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Consider the following quasi-linear PDE : $u_t + uu_x = -2u$, with the boundary condition $u(0,t) = e^{-t}$. Show, using the method of characteristics, that the solution to this boundary value problem is $u(x,t) = e^{-2t} / ( x + \sqrt{x^2 + e^{-2t}} )$.

So far I've followed the usual method when solving this and I've obtained that $t = r + s$,

$u = e^{-s-2r}$

$x = e^{-s}(1-e^{-2r})$ / 2,

but I'm not sure where to go from here to find my solution $u(x,t)$. Any help?

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What you need to finish the solution is to eliminate $s$ and $r$, writing them in terms of $t$ and $x$.

Thus \begin{align*}x^2+e^{-2t}&=\frac{1}{4}e^{-2s}(1+e^{-4r}-2e^{-2r})+e^{-2t}\\ &=\frac{1}{4}(e^{-2s}+e^{-2s-4r}-2e^{-2s-2r})+e^{-2t}\\ &=\frac{1}{4}(e^{-2s}+u^2+2e^{-2t}\\ &=\frac{1}{4}e^{-2s}(1+e^{-2r})^2\end{align*}

Hence $\sqrt{x^2+e^{-2t}}=\frac{1}{2}(e^{-s}+u)$. Also $2x=e^{-s}-u$, so eliminating the $s$ variable gives $$u=\sqrt{x^2+e^{-2t}}-x$$ (This is equal to the given solution.)

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  • $\begingroup$ yes, this is what i figured out thankyou! $\endgroup$ – sam Mar 9 at 8:21
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Forming the characteristics :

$$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}x}{u} = \frac{\mathrm{d}u}{-2u}$$

Taking the last two :

$$\frac{\mathrm{d}x}{u} = \frac{\mathrm{d}u}{-2u} \Leftrightarrow -2\mathrm{d}x = \frac{\mathrm{d}u}{u} \implies -2x = u + u_1 \implies u_1 \equiv u + 2x$$

Now, taking the first and the last :

$$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}u}{-2u} \implies t = -\frac{1}{2} \ln u + u_2\implies u_2 \equiv t + \frac{1}{2}\ln u $$

We have finally yield our two integral curves $u_1$ and $u_2$, but we see that none of them can have clear expressions of $u(x,t)$.

The solution to the PDE will be given as a smooth function $F$, such that :

$$F(u_1,u_2) = 0 \implies F\left( u(x,t) + 2x, \frac{1}{2} \ln u(x,t) + t \right)=0$$

Now, let's take account of the boundary condition $u(0,t) = e^{-t}$. It will then be :

$$F\bigg|_{x=0} = F\left( u(0,t) , \frac{1}{2} \ln u(0,t) + t \right)= F \left(e^{-t}, \frac{t}{2} \right) \equiv 0$$

But, since $F$ is a smooth function giving an arbitrary curve $(=0)$ then, can take :

$$\begin{cases} u(x,t) + 2x = e^{-t} \\ \frac{1}{2}\ln u(x,t) + t = \frac{t}{2}\end{cases}$$

I'll leave it up to you to finish it now.

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  • $\begingroup$ @LutzL Thanks for the heads up, mistake ! Corrected now. $\endgroup$ – Rebellos Mar 9 at 8:36
  • $\begingroup$ $u_1,u_2$ are not "integral curves" (what is that?) but constants on the characteristic curves. $\endgroup$ – LutzL Mar 9 at 8:40
  • $\begingroup$ We call the constants on characteristic curves "integral curves/surfaces" in Greece just as a common non-rigorous term. $\endgroup$ – Rebellos Mar 9 at 8:47
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$$u_t+uu_x=-2u$$ Charpit-Lagrange system of characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{-2u}$$ First characteristic equation from $\frac{dt}{1}=\frac{du}{-2u}$ : $$ue^{2t}=c_1$$ Second characteristic equation from $\frac{dx}{u}=\frac{du}{-2u}$ $$u+2x=c_2$$ General solution of the PDE on the form of implicit equation : $$ue^{2t}=F(u+2x)$$ $F$ is an arbitrary function, to be determined according to the specified condition.

Condition : $u(0,t)=e^{-t}$

$(e^{-t})e^{2t}=F(e^{-t}+0)$

Let $X=e^{-t}$ $$e^t=\frac{1}{X}=F(X)$$ So, the function $F(X)$ is determined. We put it into the above general solution where $X=u+2x$ $$ue^{2t}=\frac{1}{u+2x}$$ $$u^2+2xu-e^{-2t}=0$$ Solving for $u$ : $$u=-x\pm\sqrt{x^2+e^{-2t}}$$ The sign is determined according to the condition $u(0,t)=e^{-t}$ : $$u(x,t)=-x+\sqrt{x^2+e^{-2t}}$$ Note :

$-x+\sqrt{x^2+e^{-2t}}= e^{-2t} / ( x + \sqrt{x^2 + e^{-2t}} )$

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  • $\begingroup$ Note that $F$ is only determined for $X>0$, as that is the range of the exponential function. $\endgroup$ – LutzL Mar 9 at 8:38

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