🧬 Dynamic RNA velocity model-- (1) math solutions

Introduction

The steady‑state model’s reliance on true steady states is at odds with known biophysical behavior. The dynamic model removes this requirement to broaden RNA velocity’s applicability but inevitably introduces new assumptions that may not hold for every dataset. Effective use of the dynamic model therefore demands a clear understanding of its strengths and limitations. Our blog series toward this goal begins by delving into the mathematical foundations of the dynamic model.

Symbol definitions

Please refer to the Nature Biotechnology paper for the definitions of the symbols used in this derivation.

Dynamic Solutions for RNA Velocity Model

To derive the solution for s(t) from the given system of differential equations, we will follow a two-step process:

1. Solve the first differential equation for u(t) using the initial condition u(t₀)=u₀.
2. Substitute the expression for u(t) into the second differential equation and then solve for s(t) using the initial condition s(t₀)=s₀.

Step 1: Solve for u(t)

The first differential equation is:

$$ \frac{du(t)}{dt}=\alpha-\beta u(t) $$

Rearrange it into the standard form for a first-order linear ODE:

$$ \frac{du(t)}{dt}+\beta u(t)=\alpha $$

This is a linear first-order differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$, where P(t)=β and Q(t)=α. The integrating factor (IF) is $e^{\int P(t)dt}=e^{\int \beta dt}=e^{\beta t}$.

Multiply the entire equation by the integrating factor:

$$ e^{\beta t}\frac{du(t)}{dt}+\beta u(t)e^{\beta t}=\alpha e^{\beta t} $$

The left side is the derivative of the product $(u(t)e^{\beta t})$:

$$ \frac{d}{dt}(u(t)e^{\beta t})=\alpha e^{\beta t} $$

Integrate both sides with respect to t:

$$ \int \frac{d}{dt}(u(t)e^{\beta t})dt=\int \alpha e^{\beta t}dt $$ $$ u(t)e^{\beta t}=\frac{\alpha}{\beta}e^{\beta t}+C_1 $$

Now, solve for u(t):

$$ u(t)=\frac{\alpha}{\beta}+C_1e^{-\beta t} $$

Use the initial condition u(t₀)=u₀ to find C₁:

$$ u_0=\frac{\alpha}{\beta}+C_1e^{-\beta t_0} $$ $$ C_1e^{-\beta t_0}=u_0-\frac{\alpha}{\beta} $$ $$ C_1=(u_0-\frac{\alpha}{\beta})e^{\beta t_0} $$

Substitute C₁ back into the equation for u(t):

$$ u(t)=\frac{\alpha}{\beta}+(u_0-\frac{\alpha}{\beta})e^{\beta t_0}e^{-\beta t} $$ $$ u(t)=\frac{\alpha}{\beta}+(u_0-\frac{\alpha}{\beta})e^{-\beta(t-t_0)} $$

Step 2: Solve for s(t)

Now that we have u(t), we can solve the second differential equation:

$$ \frac{ds(t)}{dt}=\beta u(t)-\gamma s(t) $$

Substitute the expression for u(t):

$$ \frac{ds(t)}{dt}=\beta[\frac{\alpha}{\beta}+(u_0-\frac{\alpha}{\beta})e^{-\beta(t-t_0)}]-\gamma s(t) $$ $$ \frac{ds(t)}{dt}+\gamma s(t)=\alpha+(u_0\beta-\alpha)e^{-\beta(t-t_0)} $$

This is again a linear first-order differential equation. The integrating factor is $e^{\gamma t}$.

Multiply both sides by the integrating factor:

$$ \frac{d}{dt}(s(t)e^{\gamma t})=\alpha e^{\gamma t}+(u_0\beta-\alpha)e^{\gamma t}e^{-\beta(t-t_0)} $$

Integrate both sides:

$$ s(t)e^{\gamma t}=\frac{\alpha}{\gamma}e^{\gamma t}+(u_0\beta-\alpha)\int e^{\gamma t}e^{-\beta(t-t_0)}dt+C_2 $$

Simplify the integral:

$$ s(t)e^{\gamma t}=\frac{\alpha}{\gamma}e^{\gamma t}+(u_0\beta-\alpha)e^{\beta t_0}\int e^{(\gamma-\beta)t}dt+C_2 $$ $$ s(t)e^{\gamma t}=\frac{\alpha}{\gamma}e^{\gamma t}+(u_0\beta-\alpha)e^{\beta t_0}\frac{e^{(\gamma-\beta)t}}{\gamma-\beta}+C_2 $$

Solve for s(t):

$$ s(t)=\frac{\alpha}{\gamma}+(u_0\beta-\alpha)e^{\beta t_0}\frac{e^{-\beta t}}{\gamma-\beta}+C_2e^{-\gamma t} $$

Use the initial condition s(t₀)=s₀ to find C₂:

$$ s_0=\frac{\alpha}{\gamma}+(u_0\beta-\alpha)\frac{e^{-\beta t_0+\beta t_0}}{\gamma-\beta}+C_2e^{-\gamma t_0} $$ $$ C_2e^{-\gamma t_0}=s_0-\frac{\alpha}{\gamma}-(u_0\beta-\alpha)\frac{1}{\gamma-\beta} $$ $$ C_2=[s_0-\frac{\alpha}{\gamma}-(u_0\beta-\alpha)\frac{1}{\gamma-\beta}]e^{\gamma t_0} $$

The complete solution for s(t) is:

$$ s(t)=\frac{\alpha}{\gamma}+\frac{u_0\beta-\alpha}{\gamma-\beta}e^{-\beta(t-t_0)}+[s_0-\frac{\alpha}{\gamma}-\frac{u_0\beta-\alpha}{\gamma-\beta}]e^{-\gamma(t-t_0)} $$

Special case: If γ=β

If γ=β, the integral becomes:

$$ \int(βu_0-α)e^{(γ-β)t+βt_0}dt=(βu_0-α)e^{βt_0}\int dt=(βu_0-α)e^{βt_0}t $$

Then,

$$ s(t)e^{γt}=\frac{α}{γ}e^{γt}+(βu_0-α)e^{βt_0}t+C_2 $$ $$ s(t)=\frac{α}{γ}+(βu_0-α)te^{-γt+βt_0}+C_2e^{-γt} $$

Since γ=β:

$$ s(t)=\frac{α}{γ}+(γu_0-α)te^{-γ(t-t_0)}+C_2e^{-γt} $$ $$ s(t)=\frac{α}{γ}+(γu_0-α)(t_0+τ)e^{-γτ}+C_2e^{-γ(t_0+τ)} $$

This scenario is not covered by the target equation, which explicitly has a denominator (γ−β), implying γ≠β.