diff --git a/Project2/LyX/TheoryAndMethods.lyx b/Project2/LyX/TheoryAndMethods.lyx index 4f569760354d0c743292e08e2ed380051c49d3d8..6061c913ddcb2f09f2113b1ea65218ceef986a78 100644 --- a/Project2/LyX/TheoryAndMethods.lyx +++ b/Project2/LyX/TheoryAndMethods.lyx @@ -683,7 +683,7 @@ Rarefaction and Compression wave \end_layout \begin_layout Subsection -Rankine-Hugenoit Condition +Rankine-Hugoniot Condition \end_layout \begin_layout Standard @@ -696,7 +696,7 @@ Consider the PDE \end_inset . - The Rankine-Hugenoit condition yields an, + The Rankine-Hugoniot condition yields an, explicit, expression for the speed of a shock ( \begin_inset Formula $s$ @@ -724,7 +724,7 @@ with \end_layout \begin_layout Proof -The Rankine-Hugenoit Conditon +The Rankine-Hugoniot Conditon \begin_inset Formula $s=\frac{f(u_{l})-f(u_{r})}{u_{l}-u_{r}}$ \end_inset @@ -1008,7 +1008,7 @@ end{tikzpicture} \begin_inset Caption Standard \begin_layout Plain Layout -Schematic Figure of Rankine-Hugenoit condition window +Schematic Figure of Rankine-Hugoniot condition window \end_layout \end_inset @@ -1070,7 +1070,7 @@ and simplify the integrals to \begin{align} 0 & =\int_{x_{1}}^{x_{1}+\Delta x}\underbrace{(u_{l}-u_{r})}_{\text{constant}}dx+\int_{t_{1}}^{t_{1}+\Delta t}\left(f(u_{r})-f(u_{l})\right)dt\\ & =(u_{l}-u_{r})\Delta x+\left(f(u_{r})-f(u_{l})\right)\Delta t\\ -\Leftrightarrow\frac{\Delta x}{\Delta t} & =\frac{f(u_{l})-f(u_{r})}{u_{l}-u_{r}}=s +\Leftrightarrow\frac{\Delta x}{\Delta t} & =\frac{f(u_{l})-f(u_{r})}{u_{l}-u_{r}}=s\label{eq:RH} \end{align} \end_inset @@ -1078,5 +1078,455 @@ and simplify the integrals to \end_layout +\begin_layout Subsection +The Class of Riemann Problems +\end_layout + +\begin_layout Standard +A Riemann problem is a specific problem for the scalar conservation law, + inspected above, + which reads +\begin_inset Formula +\begin{equation} +\begin{cases} +u_{t}+f(u)_{x}=0\\ +u(x,t=0)=u_{0}(x) & =\begin{cases} +u_{l} & ,x<0\\ +u_{r} & ,x>0 +\end{cases} +\end{cases} +\end{equation} + +\end_inset + + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +Add the sketch plot +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +For the Riemann problem there can be found two different classes of solutions +\end_layout + +\begin_layout Enumerate + +\series bold +Similarity Solution +\series default +: + A similarity solution connects the two initial states by a continuous solution +\end_layout + +\begin_layout Enumerate + +\series bold +Shock Solution: + +\series default + A shock solution connects +\begin_inset Formula $u_{l}$ +\end_inset + + and +\begin_inset Formula $u_{r}$ +\end_inset + + by a discontinuous solution, + where the discontinuity moves with the speed s, + derived by the Rankine-Hugoniot condition +\end_layout + +\begin_layout Standard +In the following assume the flux function to be convex. + All derivates can be performed for a concave flux function in the same manner, + but will produce different results. +\end_layout + +\begin_layout Subsubsection +Similarity Solution +\end_layout + +\begin_layout Standard +A continuous similarity solution occurs under the condition +\begin_inset Formula $f'(u_{l})<f'(u_{r})$ +\end_inset + +. + For the Riemann problem the solution is undefined for +\begin_inset Formula $x=0$ +\end_inset + +, + where two different characteristics, + which will not cross. +\begin_inset Formula +\[ +x=f'(u_{l})t\qquad x=f'(u_{r})t +\] + +\end_inset + +In between these two characteristics, + there is one more valid characteristic, + namely +\begin_inset Formula $x=kt$ +\end_inset + + with +\begin_inset Formula $f'(u_{l})<k<f'(u_{r})$ +\end_inset + + for +\begin_inset Formula $k=\frac{x}{t}$ +\end_inset + +. + The scalar solution is of the form +\begin_inset Formula +\begin{equation} +u(x,t)=\begin{cases} +u_{l} & ,\frac{x}{t}\leq f'(u_{l})\\ +v(\frac{x}{t}) & ,f'(u_{l}<x<f'(u_{r})\\ +u_{r} & ,\frac{x}{t}\geq f'(u_{r}) +\end{cases}\label{eq:SimilaritySolutionAbstract} +\end{equation} + +\end_inset + +The arising question is how to calculate +\begin_inset