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Mathematics Calculus Equation Solving Wolfram Language

Change the lower limit of the integral when using DSolve?

Jacques Ou

Posted 8 years ago

Please tell me how to make the integral start from 0 rather than 1 in the result? Thanks. In[1702]:= eq71 = DSolve[{f2[k, t] - Dpw2[k, t] \[Eta][k] == \!\(\SuperscriptBox[\(w2\), \* TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[k, t], f3[k, t] - Dpw3[k, t] \[Eta][k] == \!\(\SuperscriptBox[\(w3\), \* TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[k, t]}, {w2[k, t], w3[k, t]}, t] // Simplify // Normal Out[1702]= {{w2[k, t] -> E^(-Dp t \[Eta][k]) (C[1] + \!\( \SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\( \SuperscriptBox[\(E\), \(Dp\ K[1]\ \[Eta][k]\)]\ f2[k, K[1]]\) \[DifferentialD]K[1]\)\)), w3[k, t] -> E^(-Dp t \[Eta][k]) (C[2] + \!\( \SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\( \SuperscriptBox[\(E\), \(Dp\ K[2]\ \[Eta][k]\)]\ f3[k, K[2]]\) \[DifferentialD]K[2]\)\))}}

Please tell me how to make the integral start from 0 rather than 1 in the result? Thanks.

In[1702]:= eq71 = DSolve[{f2[k, t] - Dp*w2[k, t] \[Eta][k] == 
\!\(\*SuperscriptBox[\(w2\), \*
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[k, t], f3[k, t] - Dp*w3[k, t] \[Eta][k] == 
\!\(\*SuperscriptBox[\(w3\), \*
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[k, t]}, {w2[k, t], w3[k, t]}, t] // 
   Simplify // Normal

Out[1702]= {{w2[k, t] -> E^(-Dp t \[Eta][k]) (C[1] + \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\(
\*SuperscriptBox[\(E\), \(Dp\ K[1]\ \[Eta][k]\)]\ f2[k, 
          K[1]]\) \[DifferentialD]K[1]\)\)), 
  w3[k, t] -> E^(-Dp t \[Eta][k]) (C[2] + \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\(
\*SuperscriptBox[\(E\), \(Dp\ K[2]\ \[Eta][k]\)]\ f3[k, 
          K[2]]\) \[DifferentialD]K[2]\)\))}}

enter image description here

POSTED BY: Jacques Ou

2 Replies

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Jacques Ou

Posted 8 years ago

I think the problem can be solved by giving initial conditions. In[2345]:= eq71 = DSolve[{f2[k, t] - Dpw2[k, t] \[Eta][k] == \!\(\SuperscriptBox[\(w2\), \* TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[k, t], f3[k, t] - Dpw3[k, t] \[Eta][k] == \!\(\SuperscriptBox[\(w3\), \* TagBox[ RowBox[{"(", RowBox[{"0", ",", "1"}], ")"}], Derivative], MultilineFunction->None]\)[k, t], w2[k, 0] == eq71, w3[k, 0] == eq72}, {w2[k, t], w3[k, t]}, t] // Simplify // Normal Out[2345]= {{w2[k, t] -> (1/((lh - rb) Sqrt[\[Eta][k]])) E^(-Dp t \[Eta][ k]) (2 pinf (Sin[lh Sqrt[\[Eta][k]]] - Sin[rb Sqrt[\[Eta][k]]]) + (-lh + rb) (\!\( \SubsuperscriptBox[\(\[Integral]\), \(1\), \(0\)]\(\( \SuperscriptBox[\(E\), \(Dp\ K[1]\ \[Eta][k]\)]\ f2[k, K[1]]\) \[DifferentialD]K[1]\)\)) Sqrt[\[Eta][ k]] + (lh - rb) (\!\( \SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\( \SuperscriptBox[\(E\), \(Dp\ K[1]\ \[Eta][k]\)]\ f2[k, K[1]]\) \[DifferentialD]K[1]\)\)) Sqrt[\[Eta][k]]), w3[k, t] -> (1/((lh - rb) Sqrt[\[Eta][k]])) E^(-Dp t \[Eta][ k]) (2 pinf (-Cos[lh Sqrt[\[Eta][k]]] + Cos[rb Sqrt[\[Eta][k]]]) + (-lh + rb) (\!\( \SubsuperscriptBox[\(\[Integral]\), \(1\), \(0\)]\(\( \SuperscriptBox[\(E\), \(Dp\ K[2]\ \[Eta][k]\)]\ f3[k, K[2]]\) \[DifferentialD]K[2]\)\)) Sqrt[\[Eta][ k]] + (lh - rb) (\!\( \SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\( \SuperscriptBox[\(E\), \(Dp\ K[2]\ \[Eta][k]\)]\ f3[k, K[2]]\) \[DifferentialD]K[2]\)\)) Sqrt[\[Eta][k]])}}

