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[doc] Add Navier Stokes inverse problem and Update Dimensional Analysis #16

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[doc] Add Navier Stokes inverse problem and Update Dimensional Analysis #16

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a22453e
Fix navier stokes
Routhleck Dec 18, 2024
ced7922
Fix bugs
Routhleck Dec 18, 2024
1e074ef
Update Navier_Stokes_inverse.ipynb
Routhleck Dec 18, 2024
72bd120
Update Navier_Stokes_inverse.ipynb
Routhleck Dec 18, 2024
eea7bc8
Update Dimensional Analysis
Routhleck Dec 18, 2024
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94 changes: 93 additions & 1 deletion docs/unit-examples-forward/Burgers_RAR.ipynb
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Expand Up @@ -22,7 +22,99 @@
"$$\n",
"u(-1,t)=u(1,t)=0, \\quad u(x,0) = - \\sin(\\pi x).\n",
"$$\n",
"\n"
"\n",
"## Dimensional Analysis\n",
"\n",
"### Step 1: Assign Dimensions to Variables\n",
"\n",
"1. **Spatial Coordinate $x$:**\n",
" - The dimension of $x$ is length:\n",
" $$\n",
" [x] = L.\n",
" $$\n",
"\n",
"2. **Time $t$:**\n",
" - The dimension of time is:\n",
" $$\n",
" [t] = T.\n",
" $$\n",
"\n",
"3. **Velocity $u$:**\n",
" - Velocity has dimensions of length per unit time:\n",
" $$\n",
" [u] = L / T.\n",
" $$\n",
"\n",
"4. **Viscosity $\\nu$:**\n",
" - The term $\\nu \\frac{\\partial^2 u}{\\partial x^2}$ involves the second spatial derivative of velocity, which must have the same dimensions as the time derivative $\\frac{\\partial u}{\\partial t}$.\n",
"\n",
"---\n",
"\n",
"### Step 2: Analyze the Dimensions of Each Term\n",
"\n",
"1. **Time Derivative Term:**\n",
" - The time derivative $\\frac{\\partial u}{\\partial t}$ has dimensions:\n",
" $$\n",
" \\left[\\frac{\\partial u}{\\partial t}\\right] = \\frac{[u]}{[t]} = \\frac{L / T}{T} = \\frac{L}{T^2}.\n",
" $$\n",
"\n",
"2. **Advection Term:**\n",
" - The advection term $u \\frac{\\partial u}{\\partial x}$ involves the spatial derivative of velocity:\n",
" $$\n",
" \\left[u \\frac{\\partial u}{\\partial x}\\right] = [u] \\cdot \\frac{[u]}{[x]} = \\frac{L}{T} \\cdot \\frac{L / T}{L} = \\frac{L}{T^2}.\n",
" $$\n",
"\n",
"3. **Diffusion Term:**\n",
" - The diffusion term $\\nu \\frac{\\partial^2 u}{\\partial x^2}$ involves the second spatial derivative of velocity:\n",
" $$\n",
" \\left[\\frac{\\partial^2 u}{\\partial x^2}\\right] = \\frac{[u]}{[x]^2} = \\frac{L / T}{L^2} = \\frac{1}{L T}.\n",
" \n",
" $$\n",
" - Therefore, the diffusion term has dimensions:\n",
" $$\n",
" \\left[\\nu \\frac{\\partial^2 u}{\\partial x^2}\\right] = [\\nu] \\cdot \\frac{1}{L T} = \\frac{L}{T^2}.\n",
" $$\n",
"\n",
"---\n",
"\n",
"### Step 3: Determine the Dimensions of $\\nu$\n",
"\n",
"- The diffusion term $\\nu \\frac{\\partial^2 u}{\\partial x^2}$ must have the same dimensions as the time derivative $\\frac{\\partial u}{\\partial t}$:\n",
" $$\n",
" [\\nu] \\cdot \\frac{1}{L T} = \\frac{L}{T^2} \\implies [\\nu] = \\frac{L^2}{T}.\n",
" $$\n",
"- Therefore, the viscosity $\\nu$ has dimensions of kinematic viscosity:\n",
" $$\n",
" [\\nu] = \\frac{L^2}{T}.\n",
" $$\n",
"\n",
"---\n",
"\n",
"### Step 4: Summary of Dimensions\n",
"\n",
"| Variable/Parameter | Physical Meaning | Dimensions |\n",
"|------------------------|-----------------------------------|-----------------------|\n",
"| $x$ | Spatial coordinate | $L$ |\n",
"| $t$ | Time | $T$ |\n",
"| $u$ | Velocity | $L / T$ |\n",
"| $\\nu$ | Kinematic viscosity | $L^2 / T$ |\n",
"\n",
"---\n",
"\n",
"### Step 5: Initial and Boundary Conditions\n",
"\n",
"1. **Boundary Conditions:**\n",
" - The boundary conditions $u(-1,t) = u(1,t) = 0$ are given in meters per second:\n",
" $$\n",
" [u(-1,t)] = [u(1,t)] = L / T.\n",
" $$\n",
"\n",
"2. **Initial Condition:**\n",
" - The initial condition $u(x,0) = -\\sin(\\pi x)$ is given in meters per second:\n",
" $$\n",
" [u(x,0)] = L / T.\n",
" $$\n",
" - The term $\\sin(\\pi x)$ is dimensionless because $x$ is in meters, and $\\pi$ is a dimensionless constant."
]
},
{
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127 changes: 126 additions & 1 deletion docs/unit-examples-forward/Laplace_disk.ipynb
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Expand Up @@ -24,7 +24,132 @@
"y(r, \\theta +2\\pi) = y(r, \\theta).\n",
"$$\n",
"\n",
"The reference solution is $y=r\\cos(\\theta)$."
