Õ 83-25 Þ ÛÐ Þ Ð ÚÔÜØ Þ ÝÒ Þ Ô ÜÞØ ¹ 3 Ñ Ð ÜÞ u x + u y = x u u(x, 0) = e x2 ÝÒ Þ Ü ÞØ º½ dt =, dt = x = t + c, y = t + c 2 We can choose c to be zero without loss of generality Note that each characteristic (the lines y = x + c 2 ) intersects the line y = 0, on which the initial data is given, exactly once, the lution will be well-defined in the whole plane with general lution dt = t u u = t + c 3 e t The initial condition is given on y = 0, ie at t = c 2 and then we have u = e x2 = e c2 2 Thus e c2 2 = c2 + c 3 e c 2, and The lution on the characteristic is c 3 = e c 2 ( + c2 + e c2 2) u = t + e (t+c 2) ( + c 2 + e c2 2) It just remains to write the lution in terms of x and y: we have x = t y = t + c 2 t = x c 2 = y x the lution is u = x + e y ( + (y x) + e (y x)2) It is a good idea to check!
u x + xu y = y 2xu u(0, y) = ÝÒ Þ Ü ÞØ º¾ dt =, dt = x x = t + c, y = 2 (t + c ) 2 + c 2 We can choose c to be zero without loss of generality Note that each characteristic (the parabolas y = 2 x2 + c 2 ) intersects the line x = 0, on which the initial data is given, exactly once, the lution will be well-defined in the whole plane Writing this as dt = 2 t2 + c 2 2tu dt + 2tu = 2 t2 + c 2 we see there is an integrating factor e t2 and d ( e t 2 u ) ( ) = dt 2 t2 + c 2 e t2, u = e t2 (c 3 + t ( ) ) 2 s2 + c 2 e s2 ds The initial condition is given on x = 0, ie at t = 0 and then we have u = y = c 3 Thus c 3 should be taken to be and the lution on the characteristic is u = e t2 ( + t ( ) ) 2 s2 + c 2 e s2 ds It just remains to write the lution in terms of x and y: we have x = t y = 2 t2 + c 2 t = x c 2 = y 2 x2 the lution is x ( ) ) u = e ( x2 + 2 (s2 x 2 ) + y e s2 ds It is a good idea to check!
xu x + yu y = u(, y) = y ÝÒ Þ Ü ÞØ º Ó ÜÞØ ÐÝ Ü Ñ Þ Ò dt = x, dt = y x = c e t, y = c 2 e t The characteristics are half lines through (0, 0) e t is always positive the characteristics never reach (0, 0) (There is al a characteristic which is a single point the point (0, 0) corresponding to c = c 2 = 0) The initial data is specified on the line x = This intersects (once) with each of the characteristics in the half-plane x > 0, we expect a lution in this region Given that x > 0 we can choose c = without loss of generality with lution dt =, u = t + c 3 The initial data is specified at x =, ie t = 0 At t = 0 we have y = c 2 and u = t 3 to impose the condition u(, y) = y we must take c 3 = c 2 Thus the lution on the characteristic is u = t + c 2 It just remains to write the lution in terms of x and y: we have x = e t y = c 2 e t t = log x c 2 = y x the lution is u = log x + y x It is easy to check this satisfies both the equation and the initial condition It is defined everywhere except on the y axis (x = 0) But it is only the lution of our problem in the region x > 0 No characteristics connect the initial data and the region x < 0
Ñ º ÝÒÒ Ö Û p ÜÝ xu x + yu y = pu ÝÒ Þ ÞÒ ÛÒ u(x, y) ÚÛÔ Ø º Ð ÖÒ ÐÖ u(x, y) = y ØÝ ÔÞ Þ ÞÒ ÛÒ Ñ u(x, y) ¹Ý Ó ÞÔ Ð Ó Ý Þ ÛÐ Ý u(x, y) ÐÝ ÞÜ Ñ Þ Ò ºu(x, y) Þ ÚÒ x 2 + y 2 = ºp ÐÝ ÜÝØ ÜÖ dt = x, dt = y x = c e t, y = c 2 e t The characteristics are half lines through (0, 0) e t is always positive the characteristics never reach (0, 0) (There is al a characteristic which is a single point the point (0, 0) corresponding to c = c 2 = 0) Each half-line intersects (once) with the circle x 2 + y 2 = we expect a lution on the whole plane except at the point (0, 0) dt = pu, with lution u = c 3 e pt The initial data is specified on the circle x 2 + y 2 = which a characteristic intersects when (c 2 + c 2 2)e 2t =, ie when e t = (c 2 +c2 2 )/2 On the circle we want to impose u(x, y) = y ie c 3 e pt = c 2 e t c 3 = c 2 (c 2 + c 2 2) (p )/2 Thus the lution on a characteristic is u = c 2 (c 2 + c 2 2) (p )/2 e pt It just remains to write the lution in terms of x and y: we have x = c e t (c 2 + c2 2 )/2 e t = x 2 + y 2 the lution is y = c 2 e t c 2 = (c 2 +c2 2 )/2 u = y(x 2 + y 2 ) (p )/2 y x 2 +y 2 It is easy to check this satisfies both the equation and the initial condition What about the domain of definition of the lution? It certainly is defined on the whole plane except possibly at (0, 0) At (0, 0) there are several possibilities: ( ) If p < then the lution is not defined ( ) If p =, 3, 5, the lution is defined and the lution is infinitely differentiable at (0, 0) ( ) If p takes other values bigger than the lution is defined, but it is only differentiable a finite number of times (in fact not at all if < p < 3)
Ö Ð Ó ÜÞØ Ñ Û Ñ º yu x xu y = u? u(x, 0) = x, x > 0 ÛÒÔ dt = y, dt = x x = c sin t, y = c cost (There is another constant of integration which we can ignore) The characteristics are circles centered at (0, 0) We are given an initial condition on the positive x- axis (not including the origin) This intersects every characteristic once except for the characteristic consisting of the single point (0, 0) But the characteristics are closed curves this is not enough to guarantee a lution with lution dt = u, u = c 2 e t The initial data is specified on the half line y = 0, x > 0 The characteristic intersects the initial line when t = 2 π, we need to choose c 2 such that c 2 e π/2 = c In other words, the lution on the characterstic is u = c e t π/2 It just remains to write the lution in terms of x and y: we have Thus the lution is u = x = c sin t y = c 2 cos t t = arctan ( x y ) c = x 2 + y 2 x 2 + y 2 e arctan ( x y) 2 π = x 2 + y 2 e arctan( x) y It is easy to check this satisfies both the equation and the initial condition But the definition of arctan means this formula is multivalued : as we go round the point (0, 0) (in an anticlockwise direction) the arctan increases by 2π This is a result of the closed trajectories In the normal sense of the word, there is no lution