1、求解以下各偏微分方程式： (1) uyy + 9 u = 0(2) uy = ux(3) uxy = ux(4) uxy = 0
2、求解以下各偏微分方程式： (1) uxy = 1 ; u (0, 0) = 0(2) y uy = 2 ; u (1, 2) = 4(3) 2 ux + 3 uy = 0 ; u (1, 4) = 7(4) uxx = 0 ; u (1, 1) = 1
- 教書的Lv 69 years agoFavorite Answer
1.We find general solutions to a single pde:
(1) [consider as an ode for u=u(y)]--> u(x,y)=A(x) cos3y +B(x) sin3y, with A(x), and B(x) arbitrary functions.
(2) [using method of characteristic to this 1st order pde] --> u(x,y)=F(x+y), where Fis any function of one variable.
(3) [ referring as an ode for u_x first]--> u_x(x,y)=c(x)e^y ; --> [integration w.r.t. x] --> u(x,y)= C(x)e^y+D(x), where C and D are arb functions of x.
(4) u_xy=0 [integral w.r.t y]--> u_x=f(x) ; [integral w.r.t. y]--> u(x,y)=F(x)+G(y) , where F and G are arb functions of one variable.
2. In this group a pde is attached by a condition each. Unfortunately, that condition is not qualified as "initial condition" , so the solution is not uniquely determined.
(1) mimicking 1(4) , u(x,y)=xy+G(x), for arb G is its general solution; put u(0,0)=0, there are still infinitely many choices for choices of G.
(2) u(x,y)=2lny+A(x) is the general solution; Choose A properly so u(1,2)=4 will do.
(3) u(x,y)=F(3x-2y) is the general solution; so pick F so that F(-5)=7 will be OK.
(4) General solution is u(x,y)=xA(y)+B(y) for arb A, B ; choose A(1)+B(1)=1 will do.