2.1.76 \(u_t + u_x = 0\) with \(u(x,1)=\frac {x}{1+x^2}\)

problem number 76

Added May 27, 2019.

From UMN Math 5587 HW2, Fall 2016, problem 3.

Solve for \(u(x,t)\) with \(x>0,t>0\) and initial conditions not zero \(u(x,1)=\frac {x}{1+x^2}\) \begin {align*} u_t + u_x = 0 \end {align*}

Mathematica


\[\left \{\left \{u(x,t)\to \frac {-t+x+1}{t^2-2 t (x+1)+x^2+2 x+3}\right \}\right \}\]

Maple


\[u \left (x , t\right ) = \frac {-t +x +1}{\left (-t +x +1\right )^{2}+1}\]

Hand solution

Solve \(u_{t}+u_{x}=0\) with initial conditions \(u\left ( x,1\right ) =\frac {x}{1+x^{2}}\). The characteristic equations are\begin {align} \frac {dt}{ds} & =1\tag {1}\\ \frac {dx}{ds} & =1\tag {2}\\ \frac {du}{ds} & =0\tag {3} \end {align}

At \(s=0\) we have \(x\left ( 0\right ) =\xi ,t\left ( 1\right ) =1,u\left ( 0\right ) =\frac {\xi }{1+\xi ^{2}}\).  Solving (1) gives \(t=s+t\left ( 1\right ) =s+1\). Solving (2) gives \(x=s+x\left ( 0\right ) =s+\xi \). From these solutions we solve for \(\xi \), which gives \(\xi =x-s=x-\left ( t-1\right ) \). Hence \[ \xi =x-t+1 \] Equation (3) gives \begin {align*} u & =u\left ( 0\right ) \\ & =\frac {\xi }{1+\xi ^{2}} \end {align*}

Hence \[ u\left ( x,y\right ) =\frac {x-t+1}{1+\left ( x-t+1\right ) ^{2}}\]

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