The function is
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
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?
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
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)
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]
Please do NOT post code as images. Follow forum guidelines: https://wolfr.am/READ-1ST
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.
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])
