# Adjoint ODE

### Question: Minimize $$F(x, p)$$ 

$$
F(x, p) = \int\_{0}^{T}f(x,p,t)dt \tag1
$$

Subject to

$$
g(x\_0,p) = x\_0 - p =0 \tag2
$$

-> If given $$x\_0$$, we can compute $$p$$ at $$t=0$$ through $$g(x\_0, p)$$, then substitue $$x\_t$$ and $$p$$ in $$h(x\_t, \dot{x}\_t, p, t)$$ and can compute $$\dot{x}\_t$$ 

{% hint style="info" %}
Why is $$\frac{\partial{x}}{\partial p}$$ difficult to calculate? Because $$fn(x,p)$$ is unknown, need to solve ALL possible $$x$$ and $$p$$ using ODE
{% endhint %}

Apply Lagrangian function $$\mathcal{L} (x, \lambda)= f(x)- \lambda g (x)$$ and combine (1) (2) (3) in one loss function

$$
Loss = \int\_{0}^{T}\[f(x,p,t) + \lambda^Th(x\_t, \dot{x}\_t, p, t)]dt + u^T g(x\_0,p) \tag4
$$

Substitute (2) (3)

$$
Loss = \int\_{0}^{T}\[f(x,p,t) + \lambda^T0]dt + u^T 0 =  \int\_{0}^{T}f(x,p,t)dt = F(x, p) \tag5
$$

{% hint style="info" %}
Why does integral equals loss? Why is this loss function? Minimize loss?
{% endhint %}

![integral equals loss](https://1490490238-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-Ma_JYIRuxx8PCv6UaBM%2F-MgGli4PKyqiuNApKBiD%2F-MgGwGfuVxcQCQs1kf4y%2FUntitled%20design.png?alt=media\&token=ca3a85c8-6b9b-44d7-b930-ee705cf7bf86)

$$
\frac{\partial{L}}{\partial p} = \frac{\partial{F}}{\partial p} \tag6
$$

{% hint style="info" %}
Why calculate $$\frac{\partial{L}}{\partial p}$$during backprop? To use Newton's Method to approximate f(x) at a given point
{% endhint %}

![approximation using updated derivative](https://1490490238-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-Ma_JYIRuxx8PCv6UaBM%2F-MgGwjPRV2rCR6lhWYgZ%2F-MgGxUCwIDyjAF7kvJiQ%2Fimage.png?alt=media\&token=2d54510d-cf02-44d3-813f-40bf821b8068)

$$
\frac{\partial{L}}{\partial p} = \frac{\partial{F}}{\partial p} =  \int\_{0}^{T}\[\frac{\partial{f}}{\partial x}\cdot\frac{\partial{x}}{\partial p}+\frac{\partial{f}}{\partial p}+\lambda^T(\frac{\partial{h}}{\partial x}\cdot\frac{\partial{x}}{\partial p}+\frac{\partial{h}}{\partial \dot{x}}\cdot\frac{\partial{\dot{x}}}{\partial p}+\frac{\partial{h}}{\partial p})]dt+u^T(\frac{\partial{g}}{\partial x\_0}\cdot\frac{\partial{x\_0}}{\partial p}+\frac{\partial{g}}{\partial p})
$$

To avoid compute $$\frac{\partial{\dot{x}}}{\partial p}$$ , we apply integration by parts $$\int u dv = u v - \int v du$$ 

$$
\frac{\partial{L}}{\partial p} = \frac{\partial{L}}{\partial p}
$$