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I heard that the following SDE can be solved analitically by substitution:

$dX(t) = - \left[ \sin (2 X(t) ) + \frac{1}{4} \sin (4 X(t) ) \right] dt + \sqrt{2} \cos^2 X(t) dB(t),$ $X(0) = 1, \; t \in [0, \tau)$

But is isn't as simple as it sounds. Any ideas? Of course $\mathbb{B}$ is a brownian motion.

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Did you try $Y= \tan X$? – Fabian Jun 7 '14 at 11:03
up vote 0 down vote accepted

Motivation: Only few stochastic differential equations (SDEs) can be solved explicitely; one of them is a linear SDE with determinstic coefficients, i.e. an SDE of the form

$$dZ_t = (\alpha(t)+\beta(t) Z_t) \, dB_t + (\gamma(t)+\delta(t) Z_t) \, dt. \tag{1}$$

That's why one usually tries to transform SDEs into a linear SDE. For SDEs of the form

$$dX_t = b(X_t) \, dt + \sigma(X_t) \, dB_t, \tag{2}$$

i.e. SDEs where the coefficient do not depend on the time $t$, there are necessary and sufficient criterions for a transformation into a linear SDE (see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Section 19.4, 2nd edition).

Roughly speaking, the only two possibilities two transform an SDE $(2)$ into a linear SDE $(1)$ are the following two substitutions: Set

$$Z_t := f(X_t)$$

where either

$$f(x) :=\gamma \int_0^x \frac{1}{\sigma(y)} \, dy \tag{3} $$


$$f(x) := \exp \left( \gamma \cdot \int_0^x \frac{1}{\sigma(y)} \, dy \right) \tag{4}$$

for some suitable constant $\gamma>0$. (In fact, there are criterions in terms of the derivatives when to use which one; but they are quite messy.)

So here we go. The SDE

$$dX_t = - \left(\sin(2X_t) + \frac{1}{4} \sin(4X_t)\right) \, dt + \sqrt{2} \cos^2 X_t \, dB_t \tag{5}$$

is obviously of the form $(5)$ where $\sigma(y) = \sqrt{2} \cos^2(y)$. If we try substitution $(3)$, then

$$Z_t := f(X_t) \stackrel{(3)}{=} \gamma \int_0^{X_t} \frac{1}{\sqrt{2} \cos^2(y)} \, dy = \gamma' \tan X_t$$

where $\gamma' := \frac{\gamma}{\sqrt{2}}$. Applying Itô's formula yields

$$\begin{align*} Z_t-Z_0 &= \gamma' \int_0^t \frac{1}{\cos^2(X_s)} \, dX_s + \gamma \int_0^t \frac{\sin X_s}{\cos^3 X_s} \, d\langle X \rangle_s \\ &\stackrel{(5)}{=} \sqrt{2}\gamma' \int_0^t dB_s - \gamma' \int_0^t \left(\frac{\sin(2X_s)}{\cos^2(X_s)}+ \frac{\sin(4X_s)}{4\cos^2(X_s)} \right) \, ds + 2\gamma' \int_0^t \frac{\sin X_s}{\cos^3 X_s} \cos^4 X_s \, ds. \end{align*}$$

In the last step we used that the quadratic variation $\langle X \rangle_s$ equals, by $(5)$,

$$d \langle X \rangle_t = 2 \cos^4 X_t \, dt.$$

Using the well-known formulas

$$\sin(2x) = 2 \sin x \cdot \cos x \qquad \sin (4x) = 8 \sin x \cdot \cos^3(x) - 4 \sin x \cdot \cos x$$

we get

$$Z_t - Z_0 = \sqrt{2}\gamma' \int_0^t dB_s -\gamma' \int_0^t Z_s \, ds.$$

This means that we have transformed the SDE $(5)$ into a linear SDE of the form $(1)$. This SDE can be solved explicitely, and therefore we find an explicit expression for the solution

$$X_t = \arctan \left(\frac{Z_t}{\gamma'} \right).$$

(Here $\gamma'>0$ is an arbitrary constant; we may choose $\gamma'=1$.)

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Thanks, I didn't know those (3) and (4) formulas and I was trying to solve the SDE with silly substitutions such as $\log \tan X_t$ or $\exp(t \cdot \tan X(t))$. – Robert Jun 9 '14 at 17:38

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