Skip to main content

differential equations - DSolve and coupled linear first order PDEs


Does any one know a trick to make DSolve find solution to this coupled linear first order PDE system: (these are Cauchy-Riemann PDE equations, but with one of them having one of the dependent variables as well).


ClearAll[F1,F2,x,y];

ode1 = D[F1[x,y],y]-D[F2[x,y],x] == 0
ode2 = D[F1[x,y],x]+D[F2[x,y],y] == y (*y here causes the problem*)

DSolve[{ode1,ode2},{F1[x,y],F2[x,y]},{x,y}]


Mathematica graphics


This can be solved in Maple:


restart;
#infolevel[pdsolve]:=3;
eq1:= diff(F1(x,y),y)-diff(F2(x,y),x) = 0;
eq2:= diff(F1(x,y),x)+diff(F2(x,y),y) = y;
pdsolve({eq1,eq2},{F1(x,y),F2(x,y)});

Solution it gives is



F1(x, y) = _F1(y-I*x)+_F2(y+I*x)
F2(x, y) = I*_F1(y-I*x)-I*_F2(y+I*x)+(1/2)*y^2+_C1

Screen shot:


Mathematica graphics


If the RHS of the second equation is not y but a constant or some other parameter, then Mathematica can now solve it:


ClearAll[F1,F2,x,y,m];
ode1 = D[F1[x,y],y]-D[F2[x,y],x] == 0
ode2 = D[F1[x,y],x]+D[F2[x,y],y] == m
DSolve[{ode1,ode2},{F1[x,y],F2[x,y]},{x,y}]


Mathematica graphics


Is this a known limitation of DSolve or is there a trick or some other method to get the same solution as in Maple?


Using version 11.2 on windows 7.



Answer



The following substitution eliminates the right side of ode2, and DSolve then can solve the resulting equations.


ode3 = Unevaluated[D[F1[x, y], y] - D[F2[x, y], x] == 0] /. F2[x, y] -> G2[x, y] + y^2/2
(* D[F1[x, y], y] - D[G2[x, y], x] == 0 *)

ode4 = Simplify[Unevaluated[D[F1[x, y], x] + D[F2[x, y], y] == y] /.

F2[x, y] -> G2[x, y] + y^2/2]
(* D[F1[x, y], x] + D[G2[x, y], y] == 0 *)

DSolve[{ode3, ode4}, {F1[x, y], G2[x, y]}, {x, y}] // Flatten
(* {F1[x, y] -> I C[1][I x + y] - I C[2][-I x + y],
G2[x, y] -> C[1][I x + y] + C[2][-I x + y]} *)

Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...