Formula $v(\frac{x}{t})=u(x,t)$ +\end_inset + +. + For this, + insert the partial derivatives +\begin_inset Formula +\begin{align} +u_{t}(x,t) & =\frac{\partial}{\partial t}v\left(\frac{x}{t}\right)=v'\left(\frac{x}{t}\right)\frac{\partial}{\partial t}\left(\frac{x}{t}\right)=\frac{-x}{t^{2}}v'\left(\frac{x}{t}\right)\\ +u_{x}(x,t) & =\frac{\partial}{\partial x}v\left(\frac{x}{t}\right)=v'(\left(\frac{x}{t}\right)\frac{\partial}{\partial x}\left(\frac{x}{t}\right)=\frac{1}{t}v'\left(\frac{x}{t}\right)\\ +f(u)_{x} & =f'(u)u_{x}=f'\left(v\left(\frac{x}{t}\right)\right)v'\left(\frac{x}{t}\right)\frac{1}{t} +\end{align} + +\end_inset + + into the partial differential equation +\begin_inset Formula +\begin{align} +0 & =u_{t}+f(u)_{x}\\ + & =\frac{-x}{t^{2}}v'\left(\frac{x}{t}\right)+f'\left(v\left(\frac{x}{t}\right)\right)v'\left(\frac{x}{t}\right)\frac{1}{t}\quad\bigg|\cdot t\\ + & =v'\left(\frac{x}{t}\right)\left[\frac{-x}{t}+f'\left(v\left(\frac{x}{t}\right)\right)\right] +\end{align} + +\end_inset + +This equation can be fulfilled in two ways, + either let +\begin_inset Formula $v'\left(\frac{x}{t}\right)=0$ +\end_inset + +. + However, + this would imply +\begin_inset Formula $v\left(\frac{x}{t}\right)=u(x,t)$ +\end_inset + + to be constant and can only be valid if +\begin_inset Formula $u_{l}=u_{r}$ +\end_inset + +. + The second possibility to fulfill the equation is to ask for +\begin_inset Formula +\begin{align} +\frac{-x}{t}+f'\left(v\left(\frac{x}{t}\right)\right) & =0\\ +\Leftrightarrow f'\left(v\left(\frac{x}{t}\right)\right) & =\frac{x}{t}\\ +\Leftrightarrow\left(f'\right)^{-1}\left(\frac{x}{t}\right) & =\left(f'\right)^{-1}\left(f\left(v\left(\frac{x}{t}\right)\right)\right)=v\left(\frac{x}{t}\right)\label{eq:v_x/t} +\end{align} + +\end_inset + +Where +\begin_inset Formula $\left(f'\right)^{-1}$ +\end_inset + + is the inverse of the derivative of the flux function. + Note that this inverse exists, + as the flux function is convex and therefore the second derivative is positive. + Now, + insert the solution +\begin_inset CommandInset ref +LatexCommand ref +reference "eq:v_x/t" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + into the general solution +\begin_inset CommandInset ref +LatexCommand ref +reference "eq:SimilaritySolutionAbstract" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + to get the solution +\begin_inset Formula +\begin{equation} +u(x,t)=\begin{cases} +u_{l} & ,\frac{x}{t}\leq f'(u_{l})\\ +\left(f'\right)^{-1}\left(\frac{x}{t}\right) & ,f'(u_{l})<\frac{x}{t}<f'(u_{r})\\ +u_{r} & ,\frac{x}{t}\geq f'(u_{r}) +\end{cases} +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Shock Solution +\end_layout + +\begin_layout Standard +Contrary to the similarity solution, + the shock solution yields a discontinuous solution to the Riemann problem under the condition +\begin_inset Formula $f'(u_{l})>f'(u_{r})$ +\end_inset + +. + For the shock solution two characteristics cross, + yielding a discontinuous solution with a jump. + This jump moves with the shock speed +\begin_inset Formula $s$ +\end_inset + +, + derived by the Rankine-Hugoniot condition (equation +\begin_inset CommandInset ref +LatexCommand ref +reference "eq:RH" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +). + The final solution reads +\begin_inset Formula +\begin{equation} +u(x,t)=\begin{cases} +u_{l} & ,\frac{x}{t}\leq s\\ +u_{r} & ,x<\frac{x}{t} +\end{cases}\qquad\text{with: }s=\frac{f(u_{l})-f(u_{r})}{u_{l}-u_{r}} +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Weak Solutions +\end_layout + +\begin_layout Standard +Weak solutions can simplify mathematical problems by introducing some test function +\begin_inset Formula $\Phi$ +\end_inset + +. + The original problem will be multiplied by the test function and integrated over the domain +\begin_inset Formula $\Omega$ +\end_inset + +. + Weak solutions have two important properties +\end_layout + +\begin_layout Enumerate + +\series bold +No Derivatives: + +\series default + Weak solutions involve no derivatives in +\begin_inset