Examples

Partial differential equations

Let's consider the following 2D partial differential equation

\[\Delta u(x,y) = f(x,y),\]

where $\Delta$ is the Laplacian operator, $u(x,y)$ is the unknown function, and $f(x,y)$ is a given function. In this example we assume $f(x,y)=0$ using Dirichlet-Dirichlet boundary conditions given by $u(0,y) = \cos(\pi y)$, $u(1,y) = \sin(y)$.

We want to solve this equation using quantics tensor trains (QTTs). We start by defining the dimensions and the resolution of the grid. Lets say we want $2^{10}$ points in each dimension, which gives us a grid of $1024 \times 1024$ points on a $[0,1]\times [0,1]$ grid.

using TensorTrainNumerics
using CairoMakie

cores = 10
a = 0.0
b = 1.0

1.0

We follow the same convention as in this paper where we define the finite difference operator using the following inputs

h = (b-a)/(2^cores)
p = 1.0
s = 0.0
v = 0.0
α = h^2*v-2*p
β = p + h*s/2
γ = p-h*s/2

Δ = toeplitz_to_qtto(α, β, γ, cores)

TToperator{Float64, 10}(10, [[1.0 0.0; 0.0 1.0;;;; 0.0 0.0; 1.0 0.0;;;; 0.0 1.0; 0.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [-2.0 1.0; 1.0 -2.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;;;]], (2, 2, 2, 2, 2, 2, 2, 2, 2, 2), [1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

To get the 2D Laplacian operator, we need to take the Kronecker product of the 1D Laplacian operator with the identity.

A = Δ ⊗ id_tto(cores) + id_tto(cores) ⊗ Δ

TToperator{Float64, 20}(20, [[1.0 0.0; 0.0 1.0;;;; 0.0 0.0; 1.0 0.0;;;; 0.0 1.0; 0.0 0.0;;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [-2.0 1.0; 1.0 -2.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 1.0 0.0; 0.0 1.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;;; 0.0 0.0; 0.0 0.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0], [1.0 0.0; 0.0 1.0;;; -2.0 1.0; 1.0 -2.0;;; 0.0 1.0; 0.0 0.0;;; 0.0 0.0; 1.0 0.0;;;;]], (2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), [1, 4, 4, 4, 4, 4, 4, 4, 4, 4  …  4, 4, 4, 4, 4, 4, 4, 4, 4, 1], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

To build the boundary vector we take the Kronecker product with the QTT basis vectors and define some random initial guess for the solution.

b = qtt_cos(cores) ⊗ qtt_basis_vector(cores, 1) + qtt_sin(cores) ⊗ qtt_basis_vector(cores, 2^cores)
initial_guess = rand_tt(b.ttv_dims, b.ttv_rks)

TTvector{Float64, 20}(20, [[-1.3068748448925214; -0.05860606952522715;;; 0.9307247391472059; 0.0810194713691914;;; 0.8875584702557205; 0.4989957495831991;;; -0.2503177494148563; -0.011242913337357866], [0.6591851852747405 -1.1234788451905564 -0.42336658831143864 -1.600065960868221; 0.167404325204573 -0.9005010322405455 -0.46541737139123196 -0.41277642532621767;;; 1.6149785515101998 -0.44098292739821704 -0.8515936280895996 0.7650517011866605; -0.7639692266194342 1.0330523849118785 -0.17058737996740808 0.7215382453692442;;; -1.3611210814849852 -0.11640715878135015 0.9327450613451204 -0.5302478539526122; -0.6221207806159756 -0.9603927028401423 -0.7077473006458824 0.49201063461378025;;; 0.12803697029706043 -0.07677045622199469 -0.306034113278485 0.9918475624762048; -0.22345590144489466 0.6026762735791237 0.5453182272538503 0.1299586615818743], [-0.3022344052975696 0.08524736075443246 -1.524445517781313 0.004516378332865016; -0.05184766219081393 -0.5191481868635125 -0.18050556580175295 -1.2195721168523193;;; 0.125641215716426 -0.6316507124961666 -0.547424175831757 0.08751207610712115; 1.986060425511894 -0.5506277702657342 -0.4110885936978945 0.7706863682101712;;; 1.3098643980810383 -0.8014564656770672 -1.250947988120645 -1.5096813666899915; 0.1356273658867086 -0.11750322658601814 -1.17146704957152 2.424170024039988;;; -0.004623444589109873 1.6701243879129883 -1.205048929013107 1.0443911411468334; 1.274813484487595 -0.1874834024470065 -1.3612575670774605 2.007009968248779], [-0.31499517206317 -0.3668475729016974 1.1739712280227335 -2.703493032845142; 0.36158548908781973 -1.158681842683639 1.644479239485255 1.1407344956709329;;; 0.864694049608145 -0.8946484325515596 -1.097410196013899 -0.6370646359001008; 1.714277026248039 0.8365063871505962 -0.8277201752976971 -0.4757628958304671;;; -2.1250278320564298 0.13316670036357905 -0.5764124297800585 -1.9308806930535551; -1.0420740173031091 0.27116061674788156 0.44750309984185127 0.021144302311001197;;; 1.0582054462669146 1.6648580738018628 -0.49380381838979953 -0.7015844342275477; 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We solve the linear system using DMRG

