PDEMOC7

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$xu_x + u_y = y, u(x,0)=x^2\,$

The initial curve will be parameterized by $\Gamma(s,0,s^2)\,$ .

The characteristics are

$\frac{dx}{dt}=x, \frac{dy}{dt}=1, \frac{dz}{dt}=y\,$

For $x\,$ ,

$\frac{dx}{dt}=x\,$

$x^{-1}dx = dt\,$

$\ln x = t+c_1(s)\,$

$x(s,t) = c_4(s)e^t\,$

$x(s,0) = c_4(s) = s\,$

So $x(s,t) = se^t\,$

For $y\,$ ,

$\frac{dy}{dt}=1$

$y(s,t) = t+c_2(s)\,$

$y(s,0) = c_2(s) = 0\,$

So $y(s,t) = t\,$

For $z\,$ ,

$\frac{dz}{dt}=y=t\,$

$z(s,t) = \frac{1}{2}t^2+c_s(3)\,$

$z(s,0) = c_s(0) = s^2\,$

$z(s,t) = \frac{1}{2}t^2+s^2\,$

The solution is

$z(s,t) = u(x,y) =\frac{1}{2}y^2 + (xe^{-y})^2\,$

Main Page : Partial Differential Equations : Method of Characteristics

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