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I am trying to solve: $u_x^2 + u_y^2=u$ with boundary conditions: $u(x,0)=x^2$. Unfortunately it leads to equations that makes no sense (sum of squares is $0$ and all constants are $0$). I would be grateful for a reasonable explanation how to solve that with help of characteristics.

J.E.M.S
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2 Answers2

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I don't see how the method of characteristics (as I know) could be applied to an equation that involves the derivatives in a nonlinear way.

Since $u\ge 0$, we can look for $u$ in the form $u=v^2$ which leads to a simplification: by the chain rule, $$u_x^2+u_y^2 = 4v^2(v_x^2+v_y^2)$$ so in terms of $v$ the PDE becomes $$ v_x^2+v_y^2 = \frac14 \tag{1} $$ This is a form of the Eikonal equation, the solution of which is $1/2$ times the distance function from the set $\{v=0\}$.

However, the boundary condition is incompatible with the PDE. Indeed, it asserts that $v(x,0)=x$. But according to (1), $|\nabla v|=1/2$ which implies the Lipschitz bound $$|v(a)-v(b)|\le \frac12|a-b|\tag{2}$$ The Lipschitz bound, being a pointwise inequality, must extend to the boundary values; however, $v(x,0)=x$ does not satisfy (2).

Lutz Lehmann
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  • Many thanks for that great answear, is it automatic way to use that substitution whenewere one is dealing with problem $u_{xx}+u_{yy}=f(u)$ ? The metod i am talking about involves using fallowing considerations: assume we are dealing with $F(x,u,Du)=0$ than we are looking for characteristics satysfing:

    $x'(s)=\nabla_{Du}F$,

    $u'(s)=Du\cdot\nabla_{Du}F$,

    $Du'(s)=-\nabla_{x}F-\nabla_u F\cdot Du$.

     – J.E.M.S Nov 23 '14 at 16:07
  • Substitution of the form $u=g(v)$ often helps get rid of term $u$ in 1st order equations. The choice of $g$ depends on the equation: you can try the general form and see what fits. Also, you wrote $u_{xx}+u_{yy}$ here; this is of second order and the substitution would be less successful. –  Nov 23 '14 at 16:18
  • yea I made a mistake in writting a comment ;) the question is fine. As i understand the link you added the solution is a distance function from the set ${y=0 }$. – J.E.M.S Nov 23 '14 at 16:21
  • That would be the solution if the PDE was $u_x^2+u_y^2=4u$. (Then $u(x,y)=x^2$ satisfies it.) –  Nov 23 '14 at 16:24
  • Hmm you are right, that is a solution. But how can you define $v$ to be the distance from ${v=0}$ ? – J.E.M.S Nov 23 '14 at 16:30
  • The solution of normalized eikonal equation $|\nabla f|=1$ has the property that $f(p)=\pm \operatorname{dist}(p,{f=0})$. This does not mean we know what the solution is, since ${f=0}$ is unknown. But it gives an idea of what the solution might look like. In your case, this information does not help since the boundary condition is incompatible with this structure of solution. –  Nov 23 '14 at 16:34
  • Now everything is clear, many thanks for exact explanation. – J.E.M.S Nov 23 '14 at 16:37
  • @user147263 At first step of your calculus $u\geq 0$ supposes that $u$ is real. This a-priori excludes the complex solutions. The actual wording of the question doesn't specify that $u$ must be real. – JJacquelin Jun 06 '19 at 09:15
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Let $u=v^2$ ,

Then $u_x=2vv_x$

$u_y=2vv_y$

$\therefore(2vv_x)^2+(2vv_y)^2=v^2$ with $v(x,0)=x$

$4v^2(v_x)^2+4v^2(v_y)^2=v^2$ with $v(x,0)=x$

$v_x^2+v_y^2=\dfrac{1}{4}$ with $v(x,0)=x$

$v_y^2=\dfrac{1}{4}-v_x^2$ with $v(x,0)=x$

$v_y=\pm\sqrt{\dfrac{1}{4}-v_x^2}$ with $v(x,0)=x$

$v_{xy}=\mp\dfrac{v_xv_{xx}}{\sqrt{\dfrac{1}{4}-v_x^2}}$ with $v(x,0)=x$

Let $w=v_x$ ,

Then $w_y=\mp\dfrac{ww_x}{\sqrt{\dfrac{1}{4}-w^2}}$ with $w(x,0)=1$

$w_y\pm\dfrac{ww_x}{\sqrt{\dfrac{1}{4}-w^2}}=0$ with $w(x,0)=1$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{dw}{dt}=0$ , letting $w(0)=w_0$ , we have $w=w_0$

$\dfrac{dx}{dt}=\pm\dfrac{w}{\sqrt{\dfrac{1}{4}-w^2}}=\pm\dfrac{w_0}{\sqrt{\dfrac{1}{4}-w_0^2}}$ , letting $x(0)=f(w_0)$ , we have $x=\pm\dfrac{w_0t}{\sqrt{\dfrac{1}{4}-w_0^2}}+f(w_0)=\pm\dfrac{wy}{\sqrt{\dfrac{1}{4}-w^2}}+f(w)$ , i.e. $w=F\left(x\mp\dfrac{wy}{\sqrt{\dfrac{1}{4}-w^2}}\right)$

$w(x,0)=1$ :

$F(x)=1$

$\therefore w=1$

$v_x=1$

$v(x,y)=x+g(y)$

$v_y=g_y(y)$

$\therefore1^2+(g_y(y))^2=\dfrac{1}{4}$

$(g_y(y))^2=-\dfrac{3}{4}$

$g_y(y)=\pm\dfrac{i\sqrt3}{2}$

$g_y=\pm\dfrac{i\sqrt3y}{2}+C$

$\therefore v(x,y)=x\pm\dfrac{i\sqrt3y}{2}+C$

$v(x,0)=x$ :

$C=0$

$\therefore v(x,y)=x\pm\dfrac{i\sqrt3y}{2}$

Hence $u(x,y)=\left(x\pm\dfrac{i\sqrt3y}{2}\right)^2=x^2\pm i\sqrt3xy-\dfrac{3y^2}{4}$

doraemonpaul
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  • I've just started my adventure with PDE but is it a normal situation to look for a complex valued solutions ? – J.E.M.S Nov 23 '14 at 16:10