I think the problem can be solved by giving initial conditions.

In[2345]:= eq71 = DSolve[{f2[k, t] - Dp*w2[k, t] \[Eta][k] == 
\!\(\*SuperscriptBox[\(w2\), \*
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[k, t], f3[k, t] - Dp*w3[k, t] \[Eta][k] == 
\!\(\*SuperscriptBox[\(w3\), \*
TagBox[
RowBox[{"(", 
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[k, t], w2[k, 0] == eq71, 
     w3[k, 0] == eq72}, {w2[k, t], w3[k, t]}, t] // Simplify // Normal

Out[2345]= {{w2[k, t] -> (1/((lh - rb) Sqrt[\[Eta][k]]))
   E^(-Dp t \[Eta][
      k]) (2 pinf (Sin[lh Sqrt[\[Eta][k]]] - 
         Sin[rb Sqrt[\[Eta][k]]]) + (-lh + rb) (\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(0\)]\(\(
\*SuperscriptBox[\(E\), \(Dp\ K[1]\ \[Eta][k]\)]\ f2[k, 
            K[1]]\) \[DifferentialD]K[1]\)\)) Sqrt[\[Eta][
        k]] + (lh - rb) (\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\(
\*SuperscriptBox[\(E\), \(Dp\ K[1]\ \[Eta][k]\)]\ f2[k, 
            K[1]]\) \[DifferentialD]K[1]\)\)) Sqrt[\[Eta][k]]), 
  w3[k, t] -> (1/((lh - rb) Sqrt[\[Eta][k]]))
   E^(-Dp t \[Eta][
      k]) (2 pinf (-Cos[lh Sqrt[\[Eta][k]]] + 
         Cos[rb Sqrt[\[Eta][k]]]) + (-lh + rb) (\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(0\)]\(\(
\*SuperscriptBox[\(E\), \(Dp\ K[2]\ \[Eta][k]\)]\ f3[k, 
            K[2]]\) \[DifferentialD]K[2]\)\)) Sqrt[\[Eta][
        k]] + (lh - rb) (\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\(
\*SuperscriptBox[\(E\), \(Dp\ K[2]\ \[Eta][k]\)]\ f3[k, 
            K[2]]\) \[DifferentialD]K[2]\)\)) Sqrt[\[Eta][k]])}}

enter image description here

POSTED BY: Jacques Ou

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 8 years ago

Possible You must do it manually typing.Maybe there is a another solution? sol = DSolve[{f2[k, t] - Dpw2[k, t]\[Eta][k] == D[w2[k, t], t], f3[k, t] - Dpw3[k, t]\[Eta][k] == D[w3[k, t], t]}, {w2[k, t], w3[k, t]}, t] For first one: w2[k, t] /. sol[[1]] With substitution: K[1]-1==K[3] differentiation on both sides: \[DifferentialD]K[1] == \[DifferentialD]K[3] and substitution to equation. The second equation the same method:

Possible You must do it manually typing.Maybe there is a another solution?

       sol = DSolve[{f2[k, t] - Dp*w2[k, t]*\[Eta][k] == D[w2[k, t], t], 
       f3[k, t] - Dp*w3[k, t]*\[Eta][k] == D[w3[k, t], t]}, {w2[k, t], 
       w3[k, t]}, t]

For first one:

       w2[k, t] /. sol[[1]]

With substitution:

       K[1]-1==K[3]

differentiation on both sides:

       \[DifferentialD]K[1] == \[DifferentialD]K[3]

and substitution to equation. The second equation the same method:

enter image description here

POSTED BY: Mariusz Iwaniuk

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