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I'm trying to show that the solution $u(x,t) = f'(\frac{x}{t})$ solves the PDE: $u_t + f'(u).u_x = 0 $ on a given domain. The problem also say that $f\in C^1$ with $f(0) = 0$.

My attempt: After substituting the proposed solution, I reached the point: $f''(\frac{x}{t}).(\frac{-x}{t^2}) + f'(u).u_x = f''(\frac{x}{t}).(\frac{-x}{t^2}) + f'(u).f''(\frac{x}{t}).(\frac{1}{t}).$

This means $f'(u)$ should equal to $\frac{x}{t}$, but unfortunately, I do not see that.

Any suggestions would be greatly appreciated.

Thanks

Nitin Uniyal
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BA26
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  • can you please be more clear about the assumptions and what you would like to show? – psl2Z Aug 22 '22 at 21:30
  • @psl2Z We need to show that $u(x,t) = f'(\frac{x}{t})$ solves the given PDE. Basically, what we try to do is, substitute $f'(\frac{x}{t}$ into the left-hand side of the PDE and try to get 0 (that equals the right-hand side). – BA26 Aug 22 '22 at 23:16

2 Answers2

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It is not true: take $f(s) = \frac{1}{3}s^3$. Then $f\in C^{\infty}$, $f(0) = 0$ and $u(x,t) = \frac{x^2}{t^2}$, $u_t(x,t) = -\frac{2x^2}{t^3}$, $f(u)_x(x,t) = \frac{2x^5}{t^6}$ and therefore $u_t + f(u)_x \neq 0$. In general for $f \in C^1$ only, $u$ is not differentiable. Further the assumption $f(0) = 0$ is no real assumption, because $f + c$, $c$ a constant is also fine, since only derivatives play a role.

psl2Z
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We have;

$$\partial_t u+\partial_xf(u)=0$$

$$\partial_t u=-\partial_xf(u)=-f'(u)\cdot\partial_x u$$

Subsituting $u(x,t)= f'({x\over t})$ gives us;

$$f'' \left(\small{x\over t}\right)\cdot \left({-x\over t^2}\right)=-f'(u)\cdot f'' \left(\small{x\over t}\right)\cdot \left({1 \over t}\right)$$

Some quick cancellation yields;

$$\left({x\over t^2}\right)=f'(u)\cdot \left({1 \over t}\right)$$

$$\therefore f'(u)={x\over t}$$

Volk
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  • Thanks for the reply. Unfortunately, here you have assumed LHS = RHS in the first place. Usually, we work (substitute and then simplify) on one side of the eqn. and try to get to the other side. – BA26 Aug 22 '22 at 23:19
  • @BA26 Happy to help. Would you mind telling me what exactly you need for a proper answer? I would be very surprised if you couldn't legitimately make the following assumption; $$a+b=0 \implies a=-b$$ – Volk Aug 23 '22 at 01:09
  • I think my answer in the original post led to confusion. If you plug the candidate solution to the LHS of the PDE, we get $f''(\frac{x}{t}).(\frac{-x}{t^2}) + f'(u).u_x = f''(\frac{x}{t}).(\frac{-x}{t^2}) + f'(u).f''(\frac{x}{t}).(\frac{1}{t}) = f''(\frac{x}{t}). (\frac{1}{t})(-\frac{x}{t} + f'(u))$.

    Now the question is, given $u(x,t) = f(\frac{x}{t})$, can you find an expression for $f'(u)$ explicitly? Hope this makes sense. Thanks!

     – BA26 Aug 23 '22 at 05:55