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How to make InverseLaplaceTransform

Posted 9 years ago

The function is function

I want to do its InverseLaplaceTransform on variable s. I tried this way,

InverseLaplaceTransform[function, s, t] but failed. Would you like to help me to realize it? Thanks.

with best regards, Jacques

POSTED BY: Jacques Ou
7 Replies
Posted 7 years ago

This problem has been solved completely and perfectly. I appreciate your help.

In[456]:= eq60 = (Dc^(1/2) E^((s^(1/2)*(lh - x))/Dc^(1/2)) F1)/(
 s^(1/2) (s - \[Theta]c))

Out[456]= (Sqrt[Dc] E^((
 Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/(Sqrt[s] (s - \[Theta]c))

In[457]:= eq61 = (Dc^(1/2) E^((s^(1/2)*(lh - x))/Dc^(1/2)) F1)/s^(1/2)

Out[457]= (Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/Sqrt[s]

In[458]:= eq62 = eq60/eq61

Out[458]= 1/(s - \[Theta]c)

In[459]:= eq63 = eq61 /. {(lh - x)/Dc^(1/2) -> a} /. {Sqrt[Dc]*F1 -> c}

Out[459]= (c E^(a Sqrt[s]))/Sqrt[s]

In[460]:= eq64 = eq62 /. {\[Theta]c -> b}

Out[460]= 1/(-b + s)

In[461]:= eq65 = InverseLaplaceTransform[eq63, s, t]

Out[461]= -((a c E^(-(a^2/(4 t))) Sqrt[1/(a^2 t)])/Sqrt[\[Pi]])

In[462]:= eq66 = FullSimplify[eq65, a < 0 && Dc > 0 && t > 0]

Out[462]= (c E^(-(a^2/(4 t))))/(Sqrt[\[Pi]] Sqrt[t])

In[463]:= eq67 = InverseLaplaceTransform[eq64, s, t]

Out[463]= E^(b t)

In[464]:= eq68 = ((eq67 /. (t -> (T - t)))*eq66)

Out[464]= (c E^(-(a^2/(4 t)) + b (-t + T)))/(Sqrt[\[Pi]] Sqrt[t])

In[465]:= eq69 = 
 Integrate[eq68, {t, 0, T}, Assumptions -> {a < 0, b > 0, T > 0}]

Out[465]= (c E^(-a Sqrt[b] + 
  b T) (-1 - Erf[(a - 2 Sqrt[b] T)/(2 Sqrt[T])] + 
   E^(2 a Sqrt[b]) (1 + Erf[a/(2 Sqrt[T]) + Sqrt[b T]])))/(2 Sqrt[b])

enter image description here enter image description here

POSTED BY: Jacques Ou
Posted 7 years ago

Hello, Mr. Iwaniuk, I can almost realize your method although the final results are a bit different. There is still another question: the convolution is from negative infinite to positive infinite, but the convolution we used is from zero to t.
Would you like to explain? Thanks.

eq60 = (Dc^(1/2) E^((s^(1/2)*(lh - x))/Dc^(1/2)) F1)/(
 s^(1/2) (s - \[Theta]c))

(Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/(Sqrt[s] (s - \[Theta]c))

eq61 = (Dc^(1/2) E^((s^(1/2)*(lh - x))/Dc^(1/2)) F1)/s^(1/2)

(Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[Dc]) F1)/Sqrt[s]

eq62 = eq60/eq61

1/(s - \[Theta]c)

eq63 = eq61 /. {(lh - x)/Dc^(1/2) -> a} /. {Sqrt[Dc]*F1 -> c}

(c E^(a Sqrt[s]))/Sqrt[s]

eq64 = eq62 /. {\[Theta]c -> b}

1/(-b + s)

eq65 = InverseLaplaceTransform[eq63, s, t]

