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Mathematics Calculus Wolfram Language

What is the substitution or replacement method for multiple functions and their derivatives?

Nomsa Ledwaba

Posted 2 months ago

I want to substitute functions and their derivatives in this: {LXi = xD[Xi[2][t], t] + Sqrt[x] h[t], Xi[2][t]}; LPhi = { (p[t] + f[x, t]) u + g[x, t]}; f[x, t] = C[1][t] + ((-((4 k \[Theta] h[t])/Sqrt[x]) + 6 Sqrt[x] (k h[t] + 2 Derivative[1][h][t]) + 6 x (k Derivative[1][Xi[2]][t] + (Xi[2]^\[Prime]\[Prime])[t]))/( 8 k \[Theta])) Is this correct? {LXi, LPhi} = ReplaceAll[{LXi = {xD[Xi[2][t], t] + Sqrt[x] h[t], Xi[2][t]}, LPhi = { (p[t] + 1/(8 k \[Theta]) (-(1/Sqrt[x]) 4 k \[Theta] h[t] + 6 Sqrt[x] (k (h[t])+ 2 D[h[t], t]) + 6 x (k D[Xi[2][t], t] + D[Xi[2][t], {t, 2}]))) u + g[x, t]}}, {h[t] -> E^((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3])) C[1] + E^(-((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3]))) C[2], p[t] -> C[3] - ( 3 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-3 k + 3 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) k - 8 \[Theta] + 8 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) \[Theta] + Sqrt[3] Sqrt[k (3 k + 8 \[Theta])] + Sqrt[3] E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) Sqrt[ k (3 k + 8 \[Theta])]) C[5])/(8 (3 k + 8 \[Theta])) - ( 3 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (3 Sqrt[3] k + 3 Sqrt[3] E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) k + 8 Sqrt[3] \[Theta] + 8 Sqrt[3] E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3]) \[Theta] - 3 Sqrt[k (3 k + 8 \[Theta])] + 3 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) Sqrt[ k (3 k + 8 \[Theta])]) C[6])/( 8 (3 k + 8 \[Theta]) Sqrt[k (3 k + 8 \[Theta])]), Xi[2][t] -> C[4] + (Sqrt[3] E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3])) C[ 5])/(2 Sqrt[k (3 k + 8 \[Theta])]) + ( 3 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]))^2 C[ 6])/(2 k (3 k + 8 \[Theta])), D[h[t], t] -> ( E^((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3])) Sqrt[k] Sqrt[3 k + 8 \[Theta]] C[1])/(2 Sqrt[3]) - ( E^(-((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3]))) Sqrt[k] Sqrt[3 k + 8 \[Theta]] C[2])/(2 Sqrt[3]) , D[Xi[2][t], t] -> E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) C[5] - 1/2 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3])) C[ 5] + ( Sqrt[3] (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3])) Sqrt[ k (3 k + 8 \[Theta])] C[6])/(k (3 k + 8 \[Theta])) - ( Sqrt[3] E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3]))^2 Sqrt[k (3 k + 8 \[Theta])] C[6])/( 2 k (3 k + 8 \[Theta])), D[Xi[2][t], {t, 2}] -> ( 2 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) k (3 k + 8 \[Theta]) C[5])/( Sqrt[3] Sqrt[k (3 k + 8 \[Theta])]) + ( E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) k (3 k + 8 \[Theta]) C[5])/( 2 Sqrt[3] Sqrt[k (3 k + 8 \[Theta])]) - ( 2 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) Sqrt[ k (3 k + 8 \[Theta])] C[5])/Sqrt[ 3] + ((-2 (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) k (3 k + 8 \[Theta]) + 1/2 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3]))^2 k (3 k + 8 \[Theta]) + 3/2 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) (2/3 E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) k (3 k + 8 \[Theta]) + 2/3 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3]) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[ 3])) k (3 k + 8 \[Theta]))) C[6])/( k (3 k + 8 \[Theta])) Please show a generic way of substitution if possible, without having to manually find their derivatives first before substitution.

I want to substitute functions and their derivatives in this:

{LXi = x*D[Xi[2][t], t] + Sqrt[x] h[t], 
     Xi[2][t]};   LPhi = { (p[t] + f[x, t]) u + g[x, t]};             
    f[x, t] = 
     C[1][t] + ((-((4 k \[Theta] h[t])/Sqrt[x]) + 
       6 Sqrt[x] (k h[t] + 2 Derivative[1][h][t]) + 
       6 x (k Derivative[1][Xi[2]][t] + (Xi[2]^\[Prime]\[Prime])[t]))/(
      8 k \[Theta]))

Is this correct?

