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I am solving by the method of characteristics the following equation: $$w_t+\frac{x}{T-t}w_x=0,\; t\in [0,T], x\in [0,1]$$ and some continuous initial data $w(x,0)=w_0(x)$ and I want to estimate the rate of convergence. Since I have a grid and characteristics don't pass through the grid points I have to interpolate. But I more concerned of the stability issue and I hope to show that in $L^2$. Consider only discretization in $t$ for the sake of simplicity. I use energy methods for the equation of the form: $$w_t-a(x)w_x=0, \;t\in[0,T],\; x \in [0,1]$$ \begin{align*} \frac{d||w||^2}{dt}&=(w_t,w)+(w,w_t)=(a(x)w_y,w)+(w,a(x)w_y)\\ &=(a(x)w_x,w)-(w_x,a(x)w)-(w,a_x(x)w)+a(x)ww|^1_0\\ &\leq max_{x\in [0,1]}|a_x(x)|||w||^2+a(x)ww|^1_0 \end{align*} Thus, the energy estimate boils down to the bound of $a_x$ provided boundary conditions are bounded. From the pde $a_x=\frac{1}{T-t}$ which is not bounded as $t \to T$. Therefore, I can only show that $||w(t)||_2\leq C(\epsilon,max[a_x(x)]||w_0||_2$ for $t \in [0,T-\epsilon]$.

However, one can solve the equaation for characteristics and see that $x(t)=\frac{x(0)}{T-t}$ which yields a solution at $t=T$ $w(x,T)\equiv w_0(0)$ and thus another stability estimate $||w(x,T)||_2 \leq K(\epsilon)||w(T-\epsilon,x)||_{\infty}$. Now that I have these two estimates combined, can I claim that I still have stability in $L^2$ and therefore I can expect convergence of the method of characteristics to be in $L_2$? I am confused because $||w(T-\epsilon,x)||_{\infty}$ is not really bounded by anything "from above", so all the estimates on $[0,T-\epsilon]$ get lost once I go to $L_{\infty}$ norm.

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It seems that your characteristic needs another factor of $T$: $x(t) = \frac{Tx(0)}{T-t}$ satisfies $x' = \frac{x}{T-t}$ and begins at $x(0)$. Take initial values $w_0(x) = 1$ in a short interval $0\le x\le\epsilon$, with $w_0$ dropping rapidly to $0$ in $[\epsilon,1]$. Then any solution given by the method of characteristics will be identically 1 in a region which includes $T-\epsilon<t<T$, $0\le x\le1$, because of the shape of the characteristics. This gives $\int_0^1 w(x,t)^2\,dx=1$ when $t>T-\epsilon$, and approximately $\epsilon$ when $t=0$. I don't know all the terminology, but doesn't that rule out $L^2$ stability?

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Yes, I did miss $T$ for the solution. I agree with an estimate and in fact you can show that any $L^p$ norm is $1$ in your example. But I am somewhat confused by the energy estimates as it tells me the norm just keep growing with $t$. What is the intuition then behind it? Energy methods are not applicable in this case or they just not of any use? – Medan Jan 21 '13 at 18:11
You might have assumed that $-(w_x,a(x)w)$ is negative? It doesn't have to be negative. – Bob Terrell Jan 23 '13 at 15:25
No, I did not. I put the absolute value to bound from above. Clearly by the methods of characteristics I have $||w(t)||_{\infty}\leq||w(0)||_{\infty}$, so I have trivially stability in $L^{\infty}$, doesn't it imply stability in $L^2$ as well? – Medan Jan 24 '13 at 1:08

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