Formula $u$ +\end_inset + + and +\begin_inset Formula $f(u)$ +\end_inset + + as they are moved to the test function +\end_layout + +\begin_layout Enumerate + +\series bold +Solution Space: + +\series default + The solution space of weak solutions is much larger then the solution space for strong solutions. + Weak solutions can even include more then one solution for a given problem +\end_layout + +\begin_layout Standard +For a weak solution consider a smooth test function +\begin_inset Formula $\Phi(x,t)$ +\end_inset + + with compact support (the test function is zero outside some finite box). + To derive the weak form of the problem multiply the PDE with the test function and integrate over the domain +\begin_inset Formula $\Omega=\underbrace{[x_{1},x_{2}]}_{\mathcal{R}}\times\underbrace{[t_{1},t_{2}]}_{\mathcal{R}^{+}}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +The weak form then reads: + Find +\begin_inset Formula $u$ +\end_inset + + s.t. + +\begin_inset Formula $\forall\Phi\in C_{0}^{1}(\mathcal{R}\times\mathcal{R}^{+})$ +\end_inset + +: +\begin_inset Formula +\begin{align} +0 & =\iint_{\Omega}\left[u_{t}+f(u)_{x}\right]\Phi(x,t)d\Omega\\ + & =\int_{0}^{\infty}\int_{-\infty}^{\infty}\left(u_{t}\Phi(x,t)+f(u)_{x}\Phi(x,t)\right)dxdt\\ + & =\int_{0}^{\infty}\int_{-\infty}^{\infty}\left(u\Phi_{t}(x,t)+f(u)\Phi_{x}(x,t)\right)dxdt+\int_{-\infty}^{\infty}u_{0}(x)\Phi(x,t=0)dx\label{eq:WeakFormAbstract} +\end{align} + +\end_inset + +Note that in the last step integration by parts and the property that the test function vanishes on the boundaries was utilized. + Any function +\begin_inset Formula $u(x,t)$ +\end_inset + + that fulfills the weak form (equation +\begin_inset CommandInset ref +LatexCommand ref +reference "eq:WeakFormAbstract" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +) is a weak solution to the initial given problem. + This yields another problem, + as the solution +\begin_inset Formula $u$ +\end_inset + + is not necessarily unique. + To determine if a weak solution is the correct solution to a specific problem, + the (Lax) entropy condition is introduced. +\end_layout + +\begin_layout Definition +Lax Entropy Condition +\end_layout + +\begin_layout Definition +A weak shock solution is a strong solution for a problem, + if, + and only if, + the shock satisfies +\begin_inset Formula +\begin{equation} +f'(u_{l})>s>f'(u_{r}) +\end{equation} + +\end_inset + + with the shock speed +\begin_inset Formula $s$ +\end_inset + +. + Conclude from this +\end_layout + +\begin_layout Enumerate + +\series bold +Jump Solution: + +\series default + A jump solution is a valid solution if +\begin_inset Formula $u_{l}>u_{r}$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Enumerate +The flux function is convex and yields +\begin_inset Formula $f'(u_{l})>f'(u_{r})$ +\end_inset + +. + Conclude +\begin_inset Formula $f'(u_{l})>s>f'(u_{r})$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Enumerate + +\series bold +Similarity Solution: + +\series default + The similarity solution is a valid solution if +\begin_inset Formula $u_{l}<u_{r}$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Enumerate +The flux function is convex and yields +\begin_inset Formula $f'(u_{l})<f'(u_{r})$ +\end_inset + +. + Conclude that no discontinuity of the form +\begin_inset Formula $f'(u_{l})>s>f'(u_{r})$ +\end_inset + + can occur. +\end_layout + +\end_deeper \end_body \end_document diff --git a/Project2/LyX/main.lyx b/Project2/LyX/main.lyx index eeb67936037136a6363e7b4d5d90157f104335f0..e7bdad92438b7555cb6a762664a01e3724753931 100644 --- a/Project2/LyX/main.lyx +++ b/Project2/LyX/main.lyx @@ -5,6 +5,9 @@ \save_transient_properties true \origin unavailable \textclass IEEEtran-CompSoc +\begin_preamble +\usepackage{tikz} +\end_preamble \use_default_options true \begin_modules eqs-within-sections @@ -152,7 +155,7 @@ figs-within-sections \color background background \end_branch \branch List of Variables -\selected 1 +\selected 0 \filename_suffix 0 \color background background \end_branch