def quadraticcase(t0,tend,dt,k,fun,teta,x0): #solver for the case of quadratic energy
    #fun is a linear function with time in argument; dt is the change in time
    f=sy.sympify(fun) #converting string into function with t in argument
    n=(tend-t0)/dt #n: number of time steps till tend is reached
    t_tuple=np.linspace(t0,tend,n+1) #time grid
    x_tuple=np.zeros([n+1]) #x-grid
    x_left=np.zeros([n+1])
    x_right=np.zeros([n+1])
    x_left[0]=(f(t[0])-teta)/k
    x_right[0]=(f(t[0])+teta)/k
    for i in range(1,n+2):
        x_left[i]=(f(t[i])-teta)/k
        x_right[i]=(f(t[i])+teta)/k
        if x[i-1]>x_left[i-1] or x[i-1]<x_right[i-1]:
            x[i]=x[i-1]
            elif x[i-1]<=x_left[i-1]:
                x[i]=x_left
                else:
                    x[i]=x_right