Problem 42891. Differential equation and events

Created by hubert andre

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>You have to solve the non-linear time variant differential equation:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[(cos(y)+2)*y''+atan(y)*sin(0.01*t)^2+y*sin(0.5*t)=0]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>with initial condition as input of the function</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[[y(0);y'(0)]=[a,b]]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>and return the time 'te' when 'y' crosses zero from the negatives values to the positive values.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[te=zero_crossing(a,b)]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>tip: use the 'Events' option in odeset to detect the crossing, stop integration and get the time 'te'. See how to use it here :</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://www.mathworks.com/help/matlab/math/ode-event-location.html?searchHighlight=ballode\"><w:r><w:t>http://www.mathworks.com/help/matlab/math/ode-event-location.html?searchHighlight=ballode</w:t></w:r></w:hyperlink></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Additional info:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>in the test case, the solver ode45 is used, with options 'RelTol'=1e-8 and 'AbsTol'=1e-10.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>because you don't know in advance the final time of the crossing, you can solve again further in time if no crossing is detected, starting from the point of the final time of previous integration.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

You have to solve the non-linear time variant differential equation:<pre class="language-matlab">(cos(y)+2)*y''+atan(y)*sin(0.01*t)^2+y*sin(0.5*t)=0
</pre>with initial condition as input of the function<pre class="language-matlab">[y(0);y'(0)]=[a,b]
</pre>and return the time 'te' when 'y' crosses zero from the negatives values to the positive values.<pre class="language-matlab">te=zero_crossing(a,b)
</pre>tip: use the 'Events' option in odeset to detect the crossing, stop integration and get the time 'te'. See how to use it here : <a href = "http://www.mathworks.com/help/matlab/math/ode-event-location.html?searchHighlight=ballode">http://www.mathworks.com/help/matlab/math/ode-event-location.html?searchHighlight=ballode</a>Additional info:<ul><li>in the test case, the solver ode45 is used, with options 'RelTol'=1e-8 and 'AbsTol'=1e-10.</li><li>because you don't know in advance the final time of the crossing, you can solve again further in time if no crossing is detected, starting from the point of the final time of previous integration.</li></ul>

Solve

Solution Stats

51 Solutions
16 Solvers

Last Solution submitted on Jun 04, 2023

Last 200 Solutions

Problem Comments

2 Comments

2 Comments

Tim on 14 Jun 2016

It looks like [y;y']=[a,b] would mean y(0)=a and y'(0)=b, but a first order differential equation should have one initial condition y(t0)=y0.

hubert andre on 14 Jun 2016

Thanks Tim, yes, I did a mistake in the equation, its a 2nd order.

Solution Comments

Show comments

Problem Recent Solvers16

Problem Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!