U

#### User123

##### Guest

`y0`

must be 1-dimensional" which I don't understand why is that.
Code:

```
import numpy as np
import imageio.v2 as imageio
from skimage import color
from skimage import io
import scipy.sparse as sp
from scipy.integrate import solve_ivp
im3 = imageio.imread('photo.jpg')
#convert the image to a numpy array for easier mathematical use
nparray3 = np.asarray(im3)
#convert the image to its grayscle version
grayimg3 = color.rgb2gray(nparray3)
#add noise on the gray image: The result is the initial condition of our system
noisygrayimg3 = grayimg3+1.5*np.random.rand(grayimg3.shape[0],grayimg3.shape[1])
#Reshape the initial conditions into a vector so it can be passed into the ODE solver
U0 = np.reshape(noisygrayimg3, (width3*height3, 1))
#solve the diffusion equation
#STEP 1: define the function that returns du/dt which computes the right hand side of
the heat equation
def Diffusion_rhs(t,u):
# Read an image
im3 = imageio.imread('photo.jpg')
#find out the dimsnions of the image
height3, width3, colours3 = im3.shape
#Define a grid of x and y variables and their derivatives
#x variable
x = np.linspace(0,1,width3) #patial vector
dx = x[1]-x[0] #spatial step size
onex = np.ones(width3)
Ix = np.eye(width3)
Dx = sp.spdiags([onex , -2*onex , onex], [-1, 0, 1], width3, width3) #x 1D matrix
500x500
#y variable
y = np.linspace(0,1,height3) #patial vector
dy = y[1]-y[0] #spatial step size
oney = np.ones(height3)
Iy = np.eye(height3)
Dy = sp.spdiags([oney , -2*oney , oney], [-1, 0, 1], height3, height3) #y 1D matrix
400x400
#create the 2D Laplacian
A = sp.kron(Dx,Iy) + sp.kron(Ix,Dy)
#Define the Diffusion coefficient
D = 0.005
#caluclate the right-hand side
dudt = D * A* u
return dudt
#STEP 2: define the Time variable for the integration
tspan = np.array([0, 2])
#STEP 3: solve the system
sol = solve_ivp(Diffusion_rhs, tspan, U0)
```

Any help would be appreciated. Thanks.

<p>I'm trying to solve the 2D heat equation using the finite difference method as a filtering technique for an image using the solve_ivp method. My code is shown which gives an error message "<code>y0</code> must be 1-dimensional" which I don't understand why is that.</p>

<pre><code>import numpy as np

import imageio.v2 as imageio

from skimage import color

from skimage import io

import scipy.sparse as sp

from scipy.integrate import solve_ivp

im3 = imageio.imread('photo.jpg')

#convert the image to a numpy array for easier mathematical use

nparray3 = np.asarray(im3)

#convert the image to its grayscle version

grayimg3 = color.rgb2gray(nparray3)

#add noise on the gray image: The result is the initial condition of our system

noisygrayimg3 = grayimg3+1.5*np.random.rand(grayimg3.shape[0],grayimg3.shape[1])

#Reshape the initial conditions into a vector so it can be passed into the ODE solver

U0 = np.reshape(noisygrayimg3, (width3*height3, 1))

#solve the diffusion equation

#STEP 1: define the function that returns du/dt which computes the right hand side of

the heat equation

def Diffusion_rhs(t,u):

# Read an image

im3 = imageio.imread('photo.jpg')

#find out the dimsnions of the image

height3, width3, colours3 = im3.shape

#Define a grid of x and y variables and their derivatives

#x variable

x = np.linspace(0,1,width3) #patial vector

dx = x[1]-x[0] #spatial step size

onex = np.ones(width3)

Ix = np.eye(width3)

Dx = sp.spdiags([onex , -2*onex , onex], [-1, 0, 1], width3, width3) #x 1D matrix

500x500

#y variable

y = np.linspace(0,1,height3) #patial vector

dy = y[1]-y[0] #spatial step size

oney = np.ones(height3)

Iy = np.eye(height3)

Dy = sp.spdiags([oney , -2*oney , oney], [-1, 0, 1], height3, height3) #y 1D matrix

400x400

#create the 2D Laplacian

A = sp.kron(Dx,Iy) + sp.kron(Ix,Dy)

#Define the Diffusion coefficient

D = 0.005

#caluclate the right-hand side

dudt = D * A* u

return dudt

#STEP 2: define the Time variable for the integration

tspan = np.array([0, 2])

#STEP 3: solve the system

sol = solve_ivp(Diffusion_rhs, tspan, U0)

</code></pre>

<p>Any help would be appreciated.

Thanks.</p>