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Dario Cosciandem0
Dario Coscia
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ndem0
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update plotter
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‎docs/source/_rst/tutorials/tutorial1/tutorial.rst‎

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@@ -28,8 +28,8 @@ Build a PINA problem
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Problem definition in the **PINA** framework is done by building a
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python ``class``, which inherits from one or more problem classes
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(``SpatialProblem``, ``TimeDependentProblem``, ``ParametricProblem``, …)
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depending on the nature of the problem. Below is an example. Consider the following
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simple Ordinary Differential Equation:
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depending on the nature of the problem. Below is an example: ### Simple
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Ordinary Differential Equation Consider the following:
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.. math::
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@@ -49,7 +49,7 @@ our ``Problem`` class is going to be inherited from the
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.. code:: python
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from pina.problem import SpatialProblem
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from pina import CartesianProblem
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from pina.geometry import CartesianProblem
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class SimpleODE(SpatialProblem):
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@@ -73,7 +73,7 @@ What about if our equation is also time dependent? In this case, our
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.. code:: ipython3
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from pina.problem import SpatialProblem, TimeDependentProblem
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from pina import CartesianDomain
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from pina.geometry import CartesianDomain
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class TimeSpaceODE(SpatialProblem, TimeDependentProblem):
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@@ -215,26 +215,26 @@ calling the attribute ``input_pts`` of the problem
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.. parsed-literal::
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Input points: {'x0': LabelTensor([[[0.]]]), 'D': LabelTensor([[[0.8633]],
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[[0.4009]],
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[[0.6489]],
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[[0.9278]],
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[[0.3975]],
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[[0.1484]],
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[[0.9632]],
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[[0.5485]],
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[[0.2984]],
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[[0.5643]],
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[[0.0368]],
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[[0.7847]],
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[[0.4741]],
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[[0.6957]],
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[[0.3281]],
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[[0.0958]],
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[[0.1847]],
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[[0.2232]],
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[[0.8099]],
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[[0.7304]]])}
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Input points: {'x0': LabelTensor([[[0.]]]), 'D': LabelTensor([[[0.7644]],
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[[0.2028]],
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[[0.1789]],
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[[0.4294]],
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[[0.3239]],
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[[0.6531]],
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[[0.1406]],
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[[0.6062]],
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[[0.4969]],
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[[0.7429]],
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[[0.8681]],
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[[0.3800]],
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[[0.5357]],
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[[0.0152]],
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[[0.9679]],
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[[0.8101]],
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[[0.0662]],
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[[0.9095]],
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[[0.2503]],
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[[0.5580]]])}
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Input points labels: ['x']
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@@ -271,7 +271,8 @@ If you want to track the metric by yourself without a logger, use
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.. code:: ipython3
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from pina import PINN, Trainer
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from pina import Trainer
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from pina.solvers import PINN
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from pina.model import FeedForward
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from pina.callbacks import MetricTracker
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@@ -300,12 +301,11 @@ If you want to track the metric by yourself without a logger, use
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TPU available: False, using: 0 TPU cores
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IPU available: False, using: 0 IPUs
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HPU available: False, using: 0 HPUs
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Missing logger folder: /Users/dariocoscia/Desktop/PINA/tutorials/tutorial1/lightning_logs
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.. parsed-literal::
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Epoch 1499: : 1it [00:00, 316.24it/s, v_num=0, mean_loss=5.39e-5, x0_loss=1.26e-6, D_loss=0.000106]
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Epoch 1499: : 1it [00:00, 272.55it/s, v_num=3, x0_loss=7.71e-6, D_loss=0.000734, mean_loss=0.000371]
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.. parsed-literal::
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@@ -314,7 +314,7 @@ If you want to track the metric by yourself without a logger, use
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.. parsed-literal::
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Epoch 1499: : 1it [00:00, 166.89it/s, v_num=0, mean_loss=5.39e-5, x0_loss=1.26e-6, D_loss=0.000106]
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Epoch 1499: : 1it [00:00, 167.14it/s, v_num=3, x0_loss=7.71e-6, D_loss=0.000734, mean_loss=0.000371]
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After the training we can inspect trainer logged metrics (by default
@@ -332,9 +332,9 @@ loss can be accessed by ``trainer.logged_metrics``
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.. parsed-literal::
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{'mean_loss': tensor(5.3852e-05),
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'x0_loss': tensor(1.2636e-06),
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'D_loss': tensor(0.0001)}
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{'x0_loss': tensor(7.7149e-06),
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'D_loss': tensor(0.0007),
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'mean_loss': tensor(0.0004)}
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@@ -362,7 +362,7 @@ indistinguishable. We can also plot easily the loss:
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.. code:: ipython3
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pl.plot_loss(trainer=trainer, label = 'mean_loss', logy=True)
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pl.plot_loss(trainer=trainer, label = 'mean_loss', logy=True)
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‎docs/source/_rst/tutorials/tutorial2/tutorial.rst‎