"The reference solution is $y=r\\cos(\\theta)$.\n",
"\n",
"# Dimensional Analysis for the Laplace Equation on a Disk\n",
"\n",
"## Problem Setup\n",
"\n",
"We will solve the Laplace equation in a polar coordinate system:\n",
"\n",
"$$\n",
"r\\frac{dy}{dr} + r^2\\frac{d^2y}{dr^2} + \\frac{d^2y}{d\\theta^2} = 0, \\qquad r \\in [0, 1], \\quad \\theta \\in [0, 2\\pi]\n",
"$$\n",
"\n",
"with the Dirichlet boundary condition:\n",
"\n",
"$$\n",
"y(1,\\theta) = \\cos(\\theta)\n",
"$$\n",
"\n",
"and the periodic boundary condition:\n",
"\n",
"$$\n",
"y(r, \\theta + 2\\pi) = y(r, \\theta).\n",
"$$\n",
"\n",
"The reference solution is:\n",
"\n",
"$$\n",
"y = r\\cos(\\theta).\n",
"$$\n",
"\n",
"---\n",
"\n",
"## Dimensional Analysis\n",
"\n",
"### Step 1: Assign Dimensions to Variables\n",
"\n",
"1. **Radial Coordinate $r$:**\n",
" - The dimension of $r$ is length:\n",
" $$\n",
" [r] = L.\n",
" $$\n",
"\n",
"2. **Angular Coordinate $\\theta$:**\n",
" - The dimension of $\\theta$ is dimensionless:\n",
" $$\n",
" [\\theta] = 1.\n",
" $$\n",
"\n",
"3. **Solution $y$:**\n",
" - The solution $y$ represents a physical quantity, which we assume to be in volts (V):\n",
" $$\n",
" [y] = V.\n",
" $$\n",
"\n",
"---\n",
"\n",
"### Step 2: Analyze the Dimensions of Each Term\n",
"\n",
"1. **First Derivative Term $r\\frac{dy}{dr}$:**\n",
" - The first derivative $\\frac{dy}{dr}$ has dimensions:\n",
" $$\n",
" \\left[\\frac{dy}{dr}\\right] = \\frac{[y]}{[r]} = \\frac{V}{L}.\n",
" $$\n",
" - Therefore, the term $r\\frac{dy}{dr}$ has dimensions:\n",
" $$\n",
" \\left[r\\frac{dy}{dr}\\right] = [r] \\cdot \\frac{V}{L} = L \\cdot \\frac{V}{L} = V.\n",
" $$\n",
"\n",
"2. **Second Derivative Term $r^2\\frac{d^2y}{dr^2}$:**\n",
" - The second derivative $\\frac{d^2y}{dr^2}$ has dimensions:\n",
" $$\n",
" \\left[\\frac{d^2y}{dr^2}\\right] = \\frac{[y]}{[r]^2} = \\frac{V}{L^2}.\n",
" $$\n",
" - Therefore, the term $r^2\\frac{d^2y}{dr^2}$ has dimensions:\n",
" $$\n",
" \\left[r^2\\frac{d^2y}{dr^2}\\right] = [r]^2 \\cdot \\frac{V}{L^2} = L^2 \\cdot \\frac{V}{L^2} = V.\n",
" $$\n",
"\n",
"3. **Second Derivative Term $\\frac{d^2y}{d\\theta^2}$:**\n",
" - The second derivative $\\frac{d^2y}{d\\theta^2}$ has dimensions:\n",
" $$\n",
" \\left[\\frac{d^2y}{d\\theta^2}\\right] = \\frac{[y]}{[\\theta]^2} = \\frac{V}{1^2} = V.\n",
" $$\n",
"\n",
"---\n",
"\n",
"### Step 3: Verify Dimensional Consistency\n",
"\n",
"The Laplace equation in polar coordinates is:\n",
"\n",
"$$\n",
"r\\frac{dy}{dr} + r^2\\frac{d^2y}{dr^2} + \\frac{d^2y}{d\\theta^2} = 0.\n",
"$$\n",
"\n",
"Each term in the equation has dimensions of $V$:\n",
"\n",
"- $r\\frac{dy}{dr}$: $V$\n",
"- $r^2\\frac{d^2y}{dr^2}$: $V$\n",
"- $\\frac{d^2y}{d\\theta^2}$: $V$\n",
"\n",
"Since all terms have the same dimensions, the equation is dimensionally consistent.\n",
"\n",
"---\n",
"\n",
"### Step 4: Summary of Dimensions\n",
"\n",
"| Variable/Parameter | Physical Meaning | Dimensions |\n",
"|------------------------|-----------------------------------|-----------------------|\n",
"| $r$ | Radial coordinate | $L$ |\n",
"| $\\theta$ | Angular coordinate | $1$ (dimensionless) |\n",
"| $y$ | Solution (e.g., voltage) | $V$ |\n",
"\n",
"---\n",
"\n",
"### Step 5: Initial and Boundary Conditions\n",
"\n",
"1. **Boundary Condition $y(1,\\theta) = \\cos(\\theta)$:**\n",
" - The boundary condition $y(1,\\theta) = \\cos(\\theta)$ is given in volts:\n",
" $$\n",
" [y(1,\\theta)] = V.\n",
" $$\n",
" - The term $\\cos(\\theta)$ is dimensionless because $\\theta$ is dimensionless.\n",
"\n",
"2. **Periodic Boundary Condition $y(r, \\theta + 2\\pi) = y(r, \\theta)$:**\n",
" - The periodic boundary condition ensures that the solution is periodic in $\\theta$ with period $2\\pi$.\n",
" - Since $\\theta$ is dimensionless, the condition is dimensionally consistent.\n"
]
},
{
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