Change variable to solve equation

[evinda]

Hello! (Wave)

I want to solve the equation $xu_t+uu_x=0$ with $u(x,0)=x$. There is a hint to change variables $x \mapsto x^2$.

I have tried the following.

Let $x=y^2$.

Then $u_x=\frac{du}{dy} \frac{dy}{dx}+\frac{du}{dt} \frac{dt}{dx}=u_y \frac{1}{\frac{dx}{dy}}=\frac{1}{2y} u_y$.

Thus, $xu_t+uu_x=0 \Leftrightarrow y^2 u_t+\frac{1}{2y}uu_y=0$. But the latter doesn't help us to use the method of my post [m]Solve equation[/m].

So do we have to make an other change of variables? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

Suppose we do not use the hint?
We can divide the ewuation by x to make it easier can't we? (Wondering)

 

[evinda]

Hey evinda! (Smile)

Suppose we do not use the hint?
We can divide the ewuation by x to make it easier can't we? (Wondering)

Then we have that $u_t+\frac{u}{x} u_x=0$.

$\frac{dx}{dt}=\frac{u}{x}$

$\frac{d}{dt}u(x(t),t)=0$.

Thus, $u(x(t),t)=c$.

The characteristic line that passes through the points $(x,t)$ and $(x_0,0)$ has slope

$\frac{x-x_0}{t-0}=\frac{dx}{dt}=\frac{u}{x}=\frac{x_0}{x}$$x-x_0=t \frac{x_0}{x} \Rightarrow x_0 \left( \frac{t}{x}+1\right)=x \Rightarrow x_0=\frac{x^2}{t+x}$.

So $u(x,t)=\frac{x^2}{t+x}$.

Is everything right?

 

[I like Serena]

Then we have that $u_t+\frac{u}{x} u_x=0$.

$\frac{dx}{dt}=\frac{u}{x}$

$\frac{d}{dt}u(x(t),t)=0$.

Thus, $u(x(t),t)=c$.

The characteristic line that passes through the points $(x,t)$ and $(x_0,0)$ has slope

$\frac{x-x_0}{t-0}=\frac{dx}{dt}=\frac{u}{x}=\frac{x_0}{x}$

This time it's not a line is it?
After all, $\d xt$ isn't constant although $u$ is.
Shouln't we solve $\d xt=\frac ux$, and then make sure that it satisfies the boundary condition? (Wondering)

 

[evinda]

This time it's not a line is it?
After all, $\d xt$ isn't constant although $u$ is.
Shouln't we solve $\d xt=\frac ux$, and then make sure that it satisfies the boundary condition? (Wondering)

So we substitute here $\d xt=\frac ux$ that $u(x(t),t)=x_0$ ?

If so, then is it as follows?

$$\frac{dx}{dt}=\frac{u}{x} \Rightarrow \frac{dx}{dt}=\frac{x_0}{x} \Rightarrow xdx=x_0 dt \Rightarrow \frac{x^2}{2}=x_0 t+c \Rightarrow x^2=2x_0 t+C \Rightarrow 2x_0 t=x^2-C \Rightarrow x_0=\frac{x^2-C}{2t}$$

Thus, $u(x,t)=\frac{x^2-C}{2t}$.

We have that for $t=0$, $u(x,0)=x$. But can that hold? Is it maybe as follows?

Since $\frac{1}{2t} \to \infty$ as $t \to 0$, it has to hold $C=x_0^2$. (Thinking)

 

[I like Serena]

So we substitute here $\d xt=\frac ux$ that $u(x(t),t)=x_0$ ?

If so, then is it as follows?

$$\frac{dx}{dt}=\frac{u}{x} \Rightarrow \frac{dx}{dt}=\frac{x_0}{x} \Rightarrow xdx=x_0 dt \Rightarrow \frac{x^2}{2}=x_0 t+c \Rightarrow x^2=2x_0 t+C \Rightarrow 2x_0 t=x^2-C \Rightarrow x_0=\frac{x^2-C}{2t}$$

Thus, $u(x,t)=\frac{x^2-C}{2t}$.

We have that for $t=0$, $u(x,0)=x$. But can that hold? Is it maybe as follows?

Since $\frac{1}{2t} \to \infty$ as $t \to 0$, it has to hold $C=x_0^2$.

