Find the complete solutions of the partial difference equation
(D +D' - 1)²= xy
Answers
Step-by-step explanation:
Find the complete solutions of the partial difference equation
(D +D' - 1)²= xy
Answer:
Martha L. Abell, James P. Braselton, in Differential Equations with Mathematica (Fourth Edition), 2016
10.1.1 Introduction
We begin our study of partial differential equations with an introduction of some of the terminology associated with the topic. A linear second-order partial differential equation (PDE) in the two independent variables x and y has the form
(10.1) A(x,y)
∂2u
∂x2
+B(x,y)
∂2u
∂y∂x
+C(x,y)
∂2u
∂y2
+D(x,y)
∂u
∂x
+E(x,y)
∂u
∂y
+F(x,y)u=G(x,y),
where the solution is u(x, y). If G(x, y) = 0 for all x and y, we say that the equation is homogeneous. Otherwise, the equation is nonhomogeneous.
Example 10.1.1
Classify the following partial differential equations: (a) uxx + uyy = u; (b) uux = x.
Solution
(a) This equation satisfies the form of the linear second-order partial differential equation (10.1) with A = C = 1, F = −1, and B = D = E = 0. Because G(x, y) = 0, the equation is homogeneous. (b) This equation is nonlinear, because the coefficient of ux is a function of u. It is also nonhomogeneous because G(x, y) = x.
Definition 40 Solution of a Partial Differential Equation
A solution of a partial differential equation in some region R of the space of the independent variables is a function that possesses all of the partial derivatives that are present in the PDE in some region containing R and satisfies the PDE everywhere in R.
Example 10.1.2
Show that u(x, y) = y2 − x2 and u(x,y)=eysinx are solutions to Laplace’s equation uxx + uyy = 0.
Solution
For u(x, y) = y2 − x2, ux(x, y) = −2x, uy(x, y) = 2y, uxx(x, y) = −2, and uyy(x, y) = 2, so we have that uxx + uyy = (−2) + 2 = 0, which we verify with Mathematica.
Clear[u]u[x_, y_]=y ∧2−x ∧2;D[u[x, y], {x, 2}]+D[u[x, y], {y, 2}]
0
Similarly, for u(x,y)=eysinx, we have ux=eycosx, uy=eycosx, uxx=−eysinx, and uyy=eysinx. Therefore, uxx+uyy=(−eysinx)+eysinx=0, so the equation is satisfied for both functions.
Clear[u]u[x_, y_]=Exp[y]Sin[x];D[u[x, y], {x, 2}]+D[u[x, y], {y, 2}]
0