Welcome to our community

Be a part of something great, join today!

[SOLVED] Change variable to solve equation

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
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)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,836

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,836