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Solve nonlinear equation

$$\left\{\begin{matrix} -xu_x+uu_y=y & \\ u(x,2x)=0& \end{matrix}\right.$$

using method of characteristic curves

my attempt:

for the pde we can write it as

$\frac{dx}{-x}=\frac{dy}{u}=\frac{du}{y}$

but i cant solve little please..thank you

2 Answers2

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The given partial differential equation is: $$\begin{matrix} -xu_x+uu_y=y & \\ u(x,2x)=0.& \end{matrix}$$

Finding the general solution

Using the method of characteristics we obtain:

$$\dfrac{dx}{-x}=\dfrac{dy}{u}=\dfrac{du}{y} \qquad (1)$$

We solve the right part of the previous equality: $$\dfrac{dy}{u}=\dfrac{du}{y}\implies ydy=udu \implies \frac{1}{2}y^2=\frac{1}{2}u^2-\frac{c_1}{2} \implies u =\pm\sqrt{c_1+y^2} $$

Now, use this in the left part of the equality (1):

$$\dfrac{dx}{-x}=\dfrac{dy}{u}\implies \dfrac{dx}{-x}=\dfrac{dy}{\pm\sqrt{c_1+y^2}} \implies -\ln x =\pm \ln(y+\sqrt{y^2+c_1})-\ln c_2 $$

$$\implies -\ln x = \pm\ln(y+u)-\ln c_2 \implies \ln c_2 = \ln(y+u)^{\pm1}+\ln x \implies c_2 = x\left(y+u \right)^{\pm 1}$$

Now, we know that $c_1=F[c_2]$, in which $F$ is an arbitrary function. Hence we obtain: $$u =\pm\sqrt{c_1+y^2}=\pm\sqrt{F\left[x\left(y+u \right)^{\pm 1}\right]+y^2}$$ $$\implies u^2 = F\left[x\left(y+u \right)^{\pm 1}\right]+y^2$$

Determining the arbitrary function $F$ by using $u(x,2x)=0$

Now, we look as $u(x,2x)=0$: $$\implies 0^2 = F\left[x\left(2x+0 \right)^{\pm 1}\right]+(2x)^2$$ $$\implies 0 = F\left[x\left(2x\right)^{\pm 1}\right]+4x^2.$$

We have to look at both cases $-1$ and $+1$ separately:

  • Case $-1$: $$0 = F\left[\frac{x}{2x}\right]+4x^2 \implies F[2]=-2[2x^2] \implies \text{contradiction as a constant cannot be a function!}$$
  • Case $+1$: $$0 = F[2x^2]+4x^2 \implies F[2x^2]=-2(2x^2) \implies F[u]=-2u, $$hence the solution become $$u^2=-2x(y+u)+y^2 \implies u^2+2xu+2xy-y^2.$$

Now, use the quadratic formula to solve for $u$ to obtain: $$u_{1,2}=\frac{-2x\pm\sqrt{(2x)^2-4(2xy-y^2)}}{2}$$ $$u_{1,2}=-x\pm\sqrt{x^2-(2xy-y^2)}$$ $$u_{1,2}=-x\pm\sqrt{x^2-2xy+y^2}=-x\pm \sqrt{(x-y)^2}=-x\pm|x-y|$$

Investigating the solution

We now have to check which of the solutions do satisfy $u(x,2x)=0$.

  • Case +1: $$u = -x +|x-y| \implies 0 = -x+|x|.$$ If $x>0$ we get $0=-x+x$, but if $x<0$ we get $0=-x-x$. So the solution $u=-x+|x-y|$ is valid for $x>0$.
  • Case -1: $$u = -x -|x-y| \implies 0 = -x-|x|.$$ If $x>0$ we get $0=-x-x$, but if $x<0$ we get $0=-x+x$. So the solution $u=-x-|x-y|$ is valid for $x<0$.
MrYouMath
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  • It is worth noting that this is a non-smooth solution, where the characteristics approach, but do not touch, the discontinuity. Very cool. – Zach Boyd Oct 02 '17 at 18:00
  • You loose the smooth solution $\quad u(x,y)=y-2x$ . – JJacquelin Oct 05 '17 at 08:34
  • @JJacquelin: Isn't your function included in the $u=-x-|x-y|$ case for $x-y>0$? – MrYouMath Oct 05 '17 at 13:43
  • OK. Looking more carefully, the collation of your two partial solutions $u=-x-|x-y|$ case for $x>y$ and $u=-x+|x-y|$ case for $x<y$ is equivalent to the whole solution $u=-2x+y$. The results are consistent from two different calculus. So, the OP can chose the one that suits him best. – JJacquelin Oct 05 '17 at 15:16
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$$-xu_x+uu_y=y $$ $$\dfrac{dx}{-x}=\dfrac{dy}{u}=\dfrac{du}{y}$$ First family of characteristic curves, from $\quad\dfrac{dy}{u}=\dfrac{du}{y}\quad\to\quad u^2-y^2=c_1$

$\dfrac{dy}{u}=\dfrac{du}{y}=\dfrac{dy+du}{u+y}=\dfrac{d(u+y)}{u+y}$

Second family of characteristic curves, from $\quad\dfrac{dx}{-x}=\dfrac{d(u+y)}{u+y} \quad\to\quad (u+y)x=c_2$

General solution on the form of implicit equation : $$(u+y)x=F(u^2-y^2)$$ where $F(X)$ is any differentiable function.

Condition :

$u(x,2x)=0\quad\implies\quad (0+2x)x=F(0^2-(2x)^2) \quad\to\quad 2x^2=F(-4x^2)$ $$F(X)=-\frac{1}{2}X$$ Particular solution according to the boundary condition, with $X=u^2-y^2$ :

$\quad (u+y)x=-\frac{1}{2}(u^2-y^2)\quad$ and after simplification : $\quad x=-\frac{1}{2}(u-y)$ $$u(x,y)=y-2x$$

JJacquelin
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