I want to show that the following PDE of the function $u(x,y)$ has infinitely many solutions: $$ \left\{ \begin{array}{c} u_x+u_y=2xu \\ u(x,x)=e^{x^2} \\ \end{array} \right. $$

By using the method of characteristics and choosing a curve $\Gamma(r,r,e^{r^2})$ in $u(x(r,s),y(r,s))$, I get the characteristic curve $(x(r,s),y(r,s),z(r,s))=(s+r,s+r,e^{(s+r)^2})$

I notice: $x(r,s)=y(r,s)$ $\implies$ $u(x,y)=z(r(x,y),s(x,y))=e^{x^2}=e^{y^2}$. However, this solution that I find is unique. I seem to be missing something to show that there are infinite solutions. Any tips?


3 Answers 3


$$u_x+u_y=2xu $$ The characteristic curves are solution of the differential equations : $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{2xu}$$ From $dx=dy$ a first family of characteristic curves is $\quad y-x=c_1$

From $dx=\frac{du}{2xu}$ a second family of characteristic curves is $\quad ue^{-x^2}=c_2$

The general solution of the PDE expressed on the form of implicit equation is : $$\Phi\left((y-x)\:,\:(ue^{-x^2})\right)=0$$ or, on explicit form : $$ue^{-x^2}=f(y-x) \quad\to\quad u=e^{x^2}f(y-x)$$ where $f$ is any differentiable function.

With the condition $u(x,x)=e^{x^2}=e^{x^2}f(x-x)=e^{x^2}f(0)\quad\implies\quad f(0)=1$

The solutions are : $$u(x,y)=e^{x^2}f(y-x)\quad \text{any function }f \text{ having the property }f(0)=1$$ Since they are an infinity of functions which have the property $f(0)=1$, this proves that they are an infinity of solutions for the PDE with condition $\begin{cases} u_x+u_y=2xu \\ u(x,x)=e^{x^2} \end{cases} $

EXAMPLE of solutions :

With $f(X)=C\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}$

With $f(X)=CX\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}(y-x)$

With $f(X)=CX^b\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}(y-x)^b$

With $f(X)=C\sin(X)\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}\sin(y-x)$

With $f(X)=Ce^{-bX^2}\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}e^{-b(y-x)^2}$

An so on ...

One see that they are an infinity of examples, many are easy to find. And all linear combinations of those solutions.

  • $\begingroup$ I would not call $y-x=c_1$ and $ue^{-x^2}=c_2$ characteristic curves. They are level sets of the corresponding first integrals of the PDE. Their intersections (take a level set from the first family, and a level set from the second family) are the characteristic curves. $\endgroup$
    – Olod
    Feb 12, 2017 at 13:15

Consider the Ansatz $u(x,y) = \exp\left(ax^2 + bxy + cy^2\right)$, then $a,b,c$ must satisfy \begin{align} u_x + u_y &= \left( 2ax + by \right) \exp\left( a^2 + bxy + cy^2 \right) + \left( bx + 2cy \right) \exp\left( ax^2 + bxy + cy^2 \right) \\ &=\left( 2a + b \right)x u(x,y) + \left( 2c +b \right) y u(x,y) \end{align} Thus $2a + b = 2$ and $2c + b = 0$, so $a = 1 - \frac{b}{2}$ and $ c =- \frac{b}{2}$.

This also gives $a + b + c = 1 - \frac{b}{2} + b - \frac{b}{2} = 1$, or \begin{align} u(x,x) = \exp\left(x^2 \right) \end{align}

So for all $b \in \mathbb{R}$ another solution is \begin{align} u(x,y) = \exp\left(x^2 \right) \exp\left(- \frac{b}{2}x^2 + b xy - \frac{b}{2} y^2\right) = \exp\left(x^2 \right) \exp\left(- \frac{b}{2} \left(x - y \right)^2\right) \end{align}


The method of characteristics for first-order quasilinear equations is not at its best, when the curve involved in a Cauchy problem is a characteristic curve (as is the case for your PDE). Try instead to show that the general solution of the PDE (treated as a semilinear one) is $$ u(x,y) =f(x-y) e^{x^2} $$ where $f$ is a continuously differentialbe function. Even simpler, to get the general solution, you can use the change of variables described here to transform your equation to a PDE reducible to an ODE.

It follows that there are infinitely many solutions of your Cauchy problem, say $$ u(x,y) =( (x-y)^n +1) e^{x^2} $$ where $n \in \mathbf N.$ (I'd recommend my book to see how to treat first-order semilinear and quasilinear PDEs.)


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