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

problem number 76

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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