{LXi, LPhi} = 
 ReplaceAll[{LXi = {x*D[Xi[2][t], t] + Sqrt[x] h[t], Xi[2][t]},
   LPhi = { (p[t] + 
         1/(8 k \[Theta]) (-(1/Sqrt[x]) 4 k \[Theta] h[t] + 
            6  Sqrt[x]  (k (*h[t]*)+ 2 D[h[t], t]) + 
            6 x (k D[Xi[2][t], t] + D[Xi[2][t], {t, 2}]))) u + 
      g[x, t]}}, {h[t] -> 
    E^((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3]))  C[1] + 
     E^(-((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3])))  C[2], 
   p[t] -> C[3] - (
     3  E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
       3]))  (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
        3]))  (-3 k + 
        3 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) k - 8 \[Theta] + 
        8 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) \[Theta] + 
        Sqrt[3] Sqrt[k (3 k + 8 \[Theta])] + 
        Sqrt[3] E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) Sqrt[
         k (3 k + 8 \[Theta])])  C[5])/(8 (3 k + 8 \[Theta])) - (
     3  E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
       3]))  (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
        3]))  (3 Sqrt[3] k + 
        3 Sqrt[3] E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) k + 
        8 Sqrt[3] \[Theta] + 
        8 Sqrt[3] E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
         3]) \[Theta] - 3 Sqrt[k (3 k + 8 \[Theta])] + 
        3 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]) Sqrt[
         k (3 k + 8 \[Theta])])  C[6])/(
     8 (3 k + 8 \[Theta]) Sqrt[k (3 k + 8 \[Theta])]), 
   Xi[2][t] -> 
    C[4] + (Sqrt[3]
        E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
       3]))  (-1 + E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]))  C[
      5])/(2 Sqrt[k (3 k + 8 \[Theta])]) + (
     3  E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
       3]))  (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]))^2  C[
      6])/(2 k (3 k + 8 \[Theta])), 
   D[h[t], t] -> (
     E^((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3]))  Sqrt[k]
        Sqrt[3 k + 8 \[Theta]]  C[1])/(2 Sqrt[3]) - (
     E^(-((Sqrt[k] t Sqrt[3 k + 8 \[Theta]])/(2 Sqrt[3])))  Sqrt[k]
        Sqrt[3 k + 8 \[Theta]]  C[2])/(2 Sqrt[3]) , 
   D[Xi[2][t], t] -> 
    E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3])  C[5] - 
     1/2  E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
       3]))  (-1 + E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]))  C[
      5] + (
     Sqrt[3]  (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3]))  Sqrt[
      k (3 k + 8 \[Theta])]  C[6])/(k (3 k + 8 \[Theta])) - (
     Sqrt[3]  E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
       3]))  (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
        3]))^2  Sqrt[k (3 k + 8 \[Theta])]  C[6])/(
     2 k (3 k + 8 \[Theta])), 
   D[Xi[2][t], {t, 2}] -> (
     2  E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3])
        k  (3 k + 8 \[Theta])  C[5])/(
     Sqrt[3] Sqrt[k (3 k + 8 \[Theta])]) + (
     E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
       3]))  (-1 + E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
        3]))  k  (3 k + 8 \[Theta])  C[5])/(
     2 Sqrt[3] Sqrt[k (3 k + 8 \[Theta])]) - (
     2  E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3])  Sqrt[
      k (3 k + 8 \[Theta])]  C[5])/Sqrt[
     3] + ((-2 (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
           3])) k (3 k + 8 \[Theta]) + 
        1/2 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
          3])) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
           3]))^2 k (3 k + 8 \[Theta]) + 
        3/2 E^(-((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
          3])) (2/3 E^((2 t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[3])
             k (3 k + 8 \[Theta]) + 
           2/3 E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
            3]) (-1 + E^((t Sqrt[k (3 k + 8 \[Theta])])/Sqrt[
              3])) k (3 k + 8 \[Theta])))  C[6])/(
     k (3 k + 8 \[Theta]))

Please show a generic way of substitution if possible, without having to manually find their derivatives first before substitution.

POSTED BY: Nomsa Ledwaba

1 Reply

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Michael Rogers

Michael Rogers, Emory University

Posted 2 months ago

I'm not sure I understand, and the expression is so complicated, I'm not tempted to dig deeply. But I hope I've guessed what you're after, which I will illustrate with a much simpler example. Suppose I want to substitute `y[x] = Exp[x] Sin[x]` into `y''[x] - 2 y'[x] + 2 y[x] == 0`. First of all, I wouldn't execute `y[x] = Exp[x] Sin[x]`. But if I had done so, I would `Clear` it. The trick is to substitute a `Function` for `y` and not to substitute a formula for `y[x]`. Clear[y] y''[x] - 2 y'[x] + 2 y[x] == 0 /. y -> Function[x, Exp[x] Sin[x]] // Simplify (* True ) Or: Clear[y] sol = Exp[x] Sin[x]; y''[x] - 2 y'[x] + 2 y[x] == 0 /. y -> Function[x, Evaluate[sol]] // Simplify ( True *)

POSTED BY: Michael Rogers

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