Laplace transform applied to differential equations

In <a href="/facts/Mathematics/pxTouaz4">mathematics</a>, the <a href="/facts/Laplace_transform/0Ge4MIc9">Laplace transform</a> is a powerful <a href="/facts/Integral_transform/AB2yjHJJ">integral transform</a> used to switch a function from the <a href="/facts/Time_domain/1w6mP2MU">time domain</a> to the <a href="/facts/Laplace_transform/0Ge4MIc9">s-domain</a>. The Laplace transform can be used in some cases to solve <a href="/facts/Linear_differential_equation/nLEbIIVf">linear differential equations</a> with given <a href="/facts/Initial_value_problem/PqmJfGML">initial conditions</a>.
First consider the following property of the Laplace transform:

L
          
        
        {
        
          f
          ′
        
        }
        =
        s
        
          
            L
          
        
        {
        f
        }
        −
        f
        (
        0
        )
      
    
    {\displaystyle {\mathcal {L}}\{f'\}=s{\mathcal {L}}\{f\}-f(0)}

L
          
        
        {
        
          f
          ″
        
        }
        =
        
          s
          
            2
          
        
        
          
            L
          
        
        {
        f
        }
        −
        s
        f
        (
        0
        )
        −
        
          f
          ′
        
        (
        0
        )
      
    
    {\displaystyle {\mathcal {L}}\{f''\}=s^{2}{\mathcal {L}}\{f\}-sf(0)-f'(0)}

One can prove by <a href="/facts/Mathematical_induction/KxAPqNrH">induction</a> that

L
          
        
        {
        
          f
          
            (
            n
            )
          
        
        }
        =
        
          s
          
            n
          
        
        
          
            L
          
        
        {
        f
        }
        −
        
          ∑
          
            i
            =
            1
          
          
            n
          
        
        
          s
          
            n
            −
            i
          
        
        
          f
          
            (
            i
            −
            1
            )
          
        
        (
        0
        )
      
    
    {\displaystyle {\mathcal {L}}\{f^{(n)}\}=s^{n}{\mathcal {L}}\{f\}-\sum _{i=1}^{n}s^{n-i}f^{(i-1)}(0)}

Now we consider the following differential equation:

∑
          
            i
            =
            0
          
          
            n
          
        
        
          a
          
            i
          
        
        
          f
          
            (
            i
            )
          
        
        (
        t
        )
        =
        ϕ
        (
        t
        )
      
    
    {\displaystyle \sum _{i=0}^{n}a_{i}f^{(i)}(t)=\phi (t)}

with given initial conditions

f
          
            (
            i
            )
          
        
        (
        0
        )
        =
        
          c
          
            i
          
        
      
    
    {\displaystyle f^{(i)}(0)=c_{i}}

Using the <a href="/facts/Linearity/fChtg3KC">linearity</a> of the Laplace transform it is equivalent to rewrite the equation as

∑
          
            i
            =
            0
          
          
            n
          
        
        
          a
          
            i
          
        
        
          
            L
          
        
        {
        
          f
          
            (
            i
            )
          
        
        (
        t
        )
        }
        =
        
          
            L
          
        
        {
        ϕ
        (
        t
        )
        }
      
    
    {\displaystyle \sum _{i=0}^{n}a_{i}{\mathcal {L}}\{f^{(i)}(t)\}={\mathcal {L}}\{\phi (t)\}}

obtaining

L
          
        
        {
        f
        (
        t
        )
        }
        
          ∑
          
            i
            =
            0
          
          
            n
          
        
        
          a
          
            i
          
        
        
          s
          
            i
          
        
        −
        
          ∑
          
            i
            =
            1
          
          
            n
          
        
        
          ∑
          
            j
            =
            1
          
          
            i
          
        
        
          a
          
            i
          
        
        
          s
          
            i
            −
            j
          
        
        
          f
          
            (
            j
            −
            1
            )
          
        
        (
        0
        )
        =
        
          
            L
          
        
        {
        ϕ
        (
        t
        )
        }
      
    
    {\displaystyle {\mathcal {L}}\{f(t)\}\sum _{i=0}^{n}a_{i}s^{i}-\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}f^{(j-1)}(0)={\mathcal {L}}\{\phi (t)\}}

Solving the equation for 
 
 
 
 
 
 L
 
 
 {
 f
 (
 t
 )
 }
 
 
 {\displaystyle {\mathcal {L}}\{f(t)\}}
 
 and substituting 
 
 
 
 
 f
 
 (
 i
 )
 
 
 (
 0
 )
 
 
 {\displaystyle f^{(i)}(0)}
 
 with 
 
 
 
 
 c
 
 i
 
 
 
 
 {\displaystyle c_{i}}
 
 one obtains

L
          
        
        {
        f
        (
        t
        )
        }
        =
        
          
            
              
                
                  L
                
              
              {
              ϕ
              (
              t
              )
              }
              +
              
                ∑
                
                  i
                  =
                  1
                
                
                  n
                
              
              
                ∑
                
                  j
                  =
                  1
                
                
                  i
                
              
              
                a
                
                  i
                
              
              
                s
                
                  i
                  −
                  j
                
              
              
                c
                
                  j
                  −
                  1
                
              
            
            
              
                ∑
                
                  i
                  =
                  0
                
                
                  n
                
              
              
                a
                
                  i
                
              
              
                s
                
                  i
                
              
            
          
        
      
    
    {\displaystyle {\mathcal {L}}\{f(t)\}={\frac {{\mathcal {L}}\{\phi (t)\}+\sum _{i=1}^{n}\sum _{j=1}^{i}a_{i}s^{i-j}c_{j-1}}{\sum _{i=0}^{n}a_{i}s^{i}}}}

The solution for f(t) is obtained by applying the <a href="/facts/Inverse_Laplace_transform/QgGzEnV4">inverse Laplace transform</a> to 
 
 
 
 
 
 L
 
 
 {
 f
 (
 t
 )
 }
 .
 
 
 {\displaystyle {\mathcal {L}}\{f(t)\}.}

Note that if the initial conditions are all zero, i.e.

f
          
            (
            i
            )
          
        
        (
        0
        )
        =
        
          c
          
            i
          
        
        =
        0
        
        ∀
        i
        ∈
        {
        0
        ,
        1
        ,
        2
        ,
        .
        .
        .
         
        n
        }
      
    
    {\displaystyle f^{(i)}(0)=c_{i}=0\quad \forall i\in \{0,1,2,...\ n\}}

then the formula simplifies to

f
        (
        t
        )
        =
        
          
            
              L
            
          
          
            −
            1
          
        
        
          {
          
            
              
                
                  
                    L
                  
                
                {
                ϕ
                (
                t
                )
                }
              
              
                
                  ∑
                  
                    i
                    =
                    0
                  
                  
                    n
                  
                
                
                  a
                  
                    i
                  
                
                
                  s
                  
                    i
                  
                
              
            
          
          }
        
      
    
    {\displaystyle f(t)={\mathcal {L}}^{-1}\left\{{{\mathcal {L}}\{\phi (t)\} \over \sum _{i=0}^{n}a_{i}s^{i}}\right\}}

Laplace transform applied to differential equations open-in-new

Laplace transform applied to differential equations