Solution 4. hence the Cauchy problem has a differentiable solution for all positive time y ą 0. b) Let hpsq ups, 0q, since hpsq is decreasing in r π

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1 Solution 4 1. ) Let hpsq ups, 0q, sine the initil ondition hpxq is differentile nd nonderesing, i.e. h 1 pxq ě 0, we hve the formul for the ritil time (see eqution (2.47) on p.4 in [PR]), y 1 h 1 ă P R, pxq hene the Cuhy prolem hs differentile solution for ll positive time y ą 0. ) Let hpsq ups, 0q, sine hpsq is deresing in r π 2, π 2 s, i.e. h1 psq ă 0, the solution will eome non-differentile. We hve the formul (2.46) in [PR], u x h1 1 ` yh 1, then the solution s derivte lows up t the time y 1 h 1 psq, the ritil time y is the infimum of t 1 h 1 psq u. h 1 psq hs infimum 1 t s 0, so 0 x ď π 2, h 1 psq os x π 2 ă x ă π 2, 0 x ě π 2. y inft 1 h 1 psq u 1 h 1 p0q 1. ) The disontinuity moves with speed 1 2 pu ` u`q 1 1 2p1 2q 2, therefore & 1 x ă 1 y upx, yq 2, % 2 x ą 1 y 2. is wek solution of the Cuhy prolem 1

2 2. ) The hrteristi equtions nd the prmetri initil onditions re y t pt, sq 1, x t pt, sq u 2, u t pt, sq 0, 1 s ď 0, yp0, sq 0, xp0, sq s, up0, sq 1 s α 0 ă s ă α, 0 s ě α. Solve the equtions to otin the hrteristis ps ` t, t, 1q s ď 0, px, y, uqpt, sq ps ` u 2 p0, sqt, t, up0, sqq ps ` p1 s qt, t, 1 s α α q 0 ă s ă α, ps, t, 0q s ě α. Two wys to proeed: 1st. Invert the trnsformtion pxpt, sq, ypt, sqq s ď 0 px, yq ps ` t, tq ñ pt, sq py, x yq, s ď 0 ñ x y ď 0 ô x ď y 0 ă s ă α s ě α px, yq ps ` p1 s x y qt, tq ñ pt, sq py, α α α y q, 0 ă s ă α ñ 0 ă α x y α y ă α ô y ă x ă α or α ă x ă y px, yq ps, tq Then for 0 ă y ă α, we hve py, x yq x ď y, pt, sq py, α x y α y q y ă x ă α, py, xq x ě α. Sustitutie to upt, sq, 1 x ď y, x α upx, yq y α y ă x ă α, 0 x ě α. The solution eomes disontinuous t y α. 2

3 2nd. 1 x ě y, u up0, s x u 2 yq 1 x u2 y y ă x ă α, α 0 x ě α. Solve u to get the sme result. ) Two wys to rgue tht the solution will e sigulr: 1st. The projetions of hrteristi urves px, yq-plne re the lines s ` y s ď 0, x s ` u 2 p0, sqy s ` p1 s qy 0 ă s ă α,, α s s ě α, for the s-vlues 0 ď s ď α, these lines will ollide t finite y, tully they ollide t one point px, yq pα, αq. However the funtion upx, yq preserve its initil vlue upx, 0q long these lines, hene we n t solve the Cuhy prolem eyond y α. The ritil time y α. 2nd. From the expliit form of the solution upx, yq, the solution is ontinuous for y ă α. When y α, # 1 x ă α, upx, αq 0 x ą α, whih is disontinuous in x. So we n t solve the Cuhy prolem eyond y α. The ritil time y α. ) See the 1st rgument of ). l 0 : x y; l α 2 : x α 2 ` y 2 ; l α : x α. Diret lultions show they interset t pα, αq.. ) Rewrite the eqution in the form u y ` 1 pu q x 0, nd integrte (with respet to x, for fixed y) over n ritrry intervl r, s to otin y ż upξ, yqdξ ` 1 ru p, yq u p, yqs 0.

