#### 2.1.70 $$y u_x - x u_y = e^u$$ with $$u(0,y)=y^2-1$$

problem number 70

Characteristics, with IC

Taken from ”the method of Characteristics” by Ryan C. Daileda. Page 16 http://ramanujan.math.trinity.edu/rdaileda/teach/s15/m3357/lectures/lecture_1_22_slides.pdf

Solve for $$u(x,y)$$ \begin {align*} y u_x - x u_y = e^u \end {align*}

With initial conditions $$u(0,y)=y^2-1$$.

Mathematica

$\left \{\left \{u(x,y)\to -\log \left (e^{-x^2-y^2+1}+\tan ^{-1}\left (\frac {x}{\sqrt {y^2}}\right )\right )\right \}\right \}$

Maple

$u \left (x , y\right ) = \ln \left (-\frac {1}{\arctan \left (\frac {x}{y}\right )-{\mathrm e}^{-x^{2}-y^{2}+1}}\right )$

Hand solution

Solve \begin {align*} yu_{x}-xu_{y} & =e^{u}\\ u\left ( 0,y\right ) & =y^{2}-1 \end {align*}

The characteristic equations (using $$s$$ as the parameter) are\begin {align} \frac {dx}{ds} & =y\tag {1}\\ \frac {dy}{ds} & =-x\tag {2}\\ \frac {du}{ds} & =e^{u} \tag {3} \end {align}

With initial point at $$s=0$$ which can be written as $$u\left ( 0,\xi \right ) =\xi ^{2}-1$$. The idea of this method is to use  (1,2) to obtain expressions for $$s,\xi$$. These are the unknowns. Then using these in (3) to obtain the ﬁnal solution $$u\left ( x,y\right )$$. One problem that we see right away, is that (1,2) are coupled. When this happens, we must decouple them ﬁrst. Diﬀerentiating (1) gives $$\frac {d^{2}x}{ds^{2}}=y^{\prime }=-x$$. Hence (1) becomes $$x^{\prime \prime }\left ( s\right ) +x=0$$. This has solution \begin {equation} x\left ( s\right ) =c_{1}\cos s+c_{2}\sin s \tag {4} \end {equation} But from (1), we see that $$x^{\prime }\left ( s\right ) =y$$. Therefore \begin {equation} y\left ( s\right ) =-c_{1}\sin s+c_{2}\cos s \tag {5} \end {equation} We made some progress. Found $$x\left ( s\right ) ,y\left ( s\right )$$. But need to solve for $$c_{1},c_{2}$$. This is done using the initial point, which is always at $$s=0$$. When $$s=0$$, using $$u\left ( 0,\xi \right ) =\xi ^{2}-1$$, which says $$x\left ( 0\right ) =0$$ and $$y\left ( 0\right ) =\xi$$. Using $$x\left ( 0\right ) =0$$ when $$s=0$$ then from (4) we obtain $$0=c_{1}$$. Now the solutions becomes\begin {align*} x\left ( s\right ) & =c_{2}\sin s\\ y\left ( s\right ) & =c_{2}\cos s \end {align*}

Now we use the second initial condition on $$y\left ( s\right )$$ which says $$y\left ( 0\right ) =\xi$$. Hence from the second equation above, $$\xi =c_{2}$$. Therefore the solution now becomes\begin {align} x\left ( s\right ) & =\xi \sin s\tag {6}\\ y\left ( s\right ) & =\xi \cos s \tag {7} \end {align}

This is as far as we can go. Remembering that we are after expressions for $$s$$ and $$\xi$$. Dividing (6/7) gives \begin {align} \frac {x}{y} & =\tan \left ( s\right ) \nonumber \\ s & =\arctan \left ( \frac {x}{y}\right ) \tag {8} \end {align}

Good. We obtained relation for $$s$$ in terms of $$x,y$$. What about $$\xi$$?. By equation (6) and (7) and adding them we obtain\begin {align} x^{2}\left ( s\right ) +y^{2}\left ( s\right ) & =\xi ^{2}\nonumber \\ \xi & =\sqrt {x^{2}+y^{2}} \tag {9} \end {align}

Good. (8,9) is what we wanted. Equations (1,2) have done their job. We used them to ﬁnd $$s,\xi$$. Now we move on to (3) which is $\frac {du}{ds}=e^{u}$ Solving it gives$-e^{-u}=s+C$ But at $$s=0$$ we know that $$u\left ( 0\right ) =\xi ^{2}-1$$. $$-e^{-\left ( \xi ^{2}-1\right ) }=C$$. Then the above becomes$-e^{-u}=s-e^{-\left ( \xi ^{2}-1\right ) }$ We are almost there. We just need now to go back to $$x,y$$ from $$s,\xi$$. By using (8,9) the above becomes\begin {align*} -e^{-u} & =\arctan \left ( \frac {x}{y}\right ) -e^{-\left ( x^{2}+y^{2}-1\right ) }\\ e^{-u} & =-\arctan \left ( \frac {x}{y}\right ) +e^{-\left ( x^{2}+y^{2}-1\right ) } \end {align*}

We can stop here. But if assume $$u>0$$ then the above can be simpliﬁed more \begin {align*} -u & =\ln \left ( -\arctan \left ( \frac {x}{y}\right ) +e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) \\ u\left ( x,y\right ) & =-\ln \left ( -\arctan \left ( \frac {x}{y}\right ) +e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) \\ & =-\ln \left ( -\left ( \arctan \left ( \frac {x}{y}\right ) -e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) \right ) \\ & =\ln \left ( \frac {1}{-\left ( \arctan \left ( \frac {x}{y}\right ) -e^{-\left ( x^{2}+y^{2}-1\right ) }\right ) }\right ) \end {align*}

I used document titled ”The method of Characteristics” by Ryan C. Daileda for help which is a very useful document.

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