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@@ -31,16 +31,12 @@ The problem definition
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----------------------
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The two-dimensional Poisson problem is mathematically written as:
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.. math::
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\begin{equation}
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\begin{cases}
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\Delta u = \sin{(\pi x)} \sin{(\pi y)} \text{ in } D, \\
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u = 0 \text{ on } \Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4,
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\end{cases}
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\end{equation}
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where :math:`D` is a square domain :math:`[0,1]^2`, and
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:raw-latex:`\begin{equation}
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\begin{cases}
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\Delta u = \sin{(\pi x)} \sin{(\pi y)} \text{ in } D, \\
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u = 0 \text{ on } \Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4,
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\end{cases}
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\end{equation}` where :math:`D` is a square domain :math:`[0,1]^2`, and
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:math:`\Gamma_i`, with :math:`i=1,...,4`, are the boundaries of the
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square.
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.. parsed-literal::
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Epoch 999: : 1it [00:00, 152.98it/s, v_num=9, mean_loss=0.000239, D_loss=0.000793, gamma1_loss=8.51e-5, gamma2_loss=0.000103, gamma3_loss=0.000122, gamma4_loss=9.14e-5]
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Epoch 999: : 1it [00:00, 158.53it/s, v_num=3, gamma1_loss=5.29e-5, gamma2_loss=4.09e-5, gamma3_loss=4.73e-5, gamma4_loss=4.18e-5, D_loss=0.00134, mean_loss=0.000304]
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Epoch 999: : 1it [00:00, 119.21it/s, v_num=9, mean_loss=0.000239, D_loss=0.000793, gamma1_loss=8.51e-5, gamma2_loss=0.000103, gamma3_loss=0.000122, gamma4_loss=9.14e-5]
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Epoch 999: : 1it [00:00, 105.33it/s, v_num=3, gamma1_loss=5.29e-5, gamma2_loss=4.09e-5, gamma3_loss=4.73e-5, gamma4_loss=4.18e-5, D_loss=0.00134, mean_loss=0.000304]
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Now the ``Plotter`` class is used to plot the results. The solution
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which coincides with the forcing term in the Laplace equation. The set
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of input variables to the neural network is:
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.. math::
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\begin{equation}
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[x, y, k(x, y)], \text{ with } k(x, y)=\sin{(\pi x)}\sin{(\pi y)},
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\end{equation}
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:raw-latex:`\begin{equation}
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[x, y, k(x, y)], \text{ with } k(x, y)=\sin{(\pi x)}\sin{(\pi y)},
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\end{equation}`
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where :math:`x` and :math:`y` are the spatial coordinates and
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:math:`k(x, y)` is the added feature.
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Epoch 999: : 1it [00:00, 119.36it/s, v_num=10, mean_loss=8.97e-7, D_loss=4.43e-6, gamma1_loss=1.37e-8, gamma2_loss=1.68e-8, gamma3_loss=1.22e-8, gamma4_loss=1.77e-8]
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Epoch 999: : 1it [00:00, 111.88it/s, v_num=4, gamma1_loss=2.54e-7, gamma2_loss=2.17e-7, gamma3_loss=1.94e-7, gamma4_loss=2.69e-7, D_loss=9.2e-6, mean_loss=2.03e-6]
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Epoch 999: : 1it [00:00, 95.23it/s, v_num=10, mean_loss=8.97e-7, D_loss=4.43e-6, gamma1_loss=1.37e-8, gamma2_loss=1.68e-8, gamma3_loss=1.22e-8, gamma4_loss=1.77e-8]
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Epoch 999: : 1it [00:00, 85.62it/s, v_num=4, gamma1_loss=2.54e-7, gamma2_loss=2.17e-7, gamma3_loss=1.94e-7, gamma4_loss=2.69e-7, D_loss=9.2e-6, mean_loss=2.03e-6]
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The predicted and exact solutions and the error between them are
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.. math::
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\begin{equation}
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k(x, \mathbf{y}) = \beta \sin{(\alpha x)} \sin{(\alpha y)},
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\end{equation}
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:raw-latex:`\begin{equation}
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k(x, \mathbf{y}) = \beta \sin{(\alpha x)} \sin{(\alpha y)},
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\end{equation}`
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where :math:`\alpha` and :math:`\beta` are the abovementioned
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parameters. Their implementation is quite trivial: by using the class
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Epoch 999: : 1it [00:00, 103.14it/s, v_num=14, mean_loss=1.39e-6, D_loss=6.04e-6, gamma1_loss=4.19e-7, gamma2_loss=2.8e-8, gamma3_loss=4.05e-7, gamma4_loss=3.49e-8]
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Epoch 999: : 1it [00:00, 119.29it/s, v_num=5, gamma1_loss=3.26e-8, gamma2_loss=7.84e-8, gamma3_loss=1.13e-7, gamma4_loss=3.02e-8, D_loss=2.66e-6, mean_loss=5.82e-7]
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Epoch 999: : 1it [00:00, 84.50it/s, v_num=14, mean_loss=1.39e-6, D_loss=6.04e-6, gamma1_loss=4.19e-7, gamma2_loss=2.8e-8, gamma3_loss=4.05e-7, gamma4_loss=3.49e-8]
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Epoch 999: : 1it [00:00, 85.94it/s, v_num=5, gamma1_loss=3.26e-8, gamma2_loss=7.84e-8, gamma3_loss=1.13e-7, gamma4_loss=3.02e-8, D_loss=2.66e-6, mean_loss=5.82e-7]
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Umh, the final loss is not appreciabily better than previous model (with
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Epoch 999: : 1it [00:00, 130.55it/s, v_num=17, mean_loss=1.34e-14, D_loss=6.7e-14, gamma1_loss=5.13e-17, gamma2_loss=9.68e-18, gamma3_loss=5.14e-17, gamma4_loss=9.75e-18]
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Epoch 0: : 0it [00:00, ?it/s]Epoch 999: : 1it [00:00, 131.20it/s, v_num=6, gamma1_loss=2.55e-16, gamma2_loss=4.76e-17, gamma3_loss=2.55e-16, gamma4_loss=4.76e-17, D_loss=1.74e-13, mean_loss=3.5e-14]
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Epoch 999: : 1it [00:00, 104.91it/s, v_num=17, mean_loss=1.34e-14, D_loss=6.7e-14, gamma1_loss=5.13e-17, gamma2_loss=9.68e-18, gamma3_loss=5.14e-17, gamma4_loss=9.75e-18]
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Epoch 999: : 1it [00:00, 98.81it/s, v_num=6, gamma1_loss=2.55e-16, gamma2_loss=4.76e-17, gamma3_loss=2.55e-16, gamma4_loss=4.76e-17, D_loss=1.74e-13, mean_loss=3.5e-14]
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In such a way, the model is able to reach a very high accuracy! Of

