Surface from equation

Question

I have no experience with this kind of tasks. Could you help?

Find the surface that follows $yu_x + u(x-1)u_y = 0$ and contains the curve $x=2$, $y=t$, $u=2t^2$.

Answer 1

The idea is to cook up a direction on the surface transverse to the curve and evolve it along the PDE to sweep out a surface. I'll briefly sketch it:

Given $yu_x + (x-1)uu_y=0$, so $u_x = (1-x)u\lambda$, $u_y = y\lambda$ for some $\lambda=\lambda(x,y)$. So $$ \begin{align*} \lambda_x y &= u_{yx} = u_{xy} \\ &= (1-x)u_y\lambda+(1-x)u\lambda_y \\ &= (1-x)y\lambda^2 + (1-x)u\lambda_y \end{align*} $$ and hence $$ (y,x-1)\cdot\nabla\lambda = (1-x)\lambda^2 y $$ Ignoring the problematic case $t=0$ for which $(y,x-1)$ is actually tangential to the given curve, the integral curves of $(y,x-1)$ starting at $(x,y)=(2,t)$ has $\sinh s$, $\cosh s$ and some $t$ lying around. Plugging back gives $$ \frac{\partial}{\partial s}\lambda(x(s,t),y(s,t)) =(1-x(s,t))\lambda(x(s,t),y(s,t))^2 y(s,t) $$ which you can solve for $\lambda(x(s,t),y(s,t))$ (easy to determine the value $\lambda(s=0)=4$), hence $$ u_s = u_x x_s + u_y y_s, u(s=0)=2t^2 $$ is a simple.integration.

Answer 2

The standard way so solve a quasilinear first order PDE is to set up the characteristic equations. Given

$$ a(x,y,u)u_x+b(x,y,u)u_y+c(x,y,u)u = 0 $$

we construct the characteristic curves as solutions to the system

$$ \frac{\text dx}{a} = \frac{\text dy}{b} = \frac{\text du}{c} $$

In our case, we have

$$ \frac{\text dx}{y} = \frac{\text dy}{u(x-1)} = \frac{\text du}{0} $$

We interpret $\frac{\text du}{0}$ as $u$ being a constant (informally, $\frac{\text dx}{y} = \frac{\text du}{0} \Rightarrow \frac{\text du}{\text dx} = 0 \Rightarrow u = c_1$). Then, since $u$ is a constant,

$$ \frac{\text dx}{y} = \frac{\text dy}{u(x-1)} \Rightarrow u(x-1)\text dx = y\text dy \Rightarrow u(x^2-2x) - y^2 = c_2 $$

Then, any surface with $g(c_1(x,y,u),c_2(x,y,u))=0$ for any $g$ will satisfy the original equation. Equivalently, we can write $c_1 = f(c_2)$ for any $f$. Plugging in our values, we get

$$ u = f(u(x^2-2x)-y^2) $$

is a family of surfaces which satisfy the PDE. To get the particular one which contains the curve $(x(t),y(t),u(t)) = (2,t,2t^2)$, we convert this to the condition $u(2,y) = 2y^2$ and plug in $x=2$ in our family of surfaces to solve for $f$.

$$ u(2,y) = f(u(2,y)(2^2-2\cdot2)-y^2) \text{ and } u(2,y) = 2y^2 \\ \Rightarrow 2y^2 = (-2)(-y^2) = f(-y^2) \Rightarrow f(\zeta) = -2\zeta $$

Finally, substituting this definition of $f$ into our surface equation gives the solution

$$ u = 2y^2-2u(x^2-2x) \Rightarrow u(x,y) = \frac{2y^2}{2x^2-4x+1} $$

We can check that $u(2,y) = 2y^2$ and that $yu_x(x,y) + u(x,y)(x-1)u_y(x,y) = 0$.