MHB Verify that the formula provides a solution of the pde

[evinda]

Hello! (Wave)

I want to check by direct differentation that the formula $u(x,t)=\phi(z)$, where $z$ is given implicitly by $x-z=t a(\phi(z))$, does indeed provide a solution of the pde $u_t+a(u) u_x=0$.I have tried the following, but we do not get the desired result. Have I done something wrong?$x-z=ta\phi (z)\Rightarrow \phi (z)=\frac{x-z}{ta}$

$u(x,t)=\phi (z)=\frac{x-z}{ta}$

$u_t=-\frac{x-z}{at^2}$

$u_x=\frac{1}{ta}$

$u_t+a(u)u_x=-\frac{x-z}{at^2}+a\frac{1}{ta}=-\frac{x-z}{at^2}+\frac{at}{t^2a}=\frac{-x+z+at}{at^2} $.

 

[S.G. Janssens]

Hello! (Wave)

I want to check by direct differentation that the formula $u(x,t)=\phi(z)$, where $z$ is given implicitly by $x-z=t a(\phi(z))$, does indeed provide a solution of the pde $u_t+a(u) u_x=0$.I have tried the following, but we do not get the desired result. Have I done something wrong?$x-z=ta\phi (z)\Rightarrow \phi (z)=\frac{x-z}{ta}$

You previously had the defining relation $x - z = ta(\phi(z))$, which is different. I think you need to go back to that relation and apply the implicit function theorem.

$u(x,t)=\phi (z)=\frac{x-z}{ta}$

$u_t=-\frac{x-z}{at^2}$

$u_x=\frac{1}{ta}$

$u_t+a(u)u_x=-\frac{x-z}{at^2}+a\frac{1}{ta}=-\frac{x-z}{at^2}+\frac{at}{t^2a}=\frac{-x+z+at}{at^2} $.

 

[S.G. Janssens]

Hold on, I was going to explain.

 

[S.G. Janssens]

So, I assume that $\phi$ is any smooth function of a real variable.

Define: $F : \mathbb{R^3} \to \mathbb{R}$ by
\[
F(x, t, z) := x - z - t a(\phi(z)).
\]
You want to use the relation $F(x, t, z) = 0$ to define $z$ as a function of $x$ and $t$. By the implicit function theorem, this can be done in a sufficiently small neighbourhood of any point $(x_0, t_0, z_0)$ for which it holds that $F(x_0, t_0, z_0) = 0$ and the derivative $D_zF(x_0, t_0, z_0) \neq 0$. So, near such a point we have $F(x, t, z(x, t)) = 0$, i.e.
\[
x - z(x,t) - t a(\phi(z(x,t))) = 0.
\]
Moreover, by the same theorem, you can find the derivatives of $z$ by implicit differentiation. Can you implicitly differentiate the above equality with respect to $x$ and $t$ to find the partial derivatives of $z$?

Once you have done this, note that $u_x(x,t) = \phi'(z(x,t)) D_xz(x,t)$ and $u_t(x,t) = \phi'(z(x,t)) D_tz(x,t)$, where the partials of $z$ that you just calculated appear again. Substitute for these and see if you can prove that $u$ indeed satisfies the PDE.