Partial Differential Equation: Solve and analyze the behavior: for t > 0 with data u(x,0)=x^2 1) xu_x+yu_y= 4u 2) xu_x+yu_y= 2u 3) tu_x -xu_t= 0 4) tu_x -xu t= 0TO solve this partial diff equation we can use the method of charactertics . let's define a new variable s = 2 + 4 and t as the independent variable Then, we have ds = I t ly. dt du = 4c from the first equation we have my =ds . - 1 Substituting this into the second equation we get off . = 4 0 = ) du . as = y ce ds aft. -) du ( It uy ) = uc ds = ) ( ds ) lu = ( It uy ) we can integrate both side w. 91. + tos log / u) = 410 9 / 1 + uyl + c where c is constant of integration using initial condition uccio ) = 2 2 2 2 = U L E 1 0 ) = 4 ( x + 0, of x ) = u ( X + x, 0 ) = u(2210 ) so we can write the equationas 4Now we can substituting this equation ux + uy = ( . This is first - order PDE which solved using Method of characteristic dx =1 , dy - 1 , du = ( 5 ) # dt solving for u we get ulx,yit) = ce ( 5+ /4 ) where C is constant that depends on the initial condition ulx , oj = 2 2 C = x2 = 2 2 Therefore the solution of PDE is we can analyze the behavior of the solution as t - c AS I increases , the exponetial term e(?) grows without bound . This means that the solk became unbounded t s and behaviour of the solution is unstable. 3 ) . we are given the p. .Die Ux - X UF = 0 with initial condition U ( x 10 ) = 22 To solve this equation , we will use the method of 'haractorstic, "we look for a solution from " (xIt ) = f (xt ) where f is unknow function using chain Rule Use = f'( x t ) t Ut = f ' ( xt ) x Substituting these expression into the PDE f' ( x t ) t - x f' ( x t ) x =0 Dividing by t we obtainF'( xt ) = x fl ( xt ) This is separable first-order ordinary diff equation . we can separete the ' variable and integrate both side to get f ( x t ) = c x 2 where c is constant of integration using initial condition we have f ( x . 0 ) = C x 2 = x2 which =, that c = 1 Therefor the PDE ulxIt ) = fect) = x2t This solution is valid for + 20 To analyze the behavior of solution we see that as + increes the solution grows and becames more spread out This is consistent with the fact that the PDE is tansport equation which means that describe the advection of quantity with velocity that depends on the spatial variable . In this case the velocity is X which means that quantity is advacted faster for langer value of x . solution became more spread own over time we can also compure the charaterstic of PDE, which are curve along which the solution is constant The Characteristic statify the ODE dx - X which is sol x ( + ) = x ( o jet This means that the characteristics are exponential core that decay to zero as + increase . This is Consistent with the behavior of solution , which Spread became more out over time4 ) we are given the PDE 4x - put = 0 with initial condition U ( x 10 ) = 2 2 To solve PDE, we use the Method of characterstic. let * = x(S ) and + = t (s) be panametric equation of the choraltenstic curve, where sis parameter Then we d x ds = 1 , d t= - x using the chain Rule we have du LlX s - le dt = Ux - XU t = O ds This means u is constant along the characterstic curve i.e c ( x ( t ) , t ) = u(x (os, o) = x(0)2 . 08 u ( d It ) = u ( x 0 1 0 ) = 20 ) 2 where to = X- tx Now we can use initial condition to find the value of to 4 (71 , 0 ) = 212 = 202 = ( x -tx ) 2 solving xo we get x0 = * ( 1 -t ) Therefor the soin of PDE is u (xIt ) = (x-tx ) ? analyze the behaviour of the solution we can compute the first and second devivate of u with respect tox and + 4 x = 2 ( x - t x ) ( 1 - t) g ut = - x ( x -tx ) 4 xx = 2 ( 1 - t) , 4+t = - 21 / 2 - 2 22 ) 7x h = 4 xx = - 2 1 1 -2t) At += 04 ( 41 0 ) = 2 2 Ux = 296 , HE = 0 Uft = O "At = ufg = 0 from these expression we can see that the soin is quadratic fun of x for all to The cofficiant of 2 is colficient of a inceasing linearly with t Theyyore . The graph of solution moves downwards along the -ouis and become more step as + increase we can also absewe that the solution staticlly conservation law. d u( it ) dy = of ( x - tx ) 'dx =d 1 / ]t 3 = +2 dt d t which means the the total ' area' under the graph of solution is concerned over time