Infinite number of solutions using method of characteristics

Question

I'm currently having a go at a question about the method of characteristics (and struggling!). I want to show that the Cauchy problem $u_{x} + 5x^{4}u_{y} = 2$ with $u(s,s^5)=2s+1$ has infinitely many $C^1(\mathbb{R}^2)$ solutions.

Solving for the characteristic curves gives $x_{1}(t) = t + s_{0}$, $x_{2}(t)=(t+s_{0})^5$ and $z(t) = 2(t_{0}+s_{0}) + 1$. Now since the integral curves cover the graph of $u$, there exists some positive $t_{0}$ such that $(x,y) = (x_{1}(t_{0}),x_{2}(t_{0}))$ = $(t_{0} + s_{0},(t_{0}+s_{0})^5)$.

Hence, $u(x,y)=z(t_{0}) = 2(t_{0}+s_{0}) + 1$. From this, we can clearly see that $u(x,y)=2x+1$ and $u(x,y)=2y^{1/5} + 1$ are solutions of the given PDE problem, but I have no idea how to show that we can find infinitely many solutions. I did see this post, but I don't particularly understand the top comments' solution. Any help is much appreciated!

Answer 1

Note that your second solution doesn't solve the BVP (always check solutions by substitution, when possible!). Apply the method of characteristics

• $\frac{d}{dt} x = 1$, letting $x(0) = s$ we know $x = t + s$,
• $\frac{d}{dt} y = 5x^4$, letting $y(0) = s^5$ we know $y = (t + s)^5$,
• $\frac{d}{dt} u = 2$, letting $u(0) = 2s+1$ we know $u = 2(t + s)+1$.

Solutions $u(x,y)$ are obtained by substitution of $t, s$ in the last equation. This is done by using the two equations of two unknowns given by the expression of $x, y$. However, this system $$ \left\lbrace \begin{aligned} t + s &= x\\ t+s &= \sqrt[5]{y} \end{aligned} \right. $$ has no solution, except along the curve $y - x^5 = 0$ where it has infinitely many solutions.

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Let's derive parameter-free solutions. The Lagrange-Charpit equations $$ \frac{dx}{1} = \frac{dy}{5x^4} = \frac{du}{2} $$ give the characteristic families $y - x^5 = c_1$ and $u - 2x = c_2$. The general solution reads $$ u = 2x + f(y - x^5) $$ in explicit form, where $c_2 = f(c_1)$ involves an arbitrary function $f$. The boundary condition $u(s,s^5) = 2s+1$ imposes $f(0) = 1$, which suggests that infinitely many functions $f$ are suitable.

Answer 2

$$u_x+5x^4u_y=2$$ Charpit-Lagrange system of characteristic ODEs : $$\frac{dx}{1}=\frac{dy}{5x^4}=\frac{du}{2}$$ A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{5x^4}$ : $$y-x^5=c_1$$ A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{2}$ : $$u-2x=c_2$$ The general solution of the PDE on implicit form $c_2=F(c_1)$ is : $$u-2x=F(y-x^5)$$ $$u(x,y)=2x+F(y-x^5)$$ $F$ is an arbitrary function (to be determined if some boundary conditions are correctly specified).

Condition : $\quad u(s,s^5)=2s+1$

$y-x^5=s^5-x^5=0$ $$u(s,s^5)=2s+1=2s+F(s^5-s^5)=2s+F(0)\quad\implies\quad F(0)=1$$

Of course they are an infinity of functions which are equal to $1$ when the argument is $0$. For exemple : $F(s)=\cos(s)$ or $F(s)= 1+s^4$ or $F(s)=e^{3s}$ etc.

As a consequence they are an infinity of solutions of the PDE which satisfies the condition. For example :

$u(x,y)=2x+\cos(y-x^5)$

$u(x,y)=2x+1+(y-x^5)^4$

$u(x,y)=2x+e^{3(y-x^5)}$

etc.