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> >~ ~ >w!x>y z > bwR >wRwR $}L fzy{ws~{ >wRzy{wR~ b 6 X S£lzy{®·6»E¸CÉThe function v(x,t) = u(x,t)−w(t) satisfies vt − kvxx = 0 0Let u(x;t) be a solution of u t= ku xx Show that the following facts hold (a) For constants a, x 0 and t 0, the function v(x;t) = u(ax x 0;a2t t 0) satis es v t= kv xx (b) For any constant k0, the function v(x;t) = u(x;(k0 k)t) satis es v t= k0v xx (c) The function v(x;t) = t 212 exp(x 4kt) u(x t;
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Letting w = 2xyields f(w) = sin(w=2) Therefore, u(t;x) = sin(x 3t=2) 2Let v= u y, then 3v v x= 0 Thus we have v(x;y) = f(y)e 3x, ie, u y(x;y) = f(y)e 3x, which implies u(x;y) = F(y)e 3x g(x), where both Fand gare arbitrary (di erentiable) functions 3The characteristic curves satisfy the ODE dy=dx= 1=(1 x2), which implies y= arctanx C Thus u(x;y) = f(y arctanx) We omit the easyThen the series (43) defines a smooth function u(x,t) for t > 0, which satisfies ut = uxx and limt↓0 ku(x,t)−u0(x)kL2 = 0 Proof Term by term differentiation of the series with respect to x,t has the effect only of multiplying by powers of m For t > 0 the exponential factor e−tλm = e−tm 2π2F is simply the value of u(x;y) on the \diagonal" fx= yg, this makes the existence of u impossible Problem 218 Statement (a) Show that the PDE u x = 0 has no solution which is C1 everywhere and satis es the side condition u(x;x2) = x (b) Find a solution of the problem in (a) which is valid in the rst quadrant x>0, y>0 (c) Explain the results of (a) and (b) in terms of the intersections
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(~u ~v ~u w~) ~x= (~u ~v) ~x (~u ~v) ~x It follows that both sides are equal This is (2) We could check (3) by a similar argument Here is another way (~u~v) w~= w~ (~u~v) = w~ ~u w~ ~v = ~u w~~v w~ This is (3) To prove (4), it su ces to prove the rst equality, since the fact that the rst term is equal to the third term follows by a similar derivation If = 0, then both sides are1 t) satis es v t= kv xx Solution (a) By a direct computation we have v t(x;t) = a2u t(ax xQt Ë 7 s»µ«t 8 ;pV ° Mp w ¬tSZ ¯q ó tx~ ³w ( UK } f\p æ px Ôó Ët w à » ;Mo
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