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Solve the equation 3uy + Uxy = 0. (Hint: Let v = uy.)

Solve the equation 3uy + Uxy = 0. (Hint: Let v = uy.)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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[Partial Differential Equations] How do you solve this?

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2. Solve the equation 3uy + xy = = 0. (Hint: Let v = =Uy.)
Transcribed Image Text:2. Solve the equation 3uy + xy = = 0. (Hint: Let v = =Uy.)
THE VARIABLE COEFFICIENT EQUATION The equation ux + yuy=0 (4) is linear and homogeneous but has a variable coefficient (y). We shall illustrate for equation (4) how to use the geometric method somewhat like Example 1. The PDE (4) itself asserts that the directional derivative in the direction of the vector (1, y) is zero. The curves in the xy plane with (1, y) as tangent vectors have slopes y (see Figure 3). Their equations are This ODE has the solutions dy y dx y = Ce*. (6) These curves are called the characteristic curves of the PDE (4). As C is changed, the curves fill out the xy plane perfectly without intersecting. On each of the curves u(x, y) is a constant because d ди ди -u(x, Ce*) = + Ce = Ux + yuy ay dx əx It follows that Thus u(x, Cet) = u(0, Ceº) = u(0, C) is independent of x. Putting y = Ce and C = e-*y, we have u(x, y) = u(0, e *y). (5) = 0. Figure 3 u(x, y) = f(e *y) (7) is the general solution of this PDE, where again f is an arbitrary function of only a single variable. This is easily checked by differentiation using the chain rule (see Exercise 4). Geometrically, the "picture" of the solution u(x, y) is that it is constant on each characteristic curve in Figure 3.
Transcribed Image Text:THE VARIABLE COEFFICIENT EQUATION The equation ux + yuy=0 (4) is linear and homogeneous but has a variable coefficient (y). We shall illustrate for equation (4) how to use the geometric method somewhat like Example 1. The PDE (4) itself asserts that the directional derivative in the direction of the vector (1, y) is zero. The curves in the xy plane with (1, y) as tangent vectors have slopes y (see Figure 3). Their equations are This ODE has the solutions dy y dx y = Ce*. (6) These curves are called the characteristic curves of the PDE (4). As C is changed, the curves fill out the xy plane perfectly without intersecting. On each of the curves u(x, y) is a constant because d ди ди -u(x, Ce*) = + Ce = Ux + yuy ay dx əx It follows that Thus u(x, Cet) = u(0, Ceº) = u(0, C) is independent of x. Putting y = Ce and C = e-*y, we have u(x, y) = u(0, e *y). (5) = 0. Figure 3 u(x, y) = f(e *y) (7) is the general solution of this PDE, where again f is an arbitrary function of only a single variable. This is easily checked by differentiation using the chain rule (see Exercise 4). Geometrically, the "picture" of the solution u(x, y) is that it is constant on each characteristic curve in Figure 3.
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