Lyapunov–Schmidt reduction

In mathematics, the Lyapunov–Schmidt reduction or Lyapunov–Schmidt construction is used to study solutions to nonlinear equations in the case when the implicit function theorem does not work. It permits the reduction of infinite-dimensional equations in Banach spaces to finite-dimensional equations. It is named after Aleksandr Lyapunov and Erhard Schmidt.

Problem setup

Let

$f(x,\lambda)=0 \,$

be the given nonlinear equation, $X,\Lambda,$ and $Y$ are Banach spaces ($\Lambda$ is the parameter space). $f(x,\lambda)$ is the $C^p$-map from a neighborhood of some point $(x_0,\lambda_0)\in X\times \Lambda$ to $Y$ and the equation is satisfied at this point

$f(x_0,\lambda_0)=0.$

For the case when the linear operator $f_x(x,\lambda)$ is invertible, the implicit function theorem assures that there exists a solution $x(\lambda)$ satisfying the equation $f(x(\lambda),\lambda)=0$ at least locally close to $\lambda_0$.

In the opposite case, when the linear operator $f_x(x,\lambda)$ is non-invertible, the Lyapunov–Schmidt reduction can be applied in the following way.

Assumptions

One assumes that the operator $f_x(x,\lambda)$ is a Fredholm operator.

$\ker f_x (x_0,\lambda_0)=X_1$ and $X_1$ has finite dimension.

The range of this operator $\mathrm{ran} f_x (x_0,\lambda_0)=Y_1$ has finite co-dimension and is a closed subspace in $Y$.

Without loss of generality, one can assume that $(x_0,\lambda_0)=(0,0).$

Lyapunov–Schmidt construction

Let us split $Y$ into the direct product $Y= Y_1 \oplus Y_2$, where $\dim Y_2 < \infty$.

Let $Q$ be the projection operator onto $Y_1$.

Let us consider also the direct product $X= X_1 \oplus X_2$.

Applying the operators $Q$ and $I-Q$ to the original equation, one obtains the equivalent system

$Qf(x,\lambda)=0 \,$

$(I-Q)f(x,\lambda)=0 \,$

Let $x_1\in X_1$ and $x_2 \in X_2$, then the first equation

$Qf(x_1+x_2,\lambda)=0 \,$

can be solved with respect to $x_2$ by applying the implicit function theorem to the operator

$Qf(x_1+x_2,\lambda): \quad X_2\times(X_1\times\Lambda)\to Y_1 \,$

(now the conditions of the implicit function theorem are fulfilled).

Thus, there exists a unique solution $x_2(x_1,\lambda)$ satisfying

$Qf(x_1+x_2(x_1,\lambda),\lambda)=0. \,$

Now substituting $x_2(x_1,\lambda)$ into the second equation, one obtains the final finite-dimensional equation

$(I-Q)f(x_1+x_2(x_1,\lambda),\lambda)=0. \,$

Indeed, the last equation is now finite-dimensional, since the range of $(I-Q)$ is finite-dimensional. This equation is now to be solved with respect to $x_1$, which is finite-dimensional, and parameters :$\lambda$

References

• Louis Nirenberg, Topics in nonlinear functional analysis, New York Univ. Lecture Notes, 1974.