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‎docs/source/_rst/tutorials/tutorial3/tutorial.rst‎

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@@ -25,14 +25,13 @@ The problem definition
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The problem is written in the following form:
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.. math::
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\begin{equation}
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\begin{cases}
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\Delta u(x,y,t) = \frac{\partial^2}{\partial t^2} u(x,y,t) \quad \text{in } D, \\\\
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u(x, y, t=0) = \sin(\pi x)\sin(\pi y), \\\\
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u(x, y, t) = 0 \quad \text{on } \Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4,
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\end{equation}
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:raw-latex:`\begin{equation}
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\begin{cases}
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\Delta u(x,y,t) = \frac{\partial^2}{\partial t^2} u(x,y,t) \quad \text{in } D, \\\\
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u(x, y, t=0) = \sin(\pi x)\sin(\pi y), \\\\
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u(x, y, t) = 0 \quad \text{on } \Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4,
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\end{cases}
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Epoch 999: : 1it [00:00, 62.13it/s, v_num=0, mean_loss=0.0268, D_loss=0.0397, t0_loss=0.121, gamma1_loss=0.000, gamma2_loss=0.000, gamma3_loss=0.000, gamma4_loss=0.000]
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Epoch 999: : 1it [00:00, 84.47it/s, v_num=0, gamma1_loss=0.000, gamma2_loss=0.000, gamma3_loss=0.000, gamma4_loss=0.000, t0_loss=0.0419, D_loss=0.0307, mean_loss=0.0121]
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Epoch 999: : 1it [00:00, 53.88it/s, v_num=0, mean_loss=0.0268, D_loss=0.0397, t0_loss=0.121, gamma1_loss=0.000, gamma2_loss=0.000, gamma3_loss=0.000, gamma4_loss=0.000]
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Epoch 999: : 1it [00:00, 68.69it/s, v_num=0, gamma1_loss=0.000, gamma2_loss=0.000, gamma3_loss=0.000, gamma4_loss=0.000, t0_loss=0.0419, D_loss=0.0307, mean_loss=0.0121]
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Notice that the loss on the boundaries of the spatial domain is exactly
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Epoch 999: : 1it [00:00, 48.54it/s, v_num=1, mean_loss=1.48e-8, D_loss=8.89e-8, t0_loss=0.000, gamma1_loss=2.06e-15, gamma2_loss=0.000, gamma3_loss=2.1e-15, gamma4_loss=0.000]
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Epoch 0: : 0it [00:00, ?it/s]Epoch 999: : 1it [00:00, 52.10it/s, v_num=1, gamma1_loss=1.97e-15, gamma2_loss=0.000, gamma3_loss=2.14e-15, gamma4_loss=0.000, t0_loss=0.000, D_loss=1.25e-7, mean_loss=2.09e-8]
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Epoch 999: : 1it [00:00, 43.25it/s, v_num=1, mean_loss=1.48e-8, D_loss=8.89e-8, t0_loss=0.000, gamma1_loss=2.06e-15, gamma2_loss=0.000, gamma3_loss=2.1e-15, gamma4_loss=0.000]
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Epoch 999: : 1it [00:00, 45.78it/s, v_num=1, gamma1_loss=1.97e-15, gamma2_loss=0.000, gamma3_loss=2.14e-15, gamma4_loss=0.000, t0_loss=0.000, D_loss=1.25e-7, mean_loss=2.09e-8]
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We can clearly see that the loss is way lower now. Let’s plot the

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