-((a c E^(-(a^2/(4 t))) Sqrt[1/(a^2 t)])/Sqrt[\[Pi]])

eq66 = FullSimplify[eq65, a < 0 && Dc > 0 && t > 0]

(c E^(-(a^2/(4 t))))/(Sqrt[\[Pi]] Sqrt[t])

eq67 = InverseLaplaceTransform[eq64, s, t]

E^(b t)

eq68 = ((eq67 /. (t -> (T - t)))*eq66)

(c E^(-(a^2/(4 t)) + b (-t + T)))/(Sqrt[\[Pi]] Sqrt[t])

eq69 = Integrate[eq68, {t, 0, T}]

ConditionalExpression[
 1/(2 Sqrt[b])
   c E^(-Sqrt[a^2] Sqrt[b] + 
    b T) (1 - E^(2 Sqrt[a^2] Sqrt[b]) - 
     Erf[(Sqrt[a^2] - 2 Sqrt[b] T)/(2 Sqrt[T])] + 
     E^(2 Sqrt[a^2] Sqrt[b])
       Erf[(Sqrt[a^2] + 2 Sqrt[b] T)/(2 Sqrt[T])] - 
     Erfc[ComplexInfinity] + 
     E^(2 Sqrt[a^2] Sqrt[b]) Erfc[ComplexInfinity]), Re[a^2] <= 0]            

enter image description here enter image description here

POSTED BY: Jacques Ou

Please do NOT post code as images. Follow forum guidelines: https://wolfr.am/READ-1ST

POSTED BY: Moderation Team

Hello,

1.This is not mine method only a math teory.

2 You made little mistakes - >see file attached.

3 You say : "convolution is from negative infinite to positive infinite, but the convolution we used is from zero to t"

I'm an engineer, not a mathematician.If you want of this answer give Your question to Here.

Regards Mariusz.

Attachments:
POSTED BY: Mariusz Iwaniuk

Using convolution:

  L1 = InverseLaplaceTransform[(c E^(a Sqrt[s]))/Sqrt[s] , s, t]
  L2 = InverseLaplaceTransform[1/(s - b), s, t]
  L12 = Integrate[(L1[[1]] /. t -> (t - \[Tau]))*(L2 /. t -> \[Tau]) /. 
  a -> (lh - x)/Sqrt[Dc] /. b -> ThetaC /.  c -> Sqrt[Dc]*F1, {\[Tau], 0, t}, 
  Assumptions -> {t > 0, ThetaC > 0, (lh - x)/Sqrt[Dc] < 0}]

$$-\frac{1}{2} e^{\text{ThetaC} \left(t+\frac{\text{lh}-x}{\sqrt{\text{Dc} \text{ThetaC}}}\right)} \text{F1} \sqrt{\frac{\text{Dc}}{\text{ThetaC}}} \left(-1+e^{2 \sqrt{\frac{\text{ThetaC}}{\text{Dc}}} (-\text{lh}+x)}+e^{2 \sqrt{\frac{\text{ThetaC}}{\text{Dc}}} (-\text{lh}+x)} \text{erf}\left(\frac{\text{lh}-2 t \sqrt{\text{Dc} \text{ThetaC}}-x}{2 \sqrt{\text{Dc} t}}\right)-\text{erf}\left(\frac{\text{lh}+2 t \sqrt{\text{Dc} \text{ThetaC}}-x}{2 \sqrt{\text{Dc} t}}\right)\right)$$

POSTED BY: Mariusz Iwaniuk
Posted 7 years ago

Hello, Mr. Eisenberg, Dc, F1, thetac and lh are constants while x is an independent variable. This InverseLaplaceTransform is about s.

with best regards, Jacques

POSTED BY: Jacques Ou

Are D and c separate symbols, or do you have a two-letter symbol named Dc? Similar question for lh and [Theta]c. If these are products of two separate symbols, then you need a space between them.

What is F1, etc? Which entities are constants and which are functions of s?

POSTED BY: Murray Eisenberg
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