5.2 Methods and Laplace transforms

Methods

D'Alembert's Theorem

Find missing solution: Try y(x)=old solution×u(x).

Variation of Parameters (for non-homogeneous ODEs)

Looking for solution in the form:

(y_par)=(∑_j=1^n)((u_j)(x)*(y_j)(x))

For the case where n=3:

(∑_j=1^3)((u_j^′)*(y_j))=0

(∑_j=1^3)((u_j^′)*(y_j^′))=0

(∑_j=1^3)((u_j^′)*(y_j^′*′))=g(x)

Solve this system, then integrate each (u_j^′) and multiply by (y_j) to get:

(y_par)=(∑_j=1^n)((u_j)(x)*(y_j)(x))

Alternatively:

(y_par)=(∑_j)((y_j)(x)*(∫_(x_0)^x)(g(s)(W_j)[s]/(W_)[(y_1)(s),⋯(y_n)(s)]*d(s)))

Diagonalisation

Introduce change of variables x=T*y:

T*(y^′)=A*T*y+g(t)⇒d(y)/d(t)=D*y+T(-1)*g=D*y+h

This leads to a system of n decoupled equations solved by direct integration:

(y_i^′)=(r_i)*(y_i)+(h_i)⇒(y_i)(t)=(c_i)*e((r_i)*t)+e((r_i)*t)*(∫_(t_0)^t)(e(-(r_i)*s)*(h_i)(s)*d(s))

General solution in the original variables:

x(t)=T*y(t)

If the matrix is not diagonalisable, find its Jordan form J and proceed similarly:

P(-1)*A*P=D, J=[[λ,1],[0,λ]]

Laplace Transform

Map y(x)→(F___^^^)(s) through:

(F___^^^)(s)=(∫_α^β)(K(s,x)*y(x)*d(x))

The Laplace Transform of ƒ(x) is defined for x∈[0,∞):

(F___^^^)(s)=L{ƒ}*(s)≡(∫_0^∞)(ℇ(-s*x)*ƒ(x)*d(x))=(lim_T→∞)((∫_0^T)(ℇ(-s*x)*ƒ(x)*d(x)))

The Laplace Transform is linear:

L{(c_1)*(ƒ_1)(x)+(c_2)*(ƒ_2)(x)}*(s)=(c_1)*L{(ƒ_1)(x)}+(c_2)*L{(ƒ_2)(x)}

If ƒ,(ƒ^′),⋯,ƒ(n-1) are continuous on [0,∞) and ∈E, then:

L{(ƒ^n)(x)}=sn*L{ƒ(x)}-s(n-1)*ƒ(0)-⋯-s*(ƒ^n-2)(0)-(ƒ^n-1)(x)

Let L{ƒ(x)}*(s)=(F___^^^)(s). For ƒ∈E:

• s-shift: L{ℇ(-c*x)*ƒ(x)}*(s)=(F___^^^)(s+c)

• x-shift:L{ƒ(x-c)}*(s)=ℇ(-s*c)*(F___^^^)(s) if c≧0 and ƒ(x)=0 for x<0

• Derivative:L{x*ƒ(x)}*(s)=-(F___^′^^)(s)

• Scaling:L{ƒ(c*x)}*(s)=1/c*(F___^^^)(s/c), (F___^^^)(s*c)=1/c*L{ƒ(x/c)} if c>0

Laplace transform of x*ℇx:

L{x*ℇx}*(s)=-(1/(s-1))′=1/((s-1)2)

Change variable from t to (t^′)=t-c:

L{(u_c)(t)*ƒ(t-c)}=ℇ(-c*s)*(F___^^^)(s)

The unit step function:

(u_c)(t)={[0,t<c],[1,t≧c])

Laplace transform of the unit step function c≥0:

L{(u_c)(t)}*(s)=(∫_0^∞)(ℇ(-s*t)*(u_c)(t)*d(t))=(∫_c^∞)(ℇ(-s*t)*d(t))=(e(-s*c))/s,s>0

For 'nice' functions ƒ(t):

(∫_-∞^∞)(δ(t-(t_0))*ƒ(t)*d(t))=ƒ((t_0))

L{δ(t-(t_0))}*(s)=ℇ(-s*(t_0))

L{δ(t)}*(s)=(lim_a→0)(ℇ)(-a*s)

(L^-1){(lim_a→0)((ℇ(-a*s))/(s2-a2))}=1/a*(u_0)(t)*sinh(a*t)

Laplace Proofs

L{(y^′)}*(s)=(∫_0^∞)(ℇ(-s*t)*(y^′)*d(t))=(∣_0^∞)(ℇ(-s*t)*y(t))+s*Y(s)=s*Y(s)-y(0)

With L{(-t)n*ƒ(t)}*(s)=(d^n)((F___^^^)(s))/(d(s)n), where (F___^^^)(s)=L{ƒ(t)}*(s), we get:

L{tn*ℇ(a*t)}*(s)=(-1)n(dn)/(d(s)n)*(s-a)(-1)=n!/((s-a)(n+1))

Convolution

(ƒ∗g)*(t)≡(∫_0^t)(ƒ((t_1))*g(t-(t_1))*d((t_1)))

L{ƒ}⋅L{g}=L{ƒ∗g}

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