x_dmrg = dmrg_linsolvee(A, b, initial_guess; sweep_count=50,tol=1e-15)

TTvector{Float64, 20}(20, [[-177.89431945442226; -192.0705136152335;;; -118.04623776156836; 109.33357096559355], [-0.7010067136069053 0.46615583448105274; -0.6906140076160482 -0.48106735494870856;;; -0.11921946291596984 -0.5160337521703952; 0.12742017107743098 -0.5243871357754554;;; 0.016020808564685397 0.07208062239346755; 0.019941605962111374 -0.06774691400427293;;; 0.016671916093650787 -0.008924459549495691; -0.016137313454375263 -0.011633141282616409], [-0.7061860003463494 0.30979853801058044 0.0879198255754485 -0.15577865624345497; -0.7040227500053253 -0.311627090694989 -0.046125993536888406 0.16250467433773372;;; -0.05236895919093486 -0.6307242320897017 -0.3330073105783008 -0.16144263634576553; 0.052912469315943185 -0.633955285548909 0.3980148623434077 -0.0922948309471017;;; 0.004527388532104473 0.05998941849816667 -0.445590799305159 -0.4898521663890756; 0.0064079854124869315 -0.05989429983395097 -0.4698102665690386 0.4971604883295441;;; 0.004905167852228789 0.003314903135655415 -0.39223844481267905 -0.4818157144948221; 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And we reshape the solution to a 2D array for visualization

solution = reshape(qtt_to_function(x_dmrg), 2^cores, 2^cores)
xes = collect(range(0,1,length=2^cores))
yes = collect(range(0,1,length=2^cores))
fig = Figure()
cmap = :roma
ax = Axis(fig[1, 1], title = "Laplace Solution", xlabel = "x", ylabel = "y")
hm = heatmap!(ax, xes, yes, solution; colormap = cmap)
Colorbar(fig[1, 2], hm, label = "u(x, y)")
fig

Time-stepping

We can also solve time-dependent PDEs using the QTT framework. In this exampl we will use the explicit Euler method, the implicit Euler method and the Crank-Nicolson scheme.

using TensorTrainNumerics
using CairoMakie

cores = 8
h = 1/cores^2
A = h^2*toeplitz_to_qtto(-2,1.0,1.0,cores)
xes = collect(range(0.0, 1.0, 2^cores))

u₀ = qtt_sin(cores,λ=π)
init = rand_tt(u₀.ttv_dims, u₀.ttv_rks)
steps = collect(range(0.0,10.0,1000))
solution_explicit, error_explicit = euler_method(A, u₀, steps; return_error=true)

solution_implicit, rel_implicit = implicit_euler_method(A, u₀, init, steps; return_error=true)

solution_crank, rel_crank = crank_nicholson_method(A, u₀, init, steps; return_error=true, tt_solver="mals")

fig = Figure()
ax = Axis(fig[1, 1], xlabel = "x", ylabel = "u(x)", title = "Comparison of Time-Stepping Methods")
lines!(ax, xes, qtt_to_function(solution_explicit), label = "Explicit Euler", linestyle = :solid, linewidth=3)
lines!(ax, xes, qtt_to_function(solution_implicit), label = "Implicit Euler", linestyle = :dot, linewidth=3)
lines!(ax, xes, qtt_to_function(solution_crank), label = "Crank-Nicolson", linestyle = :dash, linewidth=3)
axislegend(ax)
fig

And print the relative errors

println("Relative error for explicit Euler: ", error_explicit)
println("Relative error for implicit Euler: ", rel_implicit)
println("Relative error for Crank-Nicolson: ", rel_crank)

Relative error for explicit Euler: 2.7254393110550548e-5
Relative error for implicit Euler: 4.831405661616409e-6
Relative error for Crank-Nicolson: 4.8314364669860835e-6