4 ) We write the wek formultion in the form «ż γpyq y upξ, yqdξ ` ż γpyq upξ, yqdξ ff ` 1 ru p, yq u p, yqs 0 Differentiting the integrls with respet to y nd using the PDE itself leds to «ż γ y pyqu pyq γ y pyqu`pyq 1 γpyq ż ff pu pξ, yqq ξ dξ ` pu pξ, yqq ξ dξ We get ` 1 ru p, yq u p, yqs 0 ñ γ y pyqru pyq u`pyqs 1 u pyq u p, yq ` u p, yq u `pyq ` 1 ru p, yq u p, yqs 0. γ y pyq 1 ru`pyq u pyqs u`pyq u pyq. ) The solution of the Cuhy prolem in Ex 2 t y α γpyq 4. ) implies upx, αq # 1 x ă α, 0 x ą α, u pαq 1, u`pαq 0, then ) implies tht the disontinuity moves with speed 1. Therefore the following wek solution is omptile with the integrl lne for y ą α: 1 x ă α ` 1 py αq, upx, yq 0 x ą α ` 1 py αq. Lruspx, yq u xx ` 2u xy ` r1 qpyqsu yy, nd 1, 1, 1 qpyq, hene 1 y ă 1, δplqpx, yq r1 qpyqs 0 y ď 1, 1 y ą 1. So the eqution is hyperoli on ty ą 1u, proli on t y ď 1u, ellipti on ty ă 1u. 4

5 ) y ą 1: Lrus u xx ` 2u xy. Consider nonsingulr liner trnsformtion ξ ξ x x ` y, η x ` y. nd ωpξ, ηq upxpξ, ηq, ypξ, ηqq, then ω stisfies the eqution lrωs Aω ξξ ` 2ω ξη ` Cω ηη 0 where A C To otin the nonil form, we need to solve the eqution A ξ 2 x ` 2ξ x ` ξy 2 ξ 2 x ` 2ξ x 0, C η 2 x ` 2 ` η 2 y η 2 x ` 2 0 We n tke ξ x 2, 1, 0, 1, i.e. the trnsformtion ξ 2x y, η y trnsforms the eqution to nonil form lrωs 4ω ξη 0 y ă 1: Lrus u xx ` 2u xy ` u yy. Consider nonsingulr liner trnsformtion ξ ξ x x ` y, η x ` y. nd ωpξ, ηq upxpξ, ηq, ypξ, ηqq, then ω stisfies the eqution lrωs Aω ξξ ` 2ω ξη ` Cω ηη 0 where A C To otin the nonil form, we need to solve the eqution C η 2 x ` 2 ` η 2 y η 2 x ` 2 ` η 2 y 0 5

6 We n tke ξ x 1, 1, 1, 1, i.e. the trnsformtion ξ x ` y, trnsforms the eqution to nonil form y ă 1: η x y lrωs 4ω ξξ 0 Lrus u xx ` 2u xy ` 2u yy. Consider nonsingulr liner trnsformtion ξ ξ x x ` y, η x ` y. nd ωpξ, ηq upxpξ, ηq, ypξ, ηqq, then ω stisfies the eqution where A C lrωs Aω ξξ ` 2ω ξη ` Cω ηη 0 To otin the nonil form, we need to solve the eqution A ξ 2 x ` 2ξ x ` ξ 2 y ξ 2 x ` 2ξ x ` 2 1, ξ x ` pξ x ` q ` ξ x ` pξ x ` q ` 2 0, C η 2 x ` 2 ` η 2 y η 2 x ` 2 ` 2η 2 y 1 We n tke ξ x 1, 0, 1, 1, i.e. the trnsformtion ξ x, trnsforms the eqution to nonil form ) For the hyperoli se ty ą 1u, so η x y lrωs ω ξξ ` ω ηη 0 Lrus u xx ` 2u xy, 1, 1, 0, nd the hrteristi equtions re dy dx `? 2 2, nd dy dx? 2 0 so the hrteristis re the stright lines ty 2x ` su nd ty su. Drw the pitures... 6

7 Referenes [PR] Y. Pinhover, J. Ruinstein, An introdution to Prtil Differentil Equtions, Cmridge University Press(12. Mi 2005). 7

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