Welcome to our community

Be a part of something great, join today!

[SOLVED] Solve equation

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Hello!!! (Wave)

I want to solve the equation $u_t+u^2u_x=0$ with $u(x,0)=2+x$.

I have tried the following:

The characteristic curves for $u_t+u^2 u_x=0$ are the solutions of the ode $\frac{dx}{dt}=u^2$.

We have that $\frac{d}{dt}u(x(t),t)=0$, implying that $u(x(t),t)=c$.

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

$\frac{x-z}{t-0}=\frac{dx}{dt}=u^2(x,t)=u^2(z,0)=(2+z)^2$.

Thus, $x-z=t(2+z)^2 \Rightarrow x-z=t(4+4z+z^2)=4t+4tz+tz^2 \Rightarrow z^2t+(4t+1)z+4t-x=0$.

The discriminant is $\Delta=(4t+1)^2-4t(4t-x)=4tx+8t+1$.

$z_{1,2}=\frac{-(4t+1) \pm \sqrt{4tx+8t+1}}{2t}$ for $t \neq 0$ and $4tx+8t+1 \geq 0$.

Thus $u^2(x,t)=\left( 2+\frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$.

We want $(2+x)^2=u^2(x,0)=\lim_{t \to 0} \left( 2+\frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$.

When having the + sign, the limit is $+\infty$, so $u$ could not be a solution.

Also, we have that $\lim_{t \to 0} \left( 2+\frac{[-(4t+1) - \sqrt{4tx+8t+1}]}{2t}\right)=-2-x$.

So, $u^2(x,t)=\left( 2+\frac{[-(4t+1)- \sqrt{4tx+8t+1}]}{2t}\right)^2 $.

So $u(x,t)= \pm \left| 2- \frac{(4t+1)+ \sqrt{4tx+8t+1}}{2t}\right|$.

Is everything right or have I done something wrong? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Thus $u^2(x,t)=\left( 2+\frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$.
Hey evinda!! (Smile)

Can't we just use the boundary condition here?
That is, not square it? (Wondering)

We want $(2+x)^2=u^2(x,0)=\lim_{t \to 0} \left( 2+\frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$.

When having the + sign, the limit is $+\infty$, so $u$ could not be a solution.
If I fill in t=0, doesn't the numerator become 0, so that the result won't be infinity? (Wondering)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Hey evinda!! (Smile)

Can't we just use the boundary condition here?
That is, not square it? (Wondering)
You mean that we get that $u^2(x,t)=\left( 2+ \frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$ and from there that

$u(x,t)= \pm \left| 2+ \frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right|$ and that we apply now the initial condition?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
If so, is it right as follows?


$u(x,t)=\left\{\begin{matrix}
\pm \left| 2-\frac{(4t+1)-\sqrt{4tx+8t+1}}{2t}\right| & \\
\pm \left| 2-\frac{(4t+1)+\sqrt{4tx+8t+1}}{2t}\right| & , \text{ we reject it since it gets infinite when } t \to 0.
\end{matrix}\right.$

Then we get that $\lim_{t \to 0} \frac{(4t+1)-\sqrt{4tx+8t+1}}{2t}=-x$.

Since $u(x,0)=2+x$, we get that $u(x,t)=\left| 2-\frac{(4t+1)-\sqrt{4tx+8t+1}}{2t}\right|$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
You mean that we get that $u^2(x,t)=\left( 2+ \frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right)^2$ and from there that

$u(x,t)= \pm \left| 2+ \frac{[-(4t+1) \pm \sqrt{4tx+8t+1}]}{2t}\right|$ and that we apply now the initial condition?
Ah well, I think it doesn't really matter.
I can mostly suggest to leave out all the absolute value symbols.
After all, instead of writing $u^2=y^2 \Rightarrow u=\pm| y|$, isn't it easier to write $u=\pm y$? (Wondering)

If so, is it right as follows?

$u(x,t)=\left\{\begin{matrix}
\pm \left| 2-\frac{(4t+1)-\sqrt{4tx+8t+1}}{2t}\right| & \\
\pm \left| 2-\frac{(4t+1)+\sqrt{4tx+8t+1}}{2t}\right| & , \text{ we reject it since it gets infinite when } t \to 0.
\end{matrix}\right.$

Then we get that $\lim_{t \to 0} \frac{(4t+1)-\sqrt{4tx+8t+1}}{2t}=-x$.

Since $u(x,0)=2+x$, we get that $u(x,t)=\left| 2-\frac{(4t+1)-\sqrt{4tx+8t+1}}{2t}\right|$.
Shouldn't it be:
$$u(x,t)=2-\frac{(4t+1)-\sqrt{4tx+8t+1}}{2t} = \frac{\sqrt{1+4t(2+x)}-1}{2t}$$
Otherwise I think the boundary condition does not necessarily hold. (Thinking)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Ah well, I think it doesn't really matter.
I can mostly suggest to leave out all the absolute value symbols.
After all, instead of writing $u^2=y^2 \Rightarrow u=\pm| y|$, isn't it easier to write $u=\pm y$? (Wondering)
Yes, right... (Nod)

Shouldn't it be:
$$u(x,t)=2-\frac{(4t+1)-\sqrt{4tx+8t+1}}{2t} = \frac{\sqrt{1+4t(2+x)}-1}{2t}$$
Otherwise I think the boundary condition does not necessarily hold. (Thinking)
I see... And then we just verify that $\lim_{t \to 0} \frac{\sqrt{1+4t(2+x)}-1}{2t}=x+2$ so that we are sure that the initial condition is satisfied, right? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
Yes, right... (Nod)

I see... And then we just verify that $\lim_{t \to 0} \frac{\sqrt{1+4t(2+x)}-1}{2t}=x+2$ so that we are sure that the initial condition is satisfied, right? (Thinking)
Indeed. (Nod)
 

evinda

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