Second derivative in Keras

Question

For a custom loss for a NN I use the function . u, given a pair (t,x), both points in an interval, is the the output of my NN. Problem is I\'m stuck at how

Answer 1

Solution posted by Peter Szoldan is an excellent one. But it seems like the way keras.layers.Input() take in arguments has changed since the latest version with tf2 backend. The following simple fix will work though:

import tensorflow as tf
from tensorflow import keras
from tensorflow.keras import backend as K
import numpy as np

class CustomModel(tf.keras.Model):

    def __init__(self):
        super(CustomModel, self).__init__()
        self.input_layer = Lambda(lambda x: K.log( x + 2 ) )

    def findGrad(self,func,argm):
        return keras.layers.Lambda(lambda x: K.gradients(x[0],x[1])) ([func,argm])
    
    def call(self, inputs):
        log_layer = self.input_layer(inputs)
        gradient_layer = self.findGrad(log_layer,inputs)
        hessian_layer = self.findGrad(gradient_layer, inputs)
        return hessian_layer


custom_model = CustomModel()
x = np.array([[0.],
            [1],
            [2]])
custom_model.predict(x) 

• Going through layers: input layer-> lambda layer appylying log(x+2) -> lambda layer applying gradient -> one more lambda layer applying gradeint -> Output.
• Note that this solution is for a general custom model and if you are using functional api, it should be similar.
• If you are using tf backend, then using tf.hessians, instead of applying K.gradients twice, will work as well.