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How to solve $f\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}=0$? (where $f(x,t)$ is assumed to be in $C^\infty(\mathbb{R}\times\mathbb{R^+}\rightarrow\mathbb{R})$)

I can find a particular solution which is $f=\frac{x}{t}$. Is this the only solution? If not how can I find all the other solution?

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  • $\begingroup$ Anyway, the Method of Characteristics seems promising. You should try it out! $\endgroup$ – Berrick Caleb Fillmore Apr 18 '15 at 4:50
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    $\begingroup$ Using the Method of Characteristics, you get solutions of the form $$ f(x,t) = \dfrac{x - x_{0}}{t - t_{0}}, $$ where $ (x_{0},t_{0}) \in \mathbb{R} \times (- \infty,0] $ (if you require that $ f \in {C^{\infty}}(\mathbb{R} \times (0,\infty)) $). $\endgroup$ – Berrick Caleb Fillmore Apr 18 '15 at 5:08
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Note that if $f(x,t)$ satisfies the given condition, then $f(x-a,t-b)$ also satisfies the given condition.

Along the curve $x(t)$, $$ \frac{\mathrm{d}f}{\mathrm{d}t}=\frac{\mathrm{d}x}{\mathrm{d}t}\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}\tag{1} $$ the given condition implies that $f$ will remain constant on curves where $$ \frac{\mathrm{d}x}{\mathrm{d}t}=f\tag{2} $$ Suppose $f(x_0,t_0)=a$. Then $f(x,t)=a$ on the line where $\frac{x-x_0}{t-t_0}=a$.

Suppose we know that $f(x,0)=\phi(x)$, then $$ f(x+\phi(x)t,t)=\phi(x)\tag{3} $$


If we use $\phi(x)=x$ in $(3)$, we get $$ f(x+xt,t)=x\implies f(x,t)=\frac x{t+1}\tag{4} $$ which is a translate of your function $f(x,t)=\frac xt$.


If we use $\phi(x)=x^2$ in $(3)$, we get $$ f(x+x^2t,t)=x^2\implies f(x,t)=\left(\frac{-1+\sqrt{1+4xt}}{2t}\right)^2\tag{5} $$ Thus, we can generate different functions $f$ given different functions $\phi$.


O.L. mentions that the equation $$ f\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}=0\tag{6} $$ is called the Inviscid Burgers' Equation.

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  • $\begingroup$ Sorry for my misunderstood edit. It might be good to clarify that $\frac{dx}{dt}$ is a slope, since the inexperienced among us may get confused trying to interpret it as an ordinary derivative in context. $\endgroup$ – Slade Apr 18 '15 at 6:30
  • $\begingroup$ @Slade: At the expense of a very slight bit of generality, I have gotten rid of the auxiliary variable $s$. I hope it gets rid of the confusion. $\endgroup$ – robjohn Apr 18 '15 at 6:40
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    $\begingroup$ @robjohn +1. It might be helpful to mention that this is called the inviscid Burgers equation. $\endgroup$ – Start wearing purple Apr 18 '15 at 7:19
  • $\begingroup$ @O.L.: Thanks! I hadn't known that before. $\endgroup$ – robjohn Apr 18 '15 at 7:28

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