Yes.
It's a bit clearer if we substitute $t=0$ and $x=x_0$ in the earlier expression $2x_0 t=x^2-C$. (Mmm)

 

[evinda]

Yes.
It's a bit clearer if we substitute $t=0$ and $x=x_0$ in the earlier expression $2x_0 t=x^2-C$. (Mmm)

Ok, so then we get that $x_0^2+2 x_0 t-x^2=0$.

$x_0=-t \pm \sqrt{t^2+x^2}$.

Thus, $u(x,t)=-t \pm \sqrt{t^2+x^2}$.

Since $u(x,0)=x$, it follows that $u(x,t)=-t+\sqrt{t^2+x^2}$.

Is this right? Also, do we have to check seperately the case $x=0$ since we have divided at the initial equation by $x$ ? (Thinking)

 

[I like Serena]

Ok, so then we get that $x_0^2+2 x_0 t-x^2=0$.

$x_0=-t \pm \sqrt{t^2+x^2}$.

Thus, $u(x,t)=-t \pm \sqrt{t^2+x^2}$.

Since $u(x,0)=x$, it follows that $u(x,t)=-t+\sqrt{t^2+x^2}$.

Wait! (Wait)
Wouldn't we get that $u(x,0)=|x|$?
That's not correct, is it? (Wondering)

Is this right? Also, do we have to check seperately the case $x=0$ since we have divided at the initial equation by $x$ ? (Thinking)

I believe it suffices to observe that $u$ is well-defined for $x=0$. (Wink)

 

[evinda]

Wait! (Wait)
Wouldn't we get that $u(x,0)=|x|$?
That's not correct, is it? (Wondering)

Oh yes, right... So have I done something wrong? (Thinking)

I believe it suffices to observe that $u$ is well-defined for $x=0$. (Wink)

Why does this suffice? And also how do we deduce it?

 

[I like Serena]

Oh yes, right... So have I done something wrong?

It's just that when x changes sign, the solution changes sign as well.
We can fix it with;
$$u(x,t)=-t+x\sqrt{1+\frac{t^2}{x^2}}$$

Why does this suffice? And also how do we deduce it?

All our functions are continuously differentiable.
Since the solution is valid for $x\ne 0$, it will also be valid for $x=0$ as long as it is defined. (Thinking)

 

[evinda]

It's just that when x changes sign, the solution changes sign as well.
We can fix it with;
$$u(x,t)=-t+x\sqrt{1+\frac{t^2}{x^2}}$$

Ok.

All our functions are continuously differentiable.
Since the solution is valid for $x\ne 0$, it will also be valid for $x=0$ as long as it is defined. (Thinking)

The limit $\lim_{x \to 0} \left( -t+x\sqrt{1+\frac{t^2}{x^2}}\right)$ does not exist.

What does this mean? (Thinking)

 

[I like Serena]

The limit $\lim_{x \to 0} \left( -t+x\sqrt{1+\frac{t^2}{x^2}}\right)$ does not exist.

What does this mean? (Thinking)

Ah yes. It means my solution is not entirely correct. (Blush)
Btw, that limit does exist, doesn't it? It's $0$.
Either way, we should either additionally define $u(0,t)=0$, or we should make it $u(x,t)=-t+\operatorname{sgn}(x)\sqrt{t^2+x^2}$, where $\operatorname{sgn}$ is the sign function. (Thinking)

 

[evinda]

We know that the solution is continuous and we have that $\lim_{x \to 0} \left( -t+ x \sqrt{\frac{t^2}{x^2}+1}\right)=0$, so the solution will be

$u(x,t)=\left\{\begin{matrix}
-t+x \sqrt{\frac{t^2}{x^2}+1}, & x \neq 0\\
0,& x=0.
\end{matrix}\right.$

Right?

 

[I like Serena]

We know that the solution is continuous and we have that $\lim_{x \to 0} \left( -t+ x \sqrt{\frac{t^2}{x^2}+1}\right)=0$, so the solution will be

$u(x,t)=\left\{\begin{matrix}
-t+x \sqrt{\frac{t^2}{x^2}+1}, & x \neq 0\\
0,& x=0.
\end{matrix}\right.$

Right?

Yep! (Nod)

 

[evinda]

Yep! (Nod)

Nice... Thank you! (Smile)