2.1.30 \(u(x+y) u_x+u(x-y) u_y=x^2+y^2\) (Example 3.5.5 in Lokenath Debnath)

problem number 30

Added June 2, 2019.

From example 3.5.5, page 214 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\) \begin {align*} u (x+y) u_x+u (x-y) u_y&=x^2+y^2 \end {align*}

with \(u=0\) on \(y=2 x\)

Mathematica


\begin {align*} & \left \{u(x,y)\to -\sqrt {\frac {2}{7}} \sqrt {2 x^2+3 x y-2 y^2}\right \}\\& \left \{u(x,y)\to \sqrt {\frac {2}{7}} \sqrt {2 x^2+3 x y-2 y^2}\right \}\\& \left \{u(x,y)\to -\sqrt {\frac {2}{7}} \sqrt {2 x^2+3 x y-2 y^2}\right \}\\& \left \{u(x,y)\to \sqrt {\frac {2}{7}} \sqrt {2 x^2+3 x y-2 y^2}\right \}\\ \end {align*}

Maple


\[u \left (x , y\right ) = -\sqrt {-x^{2}+2 x y +y^{2}}\, \sqrt {\frac {2 x y +\mathit {\_F1} \left (-\frac {1}{\sqrt {-x^{2}+2 x y +y^{2}}}\right )}{-x^{2}+2 x y +y^{2}}}\]

Hand solution

Solve \[ u\left ( x+y\right ) u_{x}+u\left ( x-y\right ) u_{y}=x^{2}+y^{2}\] With \(u=0\) on \(y=2x\). This solution follows the book method, but adds more details and more steps to make it more clear.

The following method works only when we can find two equations when the sum of the coefficients is zero. We see that\begin {align} ydx+xdy-udu & =y\left ( u\left ( x+y\right ) \right ) +x\left ( u\left ( x-y\right ) \right ) -u\left ( x^{2}+y^{2}\right ) \nonumber \\ & =yux+y^{2}u+x^{2}u-yxu-ux^{2}-y^{2}u\nonumber \\ & =0\tag {1} \end {align}

But we need one more equation. We see that\begin {align} xdx-ydy-udu & =x\left ( u\left ( x+y\right ) \right ) -y\left ( u\left ( x-y\right ) \right ) -u\left ( x^{2}+y^{2}\right ) \nonumber \\ & =ux^{2}+uxy-yux+y^{2}u-ux^{2}-y^{2}u\nonumber \\ & =0\tag {2} \end {align}

Now we just Integrate (1,2). Rewriting (1) as\begin {align} ydx+xdy-udu & =0\nonumber \\ d\left ( xy-\frac {1}{2}u^{2}\right ) & =0\tag {1A} \end {align}

And rewriting (2) as\begin {align} xdx-ydy-udu & =0\nonumber \\ d\left ( \frac {1}{2}\left ( x^{2}-y^{2}-u^{2}\right ) \right ) & =0\tag {2A} \end {align}

Then integrating (1A,2A) now gives\begin {align} xy-\frac {1}{2}u^{2} & =C_{1}\tag {1C}\\ \frac {1}{2}\left ( x^{2}-y^{2}-u^{2}\right ) & =C_{2}\tag {2C} \end {align}

Since \(u=0\) on \(y=2x\), then the above becomes\begin {align*} x\left ( 2x\right ) & =C_{1}\\ \frac {1}{2}\left ( x^{2}-\left ( 2x\right ) ^{2}\right ) & =C_{2} \end {align*}

Or \begin {align*} 2x^{2} & =C_{1}\\ -\frac {3}{2}x^{2} & =C_{2} \end {align*}

Or \(\frac {C_{1}}{C_{2}}=\frac {2x^{2}}{-\frac {3}{2}x^{2}}=-\frac {4}{3}\). Hence \(-4C_{2}=3C_{1}\). Using this on (1C,2C) gives\begin {align*} -4\left ( \frac {1}{2}\left ( x^{2}-y^{2}-u^{2}\right ) \right ) & =3\left ( xy-\frac {1}{2}u^{2}\right ) \\ -2\left ( x^{2}-y^{2}-u^{2}\right ) & =3xy-\frac {3}{2}u^{2}\\ -2x^{2}+2y^{2}+2u^{2} & =3xy-\frac {3}{2}u^{2}\\ u^{2}\left ( 2+\frac {3}{2}\right ) & =3xy+2x^{2}-2y^{2}\\ \frac {7}{2}u^{2} & =3xy+2x^{2}-2y^{2} \end {align*}

Hence the solution is\[ u\left ( x,y\right ) =\pm \sqrt {\frac {2}{7}\left ( 3xy+2x^{2}-2y^{2}\right ) }\]

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