5
$\begingroup$

I am new to PDE and I am solving the following: $$u_x^2+u_y^2=u^2$$ with the curve $$\Gamma:\cos(s),\sin(s),1.$$ This is what I got so far. First, I set my PDE into the general first-order PDE: $$F(x,y,z,p,q)=p^2+q^2-z^2=0.$$ Then I obtained the following: $$\dfrac{dx}{dt}=F_p=2p$$ $$\dfrac{dy}{dt}=F_q=2q$$ $$\dfrac{dz}{dt}=pF_p+qF_q=2z^2$$ $$\dfrac{dp}{dt}=-F_x-pF_z=2pz$$ $$\dfrac{dq}{dt}=-F_y-qF_q=2qz.$$ By my initial curve $\Gamma,$ I need to find functions $$p=\phi(s)\text{ and }q=\psi(s)$$ such that $$F(f(s),g(s),h(s),\phi(s),\psi(s))=0$$ $$h'(s)=f'(s)\phi(s)+g'(s)\psi(s).$$ Using this relationship, I obtained the following functions: $$\phi(s)=\pm\cos(s)$$ $$\psi(s)=\pm\sin(s)$$ where the signs for my functions have to be the same. First, I am checking when both $\phi(s)$ and $\psi(s)$ have positive sign. I obtained the following solutions for my functions when I solved by system of ODEs above: $$x(s,t)=(2te^{\frac{2t}{1-2t}}+1)\cos(s)$$ $$y(s,t)=(2te^{\frac{2t}{1-2t}}+1)\sin(s)$$ $$z(s,t)=\dfrac{1}{1-2t}$$ $$p(s,t)=\cos(s)e^{\frac{2t}{1-2t}}$$ $$q(s,t)=\sin(s)e^{\frac{2t}{1-2t}}$$ I know that the next step is to solve for $s$ and $t$. Before I do this step, am I on the right track? Did I make any errors, and if so how can I make sure I avoid them when I solve first-order fully nonlinear PDEs in the future?

$\endgroup$
2
  • $\begingroup$ It should be $\frac{dz}{dt} = pF_p + qF_q$. I didn't check your computations but you are on the right track (: $\endgroup$
    – Chee Han
    Feb 18 '17 at 19:55
  • $\begingroup$ I just edited my question to replace $\frac{dz}{dt}$ with $pF_p+qF_q$. Thanks for pointing the error out and clarifying that I am doing this problem correctly! $\endgroup$ Feb 18 '17 at 23:35
1
$\begingroup$

To avoid the need of algebraic manipulation with $s,t$, you can use the folowing method. We have:

$$\frac{-dp}{2pz} = \frac{-dq}{2qz} \implies \frac{dp}{p} = \frac{dq}{q} \implies p = Cq$$

Now plugging into the equation we have that $C^2q^2 + q^2 = z^2 \implies q = \frac{z}{\sqrt{C^2+1}}$ and $p = \frac{Cz}{\sqrt{C^2+1}}$. Then plugging into $dz = pdx + qdy$ and integrating we have:

$$dz = \frac{Cz}{\sqrt{C^2+1}}dx + \frac{z}{\sqrt{C^2+1}}dy$$

$$\frac{dz}{z} = \frac{C}{\sqrt{C^2+1}}dx + \frac{1}{\sqrt{C^2+1}}dy$$

$$\ln|z| = \frac{Cx + y}{\sqrt{C^2+1}} + D$$

$$z = Ae^{\frac{Cx + y}{\sqrt{C^2+1}}}$$

This is the complete integral of the PDE.

Now for the particualr solution we use:

$$\frac{d}{dt}(1) = u_x\frac{dx}{dt} + u_y\frac{dy}{dt}$$

$$0 = C(-\sin t) + \cos t \implies C(t) = \cot t$$

Plugging this with the initial condition in the complete integral after some manipulation we have that

$$1 = A(t)e \implies A(t) = e^{-1}$$

Finally we get $u(x,y) = e^{\cos(t)x + \sin(t)y -1}$, where $t \in [0,2\pi]$. This is a family of solution that passes through the curve. To find a single solution that passes through the curve we need to find the envelope of the family of solutions given by $\phi(x,y,u,A(t),C(t)) = 0$. This happens when $\frac{d\phi}{dt} = 0$ and we get:

$$0 = \frac{d\phi}{dt} = \phi_A \frac{dA}{dt} + \phi_C \frac{dC}{dt} = \frac{x - Cy}{(1+C^2)^{\frac 32}} \implies x = yC$$

Plugging into the family of soltions we get the wanted solution:

$$\boxed{u(x,y) = e^{\sqrt{x^2+y^2} - 1}}$$

REMARK: From the solution we can conclude that if we write it in polar coordinates it will be independent of $\theta$. This prompts you to believe that if we introduce change of variables at the beginning then we would have an easier job. As an exercises maybe can check this.

$\endgroup$
0
$\begingroup$

Just as Stefan4024 remarked, we can find the radial solution without resort to Characteristics method, by setting $u(x,y)=u(r), $ where $r=\sqrt{x^2+y^2}.$ Then the boundary problem is equivalent to \begin{gather*} \begin{cases} u'(r)=\pm u(r),\\ u(1)=1. \end{cases} \end{gather*} Solve it, and we have $u(x,y)=u(r)=e^{\pm(r-1)}=e^{\pm(\sqrt{x^2+y^2}-1)}.$

$\endgroup$
0
$\begingroup$

This problem has nice radial symmetry. We first work out $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}=\frac{\partial u}{\partial r}\frac{x}{r}$$ So our PDE is $$\left(\frac{\partial u}{\partial r}\frac{x}{r}\right)^2+\left(\frac{\partial u}{\partial r}\frac{y}{r}\right)^2=u^2$$ $$\frac{1}{r^2}\left(\frac{\partial u}{\partial r}\right)^2(x^2+y^2)=u^2$$ Since $r=\sqrt{x^2+y^2}$, $$\frac{\partial u}{\partial r}=u$$ $$u(r)=Ae^r \text{ or equivalently }u(x,y)=Ae^{\sqrt{x^2+y^2}}.$$ Boundary conditions follow.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.