Plot[Pi-z - Sin[2  z], {z, 0, Pi}]

X = 10; Y = 10; Z = 1;
ParametricPlot[{z- + Z Cos[z], z}, {z, 0, Pi}]

ContourPlot[x y, {x, 0,1}, {y,0,1}]

tempFun2 = ContourPlot[1-z +0.2Cos[Pi x/Sqrt[2]] Sin[Pi z], {x, -Sqrt[2],Sqrt[2]}, {z, 0,1}, ColorFunction ->"TemperatureMap", AspectRatio -> Automatic]
tempFun3 = ContourPlot[1-z +.3Cos[Pi x/Sqrt[2]] Sin[Pi z], {x, -Sqrt[2],Sqrt[2]}, {z, 0,1}, ColorFunction ->"TemperatureMap", AspectRatio -> Automatic]

streamFun = ContourPlot[Sin[Pi x/Sqrt[2]] Sin[Pi z], {x, -Sqrt[2],Sqrt[2]}, {z, 0,1}, ContourShading->None, AspectRatio -> Automatic]

fig20 = Show[tempFun2, streamFun,FrameLabel ->{"x","height z"}, PlotLabel -> "Y = 0.2, Z = 0"]

fig30 = Show[tempFun2, streamFun,FrameLabel ->{"x","height z"}, PlotLabel -> "Y = 0.2, Z = 0"]

This shows the temperature (red-blue) and the stream function black loops.
The fluid moves along the black loops with a speed proportional to X (capitol X) in the Lorenz equations.
The deviation from a linear temperature gradient is proportional to Y.  This contour plot has Y = 0.2, or Y = 0.3 as indicated.

fig30 = Show[tempFun3, streamFun,FrameLabel ->{"x","height z"}, PlotLabel -> "Y = 0.3, Z = 0"]

figZ =ParametricPlot[{{1-z, z},{1-z-  1/2 /(2 Pi) Sin[2Pi z], z},{1-z- 1/(2 Pi) Sin[2Pi z], z},{1-z-  1.5/(2 Pi) Sin[2Pi z], z}}, {z, 0, 1}, AxesLabel ->{"(OverBar[T])","height z"}, AspectRatio -> .4, PlotLabel -> " (OverBar[T]) (Temp averaged over x) for Z = 0 (linear), .5, 1, and 1.5"]

tempFun35 = ContourPlot[1-z +0.3Cos[Pi x/Sqrt[2]] Sin[Pi z] -0.5 /(2 Pi) Sin[2 Pi z], {x, -Sqrt[2],Sqrt[2]}, {z, 0,1}, ColorFunction ->"TemperatureMap", AspectRatio -> Automatic];
fig35 = Show[tempFun35, streamFun,FrameLabel ->{"x","height z"}, PlotLabel -> "Y = 0.3, Z = 0.5"]

finalFigure =GraphicsGrid[{{fig20, fig30},{figZ, fig35}}, ImageSize -> 1200]

Export["/Users/jimswift/Library/CloudStorage/Dropbox/classes/MAT 667 2025.1/LorenzTemperatureDistribution.pdf", finalFigure, ImageSize -> 1200]