1、习题11.对下列偏微分方程,指出它的阶,并指出它是线性的、拟线性的还是非线性的.若是线性的,再指出它是齐次的还是非齐次的.(1) u3x + 2uuy = xy;(2) uuy 6xyux = 0;(3) uxx x2uy = sinx;(4) u3xx + u3x cosu = ex;(5) uyuyyx uxuxxy + u5 = f(x,y);(6) ut 3uux + 6uxxx = 0.解. (1)是一阶非线性偏微分方程; (2)是一阶拟线性偏微分方程; (3)是二阶线性非齐次偏微分方程; (4)是二阶非线性偏微分方程; (5)是三阶拟线性偏微分方程; (6)是三阶非线性偏微分方程.
2、2.证明二维拉普拉斯算子在极坐标系(r,)下可以写成u = 2ur2 +1rur +1r22u2.证. (方法一)极坐标系与直角坐标系之间的变换关系为x = rcos, y = rsin,或者为r = (x2 + y2)12, = arctan yx.于是rx =xr = cos,ry =yr = sin,x = yx2 + y2 = sinr ,y =xx2 + y2 =cosr .从而ux =urrx +ux =ur cosusinr ,uy =urry +uy =ur sin +ucosr ,22ux2 =2ur2 cos2 2 2ursincosr +2u2sin2 r2 +ursin
3、2 r +usin2r2 ,2uy2 =2ur2 sin2 + 2 2ursincosr +2u2cos2 r2 +urcos2 r usin2r2 .因此2ux2 +2uy2 =2ur2 +1rur +1r22u2.(方法二)因x = rcos, y = rsin,直接计算得ur =uxxr +uyyr =ux cos +uy sin,u =uxx +uyy =ux(rsin) +uyrcos,2ur2 =2ux2 cos2 + 2uxy sin2 +2uy2 sin2 ,2u2 =2ux2r2 sin2 2uxy sin2 +2uy2r2 cos2 uxrcosuyrsin.因此2ur2
4、+1rur +1r22u2 =2ux2 +2uy2.3.证明三维拉普拉斯算子在柱面坐标系(r,z)下可以写成u = 2ur2 +1rur +1r22u2 +2uz2.证.柱面坐标系与直角坐标系之间的变换关系为x = rcos, y = rsin, z = z,或者为r = (x2 + y2)12, = arctan yx, z = z.完全同上题的计算,得证.4.证明三维拉普拉斯算子在球坐标系(r,)下可以写成u = 2ur2 +2rur +1r2(2u2 + cotu +1sin2 2u2).3证.柱坐标系与直角坐标系之间的变换关系为x = rsincos, y = rsinsin, z =
5、 rcos.直接计算,得ur =uxxr + uyyr + uzzr=uxrcoscos + uyrcossinuzrsin,u =uxx + uyy + uzz=ux(rsinsin) + uyrsincos,urr =+ ,u =+ ,u =+ .因此2ur2 +2rur +1r2(2u2 + cotu +1sin2 2u2)= uxx + uyy + uzz.5.求下列线性偏微分方程的通解(其中u = u(x,y):(1) uxx + cu = 0 (提示:分c 0,= 0, 0时,对每个固定的y,解具有形式u(x) = C1 cos(cx)+C2 sin(cx)(可先写出特征方程,再给出解).然而,当y变化时, C1和C2的选择也会变化(即,它们可能是y的函数).因此, c 0时的通解为u(x,y) = f(y)cos(cx)+g(y)sin(cx),其中f,g是任意二阶可微函数.当c = 0和c 0,u(x,0) = 1, 0,un(x,0) = 1 + 1n sinnx, 0|un(x,t)u(x,t)| = sup10(1 + 1nen2t sinnx)1=supt01nen2t 1nen2 +.这表明原定解问题的解是不稳定的.所以,原定解问题的解是不适定的.
