3) Consider the non-linear initial value problem Ofu – 0zu = (du)² – (d¿u)², x, t e R, (2) u(x, 0) = 0, du(x, 0) = g(x), g :R → R given. (a) Prove that if w = w(x, t) satisfies { Ofw – aw = 0, w(x,0)...

Subject: Partial differential equations Kindly solve it as soon as possible , kindly give it a try at least Thank you!!3) Consider the non-linear initial value problem<br>Ofu – 0zu = (du)² – (d¿u)²,<br>x, t e R,<br>(2)<br>u(x, 0) = 0,<br>du(x, 0) = g(x),<br>g :R → R given.<br>(a) Prove that if w =<br>w(x, t) satisfies<br>{<br>Ofw – aw = 0,<br>w(x,0) = 1,<br>dw(x, 0) = g(x),<br>x,t e R,<br>then<br>w(x, t) = e
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Extracted text: 3) Consider the non-linear initial value problem Ofu – 0zu = (du)² – (d¿u)², x, t e R, (2) u(x, 0) = 0, du(x, 0) = g(x), g :R → R given. (a) Prove that if w = w(x, t) satisfies { Ofw – aw = 0, w(x,0) = 1, dw(x, 0) = g(x), x,t e R, then w(x, t) = e"(z,t). (b) Show that pI+t u(7, t) = log (1 + 5 9(e)ds). (c) Prove that if (3) | I9(s) ds < 2,="" then="" the="" solution="" u(x,="" t)="" of="" (2)="" is="" defined="" for="" all="" t="" e="" r.="" notice="" that="" if="" (3)="" fails,="" then="" the="" solution="" u(x,="" t)="" of="" (2)="" may="" only="" exist="" in="" a="" finite="" time="" interval="" [0,t),="" t=""> 0.

Jun 05, 2022
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