpose. For instance ϕ = \frac{x}{4} or ϕ = −\frac{y}{16}. Hence choosing the first one we get z_{p} = \frac{1}{4} x e^{y+4x}. Then the general solution of the equation is given by z = f_{1}(y) + f_{2}(y + 4x) + \frac{1}{4}xe^{y+4x}. Solution of the Cauchy problem. Find the functions f_{1} and f_{2} such that the conditions z(x, y) = 0 and z_{,x}(x, y) = 0 on the curve x = 0 on the solution z(x, y) are satisfied. Using these initial conditions we find that

f_{1}(y) + f_{2}(y) = 0,  (1)
4\frac{∂f_{2}(y)}{∂y} + \frac{1}{4}e^{y} = 0  (2)

The solution of the above equations are given by

f_{1}(y) = \frac{1}{16}e^{y} + a  (3)

f_{2}(y) = −\frac{1}{16}e^{y} − a  (4)

(5)

where a is an arbitrary constant. Then the solution of the Cauchy problem is

z(x, y) = \frac{1}{16}e^{y} − \frac{1}{16} e^{y+4x} (1 − 4x)  (6)

Note that the result doesnot contain the free constant a. It means that the solution is unique.

2. Find the solution of the following Cauchy problem: z_{,xy} = 0, where z = ρ(y) and z_{,x} = σ(y) on the y-axis (x = 0). Here ρ and σ are some differentiable functions of y ∈ R. Solution: The general solution is easy to obtain. It is given by z(x, y) = f(x) + g(y) , where f and g are some differentiable functions of their arguments. The solution of the Cauchy Problem: We observe that the Cauchy data is given on one of the characteristics (x = 0) of the equation. Hence either there is no solution or there exists infinitely many solution of this problem. Using the initial conditions we obtain

(z_{,x})_{,y} = 0 on the y axis implies σ′ = 0,  (7)
z(x, y) = ρ on the y axis implies f(0) + g(y) = ρ(y),  (8)
z_{,x}(x, y) = σ(y) on the y axis implies f′(0) = σ(y)  (9)

Since the functions ρ and σ are given functions, the condition σ′ = 0 may not be satisfied. In such cases there exist no solution of this problem. If σ(y) = f′(0) = constant then the solution is as follows

z(x, y) = f(x) − f(0) + ρ(y)  (10)

Here f(x) satisfies f′(0) = σ otherwise arbitrary. It means that in such a case there are infinitely many solutions of the problem. Meaning that the solution of the Cauchy problem is not unique.

3. Find the complete integral (general solution) of the equation z_{,xx} +z_{,yy} − 2z_{,xy} + \frac{z_{,y}−z_{,x}}{y−x} = 0 Solution: This is a parabolic pde, because Δ = S^{2} − 4RT = 0. The characteristic of the equation is simply found as ξ = y + x = constant. Choose η = y − x, then the equation reduces to

2η z_{,ηη} + z_{,η} = 0  (11)

This is an ODE with the solution z(ξ, η) = a(ξ) \sqrt{η} +b(ξ), where a and b are arbitrary (differentiable) functions of ξ. Hence the general solution is given by

z(x, y) = a(x + y) \sqrt{y − x} + b(x + y)  (12)

Note that solutions exist only in the region y > x.

4. Let R = R(x, y), P = P (x, y), S = S(x, y) and W = W (x, y) be some differentiable functions of (x, y) ∈ R^{2}. Show that the equation R r + S s + P p = W can be reduced to a single linear pde of first order. Give a procedure for determining the solution of the original equation. Illustrate your method by solving the equation xr + ys + 2p = y. Solution: Let z_{x}(x, y) = u(x, y) then the pde reduces to

R u_{,x} + S u_{,y} + P u = W,  (13)

This is a type of first order pde we studied in the previous chapters. The procedure to obtain should be as follows

\frac{dx}{R} = \frac{dy}{S} = \frac{du}{W − P u}  (14)

The solution of these equations define a curve ξ(x, y, u) = constant and η(x, y, u) = constant. Then the general solution of (??) is given by F (ξ, η) = 0, where F is an arbitrary (differentiable) function. Find u from this equation and solve z_{,x} = u(x, y). Illustration: xr + ys + 2p = y. We get ξ = \frac{1}{3}y^{3} − uy^{2}, η = \frac{x}{y}. Hence the general solution is given by

u = \frac{1}{3}y − \frac{1}{y^{2}} f(\frac{x}{y})  (15)

where f is an arbitrary (differentiable) function. The solution of the equation z_{,x} = u(x, y) is

z(x, y) = \frac{1}{3}xy − \frac{1}{y}g(\frac{x}{y}) + h(y)  (16)

where g′ = f and h is an arbitrary (differentiable) function.

5. Find the general solution of the equation xy s − xp − yq + z = 0 and determine the solution of this equation which satisfies the condition z = x^{n} and p = 0 on the curve y = x. Solution: This equation can be written as a product form (D − \frac{1}{x}) (D′ − \frac{1}{y}) z = 0. Hence z = z_{1} + z_{2} where

(D − \frac{1}{x})z_{1} = 0, (D′ − \frac{1}{y})z_{2} = 0  (17)

with the solutions z_{1} = x f(y), z_{2} = y g(x). Hence the general solution is

z(x, y) = x f(y) + y g(x)  (18)

where the f and g are arbitrary (differentiable) functions of their arguments. The solution of the Cauchy problem:

f(x) + g(x) = x^{n−1},  (19)
x g′(x) − g(x) + x^{n−1} = 0.  (20)

Solutions of the above equations are given by for n ≠ 2

f(x) = \frac{n − 1}{n − 2} x^{n−1} − a x,  (21)
g(x) = −\frac{1}{n − 2} x^{n−1} + a x  (22)

where a is an arbitrary constant. Hence the solution of the original pde is

z(x, y) = \frac{n − 1}{n − 2} xy^{n−1} − \frac{1}{n − 2} yx^{n−1}  (23)

For n = 2 the solution of f and g are given by

f(x) = x ln(x) − ax + x  (24)
g(x) = −x ln(x) + ax  (25)

Hence the solution becomes

z(x, y) = xy [ln(y) − ln(x)] + xy  (26)